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Algorithm Design and Analysis (ADA)

Algorithm Design and Analysis (ADA). 242-535 , Semester 1 2013-2014. Objective to introduce mathematical induction through examples . 1. M athematical Induction. Overview. 1 . Motivation 2 . Induction Defined 3 . Maths Notation Reminder 4 . Four Examples

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Algorithm Design and Analysis (ADA)

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  1. Algorithm Design and Analysis (ADA) 242-535, Semester 1 2013-2014 • Objective • to introduce mathematical induction through examples 1. Mathematical Induction

  2. Overview 1. Motivation 2. Induction Defined 3. Maths Notation Reminder 4. Four Examples 5. More General Induction Proofs 6. A Fun Tiling Problem

  3. 1. Motivation • Induction is used in mathematical proofs of many recursive algorithms • e.g. quicksort, binary search • Induction is used to mathematically define recursive data structures • e.g. lists, trees, graphs continued

  4. Induction is often used to derive mathematical estimates of program running time • timings based on the size of input data • e.g. time increases linearly with the number of data items processed • timings based on the number of times a loop executes

  5. 2. Induction Defined • Induction is used to solve problems such as: • is S(n) correct/true for all n values? • usually for all n >= 0 or all n >=1 • Example: • let S(n) be "n2 + 1 > 0" • is S(n) true for all n >= 1? S(n) can be much more complicated, such as a program that reads in ann value. continued

  6. How do we prove (disprove) S(n)? • One approach is to try every value of n: • is S(1) true? • is S(2) true? • ... • is S(10,000) true? • ... forever!!! Not very practical

  7. Induction to the Rescue • Induction is a technique for quickly proving S(n) true or false for all n • we only have to do two things • First show that S(1) is true • do that by calculation as before continued

  8. " stands for "implies" • We prove that S(n) S(n+1)is true • Read as "if S(n) is true then S(n+1) is true" • When S(1) is true and "S(n)  S(n+1)" is true, then S(n) is true for all n>=1. • Why? continued

  9. With S(1) and S(n)  S(n+1)then S(2) is true • S(1)  S(2) when n == 1 • With S(2) and S(n)  S(n+1)then S(3) is true • S(2)  S(3) when n == 2 • With S(3) and S(n)  S(n+1)then S(4) is true • S(3)  S(4) when n == 3 • and so on, for all n

  10. Proving S(n)  S(n+1) • We prove the implication by: • 1. Assume that S(n) is true; write it down • 2. Write down some extra maths, which depends on the problem • e.g. (n+1)! = n! * (n+1) • 3. Write down S(n+1) • 4. Combine S(n) and the maths to obtain S(n+1)

  11. Let’s do it • Prove S(n): "n2 + 1 > 0" for all n >= 1. • First task: show S(1) is true by calculation • S(1) == 12+ 1 == 2, which is > 0 • so S(1) is true • Second task: prove S(n)  S(n+1) is true continued

  12. Assume S(n) is true • S(n): n2 + 1 > 0 • Extra maths: n > 0 • Write down S(n+1) • S(n+1) == (n+1)2 + 1 == n2 + 2n + 1 +1 == (n2 + 1) + 2n + 1 continued

  13. Use S(n) and the extra maths to show that S(n+1) is true • S(n) means that n2 + 1 > 0, and the extra maths is n > 0,so (n2 + 1) + 2n + 1 > 0 • this means that S(n+1) is true • so S(n)  S(n+1) continued

  14. We have used induction to show two things: • S(1) is true • S(n)  S(n+1) is true • From these it follows that S(n) is true for all n >= 1

  15. Induction More Formally • Three pieces: • 1. A statement S(n) to be proved • the statement must be about an integer n • 2. A basis for the proof. This is the statement S(b) for some integer. Often b = 0 or b = 1. continued

  16. 3. Aninductive step for the proof. We prove the statement “S(n)  S(n+1)” • The statement S(n), used in this proof, is called the inductive hypothesis • We conclude that S(n) is true for all n >= b • S(n) might not be true for some n < b

  17. 3. Maths Notation Reminder • Summation: means 1+2+3+4+…+n • e.g. means 4+9+16+…+m2 • Product: means 1*2*3*…*n

  18. 4. Example 1 • Prove the statement S(n):for all n >= 1 • e.g. 1+2+3+4 = (4*5)/2 = 10 • Basis. S(1), n = 1so 1 = (1*2)/2 continued

  19. Inductive Step. Prove S(n)  S(n+1) • 1. Assume S(n) is true: • 2. The extra maths involve summations: • 3. We want to calculate S(n+1): (1) (2) (3) continued

  20. To prove S(n)  S(n+1), combine S(n) (equation 1) and the extra maths (equation 2) to calculate S(n+1) (equation 3). continued

  21. Substitute the right hand side (rhs) of (1) for the first operand of (2), to give: = (n2 + n + 2n + 2) /2 = (n2+3n+2)/2 which is (3) continued

  22. We have shown: • S(1) is true • S(n)  S(n+1) is true • This means that S(n) is true for all n ≥ 1 • Finished. continued

