# CS 441 Discrete Mathematics for CS Lecture 16. Counting. CS 441 Discrete mathematics for CS. Course administration

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1 CS 44 Discrete Mathematics for CS Lecture 6 Counting Milos Hauskrecht 5329 Sennott Square Course administration Midterm exam: Thursday, March 6, 24 Homework assignment 8: due after Spring break An opportunity to practice mathematical induction, and basic counting for the midterm Course web page:

2 Counting Assume we have a set of objects with certain properties Counting is used to determine the number of these objects Examples: Number of available phone numbers with 7 digits in the local calling area Number of possible match starters (football, basketball) given the number of team members and their positions Basic counting rules Counting problems may be very hard, not obvious Solution: simplify the solution by decomposing the problem Two basic decomposition rules: Product rule A count decomposes into a sequence of dependent counts ( each element in the first count is associated with all elements of the second count ) Sum rule A count decomposes into a set of independent counts ( elements of counts are alternatives ) 2

3 Product rule A count can be broken down into a sequence of dependent counts each element in the first count is associated with all elements of the second count Assume an auditorium with a seat labeled by a letter and numbers in between to 5 (e.g. A23). We want the total number of seats in the auditorium. 26 letters and 5 numbers How to count? One solution: write down all seats (objects) and count them A- A-2 A-3 A-5 B- Z-49 Z (n-) n eventually we get it Product rule A count can be broken down into a sequence of dependent counts each element in the first count is associated with all elements of the second count assume an auditorium with a seat labeled by a letter and numbers in between to 5 (e.g. A23). We want the total number of seats in the auditorium. 26 letters and 5 numbers A better solution? For each letter there are 5 numbers So the number of seats is 26*5 = 3 Product rule: number of letters * number of integers in [,5] 3

4 Product rule A count can be broken down into a sequence of dependent counts each element in the first count is associated with all elements of the second count Product rule: If a count of elements can be broken down into a sequence of dependent counts where the first count yields n elements, the second n2 elements, and kth count nk elements, by the product rule the total number of elements is: n = n*n2* * nk Product rule How many different bit strings of length 7 are there? E.g. Is it possible to decompose the count problem and if yes how? Yes. Count the number of possible assignments to bit For the first bit assignment (say ) count assignments to bit 2 or or Total assignments to first 2 bits: 2*2=4 4

5 Product rule How many different bit strings of length 7 are there? E.g. Is it possible to decompose the count problem and if yes how? Yes. Count the number of possible assignments to bit For the specific first bit count possible assignments to bit 2 For the specific first two bits count assignments to bit 3 Number of assignments to the first 3 bits: 2*2*2=8 Product rule How many different bit strings of length 7 are there? E.g. Is it possible to decompose the count problem and if yes how? Yes. Count the number of possible assignments to bit For the specific first bit count possible assignments to bit 2 For the specific first two bits count assignments to bit 3 Gives a sequence of n dependent counts and by the product rule we have: n = 2* 2 * 2 * 2 * 2 * 2 *2 = 2 7 5

6 Product rule The number of subsets of a set S with k elements. How to count them? Hint: think in terms of bitstring representation of a set? Assume each element in S is assigned a bit position. If A is a subset it can be encoded as a bitstring: if an element is in A then use else put How many different bitstrings are there? n = 2* 2* 2 = 2 k k bits Sum rule A count decomposes into a set of independent counts elements of counts are alternatives, they do not depend on each other You need to travel in between city A and B. You can either fly, take a train, or a bus. There are 2 different flights in between A and B, 5 different trains and buses. How many options do you have to get from A to B? We can take only one type of transportation and for each only one option. The number of options: n = 2+5+ Sum rule: n = number of flights + number of trains + number of buses 6

7 Sum rule A count decomposes into a set of independent counts elements of counts are alternatives Sum rule: If a count of elements can be broken down into a set of independent counts where the first count yields n elements, the second n2 elements, and kth count nk elements, by the sum rule the total number of elements is: n = n+n2+ + nk Beyond basic counting rules More complex counting problems typically require a combination of the sum and product rules. A login password: The minimum password length is 6 and the maximum is 8. The password can consist of either an uppercase letter or a digit. There must be at least one digit in the password. How many different passwords are there? 7

