Pattern For Induction

Pattern For Induction

CPSC 121Steve Wolfman

2012/08/02

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Starting an Induction Proof

Theorem: some property is true for all integers n ___.

Scratch work: P(n) some property on n.“Structural” Insight: a problem of size n breaks down in terms of problems of size __________.

Laying Out the Structure of an Induction Proof

Theorem: ____________________.Proof: by induction on n.BC: For n = ___,

, P(___) is true.More?

IH: WLOG, let n be an arbitrary integer > ____.Assume _________________________________________________________________________________.

IS: Since n > ____, ___________________________.By the IH, __________________________________.

P(n) is true.




State the theorem explicitly or circle it if it’s given.It may help to define a function like “Let c(n) = the # of nodes in KST tree Kn”.




State P(n) explicitly, regardless of whether it’s given. You’ll need P(n) over and over! On an exam, you can even use it to write long statements briefly.




State this insight explicitly. It may be right in the problem statement! Say a function is recursive. Then the recursive calls tell you how the problem breaks down. Say a definition is recursive. Then the self-reference tells you how the problem breaks down. Say there’s a summation. Then our self-referential definitions for summations (or products or …) tell you how the problem breaks down.






P(n) is true.

The theorem’s easy. You already wrote it.






P(n) is true.

The smallest relevant value.

For the same value as above.Prove the property for this case.






P(n) is true.

Ask: Does the structural insight break the next value of n down in terms of the existing base case? If not, add it as a base case. Continue until all remaining cases are “handled”.






P(n) is true.

The largest base case.

P(i) for every i your structural insight tells you that you need!






P(n) is true.

Restate your structural insight. Then, use the IH to “solve” the subproblems.

Finally, build back up to P(n).






State this insight explicitly. It may be right in the problem statement! Say a function is recursive. Then the recursive calls tell you how the problem breaks down. Say a definition is recursive. Then the self-reference tells you how the problem breaks down. Say there’s a summation. Then our self-referential definitions for summations (or products or …) tell you how the problem breaks down.






P(n) is true.


The smallest relevant value.


Ask: Does the structural insight break the next value of n down in terms of the existing base case? If not, add it as a base case. Continue until all remaining cases are “handled”.





Reminder: Sigma, Pi, Factorial, Powers:All Self-Referential

14

Examples• K1 is a tree with only a root node. For all integers n 2, Kn is a

root node with all previous Ki as subtrees. Prove Kn has 2n-1 nodes. (You may assume .)

• Prove for every positive integer n:

• Prove that the following algorithm takes no more than multiplications:

∑𝑗=1

𝑛

𝑗( 𝑗+1)=𝑛(𝑛+1)(𝑛+2)3

;; Integer Natural -> IntegerPow(a,b): if b = 0: return 1 else if b = 1: return a else: let part = Pow(a,) if b is odd: return part*part*a else: return part*part

Example 1

Example 1, Getting Started

Problem: K1 is a tree with only a root node. For all integers n 2, Kn is a root node with all previous Ki as subtrees. Prove Kn has 2n-1 nodes. (You may assume .)



Example 1, Getting StartedProblem: K1 is a tree with only a root node. For all integers n 2, Kn is a root node with all previous Ki as subtrees. Prove Kn has 2n-1 nodes.





Let c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.





Scratch work: P(n) c(Kn) = 2n-1

“Structural” Insight: a problem of size n breaks down in terms of problems of size __________.State this insight explicitly. It may be right in the problem statement!

Say a definition is recursive. Then the self-reference tells you how the problem breaks down.



Scratch work: P(n) c(Kn) = 2n-1

“Structural” Insight: For all integers n 2, Kn is a root node with all previous Ki as subtrees.

Example 1: Laying out the Structure





P(n) is true.



Let c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.Proof: by induction on n.BC: For n = ___,




P(n) is true.

The smallest relevant value. (Here: n 1; so, 1.)



Let c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.Proof: by induction on n.BC: For n = 1, TODO: prove the property., c(K1) = 21-1.More?



P(n) is true.

Ask: Does the structural insight break the next value of n (here: 2) down in terms of the existing base case? If not, add it as a base case. Continue until all remaining cases are “handled”.


Let c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.Proof: by induction on n.BC: For n = 1, TODO: prove the property., c(K1) = 21-1.



P(n) is true.

NO MORE BCs. 2 breaks down in terms of 1, which is our base case. 3 breaks down in terms of 2 (which we know we can handle) and 1. Everything subsequent breaks down in terms of previous values, which eventually reach 1.





P(n) is true.



Reminder: our structural insight is “For all integers n 2, Kn is a root node with all previous Ki as subtrees.”



