Practical Guide to Quantitative Finance Interviews Part 2

Brus Teanore the floor number is equal to or greater than X. You would like to determine X. What is the strategy that will minimize the number of drops for the worst case scenario? Solution: Suppose that we have a strategy with a maximum of throws. For the int throw of ball one, we can try the N-th floor. If the ball breaks, we can start to uy the second ball from the first floor and increase the floor number by one until the second ball breaks. At mort, there are N - 1 floors to test. So a maximure of N throws art enough to cover all possibilities. If the first ball thrown out of Noth floor does not break, we have N - 1 throws left. This time we can only increase the floor number by N - 1 for the fint ball since the second ball can only cover N - 2 floors if the first ball . If the first ball thrown out of (2-1)th flour does not , we have N - 2 throws left. So we can only increase the floor number by N = 2 for the first ball the second ball can only cover M = 1 flours if the first boll . Using such logic, we can see that the number of floors that these two balls can cover with a maximum of N throws is N+(N-|)+***+ 1 = N(N + 1) / 2 In order to cover 100 stories, we need to have N(N + 1) / 2 >= 100 integer, we have N = 14 Hasically, we start the first ball on the 14th floor, if the ball breaks, we can use the second ball to try floors 1,213 with a maximum throws of 14 (when the 13th of the If the first ball does not break, we will try the first ball on the 14 + (14 - 1) = 27th floor. If it breaks, we can use the second ball to cover floon 15, 16 ,***,26 with a total maximum throws of 14 as well. 2.6 The Pigeon Hole Principle Here is the basic version of the Pigeon Hole Principle: if you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then at least one pigeon hole has more than one pigoon. Basically it says that if you have a holes and more than n + 1 pigeons, at least 2 pigeons have to share one of the holes. The generalized version is that if you have a holes and at least sen+1 pigeons, at least m + 1 pigeons have to share one of the holes. These simple and intuitive ideas are surprisingly useful in many problems. Here we will use some examples to show their applications Hint: Aune design a strategy with ball can cover N - 1 floors, if the fireball menus Brows. If the first ball is thrown oece, the second the wire, the second bali can cove N * 2 floors 20 A Practical Guide To Quamicate Face Icarries Matching socks Your drawer contains 2 red socks, 20 yellow socks and 31 blue socks. Being a busy and absent-minded MIT studert, you just randomly grab a number of socks out of the draw and try to find a matching pair. Assume each soek has equal probability of being selected, what is the minimum number of socks you need to gran in order to guarantee a pair of socks of the same color? Solution: This question is just a variation of the even simpler version of two-color-socks problem, in which case you only need 3. When you have 3 colors (3 pigeon holesj, by the Pigeon Hole Principle, you will need to have 3 + 1 * 4 socks (4 pigeons) to guarantee that at least two socks have the same color (2 pigeons share a hole) Handshakes You are invited to a welcome party with 25 follow team members. Each of the fellow members shakes hands with you to welcome you. Since a number of people in the room haven't met each other, there's a lot of random handshaking among others as well. If you don't know the total number of handshakes, can you say with certainty that there are at least two people present who shook hands with exactly the same number of people? Solauron: There are 26 people at the party and each shakes hands with from 1-since everyone shakes hands with you to 25 people. In other words, there are 26 pigeons and 25 holes. As a result, at least two people must have shaken hands with exactly the same number of people. Have we met before? Show me that, if there are 6 people at a party, then either at least 3 people met each other before the party, or at least 3 people were strangers before the party. Solution: This question appears to be a complex one and interviewers often get puzzled by what the interviewer exactly wants. But once you start to analyze possible scenarios, the answer becomes obvious Let's say that you are the 6th person at the party. Then by generalized Pigeon Hole Principle (Do we even need that for such an intuitive conclusion?), among the remaining 5 people, we conclude that either at least 3 people met you or at least 3 people did not meet you. Now let's explore these two mutually exclusive and collectively exhaustive scenarios Case 1: Suppose that at least 3 people have met you before. 