# sol13 - COP3530 Solution 13 1. 1)Merge Sort initlal segment:

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Unformatted text preview: COP3530 Solution 13 1. 1)Merge Sort initlal segment: [21][25][49][25][93][62][72][8][37][16][54] merge to b : [21 25][25 49][62 93][8 72][16 37][54] copy to a : [21 25][25 49][62 93][8 72][16 37][54] merge to b : [21 25 25 49][8 62 72 93][16 37 54] copy to a : [21 25 25 49][8 62 72 93][16 37 54] merge to b: [8 21 25 25 49 62 72 93][16 37 54] copy to a: [8 21 25 25 49 62 72 93][16 37 54] merge to b : [8 16 21 25 25 37 49 54 62 72 93] 2)Quick Sort Step 1: [ 21 , 25 , 49 , 25 , 93 , 62 , 72 , 8 , 37 , 16 , 54] Step 2: [ 8 , 16 , (21) , 25 , 49 , 25 , 93 , 62 , 72 , 37 , 54] step 3: [(8), 16 , (21) ,(25), 49 , 25 , 93 , 62 , 72 , 37 , 54] step 4: [(8), (16), (21) ,(25), 25 , 37 ,(49), 93 , 62 , 72 , 54] step 5: [(8), (16), (21) ,(25),(25), 37 ,(49), 62 , 72 , 54 ,(93)] step 6: [(8), (16), (21) ,(25),(25),(37),(49), 54 ,(62), 72 ,(93)] step 7: [(8), (16), (21) ,(25),(25),(37),(49),(54),(62), (72),(93)] W < D ^ &  &  &  &  &   W D ^  > E E E ^ PROBLEM 3: part 1: Divide each of the matricices X and Y into 4 smaller matricies of size (n/2) x (n/2). Call thse matricies X11, X12, X21, X22, Y11, Y12, Y21, Y22. Then the resulting Z matrix can similary be split into 4 smaller matricies Z11, Z12, Z21 and Z22. Now define the following intermediate matricies: M1:=(X11 + X22) * (Y11 + Y22) M2:=(X21 + X22) * Y11 M3:=X11 * (Y12 ‐ Y22) M4:=X22 * (Y21 ‐ Y11) M5:=(X11 + X12) * Y22 M6:=(X21 ‐ X11) * (Y11 + Y12) M7:=(X12 ‐ X22) * (Y21 + Y22) From this we can construct Z as follows: Z11 = M1 + M4 ‐ M5 + M7 Z12 = M3 + M5 Z21 = M2 + M4 Z22 = M1 ‐ M2 + M3 + M6 part 2: //assume that you have square matricies matrix matrix_mult(matrix A, matrix B) { int n=A.size(); if(n==1) return new matrix(1,1,A[1][1]*B[1][1]); //a matrix w/ dimensions 1x1 and element A[1][1]*B[1][1] matrix A11=new matrix(A,1,1,n/2,n/2); matrix A12=new matrix(A,1,n/2+1,n/2,n); matrix A21=new matrix(A,n/2+1,1,n,n/2); matrix A22=new matrix(A,n/2+1,n/2+1,n,n); matrix B11=new matrix(B,1,1,n/2,n/2); matrix B12=new matrix(B,1,n/2+1,n/2,n); matrix B21=new matrix(B,n/2+1,1,n,n/2); matrix B22=new matrix(B,n/2+1,n/2+1,n,n); matrix M1=matrix_mult(A11+A22,B11+B22); matrix M2=matrix_mult(A21+A22,B11); matrix M3=matrix_mult(A11,B21‐B22); matrix M4=matrix_mult(A22,B21‐B11); matrix M5=matrix_mult(A11+A12,B22); matrix M6=matrix_mult(A21‐A11,B11+B12); matrix M7=matrix_mult(A12‐A22,B21+B22); matrix C11 = M1+M4‐M5+M7; matrix C12 = M3+M5; matrix C21 = M2+M4; matrix C22 = M1‐M2+M3+M6; return combine(C11,C12,C21,C22); } part 3: Assume that it takes f(n) operations to solve this problem for 2^n x 2^n matricies. Then using this algorithm we have that f(n)=7f(n‐1)+h*4^n, where h is a constant that depends on the number of additions performed at each iteration of the algorithm. Hence f(n)~(7+h)^n, so we get running time O(N^(log7 +h)) ~ O(N^2.8). ...
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## This note was uploaded on 01/15/2010 for the course COP 3530 taught by Professor Davis during the Fall '08 term at University of Florida.

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