IH: WLOG, let n be an arbitrary integer > 1.Assume c(Ki) = 2i-1 for all “previous i” (that is, all integers 1 i < n).


P(n) is true.





IS: Since n > 1, Kn is a root node with all previous Ki as subtrees.By the IH, for each of these subtrees, c(Ki) = 2i-1.

P(n) is true.

(Because n is an integer, n > 1 means the same as n 2.)






c(Kn) = 2n-1.

Finally, build back.

Example 1: Finishing the ProofLet c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.Proof: by induction on n.BC: For n = 1, K1 is defined to contain only 1 node. Therefore, c(K1) = 1.And, 21-1 = 20 = 1., c(K1) = 21-1.



c(Kn) = 2n-1.


Example 1: Finishing the ProofLet c(n) = the # of nodes in KST tree Kn.Theorem: c(Kn) = 2n-1 for all integers n 1.Proof: by induction on n.BC: For n = 1, K1 is defined to contain only 1 node. Therefore, c(K1) = 1.And, 21-1 = 20 = 1. , c(K1) = 21-1.


IS: Since n > 1, Kn is a root node with all previous Ki as subtrees.By the IH, for each of these subtrees, c(Ki) = 2i-1.Thus:

c (K n) (one root node plus the nodes in the subtrees)

By the IH

Where j = i – 1.

By assumption in the problem statement.

QED

Example 1, Other Things to TryProblem: K1 is a tree with only a root node. For all integers n 2, Kn is a root node with all previous Ki as subtrees. Prove Kn has 2n-1 nodes. (You may assume .)

• Prove the assumption (by induction!).• Prove the alternate insight “For n 2, Kn is Kn-1 plus an extra subtree

that is also Kn-1” (not by induction), then use this insight to prove the theorem instead.

• Figure out what the height of Kn is and prove it by induction (with either insight).

• Change the definition to “K0 is the empty tree” rather than starting with K1 and change n 2 to n 1. Reprove the theorem.

Example 2


Problem: Prove for every positive integer n,.



Example 2, Getting StartedProblem: Prove for every positive integer n, .

Theorem: For every positive integer n, .




Problem: Prove for every positive integer n, .


Scratch work: P(n) “Structural” Insight: a problem of size n breaks down in terms of problems of size __________.State this insight explicitly. It may be right in the problem statement!

Say there’s a summation. Then our self-referential definitions for summations (or products or …) tell you how the problem breaks down.

Reminder: Sigma, Pi, Factorial, Powers:All Self-Referential

38


Problem: Prove for every positive integer n, .


Scratch work: P(n) “Structural” Insight: For all integers n 2,






P(n) is true.



Theorem: For every positive integer n, .Proof: by induction on n.BC: For n = ___,




P(n) is true.

The smallest relevant value. (Here: n is positive or 1; so, 1.)



Theorem: For every positive integer n, .Proof: by induction on n.BC: For n = 1, TODO: prove the property., .More?



P(n) is true.



Theorem: For every positive integer n, .Proof: by induction on n.BC: For n = 1, TODO: prove the property., .



P(n) is true.

NO MORE BCs. 2 breaks down in terms of 1, which is our base case. 3 breaks down in terms of 2 (which we know we can handle). Everything subsequent breaks down in terms of the previous value, which eventually reaches 1.





P(n) is true.



Reminder: our structural insight is “”



IH: WLOG, let n be an arbitrary integer > 1.Assume .


P(n) is true.





IS: Since n > 1, .By the IH, .

P(n) is true.

(Because n is an integer, n > 1 means the same as n 2.)






.


Example 2: Finishing the ProofTheorem: For every positive integer n, .Proof: by induction on n.BC: For n = 1, and . , .



.

Finally, build back. At this point, we did scratch work on both sides of the formula: and until they connected.

Example 2: Finishing the Proof

Theorem: For every positive integer n, .Proof: by induction on n.BC: For n = 1, and . , .



QED

Example 2, Other Things to TryProblem: Prove for every positive integer n,.

• Find a formula for . Now, prove the formula correct by induction and, separately, by relating the new formula to the one you analysed in Example 2.

• Find a formula for . Prove it correct by induction. (Can’t figure out the formula? Type 1 + 8 + 27 + … into Wolfram Alpha hit enter.)

• Change the bounds to 0 to n (rather than 1 to n) and find new formulae/proofs.

Example 3

Example 3, Getting StartedProblem: Prove that Pow(a,b) takes no more than multiplications:;; Integer Natural -> IntegerPow(a,b): if b = 0: return 1 else if b = 1: return a else: let part = Pow(a,) if b is odd: return part*part*a else: return part*part




Let m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .





Scratch work: P(b) “Structural” Insight: a problem of size n breaks down in terms of problems of size __________.

