CSC 282 – Algorithms Daniel Stefankovic – CSB 620 TA: Girts Folkmanis – CSB 614

Quiz – problem 1

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 ”

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 ”

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]

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])

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

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 ?

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.

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

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

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

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

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

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

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)

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

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

Finding the minimum

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

Finding the k-th smallest element

) sort each 5-tuple

Finding the k-th smallest element ) sort each 5-tuple

Finding the k-th smallest element ) sort each 5-tuple

Finding the k-th smallest element ) sort each 5-tuple

Finding the k-th smallest element ) sort each 5-tuple

Finding the k-th smallest element ) sort each 5-tuple TIME = ?

Finding the k-th smallest element ) sort each 5-tuple TIME =  (n)

Finding the k-th smallest element ) find median of the middle n/5 elements TIME = ?

Finding the k-th smallest element ) find median of the middle n/5 elements TIME = T(n/5)

Finding the k-th smallest element At least ? Many elements in the array are  X

Finding the k-th smallest element At least ? Many elements in the array are  X

Finding the k-th smallest element At least 3n/10 elements in the array are  X

Finding the k-th smallest element At least 3n/10 elements in the array are  X

Finding the k-th smallest element At least 3n/10 elements in the array are  X XX XX

Finding the k-th smallest element At least 3n/10 elements in the array are  X XX XX Recurse, time ?

Finding the k-th smallest element At least 3n/10 elements in the array are  X XX XX Recurse, time  T(7n/10)

Finding the k-th smallest element XX XX recurse

Finding the k-th smallest element XX XX recurse  n) T(n/5)  n) T(7n/10)

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

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

Why 5-tuples? XX XX recurse  n)

Why 5-tuples? XX XX recurse  n) T(2n/3) T(n/3)

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

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