1. CSC 282 – Algorithms Daniel Stefankovic – CSB 620 stefanko@cs.rochester.edu TA: Girts Folkmanis – CSB 614 gfolkman@cs.rochester.edu www.cs.rochester.edu/~stefanko/Teaching/06CS282

2. Quiz – problem 1

3. Quiz – problem 2 45 l=1 r=2 m=1 B=5 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ”

4. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ”

5. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r]

6. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r] A[m]<B  (B  A  B  A[m+1..r])

7. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r] DONE

8. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r] DONE ?

9. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r] Still need to prove that the algorithm terminates.

10. Quiz – problem 2 1 l  1, r  n 2 while l < r do 3 m  (l+r)/2  4 if A[m] < B then 5 l  m+1 6 else 7 r  m 8 fi 9 od ANSWER = “ is A[l]=B ” B  A  B  A[l..r] Q = r-l Q’ = r-(  (l+r)/2  +1) Q’ =  (l+r)/2  -l

11. Quiz – problem 3 PROBLEM: rotate an array in-place this type problem given on interviews in MSFT, GOOG homework

12. 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

13. 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

14. 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

15. 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

16. 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)

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

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

19. Finding the minimum

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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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 = ?

28. 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)

29. 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 = ?

30. 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)

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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 ?

37. 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)

38. 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

39. 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)

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

41. 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

42. 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

43. 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)

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

45. 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