State this insight explicitly. It may be right in the problem statement! Say a function is recursive. Then the recursive calls tell you how the problem breaks down.



Scratch work: P(b) “Structural” Insight: For all integers b 2, Pow(a,b) is computed in terms of Pow(a,).


Theorem: ____________________.Proof: by induction on b.BC: For b = ___,


IH: WLOG, let b be an arbitrary integer > ____.Assume _________________________________________________________________________________.

IS: Since b > ____, ___________________________.By the IH, __________________________________.

P(b) is true.

The theorem’s easy. You already wrote it.We’ll also go through and replace n by b right away so we can keep using b.


Let m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: For b = ___,




P(b) is true.

The smallest relevant value. (Here: b is a natural or 0; so, 0.)



Let m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: For b = 0, TODO: prove the property., .More?



P(b) is true.

Ask: Does the structural insight break the next value of b (here: 1) down in terms of the existing base case? If not, add it as a base case. Continue until all remaining cases are “handled”.


Let m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: For b = 0, TODO: prove the property. , .For b = 1, TODO: prove the property. , .



P(b) is true.

The structural insight was “For all integers b 2, Pow(a,b) is computed in terms of Pow(a,).” NO, it doesn’t handle the 1 case; so, we add 1 as a base case. Ask: does it handle the 2 case?





P(b) is true.

NO MORE BCs. 2 breaks down in terms of 1, which is one of our base cases. 3 breaks down in terms of 1 (a base case). Everything subsequent breaks down in terms of the a value half as large, which eventually reaches 1 (via 2 or 3).





P(b) is true.



Reminder: our structural insight is “”



IH: WLOG, let b be an arbitrary integer > 1.Assume .


P(b) is true.





IS: Since b > 1, .By the IH, .

P(b) is true.

(Because b is an integer, n > 1 means the same as b 2.)Finally, build back up to P(b).





.


Example 3: Finishing the ProofLet m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: For b = 0, Pow returns 0 without any multiplications.

, .For b = 1, TODO: prove the property. , .



.


Of course, we did scratch work to get this!


, .For b = 1, Pow returns a without any multiplications.

, .



.




, .For b = 1, Pow returns a without any multiplications.

, .



.


Finally, build back. We cannot do this without referring to the original algorithm.

So, we include it, with numbered lines, on the next slide.

Pow(a,b) algorithm

;; Integer Natural -> IntegerPow(a,b):1. if b = 0: return 12. else if b = 1: return a3. else:4. let part = Pow(a,)5. if b is odd: return part*part*a6. else: return part*part

Example 3: Finishing the ProofLet m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: Dropped for space.IH: WLOG, let b be an arbitrary integer > 1. Assume .IS: Since b > 1, .By the IH, . Since b 2, however, .Thus, .For b > 1, Pow proceeds in two cases. We therefore divide our proof into two cases:Case 1 (b is even): TODO: prove Case 2 (b is odd): TODO: prove

(We proceed in cases to follow the algorithm’s structure.But… it will turn out that we can do it all in one case.)

Example 3: Finishing the ProofLet m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: Dropped for space.IH: WLOG, let b be an arbitrary integer > 1. Assume .IS: Since b > 1, .By the IH, . Since b 2, however, .Thus, .For b > 1, Pow proceeds in two cases. We therefore divide our proof into two cases:Case 1 (b is even): Then, the algorithm returns its result with one additional multiplication, and . So, it uses a total # of multiplications of:

Case 2 (b is odd): TODO: prove

by the IHsee above

Example 3: Finishing the ProofLet m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: Dropped for space.IH: WLOG, let b be an arbitrary integer > 1. Assume .IS: Since b > 1, .By the IH, . Since b 2, however, .Thus, .For b > 1, Pow proceeds in two cases. We therefore divide our proof into two cases:Case 1 (b is even): Dropped for space.Case 2 (b is odd): Then, the algorithm returns its result with one additional multiplication, and . So, it uses a total # of multiplications of:

by the IHsee above

QED

Example 3: SIMPLIFIED VERSIONLet m(b) = the # of multiplications required to compute Pow(a,b) for any integer a.Theorem: For every natural number b, .Proof: by induction on b.BC: Dropped for space.IH: WLOG, let b be an arbitrary integer > 1. Assume .IS: Since b > 1, .By the IH, . Since b 2, however, .Thus, .For b > 1, Pow uses at most two extra multiplications to compute its result (when b is odd). Note also that . Thus, the total # of multiplications is:

by the IHsee above

QED

We notice that there’s plenty of “slop”; so, we can handle both even and odd cases at once.

