Module 9: Mathematical Induction

Transcription

1 Module 9: Mathematical Induction

2 Module 9: Announcements There is a chance that this Friday s class may be cancelled. Assignment #4 is due Thursday March 16 th at 4pm. Midterm #2 Monday March 20 th, 2017 from 7pm to 9pm Read the Piazza post. Let us know if you have a conflict.

3 Learning goals: pre-class Epp 4 th section 5.1: Convert sequences to and from explicit formulas that describe the sequence. Convert sums to and from summation/σ notation. Convert products to and from product/π notation. Manipulate formulas in summation/product notation by adjusting their bounds, merging or splitting summations/products, and factoring out values.

4 Learning goals: pre-class Epp 4 th sections 5.2 to 5.4: Given a theorem to prove stated in terms of its induction variable (i.e., usually, in terms of n), write out the skeleton of an inductive proof including: the base case(s) that need to be proven, the induction hypothesis, and the inductive step that needs to be proven.

5 Learning goals: in-class By the end of this module, you should be able to: Establish properties of self-referential structures using inductive proofs that naturally build on those self-references. Critique formal inductive proofs to determine whether they are valid and where the error(s) lie if they are invalid. Prove properties of the non-negative integers (or a subset) with appropriate self-referential structure using weak or strong induction as needed.

6 The BIG questions: How can we convince ourselves that an algorithm does what it's supposed to do? How do we determine whether or not one algorithm is better than another one? Mathematical induction is a very useful tool when proving the correctness or efficiency of an algorithm. We will see several examples of this.

7 Module 9 outline Example: single-elimination tournaments. Defining and validating mathematical induction. More examples where we can use induction. A slightly different type of induction.

8 Single-elimination tournaments Problem: single-elimination tournament Teams play one another in pairs The winner of each pair advances to the next round Los Angeles Vancouver Toronto St. Louis Pittsburgh New York Isl. Montréal Buffalo Los Angeles Toronto New York Isl. Montréal Los Angeles Montréal Round 2 Round 3 Round 4 Montréal

9 Single-elimination tournaments If we have n rounds of playoffs, how many teams can participate? a.n b.2n c.n 2 d.2 n e.none of the above. Let s prove this using mathematical induction.

10 The inductive step If 2 n teams can participate in a playoff with n rounds, how many teams can participate in a playoff with 2 n+1 rounds? a. 2 n +1 b. 2 n+1 c. (n+1) 2 d. None of the above.

11 The inductive step Theorem: If 2 n teams can participate in a playoff of n rounds, then 2 n+1 teams can participate in a playoff of n+1 rounds. Let P(x, y) be true if y teams can play in x rounds. In predicate logic: n Z, P(n, 2, ) P(n + 1, 2,01 ) Let s prove this.

12 The inductive step Theorem: If 2 n teams can participate in a playoff of n rounds, then 2 n+1 teams can participate in a playoff of n+1 rounds. Proof: Consider an unspecified playoff with n rounds. Assume that 2 n teams can participate in a playoff of n rounds. We can think of a playoff with n+1 rounds as follows: Two playoffs with n rounds proceed in parallel. The two winners then complete in one more round to determine the winner. Since each playoff with n rounds has 2 n teams, a playoff with n+1 rounds has 2 n+1 teams. QED

13 The complete induction proof Theorem: For any positive integer n, at most 2 n teams can participate in a playoff of n rounds. In predicate logic, the theorem becomes: n Z, P(n, 2, ) Proof: We prove this by induction. Base case: Induction step:

14 What did we prove with induction? Theorem: n Z, P n, 2, P 1,2 P 2,4 P 3,8 P 4,16 Proof: We prove this by induction. Base case: n = 1, P(1,2) is true. Induction step: We will prove that n Z, P n, 2, P n + 1, 2,01 P(1,2) P(2,4) P 2,4 P(3,8) P 3,8 P(4,16)

15 What did we prove with induction? P(1,2) P(1,2) P(2,4) P(2,4) by modus ponens P(2,4) P(2,4) P(3,8) P(3,8) by modus ponens P(3,8) P(3,8) P(4,16) P(4,16) by modus ponens Therefore, we have shown that: P 1,2 P 2,4 P 3,8 P 4,16 n Z, P n, 2,

16 The connection to recursion For an unspecified positive integer n To prove P(n + 1, 2,01 ): We can instead prove P(n, 2, ). This is a recursive call. Induction is more or less the same as recursion! A fun example involving induction:

17 Module 9 outline Example: single-elimination tournaments. Defining and validating mathematical induction. More examples where we can use induction. A slightly different type of induction.

18 Ordering students using adjacent swaps We want to order a group of 5 students in order of day of birth We are only allowed to swap pairs of adjacent students. We claimed the maximum number of swaps for n students is n(n-1)/2.

