Greedy Algorithms:

transcript

112/04/22 CS4335 Design and Analysis of Algorithms/WANG Lusheng

Greedy Algorithms: A greedy algorithm always makes the

choice that looks best at the moment. It makes a local optimal choice in the

hope that this choice will lead to a globally optimal solution.

Greedy algorithms yield optimal solutions for many (but not all) problems.

The 0-1 Knapsack problem:

The 0-1 knapsack problem: N items, where the i-th item is worth vi dollars and

weight wi pounds. 11 p 3 p 4p 58 p 8p 88p vi and wi are integers.

3$ 6 $ 35$ 8$ 28$ 66$ We can carry at most W (integer) pounds. How to take as valuable a load as possible.

An item cannot be divided into pieces. The fractional knapsack problem: The same setting, but the thief can take fractions of items. W may not be integer.

Solve the fractional Knapsack problem:

Greedy on the value per pound vi/wi. Each time, take the item with maximum vi/wi . If exceeds W, take fractions of the item.

Example: (1, 5$), (2, 9$), (3, 12$), (3, 11$) and w=4. vi/wi : 5 4.5 4.0 3.667 First: (1, 5$), Second: (2, 9$), Third: 1/3 (3, 12$) Total W: 1, 3, 4. Can only take part of

Proof of correctness: (The hard part)

Let X = i1, i2, …ik be the optimal items taken. Consider the item j : (vj, wj) with the highest v /w. if j is not used in X (the optimal solution), get rid of some

items with total weight wj (possibly fractional items) and add item j.

(since fractional items are allowed, we can do it.) Total value is increased. Why?

One more item selected by greedy is added to X Repeat the process, X is changed to contain all items selected

by greedy WITHOUT decreasing the total value taken by the thief.

i2 w2 wj

)()()%...(

)(%)(...)()(

The 0-1 knapsack problem cannot be solved optimally by

greedy Counter example: (moderate part) W=10 2 1.8 Items found (6pounds, 12dollars), (5pounds, 9 dollar), 1.8 1. 1 (5pounds, 9 dollars), (3pounds, 3 dollars), (3 pounds, 3

dollars) If we first take (6, 12) according to greedy

algorithm, then solution is (6, 12), (3, 3) (total value is 12+3=15).

However, a better solution is (5, 9), (5, 9) with total value 18.

To show that a statement does not hold, we only have to give an example.

A subset of mutually

compatibles jobs: {c, f}

Sorting the n jobs based on fi needs O(nlog n) time

Example: Jobs (s, f): (0, 10), (3, 4), (2, 8), (1, 5), (4, 5), (4, 8), (5, 6) (7,9).

Sorting based on fi: (3, 4) (1, 5), (4, 5) (5, 6) (4,8) (2,8) (7, 9)(0,10).Selecting jobs:(3,4) (4, 5) (5,6) (7, 9)

Sort ob finish time: b, c, a, e, d, f, g, h.

Greedy algorithm Selects: b, e, h.

Depth: The maximum No. of jobs required at any time.

Depth:3

Greedy on start time

Depth:3

Greedy Algorithm:

ddepth

0 10 11

l1=0, l2=1l2=0, l1=0

l1=0, l2=1l1=9, l2=0

n1n2 … nk

Example: 1, 2, 3, 4, 5, 6, 7, 8, 9 1, 2, 6, 7, 3, 4, 5, 8, 9

We check every pair of consecutive

numbers, if there is not inversion, then the whole sequenc has no

inversion.

Example: Job ti di

1 2 22 3 43 4 64 4 85 6 10

2 5 9 13 19j1 j2 j3 j4 j5

Greedy Algorithms:

Documents