Lecture Note
Notes for Design and Analysis of Algorithms
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Princeton University -
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COS 226 | Algorithms and Data Structures Academic year:
2024
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DESIGN AND ANALYSIS
OF ALGORITHMS
[R18A0507]
LECTURE NOTES
B.TECH II YEAR – II SEM (R18)
(2019-20)
DEPARTMENT OF
COMPUTER SCIENCE AND ENGINEERING
MALLA REDDY COLLEGE OF
ENGINEERING & TECHNOLOGY
(Autonomous Institution – UGC, Govt. of India)
Recognized under 2(f) and 12 (B) of UGC ACT 1956
(Affiliated to JNTUH, Hyderabad, Approved by AICTE - Accredited by NBA & NAAC – ‘A’ Grade - ISO 9001:2015
Certified)
Maisammaguda, Dhulapally (Post Via. Hakimpet), Secunderabad – 500100, Telangana State, India II Year B. Tech. CSE –II Sem
L T/P/D
C
3 -/- /-
3
(R18A0507) DESIGN AND ANALYSIS OF ALGORITHMS
Objectives:
To analyze performance of algorithms.
To choose the appropriate data structure and algorithm design method for a
specified application.
To understand how the choice of data structures and algorithm design methods impacts
the performance of programs.
To solve problems using algorithm design methods such as the greedy method, divide
and conquer, dynamic programming, backtracking and branch and bound.
To understand the differences between tractable and intractable problems.
To introduce P and NP classes.
UNIT I:
Introduction-Algorithm definition, Algorithm Specification, Performance Analysis Space complexity,
Time complexity, Randomized Algorithms.
Divide and conquer- General method, applications - Binary search, Merge sort, Quick sort, Strassen’s
Matrix Multiplication.
UNIT II:
Disjoint set operations, union and find algorithms, AND/OR graphs, Connected Components and Spanning
trees, Bi-connected components
Backtracking-General method, applications- The 8-queen problem, sum of subsets problem, graph coloring,
Hamiltonian cycles.
UNIT III:
Greedy method- General method, applications- Knapsack problem, Job sequencing with deadlines,
Minimum cost spanning trees, Single source shortest path problem.
UNIT IV:
Dynamic Programming- General Method, applications- Chained matrix multiplication, All pairs shortest
path problem, Optimal binary search trees, 0/1 knapsack problem, Reliability design, Traveling sales person
problem.
UNIT V:
Branch and Bound- General Method, applications-0/1 Knapsack problem, LC Branch and Bound solution,
FIFO Branch and Bound solution, Traveling sales person problem.
NP-Hard and NP-Complete problems- Basic concepts, Non-deterministic algorithms, NP - Hard and
NP-Complete classes, Cook’s theorem.
TEXT BOOKS:
1. Fundamentals of Computer Algorithms, 2nd Edition, Ellis Horowitz, Sartaj Sahni and S.
Rajasekharan, Universities Press.
2. Design and Analysis of Algorithms, P. H. Dave, H. B. Dave, 2nd edition, Pearson Education. REFERENCES:
1. Algorithm Design: Foundations, Analysis and Internet examples, M. T. Goodrich and R. Tomassia,
John Wiley & sons.
2. Design and Analysis of Algorithms, S. Sridhar, Oxford Univ. Press
3. Design and Analysis of algorithms, Aho, Ullman and Hopcroft, Pearson Education.
4. Foundations of Algorithms,, R. Neapolitan and K. Naimipour, 4th edition, Jones and Bartlett
Student edition.
5. Introduction to Algorithms,3rd Edition, T. H. Cormen, C. E.Leiserson, R. L. Rivest, and C. Stein,
PHI
6. Design and Analysis of algorithm, Aho, Ullman and Hopcroft, Pearson Education, 2004.
Outcomes:
Be able to analyze algorithms and improve the efficiency of algorithms.
Ability to analyze the performance of algorithms.
Ability to choose appropriate algorithm design techniques for solving problems.
Ability to understand how the choice of data structures and the algorithm design methods
impact the performance of program. INDEX
UNIT
TOPIC
Introduction-Algorithm definition, Algorithm Specification,
Performance Analysis ,Space complexity, Time complexity
Randomized Algorithms.
Divide and conquer: General method, applications
I
Binary search
Merge sort
Strassen’s matrix multiplication
Quick sort
Disjoint set operations applicationsunion and find algorithms,
Spanning trees
AND/OR graphs,
II
Connected Components and, Bi-connected components
Backtracking-General method,
The 8-queen problem,
sum of subsets problem,
graph coloring,
Hamiltonian cycles.
Greedy method- General method, applications
III
Knapsack problem,
PAGE NO
01
04
08
09
10
14
17
19
25
25
30
31
36
42
43
48
50
51
53
54 III
Job sequencing with deadlines,
Minimum cost spanning trees,
Single source shortest path problem.
Dynamic Programming- General Method, applications
55
56
65
67
All pairs shortest path problem,
68
Traveling sales person problem.
71
0/1 knapsack problem,
73
Reliability design,
75
Optimal binary search trees,
80
Chained matrix Multiplication
84
IV
Branch and Bound: General method
87
- LC BB and FIFO BB
88
Travelling sales person problem
91
V
0/1 knapsack problem
102
NP-Hard and NP-Complete problems: non
deterministic algorithms
NP - Hard and NP Complete classes
106
Cook’s theorem
112
110 UNIT I:
Introduction- Algorithm definition, Algorithm Specification, Performance Analysis- Space
complexity, Time complexity, Randomized Algorithms.
Divide and conquer- General method, applications - Binary search, Merge sort, Quick sort,
Strassen’s Matrix Multiplication.
Algorithm:
An Algorithm is a finite sequence of instructions, each of which has a clear meaning and can be
performed with a finite amount of effort in a finite length of time. No matter what the input values may
be, an algorithm terminates after executing a finite number of instructions. In addition every algorithm
must satisfy the following criteria:
Input: there are zero or more quantities, which are externally supplied;
Output: at least one quantity is produced
Definiteness: each instruction must be clear and unambiguous;
Finiteness: if we trace out the instructions of an algorithm, then for all cases the algorithm will
terminate after a finite number of steps;
Effectiveness: every instruction must be sufficiently basic that it can in principle be carried out by a
person using only pencil and paper. It is not enough that each operation be definite, but it must also
be feasible.
In formal computer science, one distinguishes between an algorithm, and a program. A program does not
necessarily satisfy the fourth condition. One important example of such a program for a computer is its
operating system, which never terminates (except for system crashes) but continues in a wait loop until
more jobs areentered.
We represent algorithm using a pseudo language that is a combination of the constructs of a
programming language together with informal English statements.
Psuedo code for expressing algorithms:
Algorithm Specification: Algorithm can be described in three ways.
1. Natural language like English: When this way is choused care should be taken, we should
ensure that each & every statement is definite.
2. Graphic representation called flowchart: This method will work well when the algorithm
is small& simple.
3. Pseudo-code Method: In this method, we should typically describe algorithms as program,
which resembles language like Pascal & algol.
Pseudo-Code Conventions:
1. Comments begin with // and continue until the end of line.
2. Blocks are indicated with matching braces {and}.
3. An identifier begins with a letter. The data types of variables are not explicitly declared.
1 4. Compound data types can be formed with records. Here is an example,
Node. Record
{
data type – 1 data-1;
.
.
.
data type – n data – n;
node * link;
}
Here link is a pointer to the record type node. Individual data items of a record can
be accessed with and period.
5. Assignment of values to variables is done using the assignment statement.
:= ;
6. There are two Boolean values TRUE and FALSE.
Logical Operators AND, OR, NOT
Relational Operators <, <=,>,>=, =, !=
7. The following looping statements are employed.
For, while and repeat-until
While Loop:
While < condition > do
{
.
.
.
}
For Loop:
For variable: = value-1 to value-2 step step do
{
.
.
.
}
repeat-until:
repeat
.
.
.
2
until
8. A conditional statement has the following forms.
If then
If then
Else
Case statement:
Case
{
: :
.
.
.
: :
: else :
}
9. Input and output are done using the instructions read & write.
10. There is only one type of procedure:
Algorithm, the heading takes the form,
Algorithm ()
As an example, the following algorithm fields & returns the maximum of ‘n’ given
numbers:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Algorithm Max(A,n)
// A is an array of size n
{
Result := A[1];
for I:= 2 to n do
if A[I] > Result then
Result :=A[I];
return Result;
}
In this algorithm (named Max), A & n are procedure parameters. Result & I are Local
variables.
Algorithm:
1. Algorithm selection sort (a,n)
2. // Sort the array a[1:n] into non-decreasing
order. 3.{
4.
for I:=1 to n do
5.
{
6.
j:=I;
3 7.
8.
9.
10.
11.
12.
13. }
for k:=i+1 to n do
if (a[k]=(n+s)
[ n for a[],one each for n,I a& s]
Complexity of Algorithms
The complexity of an algorithm M is the function f(n) which gives the running time
and/or storage space requirement of the algorithm in terms of the size ‘n’ of the input
data. Mostly, the storage space required by an algorithm is simply a multiple of the data
size ‘n’. Complexity shall refer to the running time of the algorithm.
The function f(n), gives the running time of an algorithm, depends not only on the
size ‘n’ of the input data but also on the particular data. The complexity function f(n) for
certain cases are:
1. Best Case : The minimum possible value of f(n) is called the best case.
2.
Average Case
: The expected value of f(n).
5 3.
Worst Case : The maximum value of f(n) for any key possible input.
Asymptotic Notations:
The following notations are commonly use notations in performance analysis and
used to characterize the complexity of an algorithm:
1.
2.
3.
4.
Big–OH (O)
Big–OMEGA (Ω),
Big–THETA (Θ) and
Little–OH (o)
Big–OH O (Upper Bound)
f(n) = O(g(n)), (pronounced order of or big oh), says that the growth rate of f(n) is less
than or equal (<) that of g(n).
Big–OMEGA Ω (Lower Bound)
f(n) = Ω (g(n)) (pronounced omega), says that the growth rate of f(n) is greater than or
equal to (>) that of g(n).
6 Big–THETA Θ (Same order)
f(n) = Θ (g(n)) (pronounced theta), says that the growth rate of f(n) equals (=) the
growth rate of g(n) [if f(n) = O(g(n)) and T(n) = Θ (g(n)].
little-o notation
Definition: A theoretical measure of the execution of an algorithm, usually the time or memory needed,
given the problem size n, which is usually the number of items. Informally, saying some equation f(n) =
o(g(n)) means f(n) becomes insignificant relative to g(n) as n approaches infinity. The notation is read, "f
of n is little oh of g of n".
Formal Definition: f(n) = o(g(n)) means for all c > 0 there exists some k > 0 such that 0 ≤ f(n) < cg(n) for
all n ≥ k. The value of k must not depend on n, but may depend on c.
Different time complexities
Suppose ‘M’ is an algorithm, and suppose ‘n’ is the size of the input data. Clearly
the complexity f(n) of M increases as n increases. It is usually the rate of increase of
f(n) we want to examine. This is usually done by comparing f(n) with some standard
functions. The most common computing times are:
O(1), O(log2 n), O(n), O(n. log2 n), O(n2), O(n3), O(2n), n! and nn
Classification of Algorithms
If ‘n’ is the number of data items to be processed or degree of polynomial or the size of
the file to be sorted or searched or the number of nodes in a graph etc.
1
Next instructions of most programs are executed once or at most only a few
times. If all the instructions of a program have this property, we say that its
running time is a constant.
Log n
When the running time of a program is logarithmic, the program gets
slightly slower as n grows. This running time commonly occurs in
programs that solve a big problem by transforming it into a smaller
problem, cutting the size by some constant fraction., When n is a million,
log n is a doubled. Whenever n doubles, log n increases by a constant, but
log n does not double until n increases to n2.
When the running time of a program is linear, it is generally the case that a
n
7 small amount of processing is done on each input element. This is the
optimal situation for an algorithm that must process n inputs.
n log n
This running time arises for algorithms that solve a problem by breaking it
up into smaller sub-problems, solving then independently, and then
combining the solutions. When n doubles, the running time more than
doubles.
n2
When the running time of an algorithm is quadratic, it is practical for use
only on relatively small problems. Quadratic running times typically arise
in algorithms that process all pairs of data items (perhaps in a double nested
loop) whenever n doubles, the running time increases four fold.
n3
Similarly, an algorithm that process triples of data items (perhaps in a
triple–nested loop) has a cubic running time and is practical for use only on
small problems. Whenever n doubles, the running time increases eight fold.
2n
Few algorithms with exponential running time are likely to be appropriate
for practical use, such algorithms arise naturally as “brute–force” solutions
to problems. Whenever n doubles, the running time squares.
Numerical Comparison of Different Algorithms
The execution time for six of the typical functions is given below:
0
n2
1
n3
1
2n
2
1
2
4
8
4
4
2
8
16
64
16
8
3
24
64
512
256
16
4
64
256
4096
65,536
32
5
160
1024
32,768
4,294,967,296
64
6
384
4096
2,62,144
Note 1
128
7
896
16,384
2,097,152
Note 2
256
8
2048
65,536
1,677,216
????????
n
log2 n
n*log2n
1
0
2
Note1: The value here is approximately the number of machine instructions
executed by a 1 gigaflop computer in 5000 years.
Randomized algorithm:
An algorithm that uses random numbers to decide what to do next anywhere in its logic is called
Randomized Algorithm. For example, in Randomized Quick Sort, we use random number to pick the next
pivot (or we randomly shuffle the array). Quicksort is a familiar, commonly used algorithm in which
randomness can be useful. Any deterministic version of this algorithm requires O(n2) time to
sort n numbers for some well-defined class of degenerate inputs (such as an already sorted array), with the
specific class of inputs that generate this behavior defined by the protocol for pivot selection. However, if
the algorithm selects pivot elements uniformly at random, it has a provably high probability of finishing
in O(n log n) time regardless of the characteristics of the input. Typically, this randomness is used to
reduce time complexity or space complexity in other standard algorithms.
8 Divide and Conquer
General Method:
Divide and conquer is a design strategy which is well known to breaking down
efficiency barriers. When the method applies, it often leads to a large improvement in
time complexity. For example, from O (n2) to O (n log n) to sort the elements.
Divide and conquer strategy is as follows: divide the problem instance into two or
more smaller instances of the same problem, solve the smaller instances recursively,
and assemble the solutions to form a solution of the original instance. The recursion
stops when an instance is reached which is too small to divide. When dividing the
instance, one can either use whatever division comes most easily to hand or invest
time in making the division carefully so that the assembly is simplified.
Divide and conquer algorithm consists of two parts:
Divide
:
Conquer :
Divide the problem into a number of sub problems. The sub problems
are solved recursively.
The solution to the original problem is then formed from the solutions
to the sub problems (patching together the answers).
Traditionally, routines in which the text contains at least two recursive calls are called
divide and conquer algorithms, while routines whose text contains only one recursive
call are not. Divide–and–conquer is a very powerful use of recursion.
Control Abstraction of Divide and Conquer
A control abstraction is a procedure whose flow of control is clear but whose
primary operations are specified by other procedures whose precise meanings are left
undefined. The control abstraction for divide and conquer technique is DANDC(P),
where P is the problem to be solved.
DANDC (P)
{
if SMALL (P) then return S (p);
else
{
divide p into smaller instances p1, p2, …. Pk, k 1;
apply DANDC to each of these sub problems;
return (COMBINE (DANDC (p1) , DANDC (p2),…., DANDC (pk));
}
}
SMALL (P) is a Boolean valued function which determines whether the input size is
small enough so that the answer can be computed without splitting. If this is so
function ‘S’ is invoked otherwise, the problem ‘p’ into smaller sub problems. These
sub problems p1, p2, . . . , pk are solved by recursive application of DANDC.
9 If the sizes of the two sub problems are approximately equal then the computing
time of DANDC is:
T (n) =
g (n)
2 T(n/2) f (n)
n small
otherwise
Where, T (n) is the time for DANDC on ‘n’ inputs
g (n) is the time to complete the answer directly for small inputs and
f (n) is the time for Divide and Combine
Binary Search:
If we have ‘n’ records which have been ordered by keys so that x1 < x2 < … < xn .
When we are given a element ‘x’, binary search is used to find the corresponding
element from the list. In case ‘x’ is present, we have to determine a value ‘j’ such
that a[j] = x (successful search). If ‘x’ is not in the list then j is to set to zero (un
successful search).
In Binary search we jump into the middle of the file, where we find key a[mid], and
compare ‘x’ with a[mid]. If x = a[mid] then the desired record has been found.
If x < a[mid] then ‘x’ must be in that portion of the file that precedes a[mid], if there
at all. Similarly, if a[mid] > x, then further search is only necessary in that past of
the file which follows a[mid]. If we use recursive procedure of finding the middle key
a[mid] of the un-searched portion of a file, then every un-successful comparison of
‘x’ with a[mid] will eliminate roughly half the un-searched portion from consideration.
Since the array size is roughly halved often each comparison between ‘x’ and
a[mid], and since an array of length ‘n’ can be halved only about log2n times before
reaching a trivial length, the worst case complexity of Binary search is about log2n
low and high are integer variables such that each time through the loop either ‘x’
is found or low is increased by at least one or high is decreased by at least one. Thus
we have two sequences of integers approaching each other and eventually low will
become greater than high causing termination in a finite number of steps if ‘x’ is not
present.