19 How many swaps do we need? Let s place students from left to right. The students already placed are ordered by day of birth. We swap each new student with his/her neighbour until he/she is at the right place. The i th student may be swapped with all previous (i-1) students.,<1 So the total number of swaps is : i Theorem 1:,<1 n Z 0, : i =>? = =>? n n 1 2

20 Theorem 1:,<1 n Z 0, : i =>? = n n 1 2 Let s parse the theorem first. One way to think about it: Theorem 1: 0 = = = = = 10 Another way to think about it:,<1 Let n n 1 P n : i =>? = Theorem 1: P(1) P(2) P(3) P(4) 2

21 Theorem 1:,<1 n Z 0, : i =>? = n n 1 2 Which base case(s) do we need to prove? a. n = 0 b. n = 1 c. n = 2 d. n = 0 and n = 1 e. n = 1 and n = 2

22 Theorem 1:,<1 n Z 0, : i =>? = n n 1 2 What do we need to prove in the induction step? a. b.,<1 n Z 0 : i = =>?,<1 n Z 0, : i = =>? n n 1 2 n n 1 2 : i = c. Both of a and b are correct. d. Neither of a and b are correct., =>? n + 1 n 2, n Z 0, : i = =>? n + 1 n 2

23 What is induction proving?,<1 n n 1 Base case: P 1 P n : i = 2 Induction step: =>? P(1) P(2) P 2 P 3 P 3 P 4 Combining the two, we have proven that P(1) P 2 P 3 P 4

24 A fun video about induction feature=related

25 Theorem 1:,<1 n Z 0, : i =>? = n n 1 2 Proof: We prove the theorem by induction. Base case: n = 1 Clearly, with only 1 student, no swap is needed: which is equal to 1*(1-1)/2 = 0 Induction step: Pick an unspecified n 1. Assume that when we have n students, we need at most n(n-1)/2 swaps. Equivalently we assume that,<1 : i =>? This is called the Induction Hypothesis. = n n 1 2

26 Theorem 1:,<1 n Z 0, : i =>? Proof (continued) Induction step (continued) For n+1 students, we need at most swaps. = n n 1, : i =>? 2,<1 = : i =>? + n By the induction hypothesis, this is equal to n n n = n + 1 n swaps, as required for the induction step. Hence by the principle of M.I., the theorem holds. QED 2

27 Induction is like recursion Where is our recursive call in the induction proof? a. Base case: b. Induction step: consider an unspecified positive integer n. c. Induction step: Assume that (where we stated our induction hypothesis) d. Induction step: The algebra where we applied our induction hypothesis e. Induction step: some other algebra steps

28 Induction is like recursion I don t know how to calculate (n-2) + (n-1) + n However, if you tell me how to calculate (n-2) + (n-1) I can calculate the first sum as follows (n-2) + (n-1) + n = ( (n-2) + (n-1)) + n

29 Review questions on induction Give me 3 critical components of an induction proof. When we are completing the induction step, what is our most important goal in life? Writing an induction proof is like playing dominos. What does the base case correspond to? What does the induction step correspond to? Induction is similar to recursion. Which part of an induction proof corresponds to the recursive call?

30 Geometric series Theorem 2: H t N, : 5 = =>? = 5H These summations occur frequently when we need to determine the running time of divideand-conquer algorithms.

31 Theorem 2: H t N, : 5 = =>? = 5H What should we prove in the base case? a. t = -1 b. t = 0 c. t = 1 d. t = 2 e. t = 3

32 Theorem 2: What do we need to prove in the induction step? a. b. H t N, : 5 = =>? H t N, : 5 = =>? t N, : 5 = c. Both of a and b are correct. d. Neither of a and b are correct. H =>? = 5H = 5H = 5H H01 : 5 = =>? = 5H0I H01 t N, : 5 = =>? = 5H0I 1 5 1

33 Theorem 2: Proof: Base case: t = 0 The summation is 5 0 = 1, and Induction step: Consider an unspecified natural number t. Assume that We need to prove that H t N, : 5 = =>? = 5H = 1 Hence by the principle of M.I., the theorem holds. QED

34 Proving an inequality Theorem 3: For all integers n 4, 2 n < n! Rules for inequalities: Start from one side (say the left side) Work step by step towards the other. When dealing with <, you are allowed to make the expression larger, but never smaller. Example: if I am smaller than you, then I am still smaller than you when you stand on a bench.

