1. Administrative Sep. 20 (today) – HW1 due Sep. 21 8am – problem session Sep. 25 – HW3 (=QUIZ #1) due Sep. 27 – HW4 due Sep. 28 8am – problem session Oct. 2 Oct. 4 – QUIZ #2 (pages 45-79 of DPV)

2. Merge 2 sorted lists M ERGE INSTANCE: 2 lists x i, y i such that x 1  x 2  …  x n y 1  y 2  …  y m SOLUTION: ordered merge

3. 1 i  1, j  1 2 while i  n and j  n do 3 if x i  y j then 4 output x i, i  i + 1 5 else 6 output y j, j  j + 1 7 output remaining elements Merge 2 sorted lists

4. M ERGE -S ORT (a,l,r) if l < r then m   (l+r)/2  M ERGE -S ORT (a,I,m) M ERGE -S ORT (a,m+1,r) M ERGE (a,l,m,r) Mergesort

5. Running time? Mergesort

6. Running time? Mergesort [ … n/2 … ] [... n … ] [ … n/4 … ] Depth ?

7. Running time? Mergesort [ … n/2 … ] [... n … ] [ … n/4 … ] Depth = log n

8. Time spent on merge? Mergesort [ … n/2 … ] [... n … ] [ … n/4 … ] Depth = log n

9. Time spent on merge? Mergesort [ … n/2 … ] [... n … ] [ … n/4 … ] Depth = log n O(n) O(n.logn)

10. recurrence T(n)= T(n/2) +  (n) if n>1 T(1)=  (1) Mergesort

11. Recurrences T(n)  T(  n/2  ) + T(  n/2  ) + c.n T(n)=O(n log n) T(n)  2 T( n/2 ) + c.n We “showed” : for

12. Recurrences T(1) = O(1) Proposition: T(n)  d.n.lg n Proof: T(n)  2 T( n/2 ) + c.n  2 d (n/2).lg (n/2) + c.n = d.n.( lg n – 1 ) + cn = d.n.lg n + (c-d).n   d.n.lg n T(n)  2 T( n/2 ) + c.n

13. Recurrences T(n)  2 T( n/2 ) + c.n T(n)  T(  n/2  ) + T(  n/2  ) + c.n G(n) = T(n+2) G(n) = T(n+2)  T(  n/2  +1) + T(  n/2  +1) + c.n = G(  n/2  -1) + G(  n/2  -1) + c.n

14. Recurrences useful guess – substitute (prove) or use “Master theorem” T(n) = a T(n/b) + f(n) If f(n) = O(n c-  ) then T(n) =  (n c ) If f(n) =  (n c ) then T(n) =  (n c.log n) If f(n) =  (n c+  ) then T(n)=  (f(n)) if a.f(n/b)  d.f(n) for some d n 0 c=log b a

15. Karatsuba-Offman a=2 n/2 a 1 + a 0 b=2 n/2 b 1 + b 0 ab=(2 n/2 a 1 +a 0 )(2 n/2 b 1 +b 0 ) = 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1 ) + a 0 b 0

16. Karatsuba-Offman a=2 n/2 a 1 + a 0 b=2 n/2 b 1 + b 0 Multiply(a,b,n) if n=1 return a*b else R 1  Multiply(a 1,b 1,n/2) R 2  Multiply(a 0,b 1,n/2) R 3  Multiply(a 1,b 0,n/2) R 4  Multiply(a 0,b 0,n/2) return 2 n R 1 + 2 n/2 (R 2 +R 3 ) + R 4

17. Karatsuba-Offman Multiply(a,b,n) if n=1 return a*b else R 1  Multiply(a 1,b 1,n/2) R 2  Multiply(a 0,b 1,n/2) R 3  Multiply(a 1,b 0,n/2) R 4  Multiply(a 0,b 0,n/2) return 2 n R 1 + 2 n/2 (R 2 +R 3 ) + R 4 Recurrence?

18. Karatsuba-Offman Multiply(a,b,n) if n=1 return a*b else R 1  Multiply(a 1,b 1,n/2) R 2  Multiply(a 0,b 1,n/2) R 3  Multiply(a 1,b 0,n/2) R 4  Multiply(a 0,b 0,n/2) return 2 n R 1 + 2 n/2 (R 2 +R 3 ) + R 4 Recurrence? T(n) = 4T(n/2) + O(n)

19. Karatsuba-Offman T(n) = 4T(n/2) + O(n) T(n)=O(n 2 )

20. Karatsuba-Offman ab=(2 n/2 a 1 +a 0 )(2 n/2 b 1 +b 0 ) = 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1 ) + a 0 b 0 Can compute in less than 4 multiplications?

21. Karatsuba-Offman ab=(2 n/2 a 1 +a 0 )(2 n/2 b 1 +b 0 ) = 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1 ) + a 0 b 0 Can compute using 3 multiplications: (a 0 +a 1 )(b 0 +b 1 ) = a 0 b 0 + (a 1 b 0 + a 0 b 1 ) + a 1 b 1

22. Karatsuba-Offman Multiply(a,b,n) if n=1 return a*b else R 1  Multiply(a 1,b 1,n/2) R 2  Multiply(a 0,b 0,n/2) R 3  Multiply(a 1 +a 0,b 1 +b 0,n/2+1) R 4  R 3 – R 2 – R 1 return 2 n R 1 + 2 n/2 R 3 + R 2 Recurrence?