10 11 Example for Binary Search
Let us illustrate binary search on the following 9 elements:
Index
Elements
1
-15
2
-6
3
0
4
7
5
9
6
23
7
54
8
82
9
101
The number of comparisons required for searching different elements is as follows:
1. Searching for x = 101
low
1
6
8
9
high
9
9
9
9
mid
5
7
8
9
found
low
1
6
8
high
9
9
9
mid
5
7
8
found
low
1
6
6
7
high mid
9
5
9
7
6
6
6 not found
low
1
1
1
2
high mid
9
5
4
2
1
1
1 not found
Number of comparisons = 4
2. Searching for x = 82
Number of comparisons = 3
3. Searching for x = 42
Number of comparisons = 4
4. Searching for x = -14
Number of comparisons = 3
Continuing in this manner the number of element comparisons needed to find each of
nine elements is:
Index
Elements
Comparisons
1
-15
3
2
-6
2
3
0
3
4
7
4
5
9
1
6
23
3
7
54
2
8
82
3
9
101
4
No element requires more than 4 comparisons to be found. Summing the
comparisons needed to find all nine items and dividing by 9, yielding 25/9 or
approximately 2.77 comparisons per successful search on the average.
There are ten possible ways that an un-successful search may terminate depending
upon the value of x.
12 If x < a[1], a[1] < x < a[2], a[2] < x < a[3], a[5] < x < a[6], a[6] < x < a[7] or
a[7] < x < a[8] the algorithm requires 3 element comparisons to determine that ‘x’
is not present. For all of the remaining possibilities BINSRCH requires 4 element
comparisons. Thus the average number of element comparisons for an unsuccessful
search is:
(3 + 3 + 3 + 4 + 4 + 3 + 3 + 3 + 4 + 4) / 10 = 34/10 = 3.4
The time complexity for a successful search is O(log n) and for an unsuccessful
search is Θ(log n).
Successful searches
Θ(1),
Θ(log n),
Θ(log n)
Best
average
worst
un-successful searches
Θ(log n)
best, average and worst
Analysis for worst case
Let T (n) be the time complexity of Binary search
The algorithm sets mid to [n+1 / 2]
Therefore,
T(0)
= 0
T(n)
= 1
if x = a [mid]
= 1 + T([(n + 1) / 2] – 1)
if x < a [mid]
= 1 + T(n – [(n + 1)/2])
if x > a [mid]
Let us restrict ‘n’ to values of the form n = 2 K – 1, where ‘k’ is a non-negative
integer. The array always breaks symmetrically into two equal pieces plus middle
element.
2K – 1 - 1
2K – 1 - 1
2K
Algebraically this is
n 1
2
Giving,
2K 1 1 = 2K
2
T(0)
=
T(2 – 1)
=
1
=
1 + T(2K
=
1 + T(2
k
1
– 1
0
if x = a [mid]
- 1
– 1)
if x < a [mid]
k - 1
– 1)
if x > a [mid]
In the worst case the test x = a[mid] always fails, so
w(0) = 0
w(2k – 1) = 1 + w(2k
- 1
– 1)
This is now solved by repeated substitution:
w(2k – 1)
=
for K > 1
1 + w(2k
- 1
– 1)
13 =
1 + [1 + w(2k
=
1 + [1 + [1 + w(2k - 3 –1)]]
=
........
=
........
=
i + w(2k
- i
-2
–1)]
– 1)
For i < k, letting i = k gives w(2k –1) = K + w(0) = k
But as 2K – 1 = n, so K = log2(n + 1), so
w(n) = log2(n + 1) = O(log n)
for n = 2K–1, concludes this analysis of binary search.
Although it might seem that the restriction of values of ‘n’ of the form 2 K–1 weakens
the result. In practice this does not matter very much, w(n) is a monotonic
increasing function of ‘n’, and hence the formula given is a good approximation even
when ‘n’ is not of the form 2K–1.
Merge Sort:
Merge sort algorithm is a classic example of divide and conquer. To sort an array,
recursively, sort its left and right halves separately and then merge them. The time
complexity of merge mort in the best case, worst case and average case is O(n log n)
and the number of comparisons used is nearly optimal.
This strategy is so simple, and so efficient but the problem here is that there seems
to be no easy way to merge two adjacent sorted arrays together in place (The result
must be build up in a separate array).
The fundamental operation in this algorithm is merging two sorted lists. Because the
lists are sorted, this can be done in one pass through the input, if the output is put in
a third list.
Algorithm
Algorithm MERGESORT (low, high)
// a (low : high) is a global array to be sorted.
{
if (low < high)
{
mid := (low + high)/2
MERGESORT(low, mid);
MERGESORT(mid+1, high); //sort
MERGE(low, mid, high);
}
}
14
//finds where to split the set
//sort one subset
the other subset
// combine the results Algorithm MERGE (low, mid, high)
// a (low : high) is a global array containing two sorted subsets
// in a (low : mid) and in a (mid + 1 : high).
// The objective is to merge these sorted sets into single sorted
// set residing in a (low : high). An auxiliary array B is used.
{
h :=low; i := low; j:= mid + 1;
while ((h < mid) and (J < high)) do
{
if (a[h] < a[j]) then
{
b[i] := a[h]; h := h + 1;
}
else
{
b[i] :=a[j]; j := j + 1;
}
i := i + 1;
}
if (h > mid) then
for k := j to high do
{
b[i] := a[k]; i := i + 1;
}
else
for k := h to mid do
{
b[i] := a[K]; i := i + l;
}
for k := low to high do
a[k] := b[k];
}
Example
For example let us select the following 8 entries 7, 2, 9, 4, 3, 8, 6, 1 to illustrate
merge sort algorithm:
7, 2, 9, 4 | 3, 8, 6, 1 1, 2, 3, 4, 6, 7, 8, 9
7, 2 | 9, 4 2, 4, 7, 9
7 | 2 2, 7
7 7
2 2
3, 8 | 6, 1 1, 3, 6, 8
9 | 4 4, 9
9 9
4 4
3 | 8 3, 8
3 3
15
8 8
6 | 1 1, 6
6 6
1 1 Tree Calls of MERGESORT(1, 8)
The following figure represents the sequence of recursive calls that are produced by
MERGESORT when it is applied to 8 elements. The values in each node are the values
of the parameters low and high.
1, 8
1, 4
5, 8
1, 2
1, 1
3, 4
2, 2
3, 3
5, 6
4, 4
5, 5
7, 8
6, 6
7, 7
8, 8
Tree Calls of MERGE()
The tree representation of the calls to procedure MERGE by MERGESORT is as
follows:
1, 1, 2
3, 3, 4
5, 5, 6
1, 2, 4
7, 7, 8
5, 6, 8
1, 4, 8
Analysis of Merge Sort
We will assume that ‘n’ is a power of 2, so that we always split into even halves, so
we solve for the case n = 2k.
For n = 1, the time to merge sort is constant, which we will be denote by 1.
Otherwise, the time to merge sort ‘n’ numbers is equal to the time to do two
recursive merge sorts of size n/2, plus the time to merge, which is linear. The
equation says this exactly:
T(1) = 1
T(n) = 2 T(n/2) + n
This is a standard recurrence relation, which can be solved several ways. We will
solve by substituting recurrence relation continually on the right–hand side.
We have, T(n) = 2T(n/2) + n
16 Since we can substitute n/2 into this main equation
2 T(n/2)
=
=
2 (2 (T(n/4)) + n/2)
4 T(n/4) + n
=
=
2 T(n/4) + n
4 T(n/4) + 2n
We have,
T(n/2)
T(n)
Again, by substituting n/4 into the main equation, we see that
4T (n/4)
=
=
4 (2T(n/8)) + n/4
8 T(n/8) + n
=
=
2 T(n/8) + n
8 T(n/8) + 3n
So we have,
T(n/4)
T(n)
Continuing in this manner, we obtain:
T(n)
=
2k T(n/2k) + K. n
As n = 2k, K = log2n, substituting this in the above equation
n
2k
log2 n . n
T k
2
= n T(1) + n log n
= n log n + n
T (n) 2log 2
Representing this in O notation:
T(n) = O(n log n)
We have assumed that n = 2k. The analysis can be refined to handle cases when ‘n’
is not a power of 2. The answer turns out to be almost identical.
Although merge sort’s running time is O(n log n), it is hardly ever used for main
memory sorts. The main problem is that merging two sorted lists requires linear
extra memory and the additional work spent copying to the temporary array and
back, throughout the algorithm, has the effect of slowing down the sort considerably.
The Best and worst case time complexity of Merge sort is O(n log n).
Strassen’s Matrix Multiplication:
The matrix multiplication of algorithm due to Strassens is the most dramatic example
of divide and conquer technique (1969).
The usual way to multiply two n x n matrices A and B, yielding result matrix ‘C’ as
follows :
for i := 1 to n do
for j :=1 to n do
c[i, j] := 0;
for K: = 1 to n do
c[i, j] := c[i, j] + a[i, k] * b[k, j];
17 This algorithm requires n3 scalar multiplication’s (i.e. multiplication of single
numbers) and n3 scalar additions. So we naturally cannot improve upon.
We apply divide and conquer to this problem. For example let us considers three
multiplication like this:
A
A 11
A
21
A
22
12
B
B
11
21
B 12
B
22
C
C
11
21
C
C
12
22
Then cij can be found by the usual matrix multiplication algorithm,
C11 = A11 . B11 + A12 . B21
C12 = A11 . B12 + A12 . B22
C21 = A21 . B11 + A22 . B21
C22 = A21 . B12 + A22 . B22
This leads to a divide–and–conquer algorithm, which performs nxn matrix
multiplication by partitioning the matrices into quarters and performing eight
(n/2)x(n/2) matrix multiplications and four (n/2)x(n/2) matrix additions.
T(1)
T(n)
=
=
1
8 T(n/2)
Which leads to T (n) = O (n3), where n is the power of 2.
Strassens insight was to find an alternative method for calculating the C ij, requiring
seven (n/2) x (n/2) matrix multiplications and eighteen (n/2) x (n/2) matrix
additions and subtractions:
P = (A11 + A22) (B11 + B22)
Q=
(A21 + A22) B11
R = A11 (B12 – B22)
S = A22 (B21 - B11)
T =
(A11 + A12) B22
U = (A21 – A11) (B11 + B12)
V = (A12 – A22) (B21 + B22)
C11 = P + S – T + V
C12 = R + T
C21 = Q + S
C22 = P + R - Q + U.
This method is used recursively to perform the seven (n/2) x (n/2) matrix
multiplications, then the recurrence equation for the number of scalar multiplications
performed is:
18 T(1)
T(n)
=
=
1
7 T(n/2)
Solving this for the case of n = 2k is easy:
T(2k)
=
7 T(2k–1)
=
72 T(2k-2)
=
=
-----------
=
7i T(2k–i)
=
7k T(1)
=
7k
Put i = k
That is, T(n) =
n
7 log2
log 7
2
=
n
=
O(n
log 7
2 )
= O(2n.81)
So, concluding that Strassen’s algorithm is asymptotically more efficient than the
standard algorithm. In practice, the overhead of managing the many small matrices
does not pay off until ‘n’ revolves the hundreds.
Quick Sort
The main reason for the slowness of Algorithms like SIS is that all comparisons and
exchanges between keys in a sequence w1, w2, . . . . , wn take place between
adjacent pairs. In this way it takes a relatively long time for a key that is badly out of
place to work its way into its proper position in the sorted sequence.
Hoare his devised a very efficient way of implementing this idea in the early 1960’s
that improves the O(n2) behavior of SIS algorithm with an expected performance that
is O(n log n).
In essence, the quick sort algorithm partitions the original array by rearranging it
into two groups. The first group contains those elements less than some arbitrary
chosen value taken from the set, and the second group contains those elements
greater than or equal to the chosen value.
The chosen value is known as the pivot element. Once the array has been rearranged
in this way with respect to the pivot, the very same partitioning is recursively applied
to each of the two subsets. When all the subsets have been partitioned and
rearranged, the original array is sorted.
The function partition() makes use of two pointers ‘i’ and ‘j’ which are moved toward
each other in the following fashion:
Repeatedly increase the pointer ‘i’ until a[i] >= pivot.
Repeatedly decrease the pointer ‘j’ until a[j] <= pivot.
19
If j > i, interchange a[j] with a[i]
Repeat the steps 1, 2 and 3 till the ‘i’ pointer crosses the ‘j’ pointer. If ‘i’
pointer crosses ‘j’ pointer, the position for pivot is found and place pivot
element in ‘j’ pointer position.
The program uses a recursive function quicksort(). The algorithm of quick sort
function sorts all elements in an array ‘a’ between positions ‘low’ and ‘high’.
It terminates when the condition low >= high is satisfied. This condition
will be satisfied only when the array is completely sorted.
Here we choose the first element as the ‘pivot’. So, pivot = x[low]. Now it
calls the partition function to find the proper position j of the element
x[low] i.e. pivot. Then we will have two sub-arrays x[low], x[low+1], . . . .
. . . x[j-1] and x[j+1], x[j+2], ..... x[high].
It calls itself recursively to sort the left sub-array x[low], x[low+1], . . . . .
. . x[j-1] between positions low and j-1 (where j is returned by the
partition function).
It calls itself recursively to sort the right sub-array x[j+1], x[j+2], . . . . . .
. . . x[high] between positions j+1 and high.
20 Example
Select first element as the pivot element. Move ‘i’ pointer from left to right in search
of an element larger than pivot. Move the ‘j’ pointer from right to left in search of an
element smaller than pivot. If such elements are found, the elements are swapped.
This process continues till the ‘i’ pointer crosses the ‘j’ pointer. If ‘i’ pointer crosses ‘j’
pointer, the position for pivot is found and interchange pivot and element at ‘j’
position.
Let us consider the following example with 13 elements to analyze quick sort:
1
2
3
4
5
6
7
8
9
10
11
12
13
38
08
16
06
79
57
24
56
02
58
04
70
45
pivot
i
j
04
79
i
j
21
Remarks
swap i & j
swap i & j 02
(24
08
16
06
04
pivot
02)
57
j
i
38
(56
57
58
79
70
45)
j, i
(02
08
16
06
04)
pivot,
j
i
02
(08
16
pivot
i
j
04
16
swap pivot
&j
swap pivot
&j
24
swap pivot
&j
(06
04)
pivot,
j
i
(04)
06
06
04)
j
i
08
(16)
swap i & j
swap pivot
&j
swap pivot
&j
04
pivot,
j, i
16
pivot,
j, i
(02
04
06
08
16
24)
38
(56
57
pivot
i
j
45
57
(45)
58
j
i
56
(58
79
79
70
70
45)
57)
45
pivot,
j, i
swap i & j
swap pivot
&j
swap pivot
&j
(58
pivot
79
i
70
57
(57)
57)
j
swap i & j
79
j
i
58
(70
79)
(70
79)
pivot,
j
i
swap pivot
&j
57
pivot,
j, i
70
79
pivot,
j, i
02
04
06
08
16
24
38
(45
56
57
58
70
79)
45
56
57
58
70
79
22
swap pivot
&j Analysis of Quick Sort:
Like merge sort, quick sort is recursive, and hence its analysis requires solving a
recurrence formula. We will do the analysis for a quick sort, assuming a random pivot
(and no cut off for small files).
We will take
T (0) = T (1) = 1, as in merge sort.
The running time of quick sort is equal to the running time of the two recursive calls
plus the linear time spent in the partition (The pivot selection takes only constant
time). This gives the basic quick sort relation:
T (n) = T (i) + T (n – i – 1) + C n
-
(1)
Where, i = |S1| is the number of elements in S1.
Worst Case Analysis
The pivot is the smallest element, all the time. Then i=0 and if we ignore T(0)=1,
which is insignificant, the recurrence is:
T (n) = T (n – 1) + C n
n > 1
-
(2)
-
(3)
Using equation – (1) repeatedly, thus
T (n – 1) = T (n – 2) + C (n – 1)
T (n – 2) = T (n – 3) + C (n – 2)
------- T (2)
= T (1) + C (2)
Adding up all these equations yields
n
T (n) T (1)
i
i 2
= O (n2)
23 Best and Average Case Analysis
The number of comparisons for first call on partition: Assume left_to_right moves
over k smaller element and thus k comparisons. So when right_to_left crosses
left_to_right it has made n-k+1 comparisons. So, first call on partition makes n+1
comparisons. The average case complexity of quicksort is
T(n) = comparisons for first call on quicksort
+
{Σ 1<=nleft,nright<=n [T(nleft) + T(nright)]}n = (n+1) + 2 [T(0) +T(1) + T(2) +
----- + T(n-1)]/n
nT(n) = n(n+1) + 2 [T(0) +T(1) + T(2) + ----- + T(n-2) + T(n-1)]
(n-1)T(n-1) = (n-1)n + 2 [T(0) +T(1) + T(2) + ----- + T(n-2)] \
Subtracting both sides:
nT(n) –(n-1)T(n-1) = [ n(n+1) – (n-1)n] + 2T(n-1) = 2n + 2T(n-1)
nT(n) = 2n + (n-1)T(n-1) + 2T(n-1) = 2n + (n+1)T(n-1)
T(n) = 2 + (n+1)T(n-1)/n
The recurrence relation obtained is:
T(n)/(n+1) = 2/(n+1) + T(n-1)/n
Using the method of subsititution:
T(n)/(n+1)
=
T(n-1)/n
=
T(n-2)/(n-1) =
T(n-3)/(n-2) =
.
