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T (n) = = = 2j Then, Let T (n) be the time to multiply two n×n matrices. The approach gains us nothing: T (n) = 8log n T (1) + dn2 M2 M3 M4 M5 M6 M7 AE + BG AF + BH CE + DG CF + DH c 8T (n/2) + dn2 2j log n−1 = M1 Then I J K L i−1 Compute Divide X, Y, Z each into four (n/2)×(n/2) matrices. J L 8i T (n/2i ) + dn2 Strassen’s Algorithm Divide and Conquer Approach I K = j=0 Assume that all integer operations take O(1) time. The naive matrix multiplication algorithm then takes time O(n3 ). Can we do better?

Xj , which has fictitious nodes i, . . , j. We are interested in T0,n . = (ci,k−1 + wi,k−1 ) + (ck,j + wk,j ) + pk = ci,k−1 + ck,j + (wi,k−1 + wk,j + pk ) k−1 = ci,k−1 + ck,j + Let ci,j be the cost of Ti,j . ph + h=i+1 4 k−1 qh + h=i j j ph + h=k+1 qh + pk h=k j = ci,k−1 + ck,j + j ph + h=i+1 qh h=i = ci,k−1 + ck,j + wi,j Which xk do we choose to be the root? Pick k in the range i + 1 ≤ k ≤ j such that ci,j = ci,k−1 + ck,j + wi,j is smallest. Loose ends: • if k = i + 1, there is no left subtree • if k = j, there is no right subtree • Ti,i is the empty tree, with wi,i = qi , and ci,i = 0.

How should we pick a? It could be S[1] (average case significant in the long run) or a random element of S (the worst case doesn’t happen with particular inputs). Therefore, T (n) ≤ 2 n Binary Search n−1 T (j) + n − 1. Find the index of x in a sorted array A. r] if = r then return( ) else m := ( + r)/2 if x ≤ A[m] then return(search(A, x, , m)) else return(search(A, x, m + 1, r)) Claim that T (n) ≤ 4(n − 1). Proof by induction on n. The claim is true for n = 1. Now suppose that T (j) ≤ 4(j − 1) for all j < n.

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Advanced Calculus of real valued functions of real variable and vectored valued functions of a vector 0


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