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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. . - PowerPoint PPT Presentation

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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.W

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 item

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.Xi1 w1 i2 w2 wj...ip wp...ikwk

The 0-1 knapsack problem cannot be solved optimally by greedyCounter example: (moderate part)W=10 2 1.8Items 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

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

Greedy on start time

Depth: The maximum No. of jobs required at any time. Depth:3Greedy Algorithm: acbdegfjih

ddepth

01011l1=0, l2=1l2=0, l1=01101l1=0, l2=1l1=9, l2=0

- n1n2 nkExample: 1, 2, 3, 4, 5, 6, 7, 8, 9 1, 2, 6, 7, 3, 4, 5, 8, 9di
- di
Example: Job ti di 2 2 3 4 4 6 4 8 6 102591319j1j2j3j4j5

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