Presentation is loading. Please wait.

Presentation is loading. Please wait.

LCS Non-Dynamic Version int function lcs (x, y, i, j) begin if (i = 0) or (j = 0) return 0; else if (x[i] = y[j]) return lcs(x, y, i-1, j-1)+1; else return.

Similar presentations


Presentation on theme: "LCS Non-Dynamic Version int function lcs (x, y, i, j) begin if (i = 0) or (j = 0) return 0; else if (x[i] = y[j]) return lcs(x, y, i-1, j-1)+1; else return."— Presentation transcript:

1 LCS Non-Dynamic Version int function lcs (x, y, i, j) begin if (i = 0) or (j = 0) return 0; else if (x[i] = y[j]) return lcs(x, y, i-1, j-1)+1; else return max(lcs(x, y, i-1, j), lcs(x, y, i, j-1)); end; int x = lcs(“abdabcd”,”cabdaaxy”,7,8);

2 LCS Dynamic Programming Version int A{0..m,0..n); A{1..m,1..n) = -1; A(0,0..n) = 0; A(1..m,0) = 0; int function lcs (x, y, i, j) begin if (A[i,j] < 0) if (x[i] = y[j]) A[i,j] = lcs(x, y, i-1, j-1) + 1; else A[i,j] = max(lcs(x, y, i-1, j), lcs(x, y, i, j-1)); return A[i,j]; end;

3 LCS Dynamic Programming Version int c(0..m,0..n); void function lcs (x, y) begin m = length(x); n = length(y); c[1..m,0] = 0; c[0,0..n] = 0; for i = 1 to m for j = 1 to n if (x[i] = y[j]) c[i,j] = c[i-1,j-1] + 1; else if c[i-1,j] >= c[i,j-1] c[i,j] = c[i-1,j]; else c[i,j] = c[i,j-1]; end; // After execution, c[m,n] contains the length of the lcs

4 LCS Dynamic Programming Version int c(0..m,0..n); char b(0..m,0..n); void function lcs (x, y) begin m = length(x); n = length(y); c[1..m,0] = 0; c[0,0..n] = 0; for i = 1 to m for j = 1 to n if (x[i] = y[j]) c[i,j] = c[i-1,j-1] + 1; b[i,j] = “\”; else if c[i-1,j] >= c[i,j-1] c[i,j] = c[i-1,j]; b[i,j] = “|”; else c[i,j] = c[i,j-1]; b[i,j] = “--”;// After execution, b contains a description of the subsequence end;

5 LCS Print function print-lcs (b, x, i, j) begin if (i = 0) or (j =0) return; else if b[i,j] = “\” print-lcs (b,x, i-1, j-1); else if b[i,j] = “|” print-lcs(b, x, i-1,j) else // b[i,j] = “--” print-lcs(b, x, i, j-1) end; // Technically b is not necessary

6 Assembly Line Scheduling Non-Dynamic Version array e[1..2] = {2,4};// Entry times for each assembly line array x[1..2] = {3,2};// Exit times for each assembly line array a[1..2,1..6] = {{7,9,3,4,8,4}, {8,5,6,4,5,7}}; // Station times array t[1..2, 1..5] = {{2,3,1,3,4}, {2,1,2,2,1}}; // Transfer times

7 ALS Non-Dynamic Version int function shortest_schedule_time (int i, int j)// i is 1..2, j is 1..n begin if (j = 1) return e[i] + a[i,j]; else int f1 = shortest_schedule_time(1, j-1); int f2 = shortest_schedule_time(2, j-1); if (i = 1)// could be eliminated with a two-element array return min( f1 + a[1,j], f2 + t[2,j-1] + a[1,j]); else // i =2 return min( f2 + a[2,j], f1 + t[1,j-1] + a[2,j])); end; void main() { int f1 = shortest_schedule_time(1, n); int f2 = shortest_schedule_time(2, n); print(min(f1 + x[1], f2 + x[2])); }

8 ALS Dynamic Version #1 struct f_type = {int f1, int f2}; f_type function shortest_schedule_time (j) begin if (j = 1) return (e[1] + a[1,j], e[2] + a[2,j]); else f_type f; f = shortest_schedule_time(j-1); return (min( f.f1 + a[1,j], f.f2 + t[2,j-1] + a[1,j]), min( f.f2 + a[2,j], f.f1 + t[1,j-1] + a[2,j])); end; void main() { f_type f = shortest_schedule_time(n); print(min(f.f1 + x[1], f.f2 + x[2])); }

9 Assembly Line Scheduling Non-Dynamic Version array e[1..3] = {2,4};// Entry times for each assembly line array x[1..3] = {3,2};// Exit times for each assembly line array a[1..3,1..6] = {{7,9,3,4,8,4}, {8,5,6,4,5,7}}; // Station times array t[1..3, 1..3, 1..5] = {{2,3,1,3,4}, {2,1,2,2,1}}; // Transfer times

10 ALS Non-Dynamic Version int function shortest_schedule_time (int i, int j)// i is 1..3, j is 1..n begin if (j = 1) return e[i] + a[i,j]; else int f1 = shortest_schedule_time(1, j-1); int f2 = shortest_schedule_time(2, j-1); int f3 = shortest_schedule_time(3, j-1); if (i = 1) // could be eliminated with a two-element array return min( f1 + a[1,j], f2 + t[2,1,j-1] + a[1,j], f3 + t[3,1,j-1] + a[1,j]); else if (i = 2) return min( f2 + a[2,j], f1 + t[1,2, j-1] + a[2,j], f3 + t[3,2,j-1] + a[2,j]); else // i = 3 return min( f3 + a[3,j], f1 + t[1,3, j-1] + a[3,j], f2 + t[2,3,j-1] + a[3,j]); end; void main() { int f1 = shortest_schedule_time(1, n); int f2 = shortest_schedule_time(2, n); int f3 = shortest_schedule_time(3, n); print(min(f1 + x[1], f2 + x[2], f3 + x[3])); }

11 ALS Dynamic Version #1 struct f_type = {int f1, int f2, int f3}; f_type function shortest_schedule_time (j) begin if (j = 1) return (e[1] + a[1,j], e[2] + a[2,j], e[3] + a[3,j]); else f_type f; f = shortest_schedule_time(j-1); return (min( f.f1 + a[1,j], f.f2 + t[2,1,j-1] + a[1,j], f.f3 + t[3,1,j-1] + a[1,j]), min( f.f2 + a[2,j], f.f1 + t[1,2,j-1] + a[2,j], f.f3 + t[3,2,j-1] + a[3,j]), min( f.f3 + a[3,j], f.f1 + t[1,3,j-1] + a[3,j], f.f2 + t[2,3,j-1] + a[3,j]), end; void main() { f_type f = shortest_schedule_time(n); print(min(f.f1 + x[1], f.f2 + x[2], f.f3 + x[3])); }


Download ppt "LCS Non-Dynamic Version int function lcs (x, y, i, j) begin if (i = 0) or (j = 0) return 0; else if (x[i] = y[j]) return lcs(x, y, i-1, j-1)+1; else return."

Similar presentations


Ads by Google