9 Beyond basic counting rules Step : The password we select has either 6,7 or 8 characters. So the total number of valid passwords is by the sum rule: P = P6+P7+P8 The number of passwords of length 6,7 and 8 respectively Step 2 Assume passwords with 6 characters (either digits + upper case letters): How many are there? If we let each character to be at any position we have: P6-all = (26+) 6 =(36) 6 different passwords of length 6 Beyond basic counting rules Step 2 But we must have a password with at least one digit. How to account for it? A trick. Split the count of all passwords of length 6 into to two mutually exclusive groups: P6-all = P6-digits + P6-nodigits. P6-digits count when the password has one or more digits 2. P6-nodigits count when the password has no digits We know how to easily compute P6-all and P6-nodigits P6-all = 36 6 and P6-nodigits = 26 6 Then P6-digits = P6-all P6-nodigits 9

10 Beyond basic counting rules Step : the total number of valid passwords is by the sum rule: P = P6+P7+P8 The number of passwords of length 6,7 and 8 respectively Step 2 The number of valid passwords of length 6: P6= P6-digits = P6-all P6-nodigits = Analogically: P7= P7-digits = P7-all P7-nodigits = P8= P8-digits = P8-all P8-nodigits = Inclusion-Exclusion principle Used in counts where the decomposition yields two dependent count tasks with overlapping elements If we used the sum rule some elements would be counted twice Inclusion-exclusion principle: uses a sum rule and then corrects for the overlapping elements. We used the principle for the cardinality of the set union. A B = A + B - A B U B A

11 Inclusion-exclusion principle How many bitstrings of length 8 start either with a bit or end with? It is easy to count strings that start with : How many are there? 2 7 It is easy to count the strings that end with. How many are there? 2 6 Is it OK to add the two numbers to get the answer? No. Overcount. There are some strings that can both start with and end with. These strings are counted in twice. How to deal with it? How to correct for overlap? How many of strings were counted twice? 2 5 ( xxxxx ) Thus we can correct for the overlap simply by using: = = 6 Tree diagrams Tree: is a structure that consists of a root, branches and leaves. Can be useful to represent a counting problem and record the choices we made for alternatives. The count appears on the leaf nodes. What is the number of bit strings of length 4 that do not have two consecutive ones.

12 Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string 2

13 Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string 3

14 Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string () Tree diagrams What is the number of bit strings of length 4 that do not have two consecutive ones? Empty string () 4

15 Pigeonhole principle Assume you have a set of objects and a set of bins used to store objects. The pigeonhole principle states that if there are more objects than bins then there is at least one bin with more than one object. 7 balls and 5 bins to store them Pigeonhole principle Assume you have a set of objects and a set of bins used to store objects. The pigeonhole principle states that if there are more objects than bins then there is at least one bin with more than one object. 7 balls and 5 bins to store them At least one bin with more than ball exists. 5

16 Pigeonhole principle Assume you have a set of objects and a set of bins used to store objects. The pigeonhole principle states that if there are more objects than bins then there is at least one bin with more than one object. Theorem. If there are k+ objects and k bins. Then there is at least one bin with two or more objects. k bins Pigeonhole principle Assume you have a set of objects and a set of bins used to store objects. The pigeonhole principle states that if there are more objects than bins then there is at least one bin with more than one object. Theorem. If there are k+ objects and k bins. Then there is at least one bin with two or more objects. Proof. (by contradiction) Assume that we have k + objects and every bin has at most one element. Then the total number of elements is k which is a contradiction. End of proof 6

17 Pigeonhole principle Assume 367 people. Are there any two people who has the same birthday? How many days are in the year? 365. Then there must be at least two people with the same birthday. Generalized pigeonhole principle We can often say more about the number of objects. Say we have 5 bins and 2 objects. What is it we can say about the bins and number of elements they hold? There must be a bin with at least 3 elements. Why? Assume there is no bin with more than 3 elements. Then the max number of elements we can have in 5 bins is. We need to place 3 so at least one bin should have at least 3 elements. 7

18 Generalized pigeonhole principle Theorem. If N objects are placed into k bins then there is at least one bin containing at least N / k objects. Example. Assume people. Can you tell something about the number of people born in the same month. Yes. There exists a month in which at least / 2 = 8.3 = 9 people were born. 8

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