Induction

How do proofs look, about natural numbers? For example, here are two functions which process natural numbers; how can we prove that they are equivalent (that is, that they return the same result for all inputs, regardless of how they happen to compute that result)?

; Recall:
; Def'n of NatNum:
;  - 0
;  - add1(k), where k is a NatNum.
; 
; 
; How do programs look, which use this def'n?

;; sumA: NatNum -> NatNum
;; Return the sum of the natNums in [0,n].
;;
(define (sumA n)
  (cond [(zero?     n) 0]
        [(positive? n) (+ n (sumA (sub1 n)))]))
; 
; Calling sumA(4) calls sumA(3) calls sumA(2) ... calls sumA(0).

;; sumB: NatNum -> NatNum
;; Return the sum of the natNums in [0,n].
;;
(define (sumB n)
  (* n (add1 n) 1/2))


   ; See test cases in lect15-sum-test.ss


; Clearly sumB is a better solution than sumA --
; but is it correct?  Can we prove it correct?
; (Are there some implicit assumptions?  If so, what?)

Example 1

We want to prove these two functions are equivalent. We start by making a parameterized statement:

Let P(n) be the statement "(sumA n) = (sumB n)."

Here's the battle plan of proofs by induction:

• First we'll show P(0) (the “base case”).
• Then we'll show that ∀ k ≥ 0, P(k) → P(k+1) (the “inductive step”).

Proof of &forall n.P(n), by induction:

• Base case: (sumA 0) = 0 = 0*1/2 = (sumB n). This is P(0), as desired.

• Inductive step: For all k∈N, P(k) → P(k+1):
  We need to show that (sumA k+1) = (sumB k+1). One tool we'll have at hand is our inductive hypothesis P(k): (sumA k) = (sumB k).

  Okay, here goes: For k≥0,

        (sumA k+1) 
      = (+ k+1 (sumA (sub1 k+1)))  [by code of sumA]
      = (k+1) + (sumA k)           [by language's-implementation-of-arithmetic]
      = (k+1) + (sumB k)           [by inductive hypothesis P(k)]
      = (k+1) + (* k (add1 k) 1/2) [by code of sumB]
      = (k+1) +  k(k+1)/2          [by langauge's-implementation-of-arithmetic]
      = (k+1)*(1 + k/2)
      = (k+1)*(2/2 + k/2)
      = (k+1)*(k+2)/2
      = (* k+1 (add1 k+1) 1/2)
      = (sumB k+1)
  

  Following that entire chain of equalities, we see that we have P(k+1).

Really, this is a new inference rule (particular to N):

If you know…                       Then you can conclude…
  P(0)                            }
  forall k. (P(k) → P(k+1))     }  forall n.P(n)

By the way, a note on ∑_k=1ⁿ (k) = n(n+1)/2:

      *****
      ****
      ***
      **
      *

 S =   1   +    2  + … +  n-1  + n,
 S =   n   +  n-1  + … +    2  + 1
------------------------------------------ add both equations
2S = (n+1) + (n+1) + … + (n+1) + (n+1)   (n copies)

Example 2

board tiling, from Rosen.
P(n) = "A checkerboard 2ⁿ x 2ⁿ with one corner removed, can be exactly tiled by L-shaped tri-ominoes."

• P(1). Check.
• P(k) → P(k+1), forall k>0:
  …[do in lecture]…

Example 3

A homework exercise:
• Show: &sum_k=1ⁿ 2^k = 2^k+1-1.
• Show: &sum_k=1ⁿ b^k = (b^k+1-1)/(b-1).
E.g., when b=10, we have 1+10+100+1000 = 1111 = 9999/9.

…[do the proof in lecture]…

Reflection on this sum:

• This means that for exponentially growing terms, the last one dominates all the previous, and is a pretty good approximation to the sum.
  
• Alternatively: the next term is nearly b times bigger than the entire sum of previous terms! [More precisely, it's nearly b-1 times bigger.]
• Think of: Collecting an exponential-growth petri colonies (a 1-day old, a 2-day old, a 3-day old, …) Most of your biomass comes from the very last day, if b≥2.)
• In a full n-ary tree, more than (n-1)/n of the nodes are leaves.
  (This sum corresponds to balanced full trees, but in fact ``full'' is sufficient. To prove this stronger statement, we can't quite use mathematical induction (why not?). But happily, in the next two lectures we'll see strong induction and structural induction, each of which can show this stronger statement.)
• Again, there's a direct-method algebraic proof of this sum:

      S  = 1 + b + b2 + … + bk
     bS  =     b + b2 + … + bk + bk+1
  -------------------------------------------------------
   S-bS  = 1                    - bk+1
  

  You think "Aw man, why'd we have to do this roundabout induction method, when we just could have written that?" Aside from the fact that it's the induction we're trying to teach (and not so much the proof of this particular equation), another way to view it is to say that induction is taking the inherent fuzziness of the ``…'' out of the the algebraic sleight-of-hand. Note that this fuzziness is still there even if you use the summation notation — it's just absorbed into the notation.

