Objective. - mathematical induction, recursive definitions - arithmetic manipulations, series, products
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1 Recurrences
2 Objective running time as recursive function solve recurrence for order of growth method: substitution method: iteration/recursion tree method: MASTER method prerequisite: - mathematical induction, recursive definitions - arithmetic manipulations, series, products
3 Pipeline Proble Algorithm Running Time Recursion Solve Recursion
4 Running time will call it T(n) = number of computational steps required to run the algorithm/program for input of size n we are interested in order of growth, not exact values - for example T(n) = Θ(n 2 ) means quadratic running time - T(n) = O(n logn) means T(n) grows not faster than CONST*n*log(n) for simple problems, we know the answer right away - example: finding MAX of an array - solution: traverse the array, keep track of the max encountered - running time: one step for each array element, so n steps for
5 Running time for complex problems complex problems involve solving subproblems, usually - init/prepare/preprocess, define subproblems - solve subproblems - put subproblems results together thus T(n) cannot be computed straight forward - instead, follow the subproblem decomposition
6 Running time for complex problems often, subproblems are the same problem for a smaller input size: - for example max(array) can be solved as: - split array in array_left, array_right - solve max(array_left), max (array_right) - combine results to get global max Max(A=[a 1,a2,...,an]) if (n==1) return a1 k= n/2 max_left = Max([a1,a2,..,ak]) max_right = Max([ak+1,ak+2,..,an]) if(max_left>max_right) return max_left else return max_right
7 Running time for complex problems many problems can be solved using a divide-andconquer strategy - prepare, solve subproblems, combine results running time can be written recursively - T(n) = time(preparation) + time(subproblems) + time(combine) - for MAX recursive: T(n) = 2*T(n/2) + O(1)
8 Running time for complex problems many problems can be solved using a divide-andconquer strategy - prepare, solve subproblems, combine results running time can be written recursively - T(n) = time(preparation) + time(subproblems) + time(combine) - for MAX recursive: T(n) = 2*T(n/2) + O(1) 2 subproblems of size n/2 max(array_left) ; max(array_right)
9 Running time for complex problems many problems can be solved using a divide-andconquer strategy - prepare, solve subproblems, combine results running time can be written recursively - T(n) = time(preparation) + time(subproblems) + time(combine) - for MAX recursive: T(n) = 2*T(n/2) + O(1) 2 subproblems of size n/2 max(array_left) ; max(array_right) constant time to check the maximum out of the two max Left and Right
10 Recurrence examples T(n) = 2T(n/2) + O(1) T(n)= 2T(n/2) + O(n) - 2 subproblems of size n/2 each, plus O(n) steps to combine results T(n) = 4T(n/3) + n - 4 subproblems of size n/3 each, plus n steps to combine results T(n/4) + T(n/2) +n 2 - a subproblem of size n/3, another of size n/2; n 2 to combine want to solve such recurrences, to obtain the order of growth of function T