.
T(3)/4
=
T(2)/3
=
2/(n+1) + T(n-1)/n
2/n + T(n-2)/(n-1)
2/(n-1) + T(n-3)/(n-2)
2/(n-2) + T(n-4)/(n-3)
.
.
2/4 + T(2)/3
2/3 + T(1)/2 T(1)/2 = 2/2 + T(0)
Adding both sides:
T(n)/(n+1) + [T(n-1)/n + T(n-2)/(n-1) + -------------- + T(2)/3 + T(1)/2]
= [T(n-1)/n + T(n-2)/(n-1) + -------------- + T(2)/3 + T(1)/2] + T(0) +
[2/(n+1) + 2/n + 2/(n-1) + ----------- +2/4 + 2/3]
Cancelling the common terms:
T(n)/(n+1) = 2[1/2 +1/3 +1/4+ ------------- +1/n+1/(n+1)]
T(n) = (n+1)2[
2k n 1
1/ k
=2(n+1) [
]
=2(n+1)[log (n+1) – log 2]
=2n log (n+1) + log (n+1)-2n log 2 –log 2
T(n)= O(n log n)
24 UNIT II:
Disjoint set operations, union and find algorithms, AND/OR graphs, Connected Components and
Spanning trees, Bi-connected components
Backtracking-General method, applications- The 8-queen problem, sum of subsets problem, graph
coloring, Hamiltonian cycles.
Sets and Disjoint Set Union:
Disjoint Set Union: Considering a set S={1,2,3…10} (when n=10), then elements
can be partitioned into three disjoint sets s1={1,7,8,9},s2={2,5,10} and
s3={3,4,6}. Possible tree representations are:
1
7
8
5
9
3
10
2
4
6
In this representation each set is represented as a tree. Nodes are linked from the child to
parent rather than usual method of linking from parent to child.
The operations on these sets are:
1.
2.
3.
4.
5.
Disjoint set union
Find(i)
Min Operation
Delete
Intersect
1. Disjoint Set union:
If Si and Sj are two disjoint sets, then their union Si U Sj = all the elements x
such that x is in Si or Sj. Thus S1 U S2 ={1,7,8,9,2,5,10}.
2. Find(i):
Given the element I, find the set containing i. Thus, 4 is in set S 3, 9 is in S1.
UNION operation:
Union(i,j) requires two tree with roots i and j be joined. S 1 U S2 is
obtained by making any one of the sets as sub tree of other.
1
5
7
8
9
5
2
1
1
7
25
8
2
9
1 Simple Algorithm for Union:
Algorithm Union(i,j)
{
//replace the disjoint sets with roots i and j, I not equal to j by their
union Integer i,j;
P[j] :=i;
}
Example:
Implement following sequence of operations Union(1,3),Union(2,5),Union(1,2)
Solution:
Initially parent array contains zeros.
0
1
0
2
0
3
0
4
0
5
0
6
1. After performing union(1,3)operation Parent[3]:=1
0
1
0
2
1
3
0
4
0
5
0
6
2. After performing union(2,5)operation Parent[5]:=2
0
1
0
2
1
3
0
4
2
5
0
6
3. After performing union(1,2)operation Parent[2]:=1
0
1
1
2
1
3
0
4
1
3
2
5
26
2
5
0
6 Process the following sequence of union
operations Union(1,2),Union(2,3) Union(n-1,n)
Degenerate Tree:
n
n-1
1
The time taken for n-1 unions is O(n).
Find(i) operation: determines the root of the tree containing
element i. Simple Algorithm for Find:
Algorithm Find(i)
{
j:=i;
while(p[j]>0) do
j:=p[j]; return j;
}
Find Operation: Find(i) implies that it finds the root node of i th node, in other words it
returns the name of the set i.
Example: Consider the Union(1,3)
1
3
Find(1)=0
Find(3)=1, since its parent is 1. (i.e, root is 1)
27 Example:
Considering
1
3
2
5
Array Representation
P[i]
i
0
1
1
2
1
3
2
5
Find(5)=1
Find(2)=1
Find(3)=1
The root node represents all the nodes in the tree. Time Complexity of ‘n’ find operations is
O(n2).
To improve the performance of union and find algorithms by avoiding the creation of
degenerate tree. To accomplish this, we use weighting rule for Union(i,j).
Weighting Rule for Union(i,j)
Tree obtained with
weighted Initially
28 1
2
n
Union(1,2)
3
1
n
2
Union(1,3)
1
n
4
2
3
:
:
:
Union(1,n)
1
2
3
n
Collapsing Rule for Find(i)
29 Spanning Trees:
A spanning tree is a subset of Graph G, which has all the vertices covered with minimum possible
number of edges. Hence, a spanning tree does not have cycles and it cannot be disconnected..
By this definition, we can draw a conclusion that every connected and undirected Graph G has at
least one spanning tree. A disconnected graph does not have any spanning tree, as it cannot be
spanned to all its vertices.
We found three spanning trees off one complete graph. A complete undirected graph can have
maximum nn-2 number of spanning trees, where n is the number of nodes. In the above addressed
example, n is 3, hence 33−2 = 3spanning trees are possible.
General Properties of Spanning Tree
We now understand that one graph can have more than one spanning tree. Following are a few
properties of the spanning tree connected to graph G −
A connected graph G can have more than one spanning tree.
All possible spanning trees of graph G, have the same number of edges and vertices.
The spanning tree does not have any cycle (loops).
Removing one edge from the spanning tree will make the graph disconnected, i.e. the spanning tree
is minimally connected.
Adding one edge to the spanning tree will create a circuit or loop, i.e. the spanning tree is maximally
acyclic.
30 Mathematical Properties of Spanning Tree
Spanning tree has n-1 edges, where n is the number of nodes (vertices).
From a complete graph, by removing maximum e - n + 1 edges, we can construct a spanning tree.
A complete graph can have maximum nn-2 number of spanning trees.
Thus, we can conclude that spanning trees are a subset of connected Graph G and disconnected
graphs do not have spanning tree.
Application of Spanning Tree
Spanning tree is basically used to find a minimum path to connect all nodes in a graph. Common
application of spanning trees are −
Civil Network Planning
Computer Network Routing Protocol
Cluster Analysis
AND/OR GRAPH:
And/or graph is a specialization of hypergraph which connects nodes by sets of arcs rather than
by a single arcs. A hypergraph is defined as follows:
A hypergraph consists of: N, a set of nodes,
H, a set of hyperarcs defined by ordered pairs, in which the first implement of the pair is a node of
N and the second implement is the subset of N.
An ordinary graph is a special case of hypergraph in which all the sets of decendent nodes have a
cardinality of 1.
Hyperarcs also known as K-connectors, where K is the cardinality of the set of decendent nodes.
If K = 1, the descendent may be thought of as an OR nodes. If K > 1, the elements of the set of
decendents may be thought of as AND nodes. In this case the connector is drawn with individual
edges from the parent node to each of the decendent nodes; these individual edges are then joined
with a curved link. And/or graph for the expression P and Q -> R is follows:
31 The K-connector is represented as a fan of arrows with a single tie is shown above. The and/or
graphs consists of nodes labelled by global databases. Nodes labelled by compound databases have
sets of successor nodes. These successor nodes are called AND nodes, in order to process the
compound database to termination, all the compound databases must be processed to termination.
For example consider, consider a boy who collects stamps (M). He has for the purpose of exchange
a winning conker (C), a bat (B) and a small toy animal (A). In his class there are friends who are
also keen collectors of different items and will make the following exchanges.
1. 1 winning conker (C) for a comic (D) and a bag of sweets (S).
2. 1 winning conker (C) for a bat (B) and a stamp (M).
3. 1 bat (B) for two stamps (M, M).
4. 1 small toy animal (A) for two bats (B, B) and a stamp (M).
The problem is how to carry out the exchanges so that all his exchangable items are converted into
stamps (M). This task can be expressed more briefly as:
1. Initial state = (C, B, A)
2. Transformation rules:
a. If C then (D, S)
b. If C then (B, M)
c. If B then (M, M)
d. If A then (B, B, M)
3. The goal state is to left with only stamps (M, . . . . . . , M)
32 The figure shows that, a lot of extra work is done by redoing many of the transformations.
This repetition can be avoided by decomposing the problem into subproblems. There are
two major ways to order the components:
1. The components can either be arranged in some fixed order at the time they are
generated (or).
2. They can be dynamically reordered during processing.
The more flexible system is to reorder dynamically as the processing unfolds. It can be
represented by and/or graph. The solution to the exchange problem will be:
Swap conker for a bat and a stamp, then exchange this bat for two stamps. Swap his
own bat for two more stamps, and finally swap the small toy animal for two bats and a
stamp. The two bats can be exchanged for two stamps.
The previous exchange problem, when implemented as an and/or graph looks as
follows:
33 Example 1:
Draw an AND/OR graph for the following prepositions:
1.
2.
3.
4.
5.
6.
7.
8.
A
B
C
A ^ B -> D
A ^ C -> E
B ^ D -> F
F -> G
A ^ E -> H
34 Connected components
In graph theory, a connected component (or just component) of an undirected graph is
a subgraph in which any two vertices are connected to each other by paths, and which is
connected to no additional vertices in the super graph. For example, the graph shown in the
illustration has three connected components. A vertex with no incident edges is itself a connected
component. A graph that is itself connected has exactly one connected component, consisting of
the whole graph.
A graph with three connected components.
Biconnected Components:
Let G = (V, E) be a connected undirected graph. Consider the following definitions:
Articulation Point (or Cut Vertex): An articulation point in a connected
graph is a vertex (together with the removal of any incident edges) that, if
deleted, would break the graph into two or more pieces..
Bridge: Is an edge whose removal results in a disconnected graph.
Biconnected: A graph is biconnected if it contains no articulation points.
In a biconnected graph, two distinct paths connect each pair of vertices. A
graph that is not biconnected divides into biconnected components. This is
illustrated in the following figure:
36 Biconnected
Components
Articulation Point
Bridge
Articulation Points and Bridges
Biconnected graphs and articulation points are of great interest in the
design of network algorithms,
because these are the “critical" points, whose failure will result in the
network becoming disconnected.
Let us consider the typical case of vertex v, where v is not a leaf and v is
not the root. Let w1, w2, . . . . . . . wk be the children of v. For each child
there is a subtree of the DFS tree rooted at this child. If for some child,
there is no back edge going to a proper ancestor of v, then if we remove v,
this subtree becomes disconnected from the rest of the graph, and hence v
is an articulation point.
L (u) = min {DFN (u), min {L (w) w is a child of u}, min {DFN
(w) (u, w) is a back edge}}.
L (u) is the lowest depth first number that can be reached from ‘u’ using a
path of descendents followed by at most one back edge. It follows that, If
‘u’ is not the root then ‘u’ is an articulation point iff ‘u’ has a child ‘w’ such
that:
L (w) ≥ DFN (u)
Algorithm for finding the Articulation points:
Pseudocode to compute DFN and L.
Algorithm Art (u, v)
// u is a start vertex for depth first search. V is its parent if any in the depth first
// spanning tree. It is assumed that the global array dfn is initialized to zero and
that // the global variable num is initialized to 1. n is the number of vertices in G.
{
dfn [u] := num; L [u] := num; num
:= num + 1; for each vertex w
adjacent from u do
{
if (dfn [w] = 0) then
{
37 Art (w, u);
// w
is unvisited. L [u] := min (L [u], L
[w]);
}
else if (w v) then L [u] := min (L [u], dfn [w]);
}
}
Algorithm for finding the Biconnected Components:
Algorithm BiComp (u, v)
// u is a start vertex for depth first search. V is its parent if any in the depth first
// spanning tree. It is assumed that the global array dfn is initially zero and that the
// global variable num is initialized to 1. n is the number of vertices in G.
{
dfn [u] := num; L [u] := num; num
:= num + 1; for each vertex w
adjacent from u do
{
if ((v w) and (dfn [w] < dfn
[u])) then add (u, w)
to the top of a stack
s;
if (dfn [w] = 0) then
{
if (L [w] > dfn [u]) then
{
write (“New
bicomponent”);
repeat
{
Delete an edge from the
top of stack s; Let this
edge be (x, y);
Write (x, y);
} until (((x, y) = (u, w)) or ((x, y) = (w, u)));
}
BiComp (w, u);
// w is unvisited. L [u] := min (L [u], L [w]);
}
else if (w v) then L [u] : = min (L [u], dfn [w]);
}
}
Example:
For the following graph identify the articulation points and Biconnected components:
86
11
2 4
11
57
62
24
7
33
9
33
810
4
10
1 0
5
9
6
2
7
5
95
38
86
79
810 4
Graph
Dept h First Spanning Tree
To identify the articulation points, we use:
L (u) = min {DFN (u), min {L (w) w is a child of u}, min {DFN (w) w
is a vertex to which there is back edge from u}}
L (1) = min {DFN (1), min {L (4)}} = min {1, L (4)} = min {1, 1} = 1
L (4) = min {DFN (4), min {L (3)}} = min {2, L (3)} = min {2, 1} = 1
L (3) = min {DFN (3), min {L (10), L (9), L (2)}} =
= min {3, min {L (10), L (9), L (2)}} = min {3, min {4, 5, 1}} = 1
L (10) = min {DFN (10)} = 4
L (9) = min {DFN (9)} = 5
L (2) = min {DFN (2), min {L (5)}, min {DFN (1)}}
= min {6, min {L (5)}, 1} = min {6, 6, 1} = 1
L (5) = min {DFN (5), min {L (6), L (7)}} = min {7, 8, 6} = 6
L (6) = min {DFN (6)} = 8
L (7) = min {DFN (7), min {L (8}, min {DFN (2)}}
= min {9, L (8) , 6} = min {9, 6, 6} = 6
L (8) = min {DFN (8), min {DFN (5), DFN (2)}}
= min {10, min (7, 6)} = min {10, 6} = 6
Therefore, L (1: 10) = (1, 1, 1, 1, 6, 8, 6, 6, 5, 4)
Finding the Articulation Points:
Vertex 1: Vertex 1 is not an articulation point. It is a root node. Root is an
articulation point if it has two or more child nodes.
Vertex 2: is an articulation point as child 5 has L (5) = 6 and
DFN (2) = 6, So, the condition L (5) = DFN (2) is true.
Vertex 3: is an articulation point as child 10 has L (10) = 4 and
DFN (3) = 3, So, the condition L (10) > DFN (3) is true.
Vertex 4: is not an articulation point as child 3 has L (3) = 1 and
DFN (4) = 2, So, the condition L (3) > DFN (4) is false.
Vertex 5: is an articulation point as child 6 has L (6) = 8, and
DFN (5) = 7, So, the condition L (6) > DFN (5) is true.
Vertex 7: is not an articulation point as child 8 has L (8) = 6, and
DFN (7) = 9, So, the condition L (8) > DFN (7) is false.
39 Vertex 6, Vertex 8, Vertex 9 and Vertex 10 are leaf nodes.
Therefore, the articulation points are {2, 3, 5}.
Example:
For the following graph identify the articulation points and Biconnected components:
4
1
2
3
5
7
8
1
1
2
2
4
6
3
5
G ra p h
5
3
6
4
7
6
7
8
8
D F S s p a n ni n g T re e
L (u) = min {DFN (u), min {L (w) w is a child of u}, min {DFN (w) w
is a vertex to which there is back edge from u}}
L (1) = min {DFN (1), min {L (2)}} = min {1, L (2)} = min {1, 2} = 1
L (2) = min {DFN (2), min {L (3)}} = min {2, L (3)} = min {2, 3} = 2
L (3) = min {DFN (3), min {L (4), L (5), L (6)}} = min {3, min {6, 4, 5}} = 3
L (4) = min {DFN (4), min {L (7)} = min {6, L (7)} = min {6, 6} = 6
L (5) = min {DFN (5)} = 4
L (6) = min {DFN (6)} = 5
L (7) = min {DFN (7), min {L (8)}} = min {7, 6} = 6
L (8) = min {DFN (8), min {DFN (4)}} = min {8, 6} = 6
Therefore, L (1: 8) = {1, 2, 3, 6, 4, 5, 6, 6}
Finding the Articulation Points:
Check for the condition if L (w) > DFN (u) is true, where w is any
child of u. Vertex 1: Vertex 1 is not an articulation point.
It is a root node. Root is an articulation point if it has two or more
child nodes.
Vertex 2: is an articulation point as L (3) = 3 and DFN (2) = 2.
So, the condition is true
Vertex 3: is an articulation Point as:
I.
L (5) = 4 and DFN (3) = 3
40 II.
III.
