Dynamic Programming II: Algorithms and Applications, Lecture notes of Algorithms and Programming

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Lecture 8:

Dynamic Programming II

– Stable Matching

– Greedy algorithms

– Divide & Conquer algorithms

– Dynamic programming algorithms

– Network flow algorithms

General techniques in this course

Key steps: Dynamic programming

1. Define subproblems

2. Find recurrences

3. Solve the base cases

4. Transform recurrence into an efficient algorithm

Recap: Weighted Interval Scheduling

• Job j starts at s j

, finishes at f

j

, and has value v

j

• Two jobs compatible if they don't overlap.
• Goal: find maximum value subset of mutually compatible jobs. Time 0 1 2 3 4 5 6 7 8 9 10 11 f g h e a b c d

Recap: Dynamic Programming – Step 1

Step 1: Define subproblems

OPT(j) = value of optimal solution to the problem

consisting of job requests 1, 2, ..., j.

Recap: Dynamic Programming – Step 2

Step 2: Find recurrences

• Case 1: OPT selects job j.
  • can't use incompatible jobs { p(j) + 1, p(j) + 2, ..., j - 1 }
  • must include optimal solution to problem consisting of

remaining compatible jobs 1, 2, ..., p(j)

• Case 2: OPT does not select job j.
  • must include optimal solution to problem consisting of

remaining compatible jobs 1, 2, ..., j-

OPT(j) = v

j

max { + OPT (p(j)), OPT(j-1)}

Case 1 Case 2

Recap: Knapsack Problem

• Knapsack problem.
  • Given n objects and a "knapsack."
  • Item i weighs wi > 0 kilograms and has value v (^) i > 0.
  • Knapsack has capacity of W kilograms.
  • Goal: fill knapsack so as to maximize total value. 1 Value 18 22 28 1 Weight 5 6 6 2 7 Item 1 3 4 5 2 W = 11

Knapsack Algorithm Recap

n + 1 1 Value 18 22 28 1 Weight 5 6 6 2 7 Item 1 3 4 5 2 ∅ { 1, 2 } { 1, 2, 3 } { 1, 2, 3, 4 } { 1 } { 1, 2, 3, 4, 5 } 0 0 0 0 0 0 0 1 0 1 1 1 1 1 2 0 6 6 6 1 6 3 0 7 7 7 1 7 4 0 7 7 7 1 7 5 0 7 18 18 1 18 6 0 7 19 22 1 22 7 0 7 24 24 1 28 8 0 7 25 28 1 29 9 0 7 25 29 1 34 10 0 7 25 29 1 35 11 0 7 25 40 1 40 W + 1 W = 11 OPT: { 4, 3 } value = 22 + 18 = 40

Recap: Dynamic Programming – Step 2

Step 2: Find recurrences

• Case 1: OPT does not select item i.
  • OPT selects best of { 1, 2, …, i-1 } using weight limit w
• Case 2: OPT selects item i.
  • new weight limit = w – wi
  • OPT selects best of { 1, 2, …, i–1 } using this new weight limit

OPT(i,w) = max { v

i

+ OPT (i-1,w-w

i

), OPT(i-1,w) }

Recap: Dynamic Programming – Step 2

Step 2: Find recurrences

• Case 1: OPT does not select item i.
  • OPT selects best of { 1, 2, …, i-1 } using weight limit w
• Case 2: OPT selects item i.
  • new weight limit = w – wi
  • OPT selects best of { 1, 2, …, i–1 } using this new weight limit

If w

i

>w

OPT(i,w) = OPT (i-1,w)

otherwise

OPT(i,w) = max { v

i

+ OPT (i-1,w-w

i

), OPT(i-1,w) }

Recap: Longest increasing subsequence

Given a sequence of numbers X[1..n] find the longest increasing

subsequence (i

1

, i

2

, …, i

k

), that is a subsequence where numbers in

the sequence increase.

j i

LIS 1 1 1 1 1 1 1 1

Recap: Longest increasing subsequence

Given a sequence of numbers X[1..n] find the longest increasing

subsequence (i

1

, i

2

, …, i

k

), that is a subsequence where numbers in

the sequence increase.

j i

L 1 1 2 1 1 1 1 1

If X[i] > X[j] then L[i] + 1 = 1 + 1 = 2 is this > L[i] yes L[i] = 2 move j to next position and check again

Recap: Longest increasing subsequence

Given a sequence of numbers X[1..n] find the longest increasing

subsequence (i

1

, i

2

, …, i

k

), that is a subsequence where numbers in

the sequence increase.

j i

L 1 1 2 2 1 1 1 1

If X[i] > X[j] then L[i] + 1 = 1 + 1 = 2 is this > L[i] yes L[i] = 2 move j to next position and check again

Recap: Longest increasing subsequence

Given a sequence of numbers X[1..n] find the longest increasing

subsequence (i

1

, i

2

, …, i

k

), that is a subsequence where numbers in

the sequence increase.

j i

L 1 1 2 2 2 3 1 1

If X[i] > X[j] then L[i] + 1 = 1 + 1 = 2 is this > L[i] yes L[i] = 2 move j to next position and check again