Strong Induction, WOP and Invariant Method

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Strong Induction, WOP and Invariant Method

• Leo Cheung

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Project

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A Quick Review

• Strong induction
• Well ordering principle
• Invariant method

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Recall Strong Induction (Lecture 5, Slide 35)
Prove P(0). Then prove P(n1) assuming all of
P(0), P(1), , P(n) (instead of just
P(n)). Conclude ?n.P(n)
Strong induction
equivalent
0 ? 1, 1 ? 2, 2 ? 3, , n-1 ? n. So by the time
we got to n1, already know all of
P(0), P(1), , P(n)
Ordinary induction
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Strong Induction

• Given
• f1 1
• fk 2 ffloor(k/2)
• Prove fn lt n

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Strong Induction - Answer

• Base case
• f1 1 lt 1
• Assume fj j, 1 j k
• fk1 2 ffloor((k1)/2)
• 2 floor((k1)/2)
• k1
• Done.

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Strong Induction

• Given
• g0 12
• g1 29
• gk 5gk-1 6gk-2 for all integers k gt2
• Prove gn 5x3n 7x2n for all integers ngt0

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Strong Induction - Answer

• Base Case
• n 0, 1 Trivial
• Assume the hypothesis is true for 0,1,2,.k-1,k
• By induction, the proof is done.

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Well Ordering Principle

• Every non-empty set of positive integers contains
  a smallest element.

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Well Ordering Principle

• To prove P is true
• Assume P is false
• Then there are set of counter examples
• C c P(c) is false
• By WOP, choose the smallest element x in C
• From x, we get y lt x s.t. P(y) is also false
• Contradiction!

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Well Ordering Principle

• The equation
  has no non-zero positive integer solution

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Well Ordering Principle - Ans

• Assume the is solution for the equation
• By WOP, we pick the solution (u,v,w,x)
  (a,b,c,d) such that (a,b,c,d) has the smallest
  value of u among all the solutions to the
  equation.
• We get
• Since all the term of left hand side is even, d
  must be even, so
• where

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Well Ordering Principle - Ans

• Divide both sides by 2 we get
• Here c must be even, again we can rewrite the
  equation as
• where
• Divide both sides by 2 and we get

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Well Ordering Principle - Ans

• Apply the same technique to b and a, we will get
• where
• Observe that (a,b,c,d) is now another set of
  solution to the equation
  and a lt a
• This contradicts the claim that (a,b,c,d) is the
  solution set with the smallest value of u
• The assumption is false, hence the equation has
  no nonzero positive integer solution.

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Invariant Method

• Find properties (the invariants) that are
  satisfied throughout the whole process.
• Show that the target do not satisfy the
  properties.
• Conclude that the target is not achievable.

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Invariant Method

• Integers 1, 2, 3, 4 and 5 are written on a board.
  Tom picks any two of the numbers, deletes them,
  and writes on the board the absolute value of
  their difference. He repeats this procedure with
  the resulting 4 numbers, and so on. After he does
  it 4 times, only one number remains on the board.
  Can this number be 2?

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Invariant Method - Answer

• Observe that whatever move you made, you can only
  get an odd number in the final answer.
• Observe that the total sum of the numbers on the
  board remains odd.
• Consider a move, you pick m and n on the board,
  and write back m-n.
• Change in total sum mn-(m-n) 2n
• Which means whatever move you make, the change in
  total sum must be even
• The parity of total sum is the invariant
• Since we start with an odd total sum 15, it is
  not possible have 2 as the final number because
  it is even.

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Exercises

• Following exercises can be found in textbook
• 4.4 4, 15, 18, 22

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END