CSE 311: Foundations of Computing. Lecture 16: Recursion & Strong Induction Applications: Fibonacci & Euclid

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1 CSE 311: Foundations of Computing Lecture 16: Recursion & Strong Induction Applications: Fibonacci & Euclid

2 Midterm A week today (Monday, May 7) in class Closed book, closed notes You will get lists of inference rules & equivalences Covers material up to end of ordinary induction. Practice problems & practice midterm on the website Solutions later this week Prof. Beame will run a review session Sunday, May 6, 3:30-5:30 pm in EEB 105.

3 More Recursive Definitions Suppose that h: N R. Then we have familiar summation notation: h = h(0) h = h h for 0 There is also product notation: h = h(0) h = h( + 1) h for 0

4 Fibonacci Numbers = 0 = 1 = + for all 2

5 Strong Inductive Proofs In 5 Easy Steps 1. Let ()be.... We will show that ()is true for all integers by strong induction. 2. Base Case: Prove () 3. Inductive Hypothesis: Assume that for some arbitrary integer, (!) is true for every integer! from to 4. Inductive Step: Prove that ( + 1)is true: Use the goal to figure out what you need. Make sure you are using I.H. (that (),, () are true) and point out where you are using it. (Don t assume ( + 1)!!) 5. Conclusion: ()is true for all integers

6 Bounding Fibonacci I: < 2 for all 0 $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

7 Bounding Fibonacci I: < 2 for all 0 1. Let P(n)be f n < 2 n. We prove that P(n)is true for all integers n 0 by strong induction. 2. Base Case: f 0 =0 <1= 2 0 so P(0)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 0,P(j)is true for every integer jfrom 0to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 k+1 Case k+1 = 1: Then f 1 = sop(k+1)is true here. Case k+1 2: Then f k+1 =f k + f k-1 by definition 2 k +2 k-1 by the IH 2 k +2 k = 2 2 k = 2 k+1 sop(k+1)is true in this case. 5. Therefore by strong induction, f n 2 n for all integers n 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

8 Bounding Fibonacci I: < 2 for all 0 1. Let P(n)be f n < 2 n. We prove that P(n)is true for all integers n 0 by strong induction. 2. Base Case: f 0 =0 <1= 2 0 so P(0)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 0,P(j)is true for every integer jfrom 0to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 < 2 k+1 Case k+1 = 1: Then f 1 = sop(k+1)is true here. Case k+1 2: Then f k+1 =f k + f k-1 by definition 2 k +2 k-1 by the IH 2 k +2 k = 2 2 k = 2 k+1 sop(k+1)is true in this case. 5. Therefore by strong induction, f n 2 n for all integers n 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

9 Bounding Fibonacci I: < 2 for all 0 1. Let P(n)be f n < 2 n. We prove that P(n)is true for all integers n 0 by strong induction. 2. Base Case: f 0 =0 <1= 2 0 so P(0)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 0,P(j)is true for every integer jfrom 0to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 < 2 k+1 Case k+1 = 1: Then f 1 = sop(k+1)is true here. Case k+1 2: Then f k+1 =f k + f k-1 by definition 2 k +2 k-1 by the IH 2 k +2 k = 2 2 k = 2 k+1 sop(k+1)is true in this case. 5. Therefore by strong induction, f n 2 n for all integers n 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

10 Bounding Fibonacci I: < 2 for all 0 1. Let P(n)be f n < 2 n. We prove that P(n)is true for all integers n 0 by strong induction. 2. Base Case: f 0 =0 <1= 2 0 so P(0)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 0,P(j)is true for every integer jfrom 0to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 < 2 k+1 Case k+1 = 1: Then f 1 = 1 < 2 = 2 1 sop(k+1)is true here. Case k+1 2: Then f k+1 =f k + f k-1 by definition < 2 k +2 k-1 by the IHsincek-1 0 < 2 k +2 k = 2 2 k = 2 k+1 sop(k+1)is true in this case. These are the only cases so P(k+1) follows. 5. Therefore by strong induction, f n < 2 n for all integersn 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

11 Bounding Fibonacci II: 2 for all 2 $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

12 Bounding Fibonacci II: 2 for all 2 1. Let P(n)be f n 2 n/2-1. We prove that P(n)is true for all integers n 2 by strong induction. 2. Base Case: f 2 = f 1 + f 0 = 1 and2 2/2 1 = 2 0 = 1 so P(2)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 2,P(j)is true for every integer jfrom 2to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 (k+1)/2-1 Case k+1 = 3: Then f k+1 =f 3 = f 2 + f 1 =2 2 1/2 = 2 3/2-1 =2 (k+1)/2-1 Case k+1 4: f k+1 =f k + f k-1 by definition 2 k/ (k-1)/2-1 by the IHsincek (k-1)/ (k-1)/2-1 = 2 (k-1)/2 = 2 (k+1)/2-1 SoP(k+1)is true in both cases. 5. Therefore by strong induction, f n 2 n/2-1 for all integersn 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