35 Theorem 3: n 4, 2 n < n! Another version of the induction step: We want to prove that n?, 2,<1 < n 1! 2, < n!. Consider an unspecified n?. Induction hypothesis: assume that 2 n-1 < (n-1)! 2 n = 2(2 n-1 ) < 2(n-1)! < n(n-1)! = n! What is the smallest value of n that we should use in the induction step? (a) less than 4 (d) 5 (b) 3 (e) more than 5 (c) 4

36 Sum of inverted squares Theorem 4: An interesting mathematical fact (which doesn t Q help with the proof): Let s prove this!, n 1, : 1 i I =>1 : 1 i I =>1 2 1 n = πi 6

37 Midterm 2 Work through the posted practice problem on converting a DFA to a sequential circuit! Topics covered: Converting a DFA to a sequential circuit Predicate logic to proof structure (assign #4 Q1) Proof structure to predicate logic (assign #4 Q2) Direct proof Proof by contrapositive and/or contradiction Induction

38 Theorem 4: After applying the induction hypothesis, what inequality do we need to prove? a. b. c. d. n 1, : 1 i I 2 1 n + 1 (n + 1) I 2 1 n 2 1 n + 1 (n + 1) I 2 1 n n (n + 1) I 2 1 n 2 1 n (n + 1) I 2 1 n + 1, =>1 2 1 n

39 Midterm 2 Work through the posted practice problem on converting a DFA to a sequential circuit! Topics covered: Converting a DFA to a sequential circuit Predicate logic to proof structure (assign #4 Q1) Proof structure to predicate logic (assign #4 Q2) Direct proof Proof by contrapositive and/or contradiction Induction

40 Sum of inverted squares Theorem 4: An interesting mathematical fact (which doesn t Q help with the proof): Let s prove this!, n 1, : 1 i I =>1 : 1 i I =>1 2 1 n = πi 6

41 Theorem 4: n 1, : 1 i I What should we prove in the base case? a. n = -1 b. n = 0 c. n = 1 d. n = 2 e. n = 3, =>1 2 1 n

42 Theorem 4: What do we need to prove in the induction step? a. b. t N,, : 1 i I =>1, t N, : 1 i I =>1 n 1, : 1 i I c. Both of a and b are correct. d. Neither of a and b are correct., =>1, n : 1 i I =>1 2 1 n 2 1 n + 1, n t N, : 1 i I =>1 2 1 n + 1

43 Proving a theorem about a DFA Consider the following DFA:

44 Which strings does this DFA accept? In addition to the empty string, what other strings does this DFA accept? a. Strings whose length is divisible by 5. b. Unsigned binary integers that are prime numbers. c. Strings with alternating 1 s and 0 s d. Unsigned binary integers that are divisible by 5.

45 Which strings does this DFA accept? In addition to the empty string, what other strings does this DFA accept? a. Strings whose length is divisible by 5. b. Unsigned binary integers that are prime numbers. c. Strings with alternating 1 s and 0 s d. Unsigned binary integers that are divisible by 5.

46 Proving a theorem about this DFA Suppose that we want to prove that this DFA only accepts binary numbers that are divisible by 5. Which statement should we prove (ignoring empty strings)? a.the DFA is in state 0 if and only if s is divisible by 5. b.the DFA is in state r if and only if r is the remainder when we divide s by 5.

47 Proving a theorem about this DFA Suppose that we want to prove that this DFA only accepts binary numbers that are divisible by 5. Which statement should we prove (ignoring empty strings)? a.the DFA is in state 0 if and only if s is divisible by 5. b.the DFA is in state r if and only if r is the remainder when we divide s by 5. We cannot prove a fact about state 0 without discussing the other states.

48 Theorem: The DFA ends up in state r after reading a string s if and only if q Z, s = 5q + r. Proof: by induction on the length n of s. Base case: n = 1 This can be verified easily by looking at the DFA. Induction step: Consider an unspecified integer n 2. Induction hypothesis: assume that the DFA behaves correctly for any string with n bits. We need to show that the DFA behaves correctly for any string with n + 1 bits.

49 Theorem: The DFA ends up in state r after reading a string s if and only if q Z, s = 5q + r. Proof (continued): Suppose that s = b n-1 b n-2...b 2 b 1 b 0. After reading the bits b n-1 b n-2...b 2 b 1, the DFA is in state r. By the induction hypothesis, the binary integer b n-1 b n-2...b 2 b 1 = 5q + r for some integer q. Now, s = 2 (b n-1 b n-2...b 2 b 1 ) + b 0 = 2(5q+r) + b 0 = 10q + 2r + b 0 and so s = 5 (2q) + (2r + b 0 ). So the DFA should have the following transitions: r b 0 2r+b r b 0 2r+b

50 Theorem: The DFA ends up in state r after reading a string s if and only if q Z, s = 5q + r. Proof (continued): It does, and hence after reading all n bits of s, the DFA ends up in state r if and only if q Z, s = 5q + r. Hence by the principle of mathematical induction, the DFA ends up in state r after reading a string s if and only if q Z, s = 5q + r. QED Corollary: the DFA is in state 0 if and only if s is divisible by 5.