11 Substitution method T(n) = 4T(n/2) + n STEP1 : guess solution, order of growth T(n)= O(n 3 ) - that means there is a constant C and a starting value n 0, such that T(n) Cn 3, for any n n 0 STEP2: verify by induction - assume T(k) k 3, for k<n - induction step: prove that T(n) Cn 3, using T(k) Ck 3, for k<n T (n) = 4T ( n )+n (1) 2 n 3 apple 4c + n (2) 2 = c 2 n3 + n (3) = cn 3 ( c 2 n3 n) (4) apple cn 3 ; if c 2 n3 n>0,choose c 2 (5)
12 Substitution method STEP 3 : identify constants, in our case c=2 works so we proved T(n)=O(n 3 ) thats correct, but the result is too weak - technically we say the bound O(n 3 ) "cubic" is too lose - can prove better bounds like T(n) "quadratic" T(n)=O(n 2 ) - Our guess was wrong! (too big) lets try again : STEP1: guess T(n)=O(n 2 ) STEP2: verify by induction - assume T(k) Ck 2, for k<n - induction step: prove that T(n) Cn 2, using T(k) Ck 2, for k<n
13 Substitution method Fallacious argument T (n) = 4T ( n )+n (1) 2 n 2 apple 4c + n (2) 2 = cn 2 + n (3) = O(n 2 ) (4) apple cn 2 (5) - cant prove T(n)=O(n 2 ) this way: need same constant steps maybe its not true? Guess O(n 2 ) was too low? - or maybe we dont have the right proof idea common trick: if math doesnt work out, make a stronger assumption (subtract a lower degree term) - assume instead T(k) C 1 k 2 -C 2 k, for k<n
14 Substitution method T (n) = 4T ( n 2 )+n apple 4 n c c 2 n 2 + n = c 1 n 2 2c 2 n + n = c 1 n 2 c 2 n (c 2 n n) apple c 1 n 2 c 2 n for c 2 > 1 So we can prove T(n)=O(n 2 ), but is that asymptotically correct? - maybe we can prove a lower upper bound, like O(nlogn)? NOPE to make sure its the asymptote, prove its also the lower bound - T(n) =Ω(n 2 ) or there is a different constant d s.t. T(n) dn 2
15 Substitution method: lower bound induction step T (n) = 4T ( n 2 )+n n 2 4d + n 2 = dn 2 + n dn 2 now we know its asymptotically close, T(n)=Θ(n 2 ) hard to make the initial guess Θ(n 2 ) - need another method to educate our guess
16 Iteration method T (n) =n +4T ( n 2 ) = n + 4( n 2 +4T (n 4 )) = n +2n +42 T ( n 2 2 ) = n +2n +4 2 ( n T ( n 2 3 )) = n +2n +22 n +4 3 T ( n 2 3 ) =... = n +2n +2 2 n k 1 n +4 k T ( n 2 k ) = kx 1 i=0 2 i n +4 k T ( n 2 k );
17 Iteration method T (n) =n +4T ( n 2 ) = n + 4( n 2 +4T (n 4 )) = n +2n +42 T ( n 2 2 ) = n +2n +4 2 ( n T ( n 2 3 )) = n +2n +22 n +4 3 T ( n 2 3 ) =... = n +2n +2 2 n k 1 n +4 k T ( n 2 k ) = kx 1 i=0 2 i n +4 k T ( n 2 k ); want k = log(n), n 2 k =1
18 Iteration method T (n) =n +4T ( n 2 ) = n + 4( n 2 +4T (n 4 )) = n +2n +42 T ( n 2 2 ) = n +2n +4 2 ( n T ( n 2 3 )) = n +2n +22 n +4 3 T ( n 2 3 ) =... = n +2n +2 2 n k 1 n +4 k T ( n 2 k ) = kx 1 i=0 2 i n +4 k T ( n 2 k ); want k = log(n), n 2 k =1 = n log(n) 1 X i=0 2 i +4 log(n) T (1) = n 2log(n) 1 + n 2 T (1) 2 1 = n(n 1) + n 2 T (1) = (n 2 )
