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Greedy Algorithms

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Greedy Algorithms

15-211 Fundamental Data Structures and Algorithms

Peter Lee

March 19, 2004

- HW6 is due on April 5!
- Quiz #2 postponed until March 31
- an online quiz
- requires up to one hour of uninterrupted time with a web browser
- actually, only a 15-minute quiz

- must be completed by April 1, 11:59pm

Objects in calendar are closer than they appear

Greed is Good

- Suppose we want to give out change, using the minimal number of bills and coins.

- An easy algorithm for giving out N cents in change:
- Choose the largest bill or coin that is N.
- Subtract the value of the chosen bill/coin from N, to get a new value of N.
- Repeat until a total of N cents has been counted.

- Does this work? I.e., does this really give out the minimal number of coins and bills?

- For US currency, this simple algorithm actually works.
- Why do we call this a greedy algorithm?

- At every step, a greedy algorithm
- makes a locally optimal decision,
- with the idea that in the end it all adds up to a globally optimal solution.

- Being optimistic like this usually leads to very simple algorithms.

Over on Craig Street…

Think Globally

Act Locally

Eat Noodles

How Californian...

- What happens if we have a 12-cent coin?

- Greedy algorithms are often visualized as “hill-climbing”.
- Suppose you want to reach the summit, but can only see 10 yards ahead and behind (due to thick fog).

Which way?

- Greedy algorithms are often visualized as “hill-climbing”.
- Suppose you want to reach the summit, but can only see 10 yards ahead and behind (due to thick fog).

Which way?

- Making the locally-best guess is efficient and easy, but doesn’t always work.

- Greedy algorithms are common in computer science
- In fact, from last week…

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- Assume that every city is infinitely far away.
- I.e., every city is miles away from BWI (except BWI, which is 0 miles away).
- Now perform something similar to breadth-first search, and optimistically guess that we have found the best path to each city as we encounter it.
- If we later discover we are wrong and find a better path to a particular city, then update the distance to that city.

- For our airline-mileage problem, we can start by guessing that every city is miles away.
- Mark each city with this guess.

- Find all cities one hop away from BWI, and check whether the mileage is less than what is currently marked for that city.
- If so, then revise the guess.

- Continue for 2 hops, 3 hops, etc.

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- Algorithm initialization:
- Label each node with the distance , except start node, which is labeled with distance 0.
- D[v] is the distance label for v.

- Put all nodes into a priority queue Q, using the distances as labels.

- Label each node with the distance , except start node, which is labeled with distance 0.

- While Q is not empty do:
- u = Q.removeMin
- for each node z one hop away from u do:
- if D[u] + miles(u,z) < D[z] then
- D[z] = D[u] + miles(u,z)
- change key of z in Q to D[z]

- if D[u] + miles(u,z) < D[z] then

- Note use of priority queue allows “finished” nodes to be found quickly (in O(log N) time).

Another Greedy Algorithm

- You rob a store: find n kinds of items
- Gold dust. Wheat. Beer.

- You rob a store: find n kinds of items
- Gold dust. Wheat. Beer.

- The total inventory for the i th kind of item:
- Weight: wi pounds
- Value: vi dollars

- Knapsack can hold a maximum of Wpounds.
- Q: how much of each kind of item should you take?
(Can take fractional weight)

- Greedy solution:
- Fill knapsack with “most valuable” item until all is taken.
- Mostvaluable = vi /wi (dollars per pound)

- Then next “most valuable” item, etc.
- Until knapsack is full.

- Fill knapsack with “most valuable” item until all is taken.

- An optimization problem.
- Is iterative / Proceeds in stages.
- Has the greedy-choice property:
A greedy choice will lead to a globally optimal solution.

- An optimization problem:
- Maximize value of loot, subject to maximum weight W.(constrained optimization)

- Proceeds in stages:
- Knapsack is filled with one item at a time.

- Greedy-choice property:
A locally greedy choice will lead to a globally optimal solution.

- In steps…:
Step 1: Does the optimal solution contain the greedy choice?

Step 2: can the greedy choice always be made first?

- Consider total value, V, of knapsack.
- Knapsack must contain item h:
- Item h is the item with highest $/lb.

- Why? Because if h is not included, we can replace some other item in knapsack with an equivalent weight of h, and increase V.
- This can continue until knapsack is full, or all of h is taken.
- Therefore any optimal solution must include greedy-choice.

Let item h be the item with highest $/lb.Total inventory of h is wh pounds.Total value of h is vi dollars.

Let ki be weight of itemi in knapsack. Then total value:

If kh<wh, and kj>0 for some jh, then replace j with an equal weight of h. Let new total value = V’.

Difference in total value:

since, by definition of h,

Therefore all of item h should be taken.

- Now we want to show that we can always make the greedy choice first.
- If item h is more than what knapsack can hold, then fill knapsack completely with h.
- No other item gives higher total value.

- Otherwise, knapsack contains h and some other item. We can always make h the first choice, without changing total value V.
- Therefore greedy-choice can always be made first.

- Case I: wh W
- Fill knapsack completely with h.
- No other item gives higher total value.

- Case II: wh< W
- Let 1st choice be item i, and kth choice be h, then we can always swap our 1st and kth choices, and total value Vremains unchanged.

- Therefore greedy-choice can always be made first.

- You win the Supermarket Shopping Spree contest.
- You are given a shopping cart with capacity C.
- You are allowed to fill it with any items you want from Giant Eagle.
- Giant Eagle has items 1, 2, … n, which have values v1, v2, …, vn, and sizes s1, s2, …, sn.
- How do you (efficiently) maximize the value of the items in your cart?

- The obvious greedy strategy of taking the maximum value item that still fits in the cart does not work.
- Consider:
- Suppose item i has size si = C and value vi.
- It can happen that there are items j and k with combined size sj+skC but vj+vk > vi.

$220

$160

$180

Maximum weight = 50 lbs

$120, 30 lbs

$100, 20 lbs

$60, 10 lbs

(optimal)

item 1 item 2 item 3 knapsack

BKP has optimal substructure, but not greedy-choice property: optimal solution does not contain greedy choice.

- How can we (efficiently) solve the binary knapsack problem?
- One possible approach:
- Dynamic programming

Machine Scheduling

- We are given n tasks and an infinite supply of machines to perform them
- each task ti = [si, fi] has start time si and finish time fi

- An assignment of tasks to machines is feasible if no machine is assigned two overlapping tasks
- An assignment is optimal if it is feasible and uses the minimal number of machines

- Tasks:
- task: a b c d e f g
- start: 0 3 4 9 7 1 6
- finish: 2 7 7 11 10 5 8

- Can you invent a greedy algorithm to find an optimal schedule for these tasks?

3 ingredients needed:

- Optimization problem.
- Proceed in stages.
- Greedy-choice property:
A greedy choice will lead to a globally optimal solution.