Dynamic Programming Has nothing to do with programming in the way - - PDF document

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Dynamic Programming Has nothing to do with programming in the way - - PDF document

May 26, 2023 388 likes •467 views

Dynamic Programming Has nothing to do with programming in the way we normally use that term Dynamic Programming More like TV programming (scheduling) Filling in a table Chapter 16 Algorithm design technique CPTR 318 1 2

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SLIDE 1

1

Dynamic Programming

Chapter 16

CPTR 318

1

Dynamic Programming

  • Has nothing to do with programming in the

way we normally use that term

– More like TV programming (scheduling) – Filling in a table

  • Algorithm design technique

2

Longest Common Subsequence

  • Given two sequences

X = x0 x1 x2 … xm-1 Y = y0 y1 y2 … yn-1 find a longest sequence common to both

  • Example:

X = A B C B D A B Y = B D C A B A

– B D A B – B C A B – B C B A

3

Longest Common Subsequence

  • Given two sequences

X = x0 x1 x2 … xm-1 Y = y0 y1 y2 … yn-1 find a longest sequence common to both

  • Example:

X = A B C B D A B Y = B D C A B A

– B D A B – B C A B – B C B A

4

Longest Common Subsequence

  • Given two sequences

X = x0 x1 x2 … xm-1 Y = y0 y1 y2 … yn-1 find a longest sequence common to both

  • Example:

X = A B C B D A B Y = B D C A B A

– B D A B – B C A B – B C B A

5

Longest Common Subsequence

  • Given two sequences

X = x0 x1 x2 … xm-1 Y = y0 y1 y2 … yn-1 find a longest sequence common to both

  • Example:

X = A B C B D A B Y = B D C A B A

– B D A B – B C A B – B C B A

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SLIDE 2

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Simple Algorithm

  • Brute force

– Try all subsequences s of X to see if s is a subsequence of Y – Select a longest such subsequence

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Brute Force Pseudocode

string LCS(string X[0..m-1], string Y[0..n-1]) { string result = ""; if (|X| > 0 && |Y| > 0) { if (X[0] == Y[0]) result = X[0] + LCS(X[1..m-1], Y[1, n-1]); else result = maxlen(LCS(X, Y[1..n-1]), LCS(X[1..m-1], Y)); } return result; }

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Brute Force Pseudocode

string LCS(string X[0..m-1], string Y[0..n-1]) { string result = ""; if (|X| > 0 && |Y| > 0) { if (X[0] == Y[0]) result = X[0] + LCS(X[1..m-1], Y[1, n-1]); else result = maxlen(LCS(X, Y[1..n-1]), LCS(X[1..m-1], Y)); } return result; }

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+ here means

string concatenation

Brute Force Pseudocode

string LCS(string X[0..m-1], string Y[0..n-1]) { string result = ""; if (|X| > 0 && |Y| > 0) { if (X[0] == Y[0]) result = X[0] + LCS(X[1..m-1], Y[1, n-1]); else result = maxlen(LCS(X, Y[1..n-1]), LCS(X[1..m-1], Y)); } return result; }

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+ here means

string concatenation maxlen returns the longer of two strings

Brute Force Pseudocode

string LCS(string X[0..m-1], string Y[0..n-1]) { string result = ""; if (|X| > 0 && |Y| > 0) { if (X[0] == Y[0]) result = X[0] + LCS(X[1..m-1], Y[1, n-1]); else result = maxlen(LCS(X, Y[1..n-1]), LCS(X[1..m-1], Y)); } return result; }

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+ here means

string concatenation maxlen returns the longer of two strings “Absolute value” bars mean string length

Brute Force Pseudocode

string LCS(string X[0..m-1], string Y[0..n-1]) { string result = ""; if (|X| > 0 && |Y| > 0) { if (X[0] == Y[0]) result = X[0] + LCS(X[1..m-1], Y[1, n-1]); else result = maxlen(LCS(X, Y[1..n-1]), LCS(X[1..m-1], Y)); } return result; }

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SLIDE 3

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Brute Force Complexity

  • Need to check each subsequence of X
  • Θ(n) to check a subsequence of X
  • There are Θ(2m) subsequences of X
  • Total time Θ(n∙2m)

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Dynamic Programming

  • Overlapping subproblems

– The solution to a subproblem is recorded and can be used multiple times

  • Optimal substructure

– Optimal solutions of subproblems are used to compute the optimal solution to the overall problem

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Table of Subsequence Lengths

  • Consider length of LCS(X,Y)
  • Consider prefixes of X and Y
  • Build a table of prefix lengths

– C[i][j] = | LCS(X[0..i - 1], Y[0..j - 1]) | – C[m-1][n-1] = | LCS(X, Y) |

  • C[i][j] = C[i - 1][j - 1] + 1 , if X[i] = Y[j]
  • r max(C[i][j - 1] , C[i -1][j] ), otherwise

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Build the DP Table

Table LCS_table(string X[0..m-1], string Y[0..n-1]) { Table C[0..m][0..n]; for (int i = 0; i <= m; i++) C[i][0] = 0; for (int j = 0; j <= n; j++) C[0][j] = 0; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) if (X[i] == Y[j]) C[i + 1][j + 1] = C[i][j] + 1; else C[i + 1][j + 1] = max(C[i + 1][j], C[i][j + 1]); return C; }

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Build the DP Table

Table LCS_table(string X[0..m-1], string Y[0..n-1]) { Table C[0..m][0..n]; for (int i = 0; i <= m; i++) C[i][0] = 0; for (int j = 0; j <= n; j++) C[0][j] = 0; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) if (X[i] == Y[j]) C[i + 1][j + 1] = C[i][j] + 1; else C[i + 1][j + 1] = max(C[i + 1][j], C[i][j + 1]); return C; }

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Initialization

Build the DP Table

Table LCS_table(string X[0..m-1], string Y[0..n-1]) { Table C[0..m][0..n]; for (int i = 0; i <= m; i++) C[i][0] = 0; for (int j = 0; j <= n; j++) C[0][j] = 0; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) if (X[i] == Y[j]) C[i + 1][j + 1] = C[i][j] + 1; else C[i + 1][j + 1] = max(C[i + 1][j], C[i][j + 1]); return C; }

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Table construction

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SLIDE 4

4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

Extract the Subsequence

string back_trace(Table C, string X, string Y, int i, int j) { if (i < 0 || j < 0) return ""; else if (X[i] == Y[j]) return back_trace(C, X, Y, i - 1, j - 1) + X[i]; else if (C[i + 1][j] > C[i][j + 1]) return back_trace(C, X, Y, i, j - 1); else return back_trace(C, X, Y, i - 1, j); }

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Extract the Subsequence

string back_trace(Table C, string X, string Y, int i, int j) { if (i < 0 || j < 0) return ""; else if (X[i] == Y[j]) return back_trace(C, X, Y, i - 1, j - 1) + X[i]; else if (C[i + 1][j] > C[i][j + 1]) return back_trace(C, X, Y, i, j - 1); else return back_trace(C, X, Y, i - 1, j); }

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Initial call: back_trace(C, X, Y, m – 1, n – 1)

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

BCAB

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

BCAB

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SLIDE 5

5

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

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SLIDE 6

6

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X Y A B C B D A B B 1 1 1 1 1 1 D 1 1 1 2 2 2 C 1 2 2 2 2 2 A 1 1 2 2 2 3 3 B 1 2 2 3 3 3 4 A 1 2 2 3 3 4 4

BDAB DP Complexity

  • Θ(m∙n) to fill out the table
  • Θ(max(m , n)) to backtrace
  • Space: Θ(m∙n)

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