19 Iteration method T (n) =n +4T ( n 2 ) = n + 4( n 2 +4T (n 4 )) = n +2n +42 T ( n 2 2 ) = n +2n +4 2 ( n T ( n 2 3 )) = n +2n +22 n +4 3 T ( n 2 3 ) =... = n +2n +2 2 n k 1 n +4 k T ( n 2 k ) = = n kx 1 i=0 2 i n +4 k T ( n 2 k ); log(n) 1 X i=0 2 i +4 log(n) T (1) = n 2log(n) 1 + n 2 T (1) 2 1 = n(n 1) + n 2 T (1) = (n 2 ) want k = log(n), n 2 k =1 math can be messy - recap sum, product, series, logarithms - iteration method good for guess, but usually unreliable for an exact result - use iteration for guess, and substitution for proofs stopping condition - T(...) = T(1), solve for k
20 Iteration method: visual tree T (n) +n T ( n 2 ) T ( n T ( n 2 ) T ( n 2 ) 2 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) +2n +4n
21 Iteration method: visual tree T (n) +n T ( n 2 ) T ( n T ( n 2 ) T ( n 2 ) 2 ) +2n T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) T ( n 4 ) T (n 4 ) T (n 4 ) T (n 4 ) +4n compute the tree depth: how many levels till nodes become leaves T(1)? log(n) compute the total number of leaves T(1) in the tree (last level): 4 log(n) compute the total additional work (right side) n+2n+4n+... = n(n-1) add the work 4 log(n) + n(n-1) = Θ(n 2 )
22 Iteration Method : derivation T(n) = n 2 + T(n/2) + T(n/4) T (n) =n 2 + T ( n 2 )+T(n 4 ) = n 2 +( n 2 )2 + T ( n 4 )+T(n 8 )+(n 4 )2 + T ( n 8 )+T( n 16 ) = n n2 + T ( n 4 )+2T(n 8 )+T( n 16 ) = n n2 +( n 4 )2 + T ( n 8 )+T( n 16 ) + 2(n 8 )2 +2T ( n 16 )+2T( n 32 )+(n 16 )2 + T ( n 32 )+T( n 64 ) = n n2 +( 5 16 )2 n 2 + T ( n 8 )+3T( n 16 )+3T( n 32 )+T( n 64 )
23 Iteration Method : derivation T(n) = n 2 + T(n/2) + T(n/4) T (n) =n 2 + T ( n 2 )+T(n 4 ) = n 2 +( n 2 )2 + T ( n 4 )+T(n 8 )+(n 4 )2 + T ( n 8 )+T( n 16 ) = n n2 + T ( n 4 )+2T(n 8 )+T( n 16 ) = n n2 +( n 4 )2 + T ( n 8 )+T( n 16 ) + 2(n 8 )2 +2T ( n 16 )+2T( n 32 )+(n 16 )2 + T ( n 32 )+T( n 64 ) = n n2 +( 5 16 )2 n 2 + T ( n 8 )+3T( n 16 )+3T( n 32 )+T( n 64 )
24 Iteration Method : derivation T(n) = n 2 + T(n/2) + T(n/4) T (n) =n 2 + T ( n 2 )+T (n 4 ) = n 2 +( n 2 )2 + T ( n 4 )+T (n 8 )+(n 4 )2 + T ( n 8 )+T ( n 16 ) = n n2 + T ( n 4 )+2T (n 8 )+T ( n 16 ) = n n2 +( n 4 )2 + T ( n 8 )+T ( n 16 ) + 2(n 8 )2 +2T ( n 16 )+2T ( n 32 )+(n 16 )2 + T ( n 32 )+T ( n 64 ) = n n2 +( 5 16 )2 n 2 + T ( n 8 )+3T ( n 16 )+3T ( n 32 )+T ( n 64 ) = n n2 +( 5 16 )2 n 2 +T ( n 16 )+T( n 32 )+(n 8 )2 +3T ( n 32 )+3T( n 64 )+3( n 16 )2 + 3T ( n 64 )+3T( n 128 ) + 3( n 32 )2 + T ( n 128 )+T( n 256 )+(n 64 )2
25 Iteration Method : derivation T(n) = n 2 + T(n/2) + T(n/4) T (n) =n 2 + T ( n 2 )+T (n 4 ) = n 2 +( n 2 )2 + T ( n 4 )+T (n 8 )+(n 4 )2 + T ( n 8 )+T ( n 16 ) = n n2 + T ( n 4 )+2T (n 8 )+T ( n 16 ) = n n2 +( n 4 )2 + T ( n 8 )+T ( n 16 ) + 2(n 8 )2 +2T ( n 16 )+2T ( n 32 )+(n 16 )2 + T ( n 32 )+T ( n 64 ) = n n2 +( 5 16 )2 n 2 + T ( n 8 )+3T ( n 16 )+3T ( n 32 )+T ( n 64 ) = n n2 +( 5 16 )2 n 2 +T ( n 16 )+T( n 32 )+(n 8 )2 +3T ( n 32 )+3T( n 64 )+3( n 16 )2 + 3T ( n 64 )+3T( n 128 ) + 3( n 32 )2 + T ( n 128 )+T( n 256 )+(n 64 )2 = n n2 +( 5 16 )2 n 2 +( 5 16 )3 n 2 + T ( n 16 )+4T( n 32 )+6T( n 64 )+4T( n 128 )+ T ( n 256 )