L (6) = 5 and DFN (3) = 3 and
L (4) = 6 and
DFN (3) = 3 So, the
condition true in above
cases
Vertex 4: is an articulation point as L (7) = 6 and DFN (4) = 6.
So, the condition is true
Vertex 7: is not an articulation point as L (8) = 6 and DFN (7) = 7.
So, the condition is False
Vertex 5, Vertex 6 and Vertex 8 are leaf
nodes. Therefore, the articulation points
are {2, 3, 4}.
41 BACKTRACKING
General Method:
Backtracking is used to solve problem in which a sequence of objects is chosen from a
specified set so that the sequence satisfies some criterion. The desired solution is
expressed as an n-tuple (x1, . . . . , xn) where each xi Є S, S being a finite set.
The solution is based on finding one or more vectors that maximize, minimize, or
satisfy a criterion function P (x1, , xn). Form a solution and check at every step
if this has any chance of success. If the solution at any point seems not promising,
ignore it. All solutions requires a set of constraints divided into two categories: explicit
and implicit constraints.
Definition 1: Explicit constraints are rules that restrict each x i to take on values only
from a given set. Explicit constraints depend on the particular instance I
of problem being solved. All tuples that satisfy the explicit constraints
define a possible solution space for I.
Definition 2: Implicit constraints are rules that determine which of the tuples in the
solution space of I satisfy the criterion function. Thus, implicit
constraints describe the way in which the xi’s must relate to each other.
For 8-queens problem:
Explicit constraints using 8-tuple formation, for this problem are S= {1, 2, 3,
4, 5, 6, 7, 8}.
The implicit constraints for this problem are that no two queens can be the
same (i.e., all queens must be on different columns) and no two queens can be
on the same diagonal.
Backtracking is a modified depth first search of a tree. Backtracking algorithms
determine problem solutions by systematically searching the solution space for the
given problem instance. This search is facilitated by using a tree organization for the
solution space.
Backtracking is the procedure whereby, after determining that a node can lead to
nothing but dead end, we go back (backtrack) to the nodes parent and proceed with
the search on the next child.
A backtracking algorithm need not actually create a tree. Rather, it only needs to
keep track of the values in the current branch being investigated. This is the way we
implement backtracking algorithm. We say that the state space tree exists implicitly in
the algorithm because it is not actually constructed.
State space is the set of paths from root node to other nodes. State space tree is the
tree organization of the solution space. The state space trees are called static trees.
This terminology follows from the observation that the tree organizations are
independent of the problem instance being solved. For some problems it is
advantageous to use different tree organizations for different problem instance. In
this case the tree organization is determined dynamically as the solution space is
being searched. Tree organizations that are problem instance dependent are called
42 dynamic trees.
Terminology:
Problem state is each node in the depth first search tree.
Solution states are the problem states ‘S’ for which the path from the root node to
‘S’ defines a tuple in the solution space.
Answer states are those solution states for which the path from root node to s
defines a tuple that is a member of the set of solutions.
Live node is a node that has been generated but whose children have not yet been
generated.
E-node is a live node whose children are currently being explored. In other words, an
E-node is a node currently being expanded.
Dead node is a generated node that is not to be expanded or explored any further.
All children of a dead node have already been expanded.
Branch and Bound refers to all state space search methods in which all children of
an E-node are generated before any other live node can become the E-node.
Depth first node generation with bounding functions is called backtracking. State
generation methods in which the E-node remains the E-node until it is dead, lead to
branch and bound methods.
N-Queens Problem:
Let us consider, N = 8. Then 8-Queens Problem is to place eight queens on an 8 x 8
chessboard so that no two “attack”, that is, no two of them are on the same row,
column, or diagonal.
All solutions to the 8-queens problem can be represented as 8-tuples (x1, . . . . , x8),
where xi is the column of the ith row where the ith queen is placed.
The explicit constraints using this formulation are Si = {1, 2, 3, 4, 5, 6, 7, 8}, 1 < i <
8. Therefore the solution space consists of 88 8-tuples.
The implicit constraints for this problem are that no two xi’s can be the same (i.e., all
queens must be on different columns) and no two queens can be on the same
diagonal.
This realization reduces the size of the solution space from 88 tuples to 8! Tuples.
The promising function must check whether two queens are in the same column or
diagonal:
Suppose two queens are placed at positions (i, j) and (k, l) Then:
Column Conflicts: Two queens conflict if their xi values are identical.
Diag 45 conflict: Two queens i and j are on the same 450 diagonal if:
i – j = k – l.
This implies, j – l = i – k
Diag 135 conflict:
i + j = k + l. This implies, j – l = k – i
43 Therefore, two queens lie on the same diagonal if and only if:
j - l = i – k
Where, j be the column of object in row i for the i th queen and l be the column of
object in row ‘k’ for the kth queen.
To check the diagonal clashes, let us take the following tile configuration:
*
In this example, we have:
*
*
*
*
*
*
*
i
1
2
3
4
5
6
7
8
xi
2
5
1
8
4
7
3
6
Let us consider for the
case whether the queens on
3rd row and 8th row
are conflicting or not. In this
case (i, j) = (3, 1) and (k, l) = (8, 6). Therefore:
j - l = i – k 1 - 6 = 3 – 8
5 = 5
In the above example we have, j - l = i – k , so the two queens are attacking.
This is not a solution.
Example:
Suppose we start with the feasible sequence 7, 5, 3, 1.
*
*
*
*
Step 1:
Add to the sequence the next number in the sequence 1, 2, . . . , 8 not yet
used.
Step 2:
If this new sequence is feasible and has length 8 then STOP with a solution. If
the new sequence is feasible and has length less then 8, repeat Step 1.
Step 3:
If the sequence is not feasible, then backtrack through the sequence until we
find the most recent place at which we can exchange a value. Go back to Step
1.
44 5
6
7
Remarks
1
2
3
4
8
7
5
3
1
7
5
3
1*
2*
7
5
3
1
4
7*
5
3
1
4
2*
j - l = 7 – 2 = 5
i – k = 1 – 6 = 5
7
5
3*
1
4
6*
j - l = 3 – 6 = 3
i – k = 3 – 6 = 3
7
5
3
1
4
8
7
5
3
1
4*
8
2*
j - l = 4 – 2 = 2
i – k = 5 – 7 = 2
7
5
3
1
4*
8
6*
j - l = 4 – 6 = 2
i – k = 5 – 7 = 2
7
5
3
1
4
8
7
5
3
1
4
7
5
3
1
6
7*
5
3
1
6
2*
7
5
3
1
6
4
7
5
3
1
6
4
2
7
5
3*
1
6
4
2
7
5
3
1
6
4
2
7
5
3
1
6
4
7
5
3
1
6
8
7
5
3
1
6
8
2
7
5
3
1
6
8
2
j - l = 1 – 2 = 1
i – k = 4 – 5 = 1
Backtrack
Backtrack
j - l = 1 – 2 = 1
i – k = 7 – 6 = 1
j - l = 3 – 8 = 5
i – k =3 – 8 = 5
8*
Backtrack
Backtrack
4
SOLUTION
* indicates conflicting queens.
On a chessboard, the solution will look like:
*
*
*
*
*
*
*
*
45 4 – Queens Problem:
Let us see how backtracking works on the 4-queens problem. We start with the root
node as the only live node. This becomes the E-node. We generate one child. Let us
assume that the children are generated in ascending order. Let us assume that the
children are generated in ascending order. Thus node number 2 of figure is generated
and the path is now (1). This corresponds to placing queen 1 on column 1. Node 2
becomes the E-node. Node 3 is generated and immediately killed. The next node
generated is node 8 and the path becomes (1, 3). Node 8 becomes the E-node.
However, it gets killed as all its children represent board configurations that cannot
lead to an answer node. We backtrack to node 2 and generate another child, node 13.
The path is now (1, 4). The board configurations as backtracking proceeds is as
follows:
1
1
.
1
.
1
2
2
2
. . . .
(b)
(a)
1
3
(c)
1
(d)
1
.
2
.
1
.
3
.
.
2
2
3
.
.
(e)
.
.
(f)
(g)
.
4
(h)
The above figure shows graphically the steps that the backtracking algorithm goes
through as it tries to find a solution. The dots indicate placements of a queen, which
were tried and rejected because another queen was attacking.
In Figure (b) the second queen is placed on columns 1 and 2 and finally settles on
column 3. In figure (c) the algorithm tries all four columns and is unable to place the
next queen on a square. Backtracking now takes place. In figure (d) the second
queen is moved to the next possible column, column 4 and the third queen is placed
on column 2. The boards in Figure (e), (f), (g), and (h) show the remaining steps that
the algorithm goes through until a solution is found.
46 Complexity Analysis:
1 n n2 n3 . . . . . . . . . . . n n
n
n 1
1
n1
47 For the instance in which n = 8, the state space tree contains:
88 1 1 = 19, 173, 961 nodes
8 1
Sum of Subsets:
Given positive numbers wi, 1 ≤ i ≤ n, and m, this problem requires finding all subsets
of wi whose sums are ‘m’.
All solutions are k-tuples, 1 ≤ k ≤ n.
Explicit constraints:
xi Є {j | j is an integer and 1 ≤ j ≤ n}.
Implicit constraints:
No two xi can be the same.
The sum of the corresponding wi’s be m.
xi < xi+1 , 1 ≤ i < k (total order in indices) to avoid generating multiple
instances of the same subset (for example, (1, 2, 4) and (1, 4, 2)
represent the same subset).
A better formulation of the problem is where the solution subset is represented by an
n-tuple (x1, . . . . . , xn) such that xi Є {0, 1}.
The above solutions are then represented by (1, 1, 0, 1) and (0, 0, 1, 1).
For both the above formulations, the solution space is 2n distinct tuples.
For example, n = 4, w = (11, 13, 24, 7) and m = 31, the desired subsets are (11,
13, 7) and (24, 7).
The following figure shows a possible tree organization for two possible formulations
of the solution space for the case n = 4.
1
x 1 =1
2
x 2 =2
x 2 =3
6
x 3 =3
7
x 3 =4
12
x 2 =4
13 14
8
x 3 =4
x 1 =2
3
x 2 =3
9
x 1 =4
x 1 =3
4
x 2 =4
10
5
x 2 =4
11
S
x 3 =4
15
S
x 4 =4
16
A p o s s ib le s o lut io n s p ac e org a n is at io n f or t h e
s u m of t h e s u b s et s pro ble m.
The tree corresponds to the variable tuple size formulation. The edges are labeled
such that an edge from a level i node to a level i+1 node represents a value for xi. At
each node, the solution space is partitioned into sub - solution spaces. All paths from
the root node to any node in the tree define the solution space, since any such path
48 corresponds to a subset satisfying the explicit constraints.
The possible paths are (1), (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 4), (1, 3, 4), (2), (2,
3), and so on. Thus, the left mot sub-tree defines all subsets containing w1, the next
sub-tree defines all subsets containing w2 but not w1, and so on.
49 Graph Coloring (for planar graphs):
Let G be a graph and m be a given positive integer. We want to discover whether the
nodes of G can be colored in such a way that no two adjacent nodes have the same
color, yet only m colors are used. This is termed the m-colorabiltiy decision problem.
The m-colorability optimization problem asks for the smallest integer m for which the
graph G can be colored.
Given any map, if the regions are to be colored in such a way that no two adjacent
regions have the same color, only four colors are needed.
For many years it was known that five colors were sufficient to color any map, but no
map that required more than four colors had ever been found. After several hundred
years, this problem was solved by a group of mathematicians with the help of a
computer. They showed that in fact four colors are sufficient for planar graphs.
The function m-coloring will begin by first assigning the graph to its adjacency matrix,
setting the array x [] to zero. The colors are represented by the integers 1, 2, . . . , m
and the solutions are given by the n-tuple (x1, x2, . . ., xn), where xi is the color of
node i.
A recursive backtracking algorithm for graph coloring is carried out by invoking the
statement mcoloring(1);
Algorithm mcoloring (k)
// This algorithm was formed using the recursive backtracking schema. The graph is
// represented by its Boolean adjacency matrix G [1: n, 1: n]. All assignments of
// 1, 2, .......... , m to the vertices of the graph such that adjacent vertices are assigned
// distinct integers are printed. k is the index of the next vertex to color.
{
repeat
{
// Generate all legal assignments for x[k].
NextValue (k);
// Assign to x [k] a legal color.
If (x [k] = 0) then return;
// No new color possible
If (k = n) then
// at most m colors have been
// used to color the n vertices.
write (x [1: n]);
else mcoloring (k+1);
} until (false);
}
Algorithm NextValue (k)
// x [1] , .........x [k-1] have been assigned integer values in the range [1, m] such that
// adjacent vertices have distinct integers. A value for x [k] is determined in the range
// [0, m].x[k] is assigned the next highest numbered color while maintaining distinctness
// from the adjacent vertices of vertex k. If no such color exists, then x [k] is 0.
{
repeat
{
x [k]: = (x [k] +1) mod (m+1)
// Next highest color.
If (x [k] = 0) then return;
// All colors have been used
for j := 1 to n do
{
// check if this color is distinct from adjacent colors
if ((G [k, j] 0) and (x [k] = x [j]))
// If (k, j) is and edge and if adj. vertices have the same color.
then break;
50 }
if (j = n+1) then return;
} until (false);
// New color found
// Otherwise try to find another color.
}
Example:
Color the graph given below with minimum number of colors by
backtracking using state space tree
1
1
2
4
3
2
3
1
1
3
1
2
2
x1
3
2
3
3
1
1
2
3
2
x2
2
1
3
Gra p h
2
3
2
2
3
1
3
3
1
3
1
3
1
1
2
2
1
x4
2
A 4- n o d e gra p h a n d a l l p o s s ib le 3-c o lorin g s
Hamiltonian Cycles:
Let G = (V, E) be a connected graph with n vertices. A Hamiltonian cycle
(suggested by William Hamilton) is a round-trip path along n edges of G
that visits every vertex once and returns to its starting position. In other
vertices of G are visited in the order v1, v2, . . . . . , vn+1, then the edges
(vi, vi+1) are in E, 1 < i < n, and the vi are distinct expect for v1 and vn+1,
which are equal. The graph G1 contains the Hamiltonian cycle 1, 2, 8, 7, 6,
5, 4, 3, 1. The graph G2 contains no Hamiltonian cycle.
1
2
3
4
1
8
7
6
5
5
Graph G1
2
3
4
Graph G2
Two graphs to illustrate Hamiltonian cycle
The backtracking solution vector (x1, . . . . . xn) is defined so that xi
represents the ith visited vertex of the proposed cycle. If k = 1, then x 1 can
be any of the n vertices. To avoid printing the same cycle n times, we
require that x1 = 1. If 1 < k < n, then xk can be any vertex v that is
distinct from x1, x2, . . . , xk–1 and v is connected by an edge to kx-1. The
51
x3 vertex xn can only be one remaining vertex and it must be connected to
both xn-1 and x1.
Using NextValue algorithm we can particularize the recursive backtracking
schema to find all Hamiltonian cycles. This algorithm is started by first
initializing the adjacency matrix G[1: n, 1: n], then setting x[2: n] to zero
and x[1] to 1, and then executing Hamiltonian(2).
The traveling salesperson problem using dynamic programming asked for a
tour that has minimum cost. This tour is a Hamiltonian cycles. For the
simple case of a graph all of whose edge costs are identical, Hamiltonian
will find a minimum-cost tour if a tour exists.
Algorithm NextValue (k)
// x [1: k-1] is a path of k – 1 distinct vertices . If x[k] = 0, then no vertex has as yet been
// assigned to x [k]. After execution, x[k] is assigned to the next highest numbered vertex
// which does not already appear in x [1 : k – 1] and is connected by an edge to x [k – 1].
// Otherwise x [k] = 0. If k = n, then in addition x [k] is connected to x [1].
{
repeat
x { [k] := (x [k] +1) mod (n+1);
// Next vertex. If (x
[k] = 0) then return;
If (G [x [k – 1], x [k]] 0) then
{
// Is there an edge?
for j := 1 to k – 1 do if (x [j] = x [k]) then break;
// check for distinctness.
If (j = k) then
// If true, then the vertex is distinct. If
((k < n) or ((k = n) and G [x [n], x [1]] 0))
then return;
}
} until (false);
}
Algorithm Hamiltonian (k)
// This algorithm uses the recursive formulation of backtracking to find all the Hamiltonian
// cycles of a graph. The graph is stored as an adjacency matrix G [1: n, 1: n]. All cycles begin
// at node 1.
{
repeat
{
// Generate values for x [k].
NextValue (k);
//Assign a legal Next
value to x [k]. if (x [k] = 0) then return;
if (k = n) then write
(x [1: n]); else
Hamiltonian (k + 1)
} until (false)
52 UNIT III:
Greedy method- General method, applications- Knapsack problem, Job sequencing with
deadlines, Minimum cost spanning trees, Single source shortest path problem.
Greedy Method
GENERAL METHOD
Greedy is the most straight forward design technique. Most of the problems have n
inputs and require us to obtain a subset that satisfies some constraints. Any subset
that satisfies these constraints is called a feasible solution. We need to find a feasible
solution that either maximizes or minimizes the objective function. A feasible solution
that does this is called an optimal solution.