21 Besin Tessers If two people in this group met each other, you and the pair (3 people) met each other. If no pair among these people met each other, then these people (>= 1 people) did not meer cach other. In either sub-case, the conclusion holds. Case 2: Suppose at least 3 people have not met you before. If two people in this group did not meet each other, you and the pair 13 people) did not meet each other. If all pairs among these people knew each other, then these people (23 people) met each other. Again, in either sub-case, the conclusion holds. Ants on a square are 51 ants on a square with side length of 1. If you have a glass with a radius of 1 /7 . can you put your glass at a position on the square to guarantee that the glass encompasses at least 3 ants?" Solution To guarantee that the glass encompasses at least 3 asts, we can separate the square into 25 smaller areas. Applying the generalized Pigeon Hole Principle, we can show that at least one of the areas must have at least 3 ants. So we only need to make sure that the glass is large crough to cover any of the 25 smaller areas. Siamply separate the area into 5 * 5 smaller squarex with side length of 1/5 each will do since a cirele with radius of 177 cas cover a square with side length 1/5. Counterfeit coins I There are 5 bags with 100 coins in each bag. A coin can weigh 9 grans, 10 grams or 11 grams. Each bag contains coins of equd weight, but we do not know what type of coins a bag contains. You have a digital scale (the kind that tells the exact weight). How many times do you need to use the scale to determine which type of coin cach bag contains? Solution: If the answer for 5 hags is not obvious, let's start with the simplest version of the problem-1 bag. We only need to take one coin to weigh it. Now we can move on to 2 bags. How many coim do we need to take from bag 2 in order to determine the coin types of bag 1 and bag 29 Considering that there are three possible types for bag 1, we will need three coins from bag 2: two coins won't do. For notation simplicity, let's change the number weight for three types to -1, 0 and 1 (by removing the mean 10). If "Hier: Separate the squan 25 aller and then at least one ares los 3 in it. A circle with radioso a square with inde length up to Grand "Hier: Start with a simpler problem. What if you have two bags of cries 5 = 14M how many noin 50 you need from cach beg to find the type of coins in either bag? What the misimus offeresct in coin mamber? Then how about the bags? 21 A Practical Guide To Quantitative Finance Interviews we only use 2 coins from bag 2. the final sum for 1 coin from hag 1 and 2 coins from bag 2 ranges from 3 to 3 (7 pigeon holes). At the same time we have 7(3 * 3) possible combinations for the weights of coins in bag 1 and hag 2 (4 pigeons), So at least two combinations will yield the same final sum (97, so at least two pigeons need to share one hole), and we can not distinguish them. If we use 3 coins from bag 2, then the sum ranges from 4 to 4, which is possible to cover all 9 combinations. The following table exactly shows that all possible combinations yield different sur I coin, bag 1 0 1 4 2 0 0 1 2 4 3 Coins, Bag 2 Cland C2 represent the weights of coins from hand 2 respective Then how abour 3 bags? We are going to have 327 possible combinations. Surely an indicator ranging from-13 to 13 will cover it and we will need 9 coins from bag 3. The possible combinations are shown in the following table: Coins, flag 3 1 C2-6 C2-1 0 0 1 1 -13 4 4 5 6 7 0 1 10 2 4 11 17 13 C1, C2, and C represow the weights of enim from bag 1. 2. and I regretincés Following this logic, it is easy to see that we will need 27 coins from bag 4 and 81 coins from bag 5. So the answer is to take 1. 3. 