26 Iteration Method : tree T (n) T ( n 2 ) T ( n 4 ) T ( n 4 ) T ( n 8 ) T (n 8 ) T ( n 16 ) T ( n 8 ) T ( n 16 ) T ( n 16 ) T ( n 32 ) T ( n 16 ) T ( n 32 ) T ( n 32 ) T ( n 64 )
27 Iteration Method : tree T (n) +n 2 T ( n 2 ) T ( n 4 ) n2 T ( n 4 ) T ( n 8 ) T (n 8 ) T ( n 16 ) n 2 16 T ( n 8 ) T ( n 16 ) T ( n 16 ) T ( n 32 ) T ( n 16 ) T ( n 32 ) T ( n 32 ) T ( n 64 ) n 2 16
28 Iteration Method : tree T (n) +n 2 T ( n 2 ) T ( n 4 ) n2 T ( n 4 ) T ( n 8 ) T (n 8 ) T ( n 16 ) n 2 16 T ( n 8 ) T ( n 16 ) T ( n 16 ) T ( n 32 ) T ( n 16 ) T ( n 32 ) T ( n 32 ) T ( n 64 ) n 2 16 depth : at most log(n) leaves: at most 2 log(n) = n; computational cost nt(1) =O(n) work : n n2 +( 5 16 )2 n 2 +( 5 16 )3 n apple n 2 total Θ(n 2 ) 1X i=0 ( 5 16 )i = n2 = (n 2 )
29 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c )
30 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c ) R=a/b c, compare R with 1, or c with log b (a)
31 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c ) R=a/b c, compare R with 1, or c with log b (a)
32 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c ) R=a/b c, compare R with 1, or c with log b (a) - MergeSort T(n) = 2T(n/2) + Θ(n); a=2 b=2 c=1 case 2 ; T(n) = Θ(nlogn)
33 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c ) R=a/b c, compare R with 1, or c with log b (a) - MergeSort T(n) = 2T(n/2) + Θ(n); a=2 b=2 c=1 case 2 ; T(n) = Θ(nlogn) - Strassen's T(n) =7T(n/2) + Θ(n 2 ) ; a=7 b=2 c=2 case 1, T(n)= Θ(nlog 2 (7))
34 Master Theorem - simple simple general case T(n) = at(n/b) + Θ(n c ) R=a/b c, compare R with 1, or c with log b (a) - MergeSort T(n) = 2T(n/2) + Θ(n); a=2 b=2 c=1 case 2 ; T(n) = Θ(nlogn) - Strassen's T(n) =7T(n/2) + Θ(n 2 ) ; a=7 b=2 c=2 case 1, T(n)= Θ(nlog 2 (7)) - Binary Search T(n)=T(n/2) + Θ(1); a=1 b=2 c=0 case 2, T(n)= Θ(logn)
35 Master Theorem - why 3 cases that sum is geometric progression with base R=a/ b c it comes down to R being <1, =1, >1. So three
36 Master Theorem - why 3 cases
37 Master Theorem - why 3 cases
38 Master Theorem - why 3 cases
39 Master Theorem general case T(n) = at(n/b) + f(n) CASE 1 : CASE 2: CASE 3: f(n) =O(n log b a ) ) T (n) = (n log b a ) f(n) = (n log b a log k n) ) T (n) = (n log b a log k+1 n) f(n) = (n log b a+ ); af(n/b) f(n) < 1 ) T (n) = (f(n))
40 Master Theorem Example recurrence: T(n)=4T(n/2) +Θ(n 2 logn) Master Theorem: a=4; b=2; f(n)= n 2 logn - f(n) / n log b a = f(n)/n 2 = logn, so case 2 with k=1 solution T(n)=Θ(n 2 log 2 n)
41 Master Theorem Example T(n)=4T(n/2) + Θ(n 3 ) Master Theorem: a=4; b=2; f(n)=n 3 - f(n) / n log b a = f(n)/n 2 = n, so case 3 - check case 3 condition: - 4f(n/2)/f(n) = 4(n/2) 3 /n 3 = 1/2 < 1-ε solution T(n) = Θ(n 3 )
42 NON-Master Theorem Example T(n) = 4T(n/2) + n 2 /logn ; f(n) = n 2 /logn f(n) / n log b a = f(n)/n 2 = 1/logn - case1: - case2: - case3: no case applies - cant use Master Theorem use iteration method for guess, and substitution for a proof - see attached pdf
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