1 / 25

Induction and Invariant Method

Induction and Invariant Method. Leo Cheung. News. Register your project group at https://spreadsheets.google.com/viewform?hl=en&formkey=dG41TGJEQ19RM2VjX1R1US1ONzZqblE6MA#gid=0 Homework 1 is posted Bring your calculator to the next 3 weeks tutorial. A Quick Review. Induction

tyrell
Download Presentation

Induction and Invariant Method

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Induction and Invariant Method Leo Cheung

  2. News • Register your project group at • https://spreadsheets.google.com/viewform?hl=en&formkey=dG41TGJEQ19RM2VjX1R1US1ONzZqblE6MA#gid=0 • Homework 1 is posted • Bring your calculator to the next 3 weeks tutorial

  3. A Quick Review • Induction • Strong induction • Well ordering principle • Invariant method

  4. Induction - 1 7n – 1 is divisible by 6, for all n ≥ 1.

  5. Induction - 1 7n – 1 is divisible by 6, for all n ≥ 1. Base case: n=1 71-1=6 which is divisible by 6. Assume the statement is true for n=k, i.e. 7k-1=6m for some m When n = k+1 7k+1-1 = 7(6m+1) - 1 = 42m + 6 = 6(7m+1) So by induction 7n – 1 is divisible by 6, for all n ≥ 1.

  6. Induction - 2 For all integer n ≥ 0

  7. Induction - 2 For all integer n ≥ 0 Base cases: n=0, L.S.= 2 =R.S. Assume the statement is true for n=k, i.e When n = k+1

  8. Induction - 3 • For which positive integers n satisfy 2n+3≤2n? Prove your answer using induction.

  9. Induction - 3 • You can check the statement is wrong for n=1,2,3. • When n=4, 2(4)+3=11<16=24 • Assume 2k+3<2k (k≥4) • When n=k+1 Hence by induction 2n+3<2n for all integer n≥4

  10. Induction - 4 • Which of the powers of 9 (90 91 92 93 …) have 9 as unit digit? Proof your answer by induction.

  11. Induction - 4 • Lets prove the statement 92n and 92n+1 has unit digit 1 and 9 respectively for all integer n≥0. • Base case: Trivial • Assume 92k and 92k+1 has unit digit 1 and 9 respectively • When n=k+1 • 92(k+1)=9 x 92k+1 which has unit digit 1 • More precisely • Very similar proof for 92(k+1)+1 • Hence 92n+1 has unit digit 9 for all integer n≥0

  12. Recall Strong Induction (Lecture 6, Slide 11) Prove P(0). Then prove P(n+1) 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 n+1, already know all of P(0), P(1), …, P(n) Ordinary induction

  13. Strong Induction - 1 Given for all integer k ≥ 2 Prove fn ≤ n for all integer n ≥ 1

  14. Strong Induction - 1 Base case f1 = 1 ≤ 1 Assume fj ≤ j, for all j such that 1 ≤ j ≤ k Done.

  15. Strong Induction - 2 • Given • g0 = 12 • g1 = 29 • gk = 5gk-1 – 6gk-2 for all integers k ≥ 2 • Prove gn = 5x3n + 7x2n for all integers n ≥ 0

  16. Strong Induction - 2 • Base Case • n = 0, 1 : Trivial • Assume the hypothesis is true for 0,1,2,….k-1,k By induction, the proof is done.

  17. Strong induction - 3 • Show that a 6xn board (n≥2) can be tiled with L-shaped tiles, without gap and overlapping.

  18. Strong Induction - 3 Assume board of dimension 6x2, 6x3, 6x4 …. 6x(k-1), 6xk can be tiled When n=k+1 N=2 N=3 k+1 columns Tile 2 columns, the remaining k-1 columns can be tiled by inductive assumption

  19. Strong Induction - 4 • You get a chocolate bar with mxn squares. Your task is to split the bar into 1x1 squares. Show that you always need mn-1 breaks. You can only break one piece at a time.

  20. Strong Induction - 4 • Chocolate with A=mxn squares need A-1 breaks. • A=1: trivial • Assume choco with area j need j-1 breaks, for all 1 ≤ j ≤ k • When A=k+1 • We break the size k+1 choco to two piece of size X and Y, X+Y=k+1 • By inductive assumption • The size X pieces need X-1 breaks (1 ≤ X ≤ k) • The size Y pieces need Y-1 breaks (1 ≤ Y ≤ k) • Hence in total we need 1+(X-1)+(Y-1) = X+Y-1=k breaks

  21. 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.

  22. 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?

  23. 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 = m+n-(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.

  24. Exercises • Following exercises can be found in textbook • 4.2 : 12 13 15 • 4.3 : 10 14 16 19 24 34 • 4.4 : 4 15

  25. END

More Related