The greedy method is a simple strategy of progressively building up a solution, one
element at a time, by choosing the best possible element at each stage. At each stage,
a decision is made regarding whether or not a particular input is in an optimal solution.
This is done by considering the inputs in an order determined by some selection
procedure. If the inclusion of the next input, into the partially constructed optimal
solution will result in an infeasible solution then this input is not added to the partial
solution. The selection procedure itself is based on some optimization measure. Several
optimization measures are plausible for a given problem. Most of them, however, will
result in algorithms that generate sub-optimal solutions. This version of greedy
technique is called subset paradigm. Some problems like Knapsack, Job sequencing
with deadlines and minimum cost spanning trees are based on subset paradigm.
For the problems that make decisions by considering the inputs in some order, each
decision is made using an optimization criterion that can be computed using decisions
already made. This version of greedy method is ordering paradigm. Some problems like
optimal storage on tapes, optimal merge patterns and single source shortest path are
based on ordering paradigm.
CONTROL ABSTRACTION
Algorithm Greedy (a, n)
// a(1 : n) contains the ‘n’ inputs
{
solution := ;
// initialize the solution to empty
for i:=1 to n do
{
x := select (a);
if feasible (solution, x) then
solution := Union (Solution, x);
}
return solution;
}
Procedure Greedy describes the essential way that a greedy based algorithm will look,
once a particular problem is chosen and the functions select, feasible and union are
properly implemented.
The function select selects an input from ‘a’, removes it and assigns its value to ‘x’.
Feasible is a Boolean valued function, which determines if ‘x’ can be included into the
solution vector. The function Union combines ‘x’ with solution and updates the objective
function.
53 KNAPSACK PROBLEM:
Let us apply the greedy method to solve the knapsack problem. We are given ‘n’
objects and a knapsack. The object ‘i’ has a weight wi and the knapsack has a capacity
‘m’. If a fraction xi, 0 < xi < 1 of object i is placed into the knapsack then a profit of p i
xi is earned. The objective is to fill the knapsack that maximizes the total profit earned.
Since the knapsack capacity is ‘m’, we require the total weight of all chosen objects to
be at most ‘m’. The problem is stated as:
n
maximize
p x
i
i
i 1
n
subject to
a
i
xi M
where, 0 < xi < 1 and 1 < i < n
i 1
The profits and weights are positive numbers.
If the objects are already been sorted into non-increasing order of p[i] / w[i] then the
algorithm given below obtains solutions corresponding to this strategy.
Algorithm GreedyKnapsack (m, n)
// P[1 : n] and w[1 : n] contain the profits and weights respectively of
// Objects ordered so that p[i] / w[i]> p[i + 1] / w[i + 1].
// m is the knapsack size and x[1: n] is the solution vector.
{
for i := 1 to n do x[i] := 0.0
U := m;
for i := 1 to n do
{
if (w(i) > U) then break;
x [i] := 1.0; U := U – w[i];
}
if (i < n) then x[i] := U / w[i];
// initialize x
}
Running time:
The objects are to be sorted into non-decreasing order of pi / wi ratio. But if we
disregard the time to initially sort the objects, the algorithm requires only O(n) time.
Example:
Consider the following instance of the knapsack problem: n = 3, m = 20, (p 1, p2, p3) =
(25, 24, 15) and (w1, w2, w3) = (18, 15, 10).
54 1. First, we try to fill the knapsack by selecting the objects in some order:
x1
x2
x3
wi xi
pi xi
1/2
1/3
1/4
18 x 1/2 + 15 x 1/3 + 10 x 1/4
= 16.5
25 x 1/2 + 24 x 1/3 + 15 x 1/4 =
24.25
2. Select the object with the maximum profit first (p = 25). So, x1 = 1 and profit
earned is 25. Now, only 2 units of space is left, select the object with next largest
profit (p = 24). So, x2 = 2/15
x1
x2
x3
wi xi
pi xi
1
2/15
0
18 x 1 + 15 x 2/15 = 20
25 x 1 + 24 x 2/15 = 28.2
3. Considering the objects in the order of non-decreasing weights wi.
x1
x2
x3
wi xi
pi xi
0
2/3
1
15 x 2/3 + 10 x 1 = 20
24 x 2/3 + 15 x 1 = 31
4. Considered the objects in the order of the ratio p i / wi .
p1/w1
p2/w2
p3/w3
25/18
24/15
15/10
1.4
1.6
1.5
Sort the objects in order of the non-increasing order of the ratio pi / xi. Select the
object with the maximum pi / xi ratio, so, x2 = 1 and profit earned is 24. Now, only 5
units of space is left, select the object with next largest pi / xi ratio, so x3 = ½ and the
profit earned is 7.5.
wi xi
pi xi
x1
x2
x3
0
1
1/2
15 x 1 + 10 x 1/2 = 20
24 x 1 + 15 x 1/2 = 31.5
This solution is the optimal solution.
JOB SEQUENCING WITH DEADLINES:
When we are given a set of ‘n’ jobs. Associated with each Job i, deadline d i > 0 and
profit Pi > 0. For any job ‘i’ the profit pi is earned iff the job is completed by its
deadline. Only one machine is available for processing jobs. An optimal solution is the
feasible solution with maximum profit.
Sort the jobs in ‘j’ ordered by their deadlines. The array d [1 : n] is used to store the
deadlines of the order of their p-values. The set of jobs j [1 : k] such that j [r], 1 ≤ r ≤
k are the jobs in ‘j’ and d (j [1]) ≤ d (j[2]) ≤ . . . ≤ d (j[k]). To test whether J U {i} is
feasible, we have just to insert i into J preserving the deadline ordering and then verify
55 that d [J[r]] ≤ r, 1 ≤ r ≤ k+1.
Example:
Let n = 4, (P1, P2, P3, P4,) = (100, 10, 15, 27) and (d1 d2 d3 d4) = (2, 1, 2, 1). The
feasible solutions and their values are:
S. No
Feasible Solution
1
1,2
Procuring
sequence
2,1
Value
2
1,3
1,3 or 3,1
115
3
1,4
4,1
127
4
2,3
2,3
25
5
3,4
4,3
42
6
1
1
100
7
2
2
10
8
3
3
15
9
4
4
27
Remarks
110
OPTIMAL
The algorithm constructs an optimal set J of jobs that can be processed by their
deadlines.
Algorithm GreedyJob (d, J, n)
// J is a set of jobs that can be completed by their deadlines.
{
}
J := {1};
for i := 2 to n do
{
if (all jobs in J U {i} can be completed by their dead lines)
then J := J U {i};
}
Minimum Spanning Trees (MST):
A spanning tree for a connected graph is a tree whose vertex set is the same as the
vertex set of the given graph, and whose edge set is a subset of the edge set of the
given graph. i.e., any connected graph will have a spanning tree.
Weight of a spanning tree w (T) is the sum of weights of all edges in T. The Minimum
spanning tree (MST) is a spanning tree with the smallest possible weight.
56 G:
A gra p h G:
T hre e ( of ma n y p o s s ib le) s p a n n in g t re e s fro m gra p h G:
2
2
4
G:
3
3
5
6
1
A w e ig ht e d gra p h G:
1
T h e min i ma l s p a n n in g t re e fro m w e ig ht e d gra p h G:
Here are some examples:
To explain further upon the Minimum Spanning Tree, and what it applies to, let's
consider a couple of real-world examples:
1. One practical application of a MST would be in the design of a network. For
instance, a group of individuals, who are separated by varying distances, wish
to be connected together in a telephone network. Although MST cannot do
anything about the distance from one connection to another, it can be used to
determine the least cost paths with no cycles in this network, thereby
connecting everyone at a minimum cost.
2. Another useful application of MST would be finding airline routes. The vertices of
the graph would represent cities, and the edges would represent routes between
the cities. Obviously, the further one has to travel, the more it will cost, so MST
can be applied to optimize airline routes by finding the least costly paths with no
cycles.
To explain how to find a Minimum Spanning Tree, we will look at two algorithms: the
Kruskal algorithm and the Prim algorithm. Both algorithms differ in their methodology,
but both eventually end up with the MST. Kruskal's algorithm uses edges, and Prim’s
algorithm uses vertex connections in determining the MST.
Kruskal’s Algorithm
This is a greedy algorithm. A greedy algorithm chooses some local optimum (i.e.
picking an edge with the least weight in a MST).
Kruskal's algorithm works as follows: Take a graph with 'n' vertices, keep on adding the
shortest (least cost) edge, while avoiding the creation of cycles, until (n - 1) edges
have been added. Sometimes two or more edges may have the same cost. The order in
which the edges are chosen, in this case, does not matter. Different MSTs may result,
but they will all have the same total cost, which will always be the minimum cost.
57 The algorithm for finding the MST, using the Kruskal’s method is as follows:
Algorithm Kruskal (E, cost, n, t)
// E is the set of edges in G. G has n vertices. cost [u, v] is the
// cost of edge (u, v). ‘t’ is the set of edges in the minimum-cost spanning tree.
// The final cost is returned.
{
Construct a heap out of the edge costs using heapify;
for i := 1 to n do parent [i] := -1;
// Each vertex is in a different set.
i := 0; mincost := 0.0;
while ((i < n -1) and (heap not empty)) do
{
Delete a minimum cost edge (u, v) from the heap and
re-heapify using Adjust;
j := Find (u); k := Find (v);
if (j k) then
{
i := i + 1;
t [i, 1] := u; t [i, 2] := v;
mincost :=mincost + cost [u, v];
Union (j, k);
}
}
if (i n-1) then write ("no spanning tree");
else return mincost;
}
Running time:
The number of finds is at most 2e, and the number of unions at most n-1.
Including the initialization time for the trees, this part of the algorithm has a
complexity that is just slightly more than O (n + e).
We can add at most n-1 edges to tree T. So, the total time for operations on T is
O(n).
Summing up the various components of the computing times, we get O (n + e log e) as
asymptotic complexity
Example 1:
1
10
50
2
45
40
30
35
25
4
55
20
6
3
5
15
58 Arrange all the edges in the increasing order of their costs:
Cost
Edge
10
(1, 2)
15
(3, 6)
20
(4, 6)
25
(2, 6)
30
(1, 4)
35
(3, 5)
40
(2, 5)
45
(1, 5)
50
(2, 3)
The edge set T together with the vertices of G define a graph that has up to n
connected components. Let us represent each component by a set of vertices in it.
These vertex sets are disjoint. To determine whether the edge (u, v) creates a cycle,
we need to check whether u and v are in the same vertex set. If so, then a cycle is
created. If not then no cycle is created. Hence two Finds on the vertex sets suffice.
When an edge is included in T, two components are combined into one and a union is
to be performed on the two sets.
Edge
Cost
Spanning Forest
Edge Sets
Remarks
1
2
3
4
5
6
{1}, {2}, {3},
{4}, {5}, {6}
3
4
5
6
{1, 2}, {3}, {4},
{5}, {6}
The vertices 1 and
2 are in different
sets, so the edge
is combined
5
{1, 2}, {3,
{4}, {5}
The vertices 3 and
6 are in different
sets, so the edge
is combined
(1, 2)
10
1
2
(3, 6)
15
1
2
3
4
6},
6
(4, 6)
20
1
2
3
4
(2, 6)
25
1
5
{1, 2}, {3, 4, 6},
{5}
The vertices 4 and
6 are in different
sets, so the edge
is combined
5
{1, 2, 3, 4,
{5}
The vertices 2 and
6 are in different
sets, so the edge
is combined
6
2
6},
3
4
6
(1, 4)
30
(3, 5)
35
The vertices 1 and
4 are in the same
set, so the edge is
rejected
Reject
1
2
4
5
{1, 2, 3, 4, 5, 6}
3
6
59
The vertices 3 and
5 are in the same
set, so the edge is
combined
55
(5, 6) MINIMUM-COST SPANNING TREES: PRIM'S ALGORITHM
A given graph can have many spanning trees. From these many spanning trees, we
have to select a cheapest one. This tree is called as minimal cost spanning tree.
Minimal cost spanning tree is a connected undirected graph G in which each edge is
labeled with a number (edge labels may signify lengths, weights other than costs).
Minimal cost spanning tree is a spanning tree for which the sum of the edge labels is as
small as possible
The slight modification of the spanning tree algorithm yields a very simple algorithm for
finding an MST. In the spanning tree algorithm, any vertex not in the tree but
connected to it by an edge can be added. To find a Minimal cost spanning tree, we
must be selective - we must always add a new vertex for which the cost of the new
edge is as small as possible.
This simple modified algorithm of spanning tree is called prim's algorithm for finding an
Minimal cost spanning tree.
Prim's algorithm is an example of a greedy algorithm.
Algorithm Prim (E, cost, n, t)
// E is the set of edges in G. cost [1:n, 1:n] is the cost
// adjacency matrix of an n vertex graph such that cost [i, j] is
// either a positive real number or if no edge (i, j) exists.
// A minimum spanning tree is computed and stored as a set of
// edges in the array t [1:n-1, 1:2]. (t [i, 1], t [i, 2]) is an edge in
// the minimum-cost spanning tree. The final cost is returned.
{
Let (k, l) be an edge of minimum cost in E;
mincost := cost [k, l];
t [1, 1] := k; t [1, 2] := l;
for i :=1 to n do
// Initialize near
if (cost [i, l] < cost [i, k]) then near [i] := l;
else near [i] := k;
near [k] :=near [l] := 0;
for i:=2 to n - 1 do
// Find n - 2 additional edges for t.
{
Let j be an index such that near [j] 0 and
cost [j, near [j]] is minimum;
t [i, 1] := j; t [i, 2] := near [j];
mincost := mincost + cost [j, near [j]];
near [j] := 0
for k:= 1 to n do
// Update near[].
if ((near [k] 0) and (cost [k, near [k]] > cost [k, j]))
then near [k] := j;
}
return mincost;
}
60 Example:
Considering the following graph, find the minimal spanning tree using prim’s algorithm.
1
4
2
8
4
4
9
5
3
1
3
4
3
The cost adjacent matrix
4 9 8
4 4 1
is 9 4 3 3
8 1 3 4
3 4
The minimal spanning tree obtained as:
Vertex 1
Vertex 2
2
4
3
4
5
3
1
2
1
4
4
1
5
3
2
3
3
The cost of Minimal spanning tree = 11.
The steps as per the algorithm are as follows:
Algorithm near (J) = k means, the nearest vertex to J is k.
The algorithm starts by selecting the minimum cost from the graph. The minimum cost
edge is (2, 4).
K = 2, l = 4
Min cost = cost (2, 4) = 1
T [1, 1] = 2
T [1, 2] = 4
61 for i = 1 to 5
Near matrix
Edges added to min spanning
tree:
Begin
i=1
is cost (1, 4) < cost (1, 2)
8 < 4, No
Than near (1) = 2
i=2
is cost (2, 4) < cost (2, 2)
1 < , Yes
So near [2] = 4
i=3
is cost (3, 4) < cost (3, 2)
1 < 4, Yes
So near [3] = 4
i=4
is cost (4, 4) < cost (4, 2)
< 1, no
So near [4] = 2
i=5
is cost (5, 4) < cost (5, 2)
4 < , yes
So near [5] = 4
end
near [k] = near [l] = 0
near [2] = near[4] = 0
T [1, 1] = 2
T [1, 2] = 4
2
1
2
2
4
1
2
1
2
4
2
3
4
5
3
4
5
3
4
5
4
2
4
4
2
1
2
3
4
5
2
4
4
2
4
1
2
3
4
5
2
0
4
0
4
3
4
5
1
2
for i = 2 to n-1 (4) do
i=2
for j = 1 to 5
j=1
near(1)0 and cost(1, near(1))
2 0 and cost (1, 2) = 4
j=2
near (2) = 0
j=3
is near (3) 0
4 0 and cost (3, 4) = 3
62 j=4
near (4) = 0
J=5
Is near (5) 0
4 0 and cost (4, 5) = 4
select the min cost from the
above obtained costs, which is
3 and corresponding J = 3
min cost = 1 + cost(3, 4)
=1+3=4
T (2, 1) = 3
T (2, 2) = 4
T (2, 1) = 3
T (2, 2) = 4
2
Near [j] = 0
i.e. near (3) =0
1
0
2
0
0
3
4
4
5
for (k = 1 to n)
K=1
is near (1) 0, yes
2 0
and cost (1,2) > cost(1, 3)
4 > 9, No
K=2
Is near (2) 0, No
K=3
Is near (3) 0, No
K=4
Is near (4) 0, No
K=5
Is near (5) 0
4 0, yes
and is cost (5, 4) > cost (5, 3)
4 > 3, yes
than near (5) = 3
2
0
0
0
3
1
2
3
4
5
i=3
for (j = 1 to 5)
J=1
is near (1) 0
2 0
cost (1, 2) = 4
J=2
Is near (2) 0, No
63 J=3
Is near (3) 0, no
Near (3) = 0
J=4
Is near (4) 0, no
Near (4) = 0
J=5
Is near (5) 0
Near (5) = 3 3 0, yes
And cost (5, 3) = 3
Choosing the min cost from
the above obtaining costs
which is 3 and corresponding J
=5
T (3, 1) = 5
T (3, 2) = 3
Min cost = 4 + cost (5, 3)
=4+3=7
T (3, 1) = 5
T (3, 2) = 3
Near (J) = 0 near (5) = 0
for (k=1 to 5)
2
1
0
2
0
3
k=1
is near (1) 0, yes
and cost(1,2) > cost(1,5)
4 > , No
K=2
Is near (2) 0 no
K=3
Is near (3) 0 no
K=4
Is near (4) 0 no
K=5
Is near (5) 0 no
i=4
for J = 1 to 5
J=1
Is near (1) 0
2 0, yes
cost (1, 2) = 4
j=2
is near (2) 0, No
64
0
4
0
5 J=3
Is near (3) 0, No
Near (3) = 0
J=4
Is near (4) 0, No
Near (4) = 0
J=5
Is near (5) 0, No
Near (5) = 0
Choosing min cost from the
above it is only '4' and
corresponding J = 1
Min cost = 7 + cost (1,2)
= 7+4 = 11
0
0
0
0
0
T (4, 1) = 1
T (4, 2) = 2
1
2
3
4
5
T (4, 1) = 1
T (4, 2) = 2
Near (J) = 0 Near (1) = 0
for (k = 1 to 5)
K=1
Is near (1) 0, No
K=2
Is near (2) 0, No
K=3
Is near (3) 0, No
K=4
Is near (4) 0, No
K=5
Is near (5) 0, No
End.
