13 - 06 feb - dynamic programming

CS 321. Algorithm Analysis & Design Lecture 13

Dynamic Programming

Edit DistanceKnapsacks

Shortest Paths

Items with weights wi

and value vi

Knapsack with capacity C

Items with weights wi

and value vi

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

(What we want)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

A[i] stores the best value for the first i items.

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

A[i] stores the best value for the first i items.

A[i] = max(A(i-1), A[i-1] + vi)

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

A[i] stores the best value for the first i items.

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

A[i,j] stores the best value for the first i items.

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

that fits in a knapsack of size j.

A[i,j] stores the best value for the first i items.

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(What we want)

(Build up the array bottom to top.)

w1

v1

w2

v2

wi

vi

wn-1

vn-1

wn

vn… …

that fits in a knapsack of size j.

1 2 3 4 5

1

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

1 2 3 4 5

1

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3 0 6 7 9 9

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3 0 6 7 9 9

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3 0 6 7 9 9

4

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

1 2 3 4 5

1 0 3 3 3 3

2 0 6 6 9 9

3 0 6 7 9 9

4 4 6 10 11 13

A[i,j] = max(A(i-1,j), A[i-1,j-wi] + vi)

(2,3), (2,6), (3,7), (1,4) - weight/value pairs

Running time = # of subproblems x time/subproblem

Running time = # of subproblems x time/subproblem

nC x O(1) = O(nC) (n = #items, C = Capacity)

Running time = # of subproblems x time/subproblem

nC x O(1) = O(nC) (n = #items, C = Capacity)

Is this polynomial?

Running time = # of subproblems x time/subproblem

nC x O(1) = O(nC) (n = #items, C = Capacity)

Is this polynomial?

pseudo-polynomial

Edit Distance

Edit Distance

The cost of changing one word to another, using insertions/deletions/changes.

Edit Distance

The cost of changing one word to another, using insertions/deletions/changes.

Each operation can have a certain cost.

A C G T

A 0 1 1 3

C 2 0 1 1

G 1 2 0 2

T 3 2 1 0

SUNNY DEOL

SUNNY LEONE

SUNNY DEOL

SUNNY LEONE

SUNNY DEOL

SUNNY LEONE

SUNNY LEOL

SUNNY DEOL

SUNNY LEONE

SUNNY LEOL

SUNNY LEON

SUNNY DEOL

SUNNY LEONE

SUNNY LEOL

SUNNY LEON

SUNNY LEONE

Work out the solution for pairs x[i:] and y[j:].

(What we want)

(Build up the array bottom to top.)

(What we want)

(Build up the array bottom to top.)

c[i,j] stores the edit distance between x[1…i] and y[1…j].

(What we want)

(Build up the array bottom to top.)

c[i,j] stores the edit distance between x[1…i] and y[1…j].

c[i,j] = min(c[i-1,j] + w1, c[i,j-1] + w2, c[i-1,j-1] + w3)

(What we want)

(Build up the array bottom to top.)

c[i,j] stores the edit distance between x[1…i] and y[1…j].