Example 3, Other Things to TryProblem: Prove that Pow(a,b) takes no more than multiplications:;; Integer Natural -> IntegerPow(a,b): if b = 0: return 1 else if b = 1: return a else: let part = Pow(a,) if b is odd: return part*part*a else: return part*part

• Prove the algorithm correct.• Prove for all positive integers n that Pow(a,2n-1) takes 2(n-1)

multiplications.• Discuss why before our induction proof, we really should be saying

“WLOG, let a be an arbitrary integer.”

One More Example(Not as Good Fit to the Pattern)

Example 4: Product of Primes

Problem: Prove for any integer n 2 that n can be written as a product of primes. (For example, 7 alone is a product of primes (just one of them). 24 can be written as 2*3*2*2, among other ways, as a product of primes.)Reminder, an integer 2 or larger is a prime exactly when it has no integer factor larger than 1 but smaller than itself.An integer 2 or larger that is not prime is “composite”.


Problem: Prove for any integer n 2 that n can be written as a product of primes.




Example 4, Getting StartedProblem: Prove for any integer n 2 that n can be written as a product of primes.

Theorem: n can be written as a product of primes for all integers n 2.





Scratch work: P(n) n can be written as a product of primes“Structural” Insight: a problem of size n breaks down in terms of problems of size __________.State this insight explicitly. It may be right in the problem statement!

Say a definition is recursive. Then the self-reference tells you how the problem breaks down.



Scratch work: P(n) n can be written as a product of primes“Structural” Insight: For all composite integers n 2, n has a factor 1 < k < n; that is, n = k*q for some integer 1 < q < n. So, n can be written as (product of primes = to k)*(product of primes = to q).






P(n) is true.



Theorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For n = ___,




P(n) is true.

The smallest relevant value. (Here: n 2; so, 2.)



Theorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For n = 2, TODO: prove the property., 2 can be written as a product of primes.More?



P(n) is true.






P(n) is true.

TOTALLY UNCLEAR. Some cases aren’t composite. Some cases break down into other cases that aren’t composite. Let’s come back to it after we’re done with the inductive step!





P(n) is true.



Reminder: our structural insight is “n can be written as p*q for integers 1 < p < n and 1 < q < n.”



IH: WLOG, let n be an arbitrary composite integer > ??.Assume all “previous i” can be written as a product of primes (that is, all integers 1 < i < n).


P(n) is true.


Again, totally unclear what the largest base case is. Come back to it. Remember, our real goal is to keep n from being a base case.We do need n to be composite, however.




IS: Since n is composite, it has a factor k between 2 and n-1, where k*q = n for some (possibly different) factor q between 2 and n-1. By the IH, k and q can be written as products of primes.

P(n) is true.






n can be written as a product of primes.


Example 4: Finishing the ProofTheorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For n = 2, TODO: prove the property., 2 can be written as a product of primes.More?





How do we resolve this? Well, any composite integer actually works. Let’s drop it. But, we’ll need to handle anything not composite as a base case!

Example 4: Finishing the ProofTheorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For n = 2, TODO: prove the property., 2 can be written as a product of primes.More?

IH: WLOG, let n be an arbitrary composite integer (greater than or equal to 2).Assume all “previous i” can be written as a product of primes (that is, all integers 1 < i < n).




Let’s try the 2 base case and then work on “all primes” as base cases.

Example 4: Finishing the ProofTheorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For n = 2, 2 is itself a prime and so is a “product of primes” with just one term in the product. More?





Example 4: Finishing the ProofTheorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For all prime numbers n (greater than or equal to 2), n is itself a prime and so is a “product of primes” with just one term in the product.





Example 4: Finishing the ProofTheorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For all prime numbers n (greater than or equal to 2), n is itself a prime and so is a “product of primes” with just one term in the product.


IS: Since n is composite, it has a factor k between 2 and n-1, where k*q = n for some (possibly different) factor q between 2 and n-1. By the IH, k and q can be written as products of primes.Thus, n can be written as the product of primes formed by multiplying the product of primes equal to k by the product of primes equal to q.

QED

Example 4: What Does It Teach Us?Theorem: n can be written as a product of primes for all integers n 2.Proof: by induction on n.BC: For all prime numbers n (greater than or equal to 2), n is itself a prime and so is a “product of primes” with just one term in the product.


IS: Since n is composite, it has a factor k between 2 and n-1, where k*q = n for some (possibly different) factor q between 2 and n-1. By the IH, k and q can be written as products of primes.Thus, n can be written as the product of primes formed by multiplying the product of primes equal to k by the product of primes equal to q.

QED

Remember that we really write our induction proofs backward.(1) Figure out what the inductive step needs to assume and for what cases thestructural insight works. (2) For the IH, assume what needs to be assumed.(3) For the BC, prove every case that the IS cannot handle.

Pattern For Induction

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