The Single Source Shortest-Path Problem: DIJKSTRA'S ALGORITHMS:
In the previously studied graphs, the edge labels are called as costs, but here we think
them as lengths. In a labeled graph, the length of the path is defined to be the sum of
the lengths of its edges.
In the single source, all destinations, shortest path problem, we must find a shortest
path from a given source vertex to each of the vertices (called destinations) in the
graph to which there is a path.
Dijkstra’s algorithm is similar to prim's algorithm for finding minimal spanning trees.
Dijkstra’s algorithm takes a labeled graph and a pair of vertices P and Q, and finds the
65 shortest path between then (or one of the shortest paths) if there is more than one.
The principle of optimality is the basis for Dijkstra’s algorithms.
Dijkstra’s algorithm does not work for negative edges at all.
The figure lists the shortest paths from vertex 1 for a five vertex weighted digraph.
8
1
4
5
2
4
2
3
4
1
5
3
0
1
2
1
3
3
1
3
4
1
2
6
1
3
4
Graph
4
5
Shortest Paths
Algorithm:
Algorithm Shortest-Paths (v, cost, dist, n)
// dist [j], 1 < j < n, is set to the length of the shortest path
// from vertex v to vertex j in the digraph G with n vertices.
// dist [v] is set to zero. G is represented by its
// cost adjacency matrix cost [1:n, 1:n].
{
for i :=1 to n do
{
S [i] := false;
// Initialize S.
dist [i] :=cost [v, i];
}
S[v] := true; dist[v] := 0.0;
// Put v in S.
for num := 2 to n – 1 do
{
Determine n - 1 paths from v.
Choose u from among those vertices not in S such that dist[u] is minimum;
S[u] := true;
// Put u is S.
for (each w adjacent to u with S [w] = false) do
if (dist [w] > (dist [u] + cost [u, w]) then
// Update distances
dist [w] := dist [u] + cost [u, w];
}
}
Running time:
Depends on implementation of data structures for dist.
Build a structure with n elements
A
at most m = E times decrease the value of an item
‘n’ times select the smallest value
For array A = O (n); B = O (1); C = O (n) which gives O (n 2) total.
For heap A = O (n); B = O (log n); C = O (log n) which gives O (n + m log n)
mB
nC
total.
66 UNIT IV:
Dynamic Programming: General method, applications-Matrix chain multiplication, Optimal
binary search trees, 0/1 knapsack problem, All pairs shortest path problem, Travelling sales
person problem, Reliability design.
Dynamic Programming
Dynamic programming is a name, coined by Richard Bellman in 1955. Dynamic
programming, as greedy method, is a powerful algorithm design technique that can
be used when the solution to the problem may be viewed as the result of a sequence
of decisions. In the greedy method we make irrevocable decisions one at a time,
using a greedy criterion. However, in dynamic programming we examine the decision
sequence to see whether an optimal decision sequence contains optimal decision
subsequence.
When optimal decision sequences contain optimal decision subsequences, we can
establish recurrence equations, called dynamic-programming recurrence equations,
that enable us to solve the problem in an efficient way.
Dynamic programming is based on the principle of optimality (also coined by
Bellman). The principle of optimality states that no matter whatever the initial state
and initial decision are, the remaining decision sequence must constitute an optimal
decision sequence with regard to the state resulting from the first decision. The
principle implies that an optimal decision sequence is comprised of optimal decision
subsequences. Since the principle of optimality may not hold for some formulations
of some problems, it is necessary to verify that it does hold for the problem being
solved. Dynamic programming cannot be applied when this principle does not hold.
The steps in a dynamic programming solution are:
Verify that the principle of optimality holds
Set up the dynamic-programming recurrence equations
Solve the dynamic-programming recurrence equations for the value of the
optimal solution.
Perform a trace back step in which the solution itself is constructed.
Dynamic programming differs from the greedy method since the greedy method
produces only one feasible solution, which may or may not be optimal, while dynamic
programming produces all possible sub-problems at most once, one of which
guaranteed to be optimal. Optimal solutions to sub-problems are retained in a table,
thereby avoiding the work of recomputing the answer every time a sub-problem is
encountered
The divide and conquer principle solve a large problem, by breaking it up into smaller
problems which can be solved independently. In dynamic programming this principle
is carried to an extreme: when we don't know exactly which smaller problems to
solve, we simply solve them all, then store the answers away in a table to be used
later in solving larger problems. Care is to be taken to avoid recomputing previously
computed values, otherwise the recursive program will have prohibitive complexity.
In some cases, the solution can be improved and in other cases, the dynamic
67 programming technique is the best approach.
Two difficulties may arise in any application of dynamic programming:
1.
It may not always be possible to combine the solutions of smaller problems to
form the solution of a larger one.
2.
The number of small problems to solve may be un-acceptably large.
There is no characterized precisely which problems can be effectively solved with
dynamic programming; there are many hard problems for which it does not seen to
be applicable, as well as many easy problems for which it is less efficient than
standard algorithms.
All pairs shortest paths:
In the all pairs shortest path problem, we are to find a shortest path between every
pair of vertices in a directed graph G. That is, for every pair of vertices (i, j), we are
to find a shortest path from i to j as well as one from j to i. These two paths are the
same when G is undirected.
When no edge has a negative length, the all-pairs shortest path problem may be
solved by using Dijkstra’s greedy single source algorithm n times, once with each of
the n vertices as the source vertex.
The all pairs shortest path problem is to determine a matrix A such that A (i, j) is the
length of a shortest path from i to j. The matrix A can be obtained by solving n
single-source problems using the algorithm shortest Paths. Since each application of
this procedure requires O (n2) time, the matrix A can be obtained in O (n3) time.
The dynamic programming solution, called Floyd’s algorithm, runs in O (n 3) time. Floyd’s
algorithm works even when the graph has negative length edges (provided there are no
negative length cycles).
The shortest i to j path in G, i ≠ j originates at vertex i and goes through some
intermediate vertices (possibly none) and terminates at vertex j. If k is an
intermediate vertex on this shortest path, then the subpaths from i to k and from k
to j must be shortest paths from i to k and k to j, respectively. Otherwise, the i to j
path is not of minimum length. So, the principle of optimality holds. Let Ak (i, j)
represent the length of a shortest path from i to j going through no vertex of index
greater than k, we obtain:
Ak (i, j) = {min {min {Ak-1 (i, k) + Ak-1 (k, j)}, c (i, j)}
1 0 for all I and j and cij = if < i, j> E. Let |V| = n and assume n > 1. A
tour of G is a directed simple cycle that includes every vertex in V. The cost of a tour
is the sum of the cost of the edges on the tour. The traveling sales person problem is
to find a tour of minimum cost. The tour is to be a simple path that starts and ends
at vertex 1.
Let g (i, S) be the length of shortest path starting at vertex i, going through all
vertices in S, and terminating at vertex 1. The function g (1, V – {1}) is the length of
an optimal salesperson tour. From the principal of optimality it follows that:
g 1, V - 1
min
2 k n
c1k g k, V 1, k
--
1
--
2
Generalizing equation 1, we obtain (for i S)
g i, S
minci j
j S
g i, S j
The Equation can be solved for g (1, V – 1}) if we know g (k, V – {1, k}) for all
choices of k.
Example :
For the following graph find minimum cost tour for the traveling salesperson
problem:
1
3
2
4
0
5
The cost adjacency matrix =
6
8
71
10
15
0
9
13
0
8
9
20
10
12
0
Let us start the tour from vertex 1:
g (1, V – {1}) = min {c1k + g (k, V – {1, K})}
-
(1)
-
(2)
2 0, this equation generalizes to:
Fi (y) = max {fi-1 (y), fi-1 (y - wi) + pi}
Equation-2 can be solved for fn (m) by beginning with the knowledge fo (y) = 0 for all
y and fi (y) = - , y < 0. Then f1, f2, . . . fn can be successively computed using
equation–2.
When the wi’s are integer, we need to compute fi (y) for integer y, 0 < y < m. Since fi
(y) = - for y < 0, these function values need not be computed explicitly. Since
each fi can be computed from fi - 1 in Θ (m) time, it takes Θ (m n) time to compute
fn. When the wi’s are real numbers, fi (y) is needed for real numbers y such that 0 <
y < m. So, fi cannot be explicitly computed for all y in this range. Even when the w i’s
are integer, the explicit Θ (m n) computation of fn may not be the most efficient
computation. So, we explore an alternative method for both cases.
The fi (y) is an ascending step function; i.e., there are a finite number
of y’s, 0 = y1
< y2 < . . . . < yk, such that fi (y1) < fi (y2) < . . . . . < fi (yk); fi (y) = - , y < y1;
fi
(y) = f (yk), y > yk; and fi (y) = fi (yj), yj < y < yj+1. So, we need to compute only fi
(yj), 1 < j < k. We use the ordered set Si = {(f (yj), yj) | 1 < j < k} to represent fi
(y). Each number of Si is a pair (P, W), where P = fi (yj) and W = yj. Notice that S0 =
{(0, 0)}. We can compute Si+1 from Si by first computing:
Si1= {(P, W) | (P – p i, W – w i) Si}
Now, Si+1 can be computed by merging the pairs in Si and Si together.
Note that if
1
Si+1 contains two pairs (Pj, Wj) and (Pk, Wk) with the property that Pj < Pk and Wj >
Wk, then the pair (Pj, Wj) can be discarded because of equation-2. Discarding or
purging rules such as this one are also known as dominance rules. Dominated tuples
get purged. In the above, (Pk, Wk) dominates (Pj, Wj).
Example 1:
Consider the knapsack instance n = 3, (w1, w2, w3) = (2, 3, 4), (P1, P2, P3) = (1, 2,
5) and M = 6.
73 Solution:
Initially, fo (x) = 0, for all x and fi (x) = - if x < 0.
Fn (M) = max {fn-1 (M), fn-1 (M - wn) + pn}
F3 (6) = max (f2 (6), f2 (6 – 4) + 5} = max {f2 (6), f2 (2) + 5}
F2 (6) = max (f1 (6), f1 (6 – 3) + 2} = max {f1 (6), f1 (3) + 2}
F1 (6) = max (f0 (6), f0 (6 – 2) + 1} = max {0, 0 + 1} = 1
F1 (3) = max (f0 (3), f0 (3 – 2) + 1} = max {0, 0 + 1} = 1
Therefore, F2 (6) = max (1, 1 + 2} = 3
F2 (2) = max (f1 (2), f1 (2 – 3) + 2} = max {f1 (2), - + 2}
F1 (2) = max (f0 (2), f0 (2 – 2) + 1} = max {0, 0 + 1} = 1
F2 (2) = max {1, - + 2} = 1
Finally, f3 (6) = max {3, 1 + 5} = 6
Other Solution:
For the given data we have:
S0 = {(0, 0)}; S0 = {(1,1 2)}
S1 = (S0 U S01) = {(0, 0), (1, 2)}
X - 2 = 0 => x = 2.
X - 2 = 1 => x = 3.
y – 3 = 0 => y = 3
y – 3 = 2 => y = 5
S11 = {(2, 3), (3, 5)}
S2 = (S1 U S1 1) = {(0, 0), (1, 2), (2, 3), (3, 5)}
X
X
X
X
–
–
–
–
5
5
5
5
=
=
=
=
0
1
2
3
=>
=>
=>
=>
x
x
x
x
=
=
=
=
5.
6.
7.
8.
y–4=0
y–4=2
y–4=3
y–4=5
=> y = 4
=> y = 6
=> y = 7
=> y = 9
S21 = {(5, 4), (6, 6), (7, 7), (8, 9)}
S3 = (S2 U S21) = {(0, 0), (1, 2), (2, 3), (3, 5), (5, 4), (6, 6), (7, 7), (8, 9)}
By applying Dominance rule,
S3 = (S2 U S21) = {(0, 0), (1, 2), (2, 3), (5, 4), (6, 6)}
From (6, 6) we can infer that the maximum Profit pi xi = 6 and weight xi wi = 6
74 Reliability Design:
The problem is to design a system that is composed of several devices connected in
series. Let ri be the reliability of device Di (that is ri is the probability that device i
will function properly) then the reliability of the entire system is ri. Even if the
individual devices are very reliable (the ri’s are very close to one), the reliability of
the system may not be very good. For example, if n = 10 and ri = 0.99, i < i < 10,
then ri = .904. Hence, it is desirable to duplicate devices. Multiply copies of the
same device type are connected in parallel.
If stage i contains mimcopies of device Di. Then the probability that all mi have
mi a
i
malfunction is (1 - r ) . Hence the reliability of stage i becomes 1 – (1 - r) .
i
i
The reliability of stage ‘i’ is given by a function i (mi).
Our problem is to use device duplication. This maximization is to be carried out under
a cost constraint. Let ci be the cost of each unit of device i and let c be the maximum
allowable cost of the system being designed.
We wish to solve:
Maximize
1 i n
Subject
i mi
C m C
to
i
i
1 i n
mi > 1 and interger, 1 < i < n
75 Assume each Ci > 0, each mi must be in the range 1 < mi < ui, where
ui C Ci
n
1
CJ
Ci
The upper bound ui follows from the observation that mj > 1
An optimal solution m1, m2 ............... mn is the result of a sequence of decisions, one
decision for each mi.
Let fi (x) represent the maximum value of
mJ
1 j i
Subject to the constrains:
C
J
mJ x
and 1 < mj < uJ, 1 < j < i
1 j i
76 Example :
Design a three stage system with device types D1, D2 and D3. The costs are $30, $15
and $20 respectively. The Cost of the system is to be no more than $105. The
reliability of each device is 0.9, 0.8 and 0.5 respectively.
Solution:
We assume that if if stage I has mi devices of type i in parallel, then i (mi) =1 – (1ri)mi
Since, we can assume each ci > 0, each mi must be in the range 1 ≤ mi ≤ ui. Where:
u i C Ci
n
1
CJ
Ci
Using the above equation compute u1, u2 and u3.
u1
105 30 3015 20
30
10515 3015 20
u2
u3
We useS ji
15
70 2
30
55
15
3
105 20 3015 20 60
3
20
20
i:stage number and J: no. of devices in stage i mi
S o fo (x), x
initially fo x 1 and x 0, so, So 1, 0
Compute S1, S2 and S3 as follows:
S1 = depends on u1 value, as u1 = 2, so
S1
S , S
1
1
1
2
S2 = depends on u2 value, as u2 = 3, so
77 S2
S
2
, S 2 , S2
1
2
3
S3 = depends on u3 value, as u3 = 3, so
S3
Now
S , S
3
3
1
2
, S3
3
f (x), x
find,S11
1
f1 x 1 (1) fo
, 1 (2) f 0 ()} With devices m1 = 1 and m2 = 2
Compute 1 (1) and 1 (2) using the formula: i mi) 1 (1 ri )mi
1 1 1 1 r1m 1 = 1 – (1 – 0.9)1 = 0.9
1 2 1 1 0.92 0.99
S1 f1 x, x
1
0.9, 30
S12 0.99 , 30 30
0.99, 60
Therefore, S1 = {(0.9, 30), (0.99, 60)}
Next
findS 2
1
f (x), x
2
f2 (x) {2 1 * f1
2 1 1 1
, 2 2 * f1 , 2 3 * f1 }
rI mi
= 1 – (1 – 0.8)
= 1 – 0.2 = 0.8
1
2 2 1 1 0.8 2 0.96
2 3 1 1 0.8 3 0.992
S12 {(0.8(0.9),30 15), (0.8(0.99),60 15)} = {(0.72, 45), (0.792, 75)}
S22 {(0.96(0.9),30 15 15) , (0.96(0.99),60 15 15)}
= {(0.864, 60), (0.9504, 90)}
S32 {(0.992(0.9),30 15 1515) , (0.992(0.99),60 15 1515)}
= {(0.8928, 75), (0.98208, 105)}
S2
S
2
, S2 , S2
1
2
3
By applying Dominance rule to S2:
Therefore, S2 = {(0.72, 45), (0.864, 60), (0.8928, 75)}
78 Dominance Rule:
If Si contains two pairs (f1, x1) and (f2, x2) with the property that f1 ≥ f2 and x1 ≤ x2,
then (f1, x1) dominates (f2, x2), hence by dominance rule (f2, x2) can be discarded.