  23. Example 2 • Prove the statement S(n):for all n >= 0 • e.g. 1+2+4+8 = 16-1 • Basis. S(0), n = 0so 20 = 21 -1 continued

  24. Inductive Step. Prove S(n)  S(n+1) • 1. Assume S(n) is true: • 2. The extra maths involve summations: • 3. We want to calculate S(n+1): (1) (2) (3) continued

  25. To prove S(n)  S(n+1), combine S(n) (equation 1) and the extra maths (equation 2) to calculate S(n+1) (equation 3). continued

  26. Substitute the right hand side (rhs) of (1) for the first operand of (2), to give: = 2*(2n+1) - 1 which is (3)

  27. We have shown: • S(0) is true • S(n)  S(n+1) is true • This means that S(n) is true for all n ≥ 0 • Finished. continued

  28. Example 3 • Prove the statement S(n): n! >= 2n-1for all n >= 1 • e.g. 5! >= 24, which is 120 >= 16 • Basis. S(1), n = 1: 1! >= 20so 1 >= 1 continued

  29. Inductive Step. Prove S(n)  S(n+1) • 1. Assume S(n) is true: n! >= 2n-1 • 2. The extra maths involve factorials:(n+1)! = n! * (n+1) • 3. We want to calculate S(n+1):(n+1)! >= 2(n+1)-1>=2n (1) (2) (3) continued

  30. To prove S(n)  S(n+1), combine S(n) (equation 1) and the extra maths (equation 2) to calculate S(n+1) (equation 3). continued

  31. Substitute the right hand side (rhs) of (1) for the first operand of (2), to give: (n+1)! >= 2n-1 * (n+1)>= 2n-1 * 2since (n+1) >= 2 (n+1)! >= 2n which is (3) why?

  32. We have shown: • S(1) is true • S(n)  S(n+1) is true • This means that S(n) is true for all n ≥ 1 • Finished. continued

  33. Example 4 • Prove the statement S(n):for all n >= 1 • This proof can be used to show that. the limit of the sum:is 1 • Basis. S(1), n = 1so 1/2 = 1/2 continued

  34. Inductive Step. Prove S(n)  S(n+1) • 1. Assume S(n) is true: • 2. The extra maths involve summations: • 3. We want to calculate S(n+1): (1) (2) (3) continued

  35. To prove S(n)  S(n+1), combine S(n) (equation 1) and the extra maths (equation 2) to calculate S(n+1) (equation 3). continued

  36. Substitute the right hand side (rhs) of (1) for the first operand of (2), to give: = which is (3)

  37. We have shown: • S(1) is true • S(n)  S(n+1) is true • This means that S(n) is true for all n ≥ 1 • Finished. continued

  38. 5. More General Inductive Proofs • There can be more than one basis case. • We can usestronginduction where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n) • b is the lowest basis value

  39. Strong Induction Example • Show: • S(1) is true • S(2) is true • S(3) is true • S(n-2)  S(n+1) is true • This means that S(n) is true for all n ≥ 1 • Why? 3 base cases

  40. Example • Every integer >= 24 can be written as 5a+7b for non-negative integers a and b. • note that some integers < 24 cannot be expressed this way (e.g. 16, 23). • Let S(n) be the statement (for n >= 24) “n = 5a + 7b, for a >= 0 and b >= 0” continued

  41. Using Strong Induction • We will show: • S(24) is true • S(25) is true • S(26) is true • S(27) is true • S(28) is true • S(n-4)  S(n+1) is true • This means that S(n) is true for all n ≥ 24 • Why? 5 base cases

  42. Basis. The 5 basis cases are 24 through 28. • S(24) true because24 = (5*2) + (7*2) • S(25) true because 25 = (5*5) + (7*0) • S(26) true because 26 = (5*1) + (7*3) • S(27) true because 27 = (5*4) + (7*1) • S(28) true because 28 = (5*0) + (7*4) continued

  43. Inductive Step. Prove S(n-4)  S(n+1) • 1. Assume S(n-4) is true: n - 4 = 5a + 7b • 2. No extra maths • 3. We want to calculate S(n+1):n+1 = 5a'+7b' • To prove S(n-4)  S(n+1), use S(n-4) (equation 1) to calculate S(n+1) (equation 2). (1) (2) continued

  44. Use S(n-4): n-4 = 5a + 7b • Add 5 to both sides of the equation: • n-4 + 5 = 5a + 7b + 5 • n + 1 = 5(a + 1) + 7b • n + 1 = 5a' + 7b (a' is some new variable) • which means that S(n+1) is true

  45. We have shown: • S(24), S(25), S(26), S(27), S(28) are true • S(n-4)  S(n+1) is true • This means that S(n) is true for all n ≥ 24 • Finished. continued

  46. 6. A Tiling Problem • A right tromino is a “corner” shapemade of 3 squares: • Use induction to prove that any number of right trominos can be used to tile (cover) any n*n size board of squares • but n must be a power of 2 • e.g. board sizes can be 2*2, 4*4, 8*8, 16*16, ... • Show that a tiled board will always have 1 blank square. continued

  47. For example, a 4*4 board (22*22), with 1 blank square: 5 trominos used

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