9, 27 and 81 coins from bags 1, 2, 3, 4, and 5. respectively, to determine which type of coins each bag contains using a single weighing 2.7 Modular Arithmetic The modulo operation-denoted as xsy or a mod y finds the remainder of division of number x by another number y. For simpicility, we only consider the case where y is a positive integer. For example, 5% * 3 = 2 An intuitive property of modalo operation is 23 Bain Teen that if x_{i}% * y = x_{i}% * y_{i} then (x_{i} - x_{i})% 6Y =0. From this property we can also show thu a. (x+1)%, and (x + y - 1)% * y are all different numbers. Prisoner problem One hundred prisoners are given the chance to be set free tomorrow. They are all told that cach will be given a red er blue hat to wear. Each prisoner can see everyone else's hat but not his own. The hat colors are assigned randomly and once the hats are plaved on top of each prisoner's hand they cannot communicate with one another in any form, of else they are iremediately exccuted. The prisoners will be called out in random order and the prisoner called out will guess the color of his hat. Each prisoner declares the color of his hat so that everyone else can hear it. If a prisoner guesses correctly the color of his hat, he is set free immediately; otherwise he is executed. They are given the right to come up with a strategy among themselves to save as many prisoners as possible. What is the best strategy they can adopt and how many prisoners can they guarantee to save? Solution: At least 99 prisoners can be saved. The key lies in the first prisoner who can see everyone else's hat. He declares his hat to be red if the number of red bats he sees is odd. Otherwise he declares his hat to be blue He will have a 1/2 chance of having guessed correctly. Everyone else is able to deduct his own hat color combining the knowledge whether the number of red hats is odd among 99 prisoners (excluding the first) and the color of the other 98 prisoners (excluding the first and himself). For example, if the number of red hats is odd among the other 99 prisoners. A prisones wearing a red hat will see even number of red hats in the other 98 prisoners (excluding the first and himself) and deduce that he is wearing a red hat The two-color case it easy, in't it? What if there are 3 possible hur colors: red, blue, and white? What is the best strategy they can adopt and how many prisoners can they guarantee to save?ال Solution: The answer is still that at least 99 prisoners will be saved. The difference is that the first prisoner tow only has 1/3 chance of survival. Let's use the following scoring system: red-0, green1, and blue 2. The first prisoner counts the total score for Hist: The first prisoner can we the number of nd sesi bhachats of at her 99 priseurs. Ose color has odd unber of coasts and the other has even number of counts Hint That are in add singly as 21. Here we have 3 colors, so you may want to consίλετ A Practical Guide To Quantitative Finance Interviews the rest of 99 prisoners and calculates %3. If the remainder is 0, he announces red; if the remainder is 1, green; 2, blue. He has 1/3 chance of living, but all the rest of the prisoners con determine his own score (color) from the remainder. Let's consider a prisoner / among 99 prisoners (excluding the first prisoner). He can calculate the total score (x) of all other 98 prisoners. Since (x + 0)% 0 3. (x+1)%3, and (x + 2)% * 3 are all different, so from the remainder that the first prisoner gives (for the 99 prisoners including (), he can determine his own score (color). For example, if prisoner i sees that there are 32 red, 29 green and 37 blue in those 98 prisoners (excluding the fint and himself). The total score of those 98 prisoners is 103. If the first prisoner announces that the remainder is 2 (green), then prisoner i knows his own color is green (1) since enly 104% * 3 = 2 among 103, 104 and 105. Theoretically, a similar strategy can be extended to any number of colors. Surely that requires all prisoners to have exceptional memory and calculation capability. Division by 9 Given an arbitrary integer, come up with a rule to decide whether it is divisible by 9 and prove it. Schaion Hopefully you still remember the rules from your high school math class. Add up all the digits of the integer. If the sum is divisible by 9, then the integer is divisible by 9; otherwise the integer is not divisible by 9. But how do we prove it? Let's express the original integer as a= a_{a} * 10 ^ n +a a-1 10^ a-1 +***+a i 10^ i +d 0^ n Basically we state that if (z is a integer), then the a is divisible by 9 as well. The proof a n +a n - 1 +***+a i +a 0 =9 1 For any a= a_{a} * 10 ^ n +a a-1 10^ n-1 +***+a i 10^ prime +a n^ n let overline a =a-(a c +a c - 1 +***+a i +u b ). We have b= a_{a}(10 ^ a - 1) +a a-1 (10^ a-1 -1)+...+a i (10^ i -1)=a-9x n which is divisible by 9 since all (10 ^ k - 1), k = 1 ,***,a are divisible by 9. Because both & and 92 ure divisible by 9. by 9 as well a = b + 9x (Similarly you can also show that a= (- 1) ^ n * a_{n} +(-1)^ n-1 a n-1 +***+(-1)^ prime a i +a n =11x. is the necessary and sufficient condition for a to be divisible by 11.) 24 25