Discarding or pruning rules such as the one above is known as dominance rule.
Dominating tuples will be present in Si and Dominated tuples has to be discarded
from Si.
Case 1: if f1 ≤ f2 and x1 > x2 then discard (f1, x1)
Case 2: if f1 > f2 and x1 < x2 the discard (f2, x2)
Case 3: otherwise simply write (f1, x1)
S2 = {(0.72, 45), (0.864, 60), (0.8928, 75)}
3 1 1 1 rI
mi
= 1 – (1 – 0.5)1 = 1 – 0.5 = 0.5
3 2 1 1 0.5 2 0.75
3 3 1 1 0.5 3 0.875
S13 0.5 (0.72), 45 20, 0.5 (0.864), 60 20, 0.5 (0.8928), 75 20
S13 0.36, 65, 0.437, 80, 0.4464, 95
S23 {0.75 (0.72), 45 20 20, 0.75 (0.864), 60 20 20,
0.75 (0.8928), 75 20 20}
= {(0.54, 85), (0.648, 100), (0.6696, 115)}
S33 0.875 (0.72), 45 20 20 20, 0.875 (0.864), 60 20 20 20,
0.875 (0.8928), 75 20 20 20
S33 (0.63, 105), 1.756, 120, 0.7812, 135
If cost exceeds 105, remove that tuples
S3 = {(0.36, 65), (0.437, 80), (0.54, 85), (0.648, 100)}
The best design has a reliability of 0.648 and a cost of 100. Tracing back for the
solution through Si ‘s we can determine that m3 = 2, m2 = 2 and m1 = 1.
79 Optimal Binary Search Tree:
In computer science, an optimal binary search tree (Optimal BST), sometimes called a weight-balanced
binary tree,[1] is a binary search tree which provides the smallest possible search time (or expected search
time) for a given sequence of accesses (or access probabilities).
The no of external nodes are same in both trees.
80 The C (i, J) can be computed as:
C (i, J) = min {C (i, k-1) + C (k, J) + P (K) + w (i, K-1) + w (K, J)}
i g( y ) whenever y is a child of x, then the search is
essentially a D-search.
An LC-search coupled with bounding functions is called an LC-branch and bound
search
We associate a cost c(x) with each node x in the state space tree. It is not possible to
easily compute the function c(x). So we compute a estimate c( x ) of c(x).
Control Abstraction for LC-Search:
Let t be a state space tree and c() a cost function for the nodes in t. If x is a node in
t, then c(x) is the minimum cost of any answer node in the subtree with root x. Thus,
c(t) is the cost of a minimum-cost answer node in t.
A heuristic c(.) is used to estimate c(). This heuristic should be easy to compute and
generally has the property that if x
is either an answer node or a leaf node, then
c(x) = c( x ) .
LC-search uses c to find an answer node. The algorithm uses two functions Least() and
Add() to delete and add a live node from or to the list of live nodes, respectively.
Least() finds a live node with least c(). This node is deleted from the list of live nodes
and returned.
88 Add(x) adds the new live node x to the list of live nodes. The list of live nodes be
implemented as a min-heap.
Algorithm LCSearch outputs the path from the answer node it finds to the root node
t. This is easy to do if with each node x that becomes live, we associate a field parent
which gives the parent of node x. When the answer node g is found, the path from g
to t can be determined by following a sequence of parent values starting from the
current E-node (which is the parent of g) and ending at node t.
Listnode = record
{
Listnode * next, *parent; float cost;
}
Algorithm LCSearch(t)
{
//Search t for an answer node
if *t is an answer node then output *t and return;
E := t;
//E-node.
initialize the list of live nodes to be empty;
repeat
{
for each child x of E do
{
if x is an answer node then output the path from x to t and return;
Add (x);
//x is a new live node.
(x parent) := E;
// pointer for path to root
}
if there are no more live nodes then
{
write (“No answer node”);
return;
}
E := Least();
} until (false);
}
The root node is the first, E-node. During the execution of LC search, this list
contains all live nodes except the E-node. Initially this list should be empty.
Examine all the children of the E-node, if one of the children is an answer node, then
the algorithm outputs the path from x to t and terminates. If the child of E is not an
answer node, then it becomes a live node. It is added to the list of live nodes and its
parent field set to E. When all the children of E have been generated, E becomes a
dead node. This happens only if none of E’s children is an answer node. Continue the
search further until no live nodes found. Otherwise, Least(), by definition, correctly
chooses the next E-node and the search continues from here.
LC search terminates only when either an answer node is found or the entire state
space tree has been generated and searched.
Bounding:
A branch and bound method searches a state space tree using any search
mechanism in which all the children of the E-node are generated before another node
becomes the E-node. We assume that each answer node x has a cost c(x) associated
with it and that a minimum-cost answer node is to be found. Three common search
strategies are FIFO, LIFO, and LC. The three search methods differ only in the
selection rule used to obtain the next E-node.
89 A good bounding helps to prune efficiently the tree, leading to a faster exploration of
the solution space.
A cost function c(.) such that c( x ) < c(x) is used to provide lower bounds on
solutions obtainable from any node x. If upper is an upper bound on the cost of a
minimum-cost solution, then all live nodes x with c(x) > c( x ) > upper. The starting
value for upper can be obtained by some heuristic or can be set to .
As long as the initial value for upper is not less than the cost of a minimum-cost
answer node, the above rules to kill live nodes will not result in the killing of a live
node that can reach a minimum-cost answer node. Each time a new answer node is
found, the value of upper can be updated.
Branch-and-bound algorithms are used for optimization problems where, we deal
directly only with minimization problems. A maximization problem is easily converted
to a minimization problem by changing the sign of the objective function.
To formulate the search for an optimal solution for a least-cost answer node in a
state space tree, it is necessary to define the cost function c(.), such that c(x) is
minimum for all nodes representing an optimal solution. The easiest way to do this is
to use the objective function itself for c(.).
For nodes representing feasible solutions, c(x) is the value of the objective
function for that feasible solution.
For nodes representing infeasible solutions, c(x) = .
For nodes representing partial solutions, c(x) is the cost of the minimum-cost
node in the subtree with root x.
Since, c(x) is generally hard to compute, the branch-and-bound algorithm will use an
estimate c( x ) such that c( x ) < c(x) for all x.
FIFO Branch and Bound:
A FIFO branch-and-bound algorithm for the job sequencing problem can begin with
upper = as an upper bound on the cost of a minimum-cost answer node.
Starting with node 1 as the E-node and using the variable tuple size formulation of
Figure 8.4, nodes 2, 3, 4, and 5 are generated. Then u(2) = 19, u(3) = 14, u(4) =
18, and u(5) = 21.
The variable upper is updated to 14 when node 3 is generated. Since
c (4) and
c(5) are greater than upper, nodes 4 and 5 get killed. Only nodes 2 and 3 remain
alive.
Node 2 becomes the next E-node. Its children, nodes 6, 7 and 8 are generated.
Then u(6) = 9 and so upper is updated to 9. The cost c(7) = 10 > upper and node 7
gets killed. Node 8 is infeasible and so it is killed.
Next, node 3 becomes the E-node. Nodes 9 and 10 are now generated. Then u(9) =
8 and so upper becomes 8. The cost c(10) = 11 > upper, and this node is killed.
90 The next E-node is node 6. Both its children are infeasible. Node 9’s only child is also
infeasible. The minimum-cost answer node is node 9. It has a cost of 8.
When implementing a FIFO branch-and-bound algorithm, it is not economical to kill
live nodes with c(x) > upper each time upper is updated. This is so because live
nodes are in the queue in the order in which they were generated. Hence, nodes with
c(x) > upper are distributed in some random way in the queue. Instead, live nodes
with c(x) > upper can be killed when they are about to become E-nodes.
The FIFO-based branch-and-bound algorithm with an appropriate
called FIFOBB.
c(.) and u(.) is
LC Branch and Bound:
An LC Branch-and-Bound search of the tree of Figure 8.4 will begin with upper =
and node 1 as the first E-node.
When node 1 is expanded, nodes 2, 3, 4 and 5 are generated in that order.
As in the case of FIFOBB, upper is updated to 14 when node 3 is generated and
nodes 4 and 5 are killed as c(4) > upper and c(5) > upper.
Node 2 is the next E-node as c(2) = 0 and c(3) = 5. Nodes 6, 7 and 8 are generated
and upper is updated to 9 when node 6 is generated. So, node 7 is killed as c(7) = 10
> upper. Node 8 is infeasible and so killed. The only live nodes now are nodes 3 and
6.
Node 6 is the next E-node as c(6) = 0 < c(3) . Both its children are infeasible.
Node 3 becomes the next E-node. When node 9 is generated, upper is updated to 8
as u(9) = 8. So, node 10 with c(10) = 11 is killed on generation.
Node 9 becomes the next E-node. Its only child is infeasible. No live nodes remain.
The search terminates with node 9 representing the minimum-cost answer node.
2
3
The path = 1 3 9 = 5 + 3 = 8
Traveling Sale Person Problem:
By using dynamic programming algorithm we can solve the problem with time
complexity of O(n22n) for worst case. This can be solved by branch and bound
technique using efficient bounding function. The time complexity of traveling sale
person problem using LC branch and bound is O(n22n) which shows that there is no
change or reduction of complexity than previous method.
We start at a particular node and visit all nodes exactly once and come back to initial
node with minimum cost.
Let G = (V, E) is a connected graph. Let C(i, J) be the cost of edge . cij = if
E and let |V| = n, the number of vertices. Every tour starts at vertex 1 and
ends at the same vertex. So, the solution space is given by S = {1, , 1 | is a
91 permutation of (2, 3, . . . , n)} and |S| = (n – 1)!. The size of S can be reduced by
restricting S so that (1, i 1, i2, . . . . in-1, 1) S iff E, 0 < j < n - 1 and
i0 = in =1.
Procedure for solving traveling sale person problem:
1.
Reduce the given cost matrix. A matrix is reduced if every row and column is
reduced. A row (column) is said to be reduced if it contain at least one zero
and all-remaining entries are non-negative. This can be done as follows:
a)
b)
Row reduction: Take the minimum element from first row, subtract it
from all elements of first row, next take minimum element from the
second row and subtract it from second row. Similarly apply the same
procedure for all rows.
Find the sum of elements, which were subtracted from rows.
c)
Apply column reductions for the matrix obtained after row reduction.
Column reduction: Take the minimum element from first column,
subtract it from all elements of first column, next take minimum
element from the second column and subtract it from second column.
Similarly apply the same procedure for all columns.
d)
Find the sum of elements, which were subtracted from columns.
e)
Obtain the cumulative sum of row wise reduction and column wise
reduction.
Cumulative reduced sum = Row wise reduction sum + column wise
reduction sum.
Associate the cumulative reduced sum to the starting state as lower
bound and as upper bound.
2.
Calculate the reduced cost matrix for every node R. Let A is the reduced cost
matrix for node R. Let S be a child of R such that the tree edge (R, S)
corresponds to including edge in the tour. If S is not a leaf node, then
the reduced cost matrix for S may be obtained as follows:
a)
Change all entries in row i and column j of A to .
b)
Set A (j, 1) to .
c)
Reduce all rows and columns in the resulting matrix except for rows
and column containing only . Let r is the total amount subtracted to
reduce the matrix.
c)
3.
c S cR A i, j r, where ‘r’ is the total amount
subtracted to reduce the matrix, cR indicates the lower bound of the
ith node in (i, j) path and c S is called the cost function.
Find
Repeat step 2 until all nodes are visited.
92 Example:
Find the LC branch and bound solution for the traveling sale person problem whose
cost matrix is as follows:
20
15
The cost matrix is 3 5
19 6
16 4
30 10 11
16 4 2
2 4
18
3
7
16
Step 1: Find the reduced cost matrix.
Apply row reduction method:
Deduct
Deduct
Deduct
Deduct
Deduct
10 (which is the minimum) from all values in the 1 st row.
2 (which is the minimum) from all values in the 2nd row.
2 (which is the minimum) from all values in the 3rd row.
3 (which is the minimum) from all values in the 4th row.
4 (which is the minimum) from all values in the 5th row.
10 20
13
14
The resulting row wise reduced cost matrix = 1 3
3 15
16
12 0 3
0
2
0
12
1
0
0
0
Row wise reduction sum = 10 + 2 + 2 + 3 + 4 = 21
Now apply column reduction for the above matrix:
Deduct 1 (which is the minimum) from all values in the 1st column.
Deduct 3 (which is the minimum) from all values in the 3rd column.
10 17 0
12
11 2
The resulting column wise reduced cost matrix (A) = 0 3 0
3 12
15
11 0 0 12
1
0
2
0
Column wise reduction sum = 1 + 0 + 3 + 0 + 0 = 4
Cumulative reduced sum = row wise reduction + column wise reduction sum.
= 21 + 4 = 25.
This is the cost of a root i.e., node 1, because this is the initially reduced cost matrix.
The lower bound for node is 25 and upper bound is .
93 Starting from node 1, we can next visit 2, 3, 4 and 5 vertices. So, consider to explore
the paths (1, 2), (1, 3), (1, 4) and (1, 5).
The tree organization up to this point is as follows:
1
i=2
i=3
2
3
U =
L = 25
i = 4
i= 5
4
5
Variable ‘i’ indicates the next node to visit.
Step 2:
Consider the path (1, 2):
Change all entries of row 1 and column 2 of A to and also set A(2, 1) to .
0
15
11
11 2
0
12
0 12
0
2
0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
Then the resultant matrix is 0
15
11
11 2
0
12
0 12
0
2
0
Row reduction sum = 0 + 0 + 0 + 0 = 0
Column reduction sum = 0 + 0 + 0 + 0 = 0
Cumulative reduction (r) = 0 + 0 = 0
Therefore, as
cS cR A 1, 2 r
c S = 25 + 10 + 0 = 35
Consider the path (1, 3):
Change all entries of row 1 and column 3 of A to and also set A(3, 1) to .
94
12 2
3 0
15 3
11 0 12
0
2
0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
Then the resultant matrix is
4
0
1 2 0
3 0 2
3 0
0 12
Row reduction sum = 0
Column reduction sum = 11
Cumulative reduction (r) = 0 + 11 = 11
Therefore, as
cS cR A 1, 3 r
c S = 25 + 17 + 11 = 53
Consider the path (1, 4):
Change all entries of row 1 and column 4 of A to and also set A(4, 1) to .
12
0
11
11 0
2
3
3 12 0
0 0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
12
Then the resultant matrix is 0
11
11 0
2
3
3 12 0
0 0
Row reduction sum = 0
Column reduction sum = 0
Cumulative reduction (r) = 0 + 0 = 0
95 Therefore, as
cS cR A 1, 4 r
c S = 25 + 0 + 0 = 25
Consider the path (1, 5):
Change all entries of row 1 and column 5 of A to and also set A(5, 1) to .
12
0 3
15 3
0
11 2
0
12
0 12
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
10
Then the resultant matrix is 0
12
9 0
3 0
0 9
0 0 12
Row reduction sum = 5
Column reduction sum = 0
Cumulative reduction (r) = 5 + 0 = 0
Therefore, as
cS cR A 1, 5 r
c S = 25 + 1 + 5 = 31
The tree organization up to this point is as follows:
U =
L = 25
1
i=2
i=3
35
2
53
3
i = 4
25
i = 2
6
4
i=3
7
i= 5
5
31
i= 5
8
The cost of the paths between (1, 2) = 35, (1, 3) = 53, (1, 4) = 25 and (1, 5) = 31.
The cost of the path between (1, 4) is minimum. Hence the matrix obtained for path
(1, 4) is considered as reduced cost matrix.
96
12
A = 0
11
11 0
3 2
3 12 0
0 0
The new possible paths are (4, 2), (4, 3) and (4, 5).
Consider the path (4, 2):
Change all entries of row 4 and column 2 of A to and also set A(2, 1) to .
0
11
11 0
2
0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
Then the resultant matrix is 0
11
11 0
2
0
Row reduction sum = 0
Column reduction sum = 0
Cumulative reduction (r) = 0 + 0 = 0
Therefore, as
cS cR A 4, 2 r
c S = 25 + 3 + 0 = 28
Consider the path (4, 3):
Change all entries of row 4 and column 3 of A to and also set A(3, 1) to .
12
3
11 0
0
2
97 Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
1
Then the resultant matrix is 1
0 0
0
0
Row reduction sum = 2
Column reduction sum = 11
Cumulative reduction (r) = 2 + 11 = 13
Therefore, as
cS cR A 4, 3 r
c S = 25 + 12 + 13 = 50
Consider the path (4, 5):
Change all entries of row 4 and column 5 of A to and also set A(5, 1) to .