23. Karatsuba-Offman Multiply(a,b,n) if n=1 return a*b else R 1  Multiply(a 1,b 1,n/2) R 2  Multiply(a 0,b 0,n/2) R 3  Multiply(a 1 +a 0,b 1 +b 0,n/2+1) R 4  R 3 – R 2 – R 1 return 2 n R 1 + 2 n/2 R 3 + R 2 Recurrence? T(n) = 3T(n/2) + O(n)

24. Recurrences T(n) = a T(n/b) + f(n) If f(n) = O(n c-  ) then T(n) =  (n c ) If f(n) =  (n c ) then T(n) =  (n c.log n) If f(n) =  (n c+  ) then T(n)=  (f(n)) if a.f(n/b)  d.f(n) for some d n 0 c=log b a T(n) = 3 T(n/2) +  (n) T(n) = 2T(n/2) +  (n.log n)

25. Karatsuba-Offman T(n) = 3T(n/2) + O(n) T(n)=O(n C ) C=log 2 3  1.58

26. Finding the minimum min  A[1] for i from 2 to n do if A[i]<min then min  A[i] How many comparisons?

27. Finding the minimum How many comparisons? comparison based algorithm: The only allowed operation is comparing the elements

28. Finding the minimum

29. Finding the k-th smallest element k =  n/2  = MEDIAN

30. Finding the k-th smallest element 6 3 1 8 7 2 6 1 8 5 8 9 1 3 2 631 8 7 26 1 8589 1 32

31. 6 3 1 8 7 2 6 1 8 5 8 9 1 3 2 1) sort each 5-tuple 631 8 7 26 1 8589 1 32

32. Finding the k-th smallest element 6 3 1 8 7 2 6 1 8 5 8 9 1 3 2 1) sort each 5-tuple

33. Finding the k-th smallest element 1 3 6 8 7 2 6 1 8 5 8 9 1 3 2 1) sort each 5-tuple

34. Finding the k-th smallest element 1 3 6 8 7 2 6 1 8 5 8 9 1 3 2 1) sort each 5-tuple

35. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 8 9 1 3 2 1) sort each 5-tuple

36. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 1) sort each 5-tuple TIME = ?

37. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 1) sort each 5-tuple TIME =  (n)

38. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 2) find median of the middle n/5 elements TIME = ?

39. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 2) find median of the middle n/5 elements TIME = T(n/5)

40. Finding the k-th smallest element 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 At least ? Many elements in the array are  X

41. Finding the k-th smallest element 1 2 3 9 7 At least ? Many elements in the array are  X 1 2 5 8 6 1 3 6 8 7

42. Finding the k-th smallest element 1 2 3 9 7 At least 3n/10 elements in the array are  X 1 2 5 8 6 1 3 6 8 7

43. Finding the k-th smallest element 1 2 3 9 7 At least 3n/10 elements in the array are  X 1 2 5 8 6 1 3 6 8 7 631 8 7 26 1 8589 1 32

44. Finding the k-th smallest element At least 3n/10 elements in the array are  X 631 8 7 26 1 8589 1 32 631 3 2 21 1 8589 6 87 XX XX

45. Finding the k-th smallest element At least 3n/10 elements in the array are  X 631 8 7 26 1 8589 1 32 631 3 2 21 1 8589 6 87 XX XX Recurse, time ?

46. Finding the k-th smallest element At least 3n/10 elements in the array are  X 631 8 7 26 1 8589 1 32 631 3 2 21 1 8589 6 87 XX XX Recurse, time  T(7n/10)

47. Finding the k-th smallest element 6 3 1 8 7 2 6 1 8 5 8 9 1 3 2 631 8 7 26 1 8589 1 32 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 63 1 87 2 618 58 913 2 63 1 32 2 1 1 85 8 968 7 XX XX recurse

48. Finding the k-th smallest element 6 3 1 8 7 2 6 1 8 5 8 9 1 3 2 631 8 7 26 1 8589 1 32 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 1 3 6 8 7 1 2 5 8 6 1 2 3 9 7 63 1 87 2 618 58 913 2 63 1 32 2 1 1 85 8 968 7 XX XX recurse  n) T(n/5)  n) T(7n/10)

49. Finding the k-th smallest element T(n)  T(n/5) + T(7n/10) + O(n)

50. Finding the k-th smallest element T(n)  T(n/5) + T(7n/10) + O(n) T(n)  d.n Induction step: T(n)  T(n/5) + T(7n/10) + O(n)  d.(n/5) + d.(7n/10) + O(n)  d.n + (O(n) – dn/10)  d.n

51. Why 5-tuples? 3 1 7 6 1 5 9 1 2 317165912 1 3 7 1 5 6 1 2 9 63 1 87 2 618 58 913 2 63 1 32 2 1 1 85 8 968 7 XX XX recurse  n) 1 3 7 1 5 6 1 2 9

52. Why 5-tuples? 3 1 7 6 1 5 9 1 2 317165912 1 3 7 1 5 6 1 2 9 63 1 87 2 618 58 913 2 63 1 32 2 1 1 85 8 968 7 XX XX recurse  n) T(2n/3) 1 3 7 1 5 6 1 2 9 T(n/3)

53. Why 5-tuples? T(n)  T(n/3) + T(2n/3) +  (n)

54. Why 5-tuples? T(n)  T(n/3) + T(2n/3) +  (n) T(n)  c.n.ln n Induction step: T(n) = T(n/3) + T(2n/3) +  (n)  c.(n/3).ln (n/3) + c.(2n/3).ln (2n/3) +  (n)  c.n.ln n - c.n.((1/3)ln 3+(2/3)ln 3/2)+  (n)  c.n.ln n