51 Module 9 outline Example: single-elimination tournaments. Defining and validating mathematical induction. More examples where we can use induction. A slightly different type of induction.

52 The induction we have seen so far To solve n + 1, we make a recursive call to n (the previous case). Examples ( n n ) + 5 n+1 2 n < n! ---> 2 n+1 < (n+1)! The DFA behaves correctly for a string with n bits, then it also behaves correctly for a string with (n+1) bits. What if we need to make recursive call to one or more previous cases?

53 Carrying water across a desert We need to buy jugs of water in order to cross a desert. Unfortunately, the water jugs only come in two sizes: 5 liters and 7 liters. Prove that as long as n >= 24, we can buy n liters of water using 5-liter and 7-liter jugs only.

54 The induction step/the recursive call Which of the following ideas can be used for the induction step? If I can buy n-1 liters of water, then I can also buy n liters of water. If I can buy n-3 liters of water, then I can also buy n liters of water. If I can buy n-5 liters of water, then I can also buy n liters of water. If I can buy n-7 liters of water, then I can also buy n liters of water.

55 The induction step Version 1: P(n-5) -> P(n) Suppose that we can buy (n-5) liters of water using x 5-liter jugs and y 7-liter jugs. Then we can buy n liters of water using (x+1) 5- liter jugs and y 7-liter jugs only. Version 2: P(n-7) -> P(n) Suppose that we can buy (n-7) liters of water using x 5-liter jugs and y 7-liter jugs. Then we can buy n liters of water using x 5-liter jugs and (y+1) 7-liter jugs only.

56 The base cases Induction step: Assume that we can buy (n-5)l of water, then we can also buy nl of water. What is the smallest value of n for which we can apply the induction step? a. 24 b. 25 c. 28 d. 29 e. 30

57 The base cases Induction step: Assume that we can buy (n-5)l of water, then we can also buy nl of water. How many base cases do we need to prove? a. 1 b. 3 c. 5 d. 7 e. 9

58 A lower bound for Fibonacci numbers The Fibonacci numbers are defined as follows: F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) for any n >= 2. Theorem: for every integer n >= 1, F(n) >= 1.6 n-2.

59 The induction step/the recursive call What should we prove in the induction step? a. If F(n-1) satisfies the theorem, then F(n) also satisfies the theorem. b. If F(n-2) satisfies the theorem, then F(n) also satisfies the theorem. c. If F(n-2) and F(n-1) satisfy the theorem, then F(n) also satisfies the theorem. d. If F(n) and F(n+1) satisfy the theorem, then F(n+2) also satisfies the theorem.

60 Let s complete the induction step.

61 The base cases What is the smallest value of n for which we can apply the induction step? a. 0 b. 1 c. 2 d. 3 e. 4 What are the base cases that we need to prove?

62 Additional induction examples

63 A binary tree A binary tree is a data structure that is defined recursively A binary tree is Empty, or A node with some data, and two children that are themselves trees

64 Calculating the size of a binary tree Consider the following procedure to calculate the size of a binary tree. If the tree is null, then the size of the tree is 0. Otherwise, the size of the tree = 1 + the size of its left child + the size of its right child Theorem 6: Prove that the above procedure correctly computes the number of (non-null) nodes of the tree.

65 Calculating the size of a binary tree We prove this using mathematical induction on the size of the tree. Base case: t is null In this case t contains exactly 0 nodes. Induction step: Assume the algorithm works for trees that are smaller than t. Because the left sub-tree of t is smaller than t, the 1st recursive calls correctly returns the size of the left sub-tree of t.

66 Calculating the size of a binary tree Proof (continued) Induction step (continued) Similarly the right sub-tree of t is smaller than t, and so the 2nd recursive call correctly returns the size of the right sub-tree of t. But we return 1 + the sum of the values returned by the recursive calls. This is exactly the size of t (1 for the root, and the sum of the sizes of the two sub-trees). Hence by the principle of M.I., our algorithm computes correctly the size of every tree. QED

67 Randomized-quick-select We can find the i th smallest element in an unsorted list as follows: Pick a random element x of the list. Divide the list into three sublists: list-smaller: elements smaller than x list-equal: elements equal to x list-larger: elements larger than x Then search list-smaller if i length of list-smaller list-larger if i > length of list smaller + length of list-equal otherwise return x

68 Randomized-quick-select This algorithm is called randomized-quick-select. A student shows the expected number of steps S(n) of the algorithm on a list with n elements is: S(1) = 4c S(2) = 12c S(3) = 20c S(n) 2cn + S(Floor(3n/4)) when n 4

69 Randomized-quick-select Prove: for every n 1, S(n) 8cn.