13 Bounding Fibonacci II: 2 for all 2 1. Let P(n)be f n 2 n/2-1. We prove that P(n)is true for all integers n 2 by strong induction. 2. Base Case: f 2 = f 1 + f 0 = 1 and2 2/2 1 = 2 0 = 1 so P(2)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 2,P(j)is true for every integer jfrom 2to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 (k+1)/2-1 Case k+1 = 3: Then f k+1 =f 3 = f 2 + f 1 =2 2 1/2 = 2 3/2-1 =2 (k+1)/2-1 No need for cases for the definition here: Case f k+1 4: f k+1 =f k + f k-1 by definition k+1 = f k + f k-1 since k+1 2 Now just want 2 to k/2-1 apply +2 (k-1)/2-1 the IH by to the getp(k)and IHsincek-1 P(k-1): 2 2 (k-1)/ (k-1)/2-1 = 2 (k-1)/2 = 2 (k+1)/2-1 Problem: Though we can get P(k)since k 2, SoP(k+1)is true k-1may in both only cases. be 1so we can t conclude P(k-1) 5. Therefore Solution: by strong Separate induction, cases for f n when 2 n/2-1 for k-1=1(or all integersn k+1=3). 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

14 Bounding Fibonacci II: 2 for all 2 1. Let P(n)be f n 2 n/2-1. We prove that P(n)is true for all integers n 2 by strong induction. 2. Base Case: f 2 = f 1 + f 0 = 1 and2 2/2 1 = 2 0 = 1 so P(2)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 2,P(j)is true for every integer jfrom 2to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 (k+1)/2-1 Case k = 2: Then f k+1 =f 3 = f 2 + f 1 =2 2 1/2 = 2 3/2-1 =2 (k+1)/2-1 Case k 3: f k+1 =f k + f k-1 by definition 2 k/ (k-1)/2-1 by the IHsincek (k-1)/ (k-1)/2-1 = 2 (k-1)/2 = 2 (k+1)/2-1 SoP(k+1)is true in both cases. 5. Therefore by strong induction, f n 2 n/2-1 for all integersn 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

15 Bounding Fibonacci II: 2 for all 2 1. Let P(n)be f n 2 n/2-1. We prove that P(n)is true for all integers n 2 by strong induction. 2. Base Case: f 2 = f 1 + f 0 = 1 and2 2/2 1 = 2 0 = 1 so P(2)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 2,P(j)is true for every integer jfrom 2to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 (k+1)/2-1 Case k = 2: Then f k+1 =f 3 = f 2 + f 1 =2 2 1/2 = 2 3/2-1 =2 (k+1)/2-1 Case k 3: f k+1 =f k + f k-1 by definition 2 k/ (k-1)/2-1 by the IHsincek (k-1)/ (k-1)/2-1 = 2 (k-1)/2 = 2 (k+1)/2-1 SoP(k+1)is true in both cases. 5. Therefore by strong induction, f n 2 n/2-1 for all integersn 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

16 Bounding Fibonacci II: 2 for all 2 1. Let P(n)be f n 2 n/2-1. We prove that P(n)is true for all integers n 2 by strong induction. 2. Base Case: f 2 = f 1 + f 0 = 1 and2 2/2 1 = 2 0 = 1 so P(2)is true. 3. Inductive Hypothesis: Assume that for some arbitrary integer k 2,P(j)is true for every integer jfrom 2to k. 4. Inductive Step: Goal: Show P(k+1); that is, f k+1 2 (k+1)/2-1 Case k = 2: Then f k+1 =f 3 = f 2 + f 1 =2 2 1/2 = 2 3/2-1 =2 (k+1)/2-1 Case k 3: f k+1 =f k + f k-1 by definition 2 k/ (k-1)/2-1 by the IHsincek (k-1)/ (k-1)/2-1 = 2 (k-1)/2 = 2 (k+1)/2-1 SoP(k+1)is true in both cases. 5. Therefore by strong induction, f n 2 n/2-1 for all integersn 0. $ % = % $ & = & $ ' = $ '& + $ '( for all ' (

17 Running time of Euclid s algorithm Theorem: Suppose that Euclid s Algorithm takes steps for gcd (-, )with - > 0. Then, -. An informal way to get the idea: Consider ann stepgcd calculation starting with r n+1 =a andr n =b: r n+1 = q n r n + r n-1 r n = q n-1 r n-1 + r n-2 r 3 = q 2 r 2 + r 1 r 2 = q 1 r 1 For all k 2, r k-1 = r k+1 mod r k Nowr 1 1 and each q k must be 1. If we replace all the q K sby1 and replacer 1 by1, we can only reduce the r k s. After that reduction, r k =f k for every k.

