Counting Techniques. L172 Agenda Section 4.1: Counting Basics Sum Rule Product Rule Inclusion-Exclusion.

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Presentation transcript:

Counting Techniques

L172 Agenda Section 4.1: Counting Basics Sum Rule Product Rule Inclusion-Exclusion

L173 Counting Basics Counting techniques are important in programming design. EG: How large an array or hash table or heap is needed? EG: What is the average case complexity of quick-sort? Answers depend on being able to count. Counting is useful in the gambling arena also. EG: What should your poker strategy be?

L174 Counting Basics Set Cardinalities Interested in answering questions such as: How many bit strings of length n are there? How many ways are there to buy 13 different bagels from a shop that sells 17 types? How many bit strings of length 11 contain a streak of one type of bit of exact length 7? COMMON THEME: convert to set cardinality problems so each question above is a about counting the number of elements in some set: Q: What is the corresponding set in each case?

L175 Counting Basics Set Cardinalities A: The set to measure the cardinality of is… How many bit strings of length n are there? {bit strings of length n} How many ways are there to buy 13 different bagels from a shop that sells 17 types? {S  {1, 2, … 17} | |S | = 13 } (…arguable) How many bit strings of length 11 contain a streak of one type of bit of exact length 7? {length 11 bit strings with 0-streak of length 7}  {length 11 bit strings with 1-streak of length 7}

L176 Product Rule As counting problems can be turned into set cardinality problems, useful to express counting principles set theoretically. Product-Rule: For finite sets A, B: |A  B| = |A|  |B| Q: How many bit strings of length n are there?

L177 Product Rule A: 2 n. Proof : Let S = {bit strings of length n}. S is in 1-to-1 correspondence with Consequently the product rule implies:

L178 Cardinality of Power Set THM: |P ({1,2,3,…,n})| = 2 n Proof. The set of bit strings of length n is in 1-to-1 correspondence with the P({1,2,3,…,n}) since subsets are represented by length n bit strings.

L179 Sum Rule Next the number of length 11 bit strings with a streak of length exactly 7. Q: Which of the following should be counted:

L1710 Sum Rule Next the number of length 11 bit strings with a streak of length exactly 7. Q: Which of the following should be counted: No!, longest streak has length No! Too long No! Streak too long Yes! Yes!

L1711 Sum Rule We are trying to compute the cardinality of: {length 11 bit strings with 0-streak of length 7}  {length 11 bit strings with 1-streak of length 7} Call the first set A and the second set B. Q: Are A and B disjoint?

L1712 Sum Rule A: Yes. If had both a 0-streak and a 1-streak of length 7 each, string would have length at least 14! When counting the cardinality of a disjoint union we use: SUM RULE: If A and B are disjoint, then |A  B| = |A|+|B| By symmetry, in our case A and B have the same cardinality. Therefore the answer would be 2|A|.

L1713 Sum Rule Break up A = {length 11 bit strings with 0-streak of length exactly 7} into more cases and use sum rule: A 1 = { ***} (* is either 0 or 1) A 2 = { **} A 3 = {* *} A 4 = {** } A 5 = {*** }. Apply sum rule: |A| = |A 1 | +|A 2 | +|A 3 | +|A 4 | +|A 5 |

L1714 Sum Rule So let’s count each set. A 1 = { ***}. There are 3 *’s, each with 2 choices, so product rule gives |A 1 | = 2 3 = 8 A 2 = { **}. There are 2 *’s. Therefore, |A 2 | = 2 2 = 4 A 3 = {* *}, A 4 = {** } Similarly: |A 2 | = |A 3 | = |A 4 | = 4 A 5 = {*** }. |A 1 | = |A 5 | = 8 |A| = |A 1 | +|A 2 | +|A 3 | +|A 4 | +|A 5 | = = 28. Therefore answer is 56.

L1715 Counting Functions How many ways to match 13M to 17W ? { f :{1,2,…,13}  P {1,2,…,17} | f is 1-to-1} Use product rule thoughtfully possible output values for f (1) values remain for f (2) …………………………… i. 17-i+1 values remain for f (i ) …………………………… =5 values remain for f (13) ANS: 17·16·15 ·14 ·… ·7·6·5 = 17! / 4! Q: In general how many 1-to-1 functions from size k to size n set?

L1716 Counting Functions A: The number of 1-to-1 functions from a size k set to a size n set is n ! / (n - k) ! As long as k is no larger than n. If k > n there are no 1-to-1 functions. Q: How about general functions from size k sets to size n sets?

L1717 Counting Functions A: The number of functions from a size k set to a size n set is n k

L1718 Inclusion-Exclusion The principle of Inclusion-Exclusion generalized the sum rule the the case of non-empty intersection: INCLUSION-EXCLUSION: If A and B are sets, then |A  B | = |A|+|B |- |A  B | This says that when counting all the elements in A or B, if we just add the the sets, we have double-counted the intersection, and must therefore subtract it out.

L1719 Inclusion-Exclusion Visualize. Diagram gives proof Inclusion- Exclusion principle: A-A  B U ABAB B-A  B

L1720 Blackboard Exercises for Section 4.1 Problem (a,b,c,f). How many positive integers with exactly three decimal digits are… …divisible by 7? …odd? …have the same 3 digits? …are not divisible by 3 or 4? Problem How many ways can the letters a,b,c,d be arranged such that a is not immediately followed by b?