Note that induction doesn't provide us with any hint as to what our sums should look like, but if we have some guess then it gives us a method to prove that guess correct. The book has the formulas for (say) the sum-of-squares and sum-of-cubes. What about ∑_k=1ⁿk¹⁷, if we really needed a nice expression for this? We could guess that the answer is an 18th-degree polynomial, set up a generic polynomial with coefficients a₁₈,…a₀, and then (by plugging in 19 different values for n) come up with 19 linear equations in the 19 unknowns a₁₈,…a₀. We then solve for these coefficients. Finally, to wrap it off entirely, we could use induction to prove the formula correct (since it was intuition that an 18th-degree polynomial really would provide a solution).

Example 4, an inequality: Harmonic series (in Rosen p.243, Example #6):
H_2^k ≥ 1+(k/2).
Note that this can be re-stated as, H_n ≥ 1+lg(n)/2, though suddenly we'd want to show this for all n (not just powers-of-2).

More complicated example

If all cities connected in one direction or the other, then there is some city which is connected to all others in either one or two steps.

After some work, I found I was using the same phrases over and over, like "a city connected to all cities within one or two steps". Some select definitions make the problem not only more concise, but also more understandable:

• A graph is semi-connected if for all vertices v,w E(v,w) or E(w,v).
  (Corollary: every vertex of a semi-connected graph has a self-loop.)
• A hub h is a vertex such that for all vertices v, there is a path of length ≤ 2 from h to v.

Given a semi-connected graph G, choose an arbitrary vertex v.
Note that G-{v} is semi-connected. [Aside: Def'n of G-{v} ?]
Let h₀ be "old" hub of G-{v}.

• Case 1: E(h₀,v). Done; h₀ is still hub for all G.
• Case 2: E(v,h₀). Consider H1, set of all 1-neighbors of h₀ in G-{v},
  and H2, set of all 2-neighbors of h₀.
  Note that {h₀},H1,H2 partition G-{v}.
  • Case 2a: for some n in H1, E(n,v). Done: h₀ is still hub for all G.
  • Case 2b: v is new hub: It has an path of length one to each of H1 &union; {h₀}, and it has a path of length two to each of H2.

False proofs-by-induction

What's wrong with this inductive proof: P(n) is "n > 1000".
(After all, it's easy to show P(k) → P(k+1).)
Okay, we'll patch that problem; what's wrong with the induction proof of: P(n) is "n>1000 or n=0".

False proof: All horses are the same color.

Proof by induction for n>1. let P(n) be "for a set of n horses, they all have the same color".

• P(1): yep, in a singleton set {h1}, every horse is h1's color.

• P(k) → P(k+1), ∀ k ≥ 1:
  We need to show P(k+1). Start with any set of k+1 horses; we'll show every member of H = { h₁, …, h_k, h_k+1 } has the same color as h₁.

  First, h₁ through h_k all have the same color as h₁ (by inductive hypothesis regarding the set H - {h_k+1}). It only remains to show that h_k+1 is the same color.
  But again by applying P(k) to the k-element set H-{h_k} = {h₁, …, h_k-1, h_k+1 }, all have the same color as h₁ (which includes h_k+1).

  Since "has the same color as" is transitive, all elements of H are the same color as h₁. Q.E.D.

The question is, what is wrong with the proof? (We know the conclusion is wrong; we are asking, why is this proof not really following the rule of induction. If it were, then we'd have an example that induction isn't actually a valid rule of inference!)

Q&A

• Q: When we make a parameterized statement P(k), is the range of values for k part of P?
  A: No.
  • Right: Let P(n) be “In a game of Nim w/ n stones, player 1 has a win iff remainder(n,4)=0.”
    We will show P(n) holds for all n≥0.
  • Wrong: Let P(n) be “In a game of Nim w/ n stones, player 1 has a win iff remainder(n,4)=0, for all n≥0.”

• Q: How formal should this step in showing that B ≤ C be?

  A: For inequalities you can state "B ≤ C" as obvious when the right-hand-side C is:

  • B + some-positive-amount, or
  • like B but with a (bunch of) subterms (each) replaced with larger amounts.

  Thus like in today's class example with the harmonic-number sum:

    1 + 1/2 + 1/3 + … + 1/2k  + 1/(2k+1) + 1/(2k+2) + … + 1/(2k+2k)
  ≥ 1 + 1/2 + 1/3 + … + 1/2k  + 1/2k+1    + 1/2k+1   + … + 1/2k+1
  

  is fine, with the explanation "Replacing some terms by larger ones". Also, indenting can go a long way in making it clear how you're doing the replacement.

  Similarly, it's fine to say

  blah ≤ blah + (a² - 1)

  (in this case, need to take care that a²-1 is non-negative, which means you might need to mention why a isn't zero.)

  Tip: when using < vs ≤, use the strongest statement possible — e.g. use < when the inequality really is strict. (So in the harmonic example above, is the "<" actually true for k=0?)