Brief intro to knapsack problem

1. BRIEF INTRO TO
KNAPSACK PROBLEM

2. Knapsack
 A 1998 study of the Stony Brook University Algorithm
Repository showed that, out of 75 algorithmic problems,
the knapsack problem was the 18th most popular and the
4th most needed after kd-trees, suffix trees, and the bin
packing problem

3. Knapsack problem
 Given a set of items, each with a mass and a value,
determine the number of each item to include in a
collection so that the total weight is less than or equal to a
given limit and the total value is as large as possible.
 It derives its name from the problem faced by someone
who is constrained by a fixed-size knapsack and must fill
it with the most valuable items.

4. Given some items, pack the knapsack to get
the maximum total value. Each item has some
weight and some value. Total weight that we can
carry is no more than some fixed number W.
So we must consider weights of items as well as
their values.
Item # Weight Value
1 1 8
2 3 6
3 5 5
Knapsack problem

5. which boxes should be chosen to maximize
the amount of money while still keeping
the overall weight under or equal to 15 kg?

6. 
Answer: 3 yellow boxes and 3 grey boxes .

7. Problem
 A thief breaks into a house. Around the thief are
various objects: a diamond ring, a silver candelabra, a
Bose Wave Radio (tm), a large portrait of Elvis Presley
painted on a black velvet background (a "velvet-elvis"),
and a large tiffany crystal vase. The thief has a
knapsack that can only hold a certain capacity. Each of
the items has a value and a size, and cannot hold all of
the items in the knapsack.

8. Item Size Value
1-Ring 1 15
2-Candelabra 5 10
3-Radio 3 9
4-Elvis 4 5
The problem is, which items should the thief take? If the
knapsack were large enough, the thief could take all of the
items and run. But, that is not the case (the problem states
that the knapsack cannot hold all of the items).
There are three types of "thieves" that we shall consider: a greedy
thief, a foolish and slow thief, and a wise thief. Each of these thieves
have a knapsack that can old a total size of 8.

9. The greedy thief breaks into the window, and sees the items.
He makes a mental list of the items available, and grabs the
most expensive item first. The ring goes in first, leaving a
capacity of 7, and a value of 15. Next, he grabs the candelabra,
leaving a knapsack of size 2 and a value of 25. No other items
will fit in his knapsack, so he leaves.
The foolish and slow thief climbs in the window, and sees the
items. This thief was a programmer, downsized as part of the
"dot-bomb" blowout. Possessing a solid background in Boolean
logic, he figures that he can simply compute all combinations of
the objects and choose the best. So, he starts going through the
binary combinations of objects - all 2^5 of them. While he is still
drawing the truth table, the police show up, and arrest him.
Although his solution would certainly have given him the best
answer, it just took long to compute.

10.  The wise thief appears, and observes the items. He
notes that an empty knapsack has a value of 0. Further,
he notes that a knapsack can either contain each item,
or not. Further, his decision to include an item will be
based on a quick calculation - either the knapsack with
some combination of the previous items will be worth
more, or else the knapsack of a size that will fit the
current item was worth more. So, he does this quick
computation, and figures out that the best knapsack
he can take is made up of items 1,3, and 4, for a total
value of 29.

11. Knapsack problem
There are two versions of the problem:
1. “0-1 knapsack problem”
 Items are indivisible; you either take an item or not.
Some special instances can be solved with dynamic
programming
 Exhibit No greedy choice property.
No greedy algorithm exists.
 Exhibit optimal substructure property.
 Only dynamic programming algorithm exists.
 .

12. 1. “Fractional knapsack problem”
 Items are divisible: you can take any fraction of an item
 Exhibit greedy choice property.
 Greedy algorithm exists.
 Exhibit optimal substructure property

13. 0-1 Knapsack Problem
 The thief wants to steal n items
 The ith item weighs wi and has value vi
 Take the most valuable load, limit of W pounds
 This is called the 0-1 version of the problem because
there are no fractions.
The thief must take the whole item or leave it behind.

14.  Both the 0-1 Knapsack Problem and the Fractional
Knapsack Problem have optimal substructure. If we
remove item j from the optimal load, the remaining
problem must be optimal for the remaining W -
wj pounds for the overall solution to be optimal.
 The greedy solution for the Fractional Knapsack
Problem does not work here

15.  Example:
Item Value Weight
1 60 10
2 100 20
3 120 30
W = 50
The greedy solution would select 1, leaving 40 pounds.
At this point if we pick items 1 and 2 the total value is
160.
If we pick items 1 and 3 the total value is 180.
The optimal solution is to pick items 2 and 3. The total
value of this solution is 220

16. Example: Fractional Knapsack
Problem
A thief considers taking W pounds of loot. The loot is in the
form of n items, each with weight wi and value vi. Any amount
of an item can be put in the knapsack as long as the weight
limit W is not exceeded.

17.  Question:
 How much of which items should the thief take to maximize the
value of the loot?
 Greedy Algorithm: Take as much of the item with the
highest value per pound (vi/wi) as you can. If you run out of that
item, take from the next highest (vi/wi) item. Continue until
knapsack is full.

18. Thank
You

19. QUESTIONS ANSWER
SESSION?