18 Running time of Euclid s algorithm Theorem: Suppose that Euclid s Algorithm takes steps for gcd (-, )with - > 0. Then, -. We go by strong induction on n. Let P(n)be gcd(a,b)witha b>0 takes nsteps a f n+1 for all n 1. Base Case: n=1 If Euclid s Algorithm on a, b witha b > 0 takes 1step, then a=q 1 bfor some q 1 and a b 1=f 2 andp(1) holds Induction Hypothesis: Suppose that for some integerk 1, P(j) is true for all integersj s.t.1 j k Inductive Step: We want to show: ifgcd(a,b) witha b > 0takes k+1 steps, thena f k+2.

19 Running time of Euclid s algorithm Induction Hypothesis: Suppose that for some integerk 1, P(j) is true for all integersj s.t.1 j k Inductive Step: We want to show: ifgcd(a,b) witha b>0 takes k+1 steps, thena f k+2. Now if k+1=2,then Euclid s algorithm on aand bcan be written as a = q 2 b + r 1 b = q 1 r 1 andr 1 > 0. Also, since a b>0 we must haveq 2 1 and b 1. Soa = q 2 b + r 1 b + r = 2 = f 3 = f k+2 as required.

20 Running time of Euclid s algorithm Induction Hypothesis: Suppose that for some integerk 1, P(j) is true for all integersj s.t.1 j k Inductive Step: We want to show: ifgcd(a,b) witha b>0 takes k+1 steps, thena f k+2. Next suppose that k+1 3so for the first 3 steps of Euclid s algorithm on aand bwe have a = q k+1 b + r k b = q k r k + r k-1 r k = q k-1 r k-1 + r k-2 and there are k-2more steps after this. Note that this means that thegcd(b,r k )takes ksteps and gcd(r k,r k-1 )takes k-1steps and b>r k >r k-1. So since k, k-1 1 by the IH we haveb f k+1 and r k f k. Also, since a b we must haveq k+1 1. Soa = q k+1 b + r k b + r k f k+1 +f k = f k+2 as required.

21 Running time of Euclid s algorithm Theorem: Suppose that Euclid s Algorithm takes steps for gcd (-, )with - > 0. Then, -. Why does this help us bound the running time of Euclid s Algorithm? We already proved that 2 so 2 () Therefore: if Euclid s Algorithm takes steps for gcd (-, )with - > 0 then - 2 () so ( 1)/2 log - or 1 + 2log - i.e., # of steps twice the # of bits in -.

22 Recursive Definition of Sets Recursive Definition Basis Step: 0 S Recursive Step: If x S, then x + 2 S Exclusion Rule: Every element in S follows from basis steps and a finite number of recursive steps.

23 Recursive Definitions of Sets Basis: 6 S, 15 S Recursive: If x,y S, then x+y S Basis: [1, 1, 0] S, [0, 1, 1] S Recursive: If [x, y, z] S, then [αx,αy, αz] S for all α R If [x 1, y 1, z 1 ] S and [x 2, y 2, z 2 ] S, then [x 1 + x 2, y 1 + y 2, z 1 + z 2 ] S. Powers of 3:

24 Recursive Definitions of Sets Basis: 6 S, 15 S Recursive: If x,y S, then x+y S Basis: [1, 1, 0] S, [0, 1, 1] S Recursive: If [x, y, z] S, then [αx,αy, αz] Sfor all α R If [x 1, y 1, z 1 ] S and [x 2, y 2, z 2 ] S, then [x 1 + x 2, y 1 + y 2, z 1 + z 2 ] S. Powers of 3: Basis: 1 S Recursive: If x S, then 3x S.

25 Recursive Definitions of Sets: General Form Recursive definition Basis step: Some specific elements are in 5 Recursive step: Given some existing named elements in 5some new objects constructed from these named elements are also in 5. Exclusion rule: Every element in 5 follows from basis steps and a finite number of recursive steps

26 Strings An alphabetσis any finite set of characters The set Σ*of stringsover the alphabet Σis defined by Basis: ℇ Σ (ℇis the empty string) Recursive: if 7 Σ*, - Σ, then 7- Σ*

27 Palindromes Palindromes are strings that are the same backwards and forwards Basis: ℇis a palindrome and any - Σis a palindrome Recursive step: If 8is a palindrome then -8-is a palindrome for every - Σ

28 All Binary Strings with no 1 s before 0 s

29 All Binary Strings with no 1 s before 0 s Basis: ℇ S Recursive: If x S,then 0x S If x S,then x1 S

30 Function Definitions on Recursively Defined Sets Length: len(ℇ) = 0 len(wa) = 1 + len(w) for w Σ *, a Σ Reversal: ℇ R = ℇ (wa) R = aw R for w Σ *, a Σ Concatenation: x ℇ= x for x Σ * x wa = (x w)a for x Σ *, a Σ

Fall 2015 Lecture 14: Modular congruences. cse 311: foundations of computing

Fall 2015 Lecture 14: Modular congruences cse 311: foundations of computing If a and b are positive integers, then gcd a, b = gcd (b, a mod b) Useful GCD Fact Proof: By definition a = a div b b + (a mod