12
0 3
0
11
0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
1 0
Then the resultant matrix is 0 3
0 0
Row reduction sum = 11
Column reduction sum = 0
Cumulative reduction (r) = 11+0 = 11
Therefore, as
c S cR A 4, 5 r
c S = 25 + 0 + 11 = 36
98 The tree organization up to this point is as follows:
U =
L = 25
1
i=2
i=3
35
2
53
3
i = 4
25
i = 2
i=3
9
4
31
5
i= 5
i=3
6
28
i= 5
7
8
50
36
i=5
10
The cost of the paths between (4, 2) = 28, (4, 3) = 50 and (4, 5) = 36. The cost of
the path between (4, 2) is minimum. Hence the matrix obtained for path (4, 2) is
considered as reduced cost matrix.
A = 0
11
11 0
2
0
The new possible paths are (2, 3) and (2, 5).
Consider the path (2, 3):
Change all entries of row 2 and column 3 of A to and also set A(3, 1) to .
11
2
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
99
Then the resultant matrix is
0
0
Row reduction sum = 2
Column reduction sum = 11
Cumulative reduction (r) = 2 + 11 = 13
Therefore, as
cS cR A 2, 3 r
c S = 28 + 11 + 13 = 52
Consider the path (2, 5):
Change all entries of row 2 and column 5 of A to and also set A(5, 1) to .
0
0
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
Then the resultant matrix is 0
Row reduction sum = 0
Column reduction sum = 0
Cumulative reduction (r) = 0 + 0 = 0
Therefore, as
0
cS cR A 2, 5 r
c S = 28 + 0 + 0 = 28
The tree organization up to this point is as follows:
100 U =
L = 25
1
i=2
35
2
3
53
5
31
i= 5
i=3
6
i=3
52
4
25
i = 2
28
i= 5
i = 4
i=3
7
8
36
50
i=5
9
10
28
i= 3
11
The cost of the paths between (2, 3) = 52 and (2, 5) = 28. The cost of the path
between (2, 5) is minimum. Hence the matrix obtained for path (2, 5) is considered
as reduced cost matrix.
A = 0
0
The new possible paths is (5, 3).
Consider the path (5, 3):
Change all entries of row 5 and column 3 of A to and also set A(3, 1) to .
Apply row and column reduction for the rows and columns whose rows and
columns are not completely .
Then the resultant matrix is
Row reduction sum = 0
Column reduction sum = 0
Cumulative reduction (r) = 0 + 0 = 0
Therefore, as
cS cR A 5, 3 r
c S = 28 + 0 + 0 = 28
The overall tree organization is as follows:
101 U =
L = 25
1
i=2
i=3
35
2
53
3
i=3
5
31
i= 5
i=3
6
7
8
50
36
i=5
9
52
4
25
i = 2
28
i= 5
i = 4
10
28
i=3
11
28
The path of traveling sale person problem is:
1
4
2
5
3
1
The minimum cost of the path is: 10 + 6 +2+ 7 + 3 = 28.
0/1 Knapsack Problem
Consider the instance: M = 15, n = 4, (P1, P2, P3, P4) = (10, 10, 12, 18) and
(w1, w2, w3, w4) = ( 2, 4, 6, 9).
0/1 knapsack problem can be solved by using branch and bound technique. In this
problem we will calculate lower bound and upper bound for each node.
Place first item in knapsack. Remaining weight of knapsack is 15 – 2 = 13. Place
next item w2 in knapsack and the remaining weight of knapsack is 13 – 4 = 9. Place
next item w3 in knapsack then the remaining weight of knapsack is 9 – 6 = 3. No
fractions are allowed in calculation of upper bound so w4 cannot be placed in
knapsack.
Profit
= P1 + P2 + P3 = 10 + 10 + 12
So, Upper bound = 32
To calculate lower bound we can place w4 in knapsack since fractions are allowed in
calculation of lower bound.
3
Lower bound = 10 + 10 + 12 + ( X 18) = 32 + 6 = 38
9
Knapsack problem is maximization problem but branch and bound technique is
applicable for only minimization problems. In order to convert maximization problem
into minimization problem we have to take negative sign for upper bound and lower
bound.
Therefore, Upper bound (U) = -32
Lower bound (L) = -38
We choose the path, which has minimum difference of upper bound and lower bound.
If the difference is equal then we choose the path by comparing upper bounds and
we discard node with maximum upper bound.
102 U = - 32
L = -38
1
x1 = 1
U = - 32
L = -38
x1 = 0
2
U = - 22
L = -32
3
Now we will calculate upper bound and lower bound for nodes 2, 3.
For node 2, x1= 1, means we should place first item in the knapsack.
U = 10 + 10 + 12 = 32, make it as -32
3
L = 10 + 10 + 12 + x 18 = 32 + 6 = 38, make it as -38
9
For node 3, x1 = 0, means we should not place first item in the knapsack.
U = 10 + 12 = 22, make it as -22
L = 10 + 12 +
5
9
x 18 = 10 + 12 + 10 = 32, make it as -32
Next, we will calculate difference of upper bound and lower bound for nodes 2, 3
For node 2, U – L = -32 + 38 = 6
For node 3, U – L = -22 + 32 = 10
Choose node 2, since it has minimum difference value of 6.
U = - 32
L = -38
1
x1 = 1
U = - 32
L = - 38
x2 = 1
U = - 32
L = - 38
4
x1 = 0
2
3
U = - 22
L = -32
x2 = 0
5
U = - 22
L = -36
Now we will calculate lower bound and upper bound of node 4 and 5. Calculate
difference of lower and upper bound of nodes 4 and 5.
For node 4, U – L = -32 + 38 = 6
For node 5, U – L = -22 + 36 = 14
Choose node 4, since it has minimum difference value of 6.
103 U = - 32
L = -38
1
x1 = 1
U = - 32
L = -38
2
x2 = 0
4
x3 = 1
U = -32
L = -38
U = - 22
L = -32
3
x2 = 1
U = -32
L = -38
x1 = 0
U = - 22
L = -36
5
x3 = 0
6
U = -38
L = -38
7
Now we will calculate lower bound and upper bound of node 8 and 9. Calculate
difference of lower and upper bound of nodes 8 and 9.
For node 6, U – L = -32 + 38 = 6
For node 7, U – L = -38 + 38 = 0
Choose node 7, since it is minimum difference value of 0.
U = - 32
L = -38
1
x1 = 1
U = - 32
L = -38
2
3
4
x3 = 1
U = - 32
L = -38
U = - 22
L = -32
x2 = 0
x2 = 1
U = - 32
L = -38
x1 = 0
U = - 22
L = -36
5
x3 = 0
6
7
x4 = 1
U = - 38
L = -38
8
U = - 38
L = -38
x4 = 0
9
U = - 20
L = -20
Now we will calculate lower bound and upper bound of node 4 and 5. Calculate
difference of lower and upper bound of nodes 4 and 5.
For node 8, U – L = -38 + 38 = 0
For node 9, U – L = -20 + 20 = 0
Here the difference is same, so compare upper bounds of nodes 8 and 9. Discard the
node, which has maximum upper bound. Choose node 8, discard node 9 since, it has
maximum upper bound.
Consider the path from 1 2 4 7 8
X1 = 1
X2 = 1
X3 = 0
104 X4 = 1
The solution for 0/1 Knapsack problem is (x1, x2, x3, x4) = (1, 1, 0, 1)
Maximum profit is:
Pi xi = 10 x 1 + 10 x 1 + 12 x 0 + 18 x 1
= 10 + 10 + 18 = 38.
Portion of state space tree using FIFO Branch and Bound for above problem:
As follows:
105 NP-Hard and NP-Complete problems
Deterministic and non-deterministic algorithms
Deterministic: The algorithm in which every operation is uniquely defined is called
deterministic algorithm.
Non-Deterministic: The algorithm in which the operations are not uniquely defined but
are limited to specific set of possibilities for every operation, such an algorithm is called
non-deterministic algorithm.
The non-deterministic algorithms use the following functions:
1. Choice: Arbitrarily chooses one of the element from given set.
2. Failure: Indicates an unsuccessful completion
3. Success: Indicates a successful completion
A non-deterministic algorithm terminates unsuccessfully if and only if there exists no
set of choices leading to a success signal. Whenever, there is a set of choices that leads to
a successful completion, then one such set of choices is selected and the algorithm
terminates successfully.
In case the successful completion is not possible, then the complexity is O(1). In case of
successful signal completion then the time required is the minimum number of steps
needed to reach a successful completion of O(n) where n is the number of inputs.
The problems that are solved in polynomial time are called tractable problems and the
problems that require super polynomial time are called non-tractable problems. All
deterministic polynomial time algorithms are tractable and the non-deterministic
polynomials are intractable.
106 Satisfiability Problem:
The satisfiability is a boolean formula that can be constructed using the
following literals and operations.
1. A literal is either a variable or its negation of the variable.
2. The literals are connected with operators ˅, ˄͢, ⇒ , ⇔
3. Parenthesis
The satisfiability problem is to determine whether a Boolean formula is true
for some assignment of truth values to the variables. In general, formulas are
expressed in Conjunctive Normal Form (CNF).
A Boolean formula is in conjunctive normal form iff it is represented by
( xi ∨ xj ∨ xk1 ) ∧ ( xi ∨ x 1j ∨ xk )
A Boolean formula is in 3CNF if each clause has exactly 3 distinct literals.
Example:
The non-deterministic algorithm that terminates successfully iff a given
formula E(x1,x2,x3) is satisfiable.
107 Reducability:
A problem Q1 can be reduced to Q2 if any instance of Q1 can be easily rephrased as an
instance of Q2. If the solution to the problem Q2 provides a solution to the problem Q1,
then these are said to be reducable problems.
Let L1 and L2 are the two problems. L1 is reduced to L2 iff there is a way to solve L1 by
a deterministic polynomial time algorithm using a deterministic algorithm that solves L2
in polynomial time and is denoted by L1α L2.
If we have a polynomial time algorithm for L2 then we can solve L1 in polynomial time.
Two problems L1 and L2 are said to be polynomially equivalent iff L1α L2 and L2 α L1.
Example: Let P1 be the problem of selection and P2 be the problem of sorting. Let the
input have n numbers. If the numbers are sorted in array A[ ] the ith smallest element of
the input can be obtained as A[i]. Thus P1 reduces to P2 in O(1) time.
Decision Problem:
Any problem for which the answer is either yes or no is called decision problem. The
algorithm for decision problem is called decision algorithm.
Example: Max clique problem, sum of subsets problem.
Optimization Problem: Any problem that involves the identification of an optimal value
(maximum or minimum) is called optimization problem.
Example: Knapsack problem, travelling salesperson problem.
In decision problem, the output statement is implicit and no explicit statements are
permitted.
The output from a decision problem is uniquely defined by the input parameters and
algorithm specification.
Many optimization problems can be reduced by decision problems with the property that
a decision problem can be solved in polynomial time iff the corresponding optimization
problem can be solved in polynomial time. If the decision problem cannot be solved in
polynomial time then the optimization problem cannot be solved in polynomial time.
108 Class P:
P: the class of decision problems that are solvable in O(p(n)) time, where p(n) is a
polynomial of problem’s input size n
Examples:
• searching
• element uniqueness
• graph connectivity
• graph acyclicity
• primality testing
Class NP
NP (nondeterministic polynomial): class of decision problems whose proposed
solutions can be verified in polynomial time = solvable by a nondeterministic
polynomial algorithm
A nondeterministic polynomial algorithm is an abstract two-stage procedure that:
• generates a random string purported to solve the problem
• checks whether this solution is correct in polynomial time
By definition, it solves the problem if it’s capable of generating and verifying a
solution on one of its tries
Example: CNF satisfiability
Problem: Is a boolean expression in its conjunctive normal form (CNF) satisfiable, i.e.,
are there values of its variables that makes it true? This problem is in NP.
Nondeterministic algorithm:
• Guess truth assignment
• Substitute the values into the CNF formula to see if it evaluates to true
What problems are in NP?
• Hamiltonian circuit existence
• Partition problem: Is it possible to partition a set of n integers into two
disjoint subsets with the same sum?
• Decision versions of TSP, knapsack problem, graph coloring, and many other
combinatorial optimization problems. (Few exceptions include: MST, shortest
paths)
• All the problems in P can also be solved in this manner (but no guessing is
necessary), so we have:
P NP
• Big question: P = NP ?
109 NP HARD AND NP COMPLETE
Polynomial Time algorithms
Problems whose solutions times are bounded by polynomials of small degree are called
polynomial time algorithms
Example: Linear search, quick sort, all pairs shortest path etc.
Non- Polynomial time algorithms
Problems whose solutions times are bounded by non-polynomials are called nonpolynomial time algorithms
Examples: Travelling salesman problem, 0/1 knapsack problem etc
It is impossible to develop the algorithms whose time complexity is polynomial for
non-polynomial time problems, because the computing times of non-polynomial are
greater than polynomial. A problem that can be solved in polynomial time in one model
can also be solved in polynomial time.
NP-Hard and NP-Complete Problem:
Let P denote the set of all decision problems solvable by deterministic algorithm in
polynomial time. NP denotes set of decision problems solvable by nondeterministic
algorithms in polynomial time. Since, deterministic algorithms are a special case of
nondeterministic algorithms, P ⊆ NP. The nondeterministic polynomial time problems
can be classified into two classes. They are
1. NP Hard and
2. NP Complete
NP-Hard: A problem L is NP-Hard iff satisfiability reduces to L i.e., any
nondeterministic polynomial time problem is satisfiable and reducable then the problem
is said to be NP-Hard.
Example: Halting Problem, Flow shop scheduling problem
NP-Complete: A problem L is NP-Complete iff L is NP-Hard and L belongs to NP
(nondeterministic polynomial).
A problem that is NP-Complete has the property that it can be solved in polynomial time
iff all other NP-Complete problems can also be solved in polynomial time. (NP=P)
110 If an NP-hard problem can be solved in polynomial time, then all NP- complete problems
can be solved in polynomial time. All NP-Complete problems are NP-hard, but some NPhard problems are not known to be NP- Complete.
Normally the decision problems are NP-complete but the optimization problems are NPHard.
However if problem L1 is a decision problem and L2 is an optimization problem, then it is
possible that L1α L2.
Example:
Knapsack
decision problem
can be reduced to knapsack
optimization problem.
There are some NP-hard problems that are not NP-Complete.
Relationship between P,NP,NP-hard, NP-Complete
Let P, NP, NP-hard, NP-Complete are the sets of all possible decision problems that are
solvable in polynomial time by using deterministic algorithms, non-deterministic
algorithms, NP-Hard and NP-complete respectively. Then the relationship between P,
NP, NP-hard, NP-Complete can be expressed using Venn diagram as:
Problem conversion
A decision problem D1 can be converted into a decision problem D2 if there is an
algorithm which takes as input an arbitrary instance I1 of D1 and delivers as output an
instance I2 of D2such that I2 is a positive instance of D2 if and only if I1 is a positive
instance of D1. If D1 can be converted into D2, and we have an algorithm which solves
D2, then we thereby have an algorithm which solves D1. To solve an instance I of D1,
we first use the conversion algorithm to generate an instance I0 of D2, and then use the
algorithm for solving D2 to determine whether or not I0 is a positive instance of D2. If it
is, then we know that I is a positive instance of D1, and if it is not, then we know that I is
a negative instance of D1. Either way, we have solved D1 for that instance. Moreover, in
this case, we can say that the computational complexity of D1 is at most the sum of the
computational complexities of D2 and the conversion algorithm. If the conversion
algorithm has polynomial complexity, we say that D1 is at most polynomially harder than
D2. It means that the amount of computational work we have to do to solve D1, over and
111 above whatever is required to solve D2, is polynomial in the size of the problem instance.
In such a case the conversion algorithm provides us with a feasible way of solving D1,
given that we know how to solve D2.
Given a problem X, prove it is in NP-Complete.
1. Prove X is in NP.
2. Select problem Y that is known to be in NP-Complete.
3. Define a polynomial time reduction from Y to X.
4. Prove that given an instance of Y, Y has a solution iff X has a solution.
Cook’s theorem:
Cook’s Theorem implies that any NP problem is at most polynomially harder than SAT.
This means that if we find a way of solving SAT in polynomial time, we will then be in a
position to solve any NP problem in polynomial time. This would have huge practical
repercussions, since many frequently encountered problems which are so far believed to
be intractable are NP. This special property of SAT is called NP-completeness. A
decision problem is NP-complete if it has the property that any NP problem can be
converted into it in polynomial time. SAT was the first NP-complete problem to be
recognized as such (the theory of NP-completeness having come into existence with the
proof of Cook’s Theorem), but it is by no means the only one. There are now literally
thousands of problems, cropping up in many different areas of computing, which have
been proved to be NP- complete.
In order to prove that an NP problem is NP-complete, all that is needed is to show that
SAT can be converted into it in polynomial time. The reason for this is that the sequential
composition of two polynomial-time algorithms is itself a polynomial-time algorithm,
since the sum of two polynomials is itself a polynomial.
Suppose SAT can be converted to problem D in polynomial time. Now take any NP
problem D0. We know we can convert it into SAT in polynomial time, and we know we
can convert SAT into D in polynomial time. The result of these two conversions is a
polynomial-time conversion of D0 into
D. since D0 was an arbitrary NP problem, it follows that D is NP-complete
112
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