1. 241-303 Discrete Maths: Induction/1 1 Discrete Maths Objective – –to introduce mathematical induction through examples 241-303, Semester 1 2014-2015 1. Mathematical Induction

2. 241-303 Discrete Maths: Induction/1 2 Overview 1. Motivation 2. Induction Defined 3.Maths Notation Reminder 4.Four Examples 5.More General Induction Proofs 6.A Fun Tiling Problem 7.Further Information

3. 241-303 Discrete Maths: Induction/1 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. 241-303 Discrete Maths: Induction/1 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. 241-303 Discrete Maths: Induction/1 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 "n 2 + 1 > 0" – –is S(n) true for all n >= 1? continued S(n) can be much more complicated, such as a program that reads in a n value.

6. 241-303 Discrete Maths: Induction/1 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. 241-303 Discrete Maths: Induction/1 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. 241-303 Discrete Maths: Induction/1 8 Second, assume that S(n) is true, and use it to show that S(n+1) is true – –mathematically, we show that S(n) --> S(n+1) Now we know S(n) is true for all n>=1. Why? continued "--> stands for "implies"

9. 241-303 Discrete Maths: Induction/1 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. 241-303 Discrete Maths: Induction/1 10 Let’s do it Prove S(n): "n 2 + 1 > 0" for all n >= 1. First task: show S(1) is true – –S(1) == 1 2 + 1 == 2, which is > 0 – –so S(1) is true Second task: show S(n+1) is true by assuming S(n) is true continued

11. 241-303 Discrete Maths: Induction/1 11 Assume S(n) is true, so n 2 +1 > 0 Prove S(n+1) – –S(n+1) == (n+1) 2 + 1 == n 2 + 2n + 1 +1 == (n 2 + 1) + 2n + 1 – –since n 2 + 1 > 0, then (n 2 + 1) + 2n + 1 > 0 – –so S(n+1) is true, by assuming S(n) is true – –so S(n) --> S(n+1) continued

12. 241-303 Discrete Maths: Induction/1 12 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

13. 241-303 Discrete Maths: Induction/1 13 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

14. 241-303 Discrete Maths: Induction/1 14 3. An inductive step for the proof. We prove the statement “S(n) --> S(n+1)” for all n. – –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

15. 241-303 Discrete Maths: Induction/1 15 3. Maths Notation Reminder Summation: means 1+2+3+4+…+n – –e.g.means 4+9+16+…+m 2 Product: means 1*2*3*…*n

16. 241-303 Discrete Maths: Induction/1 16 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 = 1 so 1 = (1*2)/2 continued (1)

17. 241-303 Discrete Maths: Induction/1 17 Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued

18. 241-303 Discrete Maths: Induction/1 18 Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = (n 2 + n + 2n + 2) /2 = (n 2 +3n+2)/2 which is (2)

19. 241-303 Discrete Maths: Induction/1 19 Example 2 Prove the statement S(n): for all n >= 0 – –e.g. 1+2+4+8 = 16-1 Basis. S(0), n = 0 so 2 0 = 2 1 -1 continued (1)

20. 241-303 Discrete Maths: Induction/1 20 Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued

21. 241-303 Discrete Maths: Induction/1 21 Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = 2*(2 n+1 ) - 1 which is (2) +1

22. 241-303 Discrete Maths: Induction/1 22 Example 3 Prove the statement S(n): n! >= 2 n-1 for all n >= 1 – –e.g. 5! >= 2 4, which is 120 >= 16 Basis. S(1), n = 1:1! >= 2 0 so 1 >= 1 continued (1)

23. 241-303 Discrete Maths: Induction/1 23 Induction. Assume S(n) to be true. Prove S(n+1), which is: (n+1)! >= 2 (n+1)-1 >= 2 n (2) Notice that: (n+1)! = n! * (n+1)(3) continued

24. 241-303 Discrete Maths: Induction/1 24 Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: (n+1)! >= 2 n-1 * (n+1) >= 2 n-1 * 2 since (n+1) >= 2 (n+1)! >= 2 n which is (2) why?

25. 241-303 Discrete Maths: Induction/1 25 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 = 1 so 1/2 = 1/2 continued (1)

26. 241-303 Discrete Maths: Induction/1 26 Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued

27. 241-303 Discrete Maths: Induction/1 27 Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = which is (2)

28. 241-303 Discrete Maths: Induction/1 28 5. More General Inductive Proofs There can be more than one basis case. We can do a complete induction (or “strong” induction) where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n) – –b is the lowest basis value

29. 241-303 Discrete Maths: Induction/1 29 Example We claim that 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

30. 241-303 Discrete Maths: Induction/1 30 Basis. The 5 basis cases are 24 through 28. – –24 = (5*2) + (7*2) – –25 = (5*5) + (7*0) – –26 = (5*1) + (7*3) – –27 = (5*4) + (7*1) – –28 = (5*0) + (7*4) continued

31. 241-303 Discrete Maths: Induction/1 31 Induction: Let n+1 >= 29. Then n-4 >= 24, the lowest basis case. – –Thus S(n-4) is true, and we can write n - 4 = 5a + 7b – –Thus, n+1 = 5(a+1) + 7b, proving S(n+1)

32. 241-303 Discrete Maths: Induction/1 32 6. A Fun Tiling Problem A right tromino is a “corner” shape: Use induction to proof that right trominos can be used to tile (cover) any n*n size board, where n is a power of 2 and the board has 1 square missing. continued

33. 241-303 Discrete Maths: Induction/1 33 For example, a 4*4 board (2 2 *2 2 ), minus 1 square:

34. 241-303 Discrete Maths: Induction/1 34 7. Further Information DM: section 1.6