Dynamic Programming knapsack 0 1

Dynamic Programming …
Continued
0-1 Knapsack Problem
Dr. AMIT KUMAR @JUET

Knapsack 0-1 Problem
• The goal is to maximize
the value of a knapsack
that can hold at most W
units (i.e. lbs or kg) worth
of goods from a list of
items I0, I1, … In-1.
– Each item has 2
attributes:
1) Value – let this be vi for
item Ii
2) Weight – let this be wi for
item Ii

• The difference
between this problem
and the fractional
knapsack one is that
you CANNOT take a
fraction of an item.
– You can either take it or
not.
– Hence the name
Knapsack 0-1 problem.

• Brute Force
– The naïve way to solve this problem is to cycle
through all 2n subsets of the n items and pick the
subset with a legal weight that maximizes the
value of the knapsack.
– We can come up with a dynamic programming
algorithm that will USUALLY do better than this
brute force technique.

• As we did before we are going to solve the
problem in terms of sub-problems.
– So let’s try to do that…
• Our first attempt might be to characterize a sub-
problem as follows:
– Let Sk be the optimal subset of elements from {I0, I1, …, Ik}.
• What we find is that the optimal subset from the elements
{I0, I1, …, Ik+1} may not correspond to the optimal subset of
elements from {I0, I1, …, Ik} in any regular pattern.
– Basically, the solution to the optimization problem for
Sk+1 might NOT contain the optimal solution from
problem Sk. Dr. AMIT KUMAR @JUET

• Let’s illustrate that point with an example:
Item Weight Value
I0 3 10
I1 8 4
I2 9 9
I3 8 11
• The maximum weight the knapsack can hold is 20.
• The best set of items from {I0, I1, I2} is {I0, I1, I2}
• BUT the best set of items from {I0, I1, I2, I3} is {I0, I2, I3}.
– In this example, note that this optimal solution, {I0, I2, I3}, does
NOT build upon the previous optimal solution, {I0, I1, I2}.
• (Instead it build's upon the solution, {I0, I2}, which is really the
optimal subset of {I0, I1, I2} with weight 12 or less.)

Knapsack 0-1 problem
• So now we must re-work the way we build upon previous sub-
problems…
– Let B[k, w] represent the maximum total value of a subset Sk with
weight w.
– Our goal is to find B[n, W], where n is the total number of items and
W is the maximal weight the knapsack can carry.
• So our recursive formula for subproblems:
B[k, w] = B[k - 1,w], if wk > w
= max { B[k - 1,w], B[k - 1,w - wk] + vk}, otherwise
• In English, this means that the best subset of Sk that has total
weight w is:
1) The best subset of Sk-1 that has total weight w, or
2) The best subset of Sk-1 that has total weight w-wk plus the item kDr. AMIT KUMAR @JUET

Knapsack 0-1 Problem –
Recursive Formula
• The best subset of Sk that has the total weight
w, either contains item k or not.
• First case: wk > w
– Item k can’t be part of the solution! If it was the
total weight would be > w, which is unacceptable.
• Second case: wk ≤ w
– Then the item k can be in the solution, and we
choose the case with greater value.Dr. AMIT KUMAR @JUET

Knapsack 0-1 Algorithm
for w = 0 to W { // Initialize 1st row to 0’s
B[0,w] = 0
}
for i = 1 to n { // Initialize 1st column to 0’s
B[i,0] = 0
}
for i = 1 to n {
for w = 0 to W {
if wi <= w { //item i can be in the solution
if vi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
}
else B[i,w] = B[i-1,w] // wi > w
}
}

• Let’s run our algorithm on the following data:
– n = 4 (# of elements)
– W = 5 (max weight)
– Elements (weight, value):
(2,3), (3,4), (4,5), (5,6)

Knapsack 0-1 Example
// Initialize the base cases
for w = 0 to W
B[0,w] = 0
for i = 1 to n
B[i,0] = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0
2 0
3 0
4 0

if wi <= w //item i can be in the solution
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0
2 0
3 0
4 0
i = 1
vi = 3
wi = 2
w = 1
w-wi = -1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0
2 0
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0
2 0
3 0
4 0
i = 1
vi = 3
wi = 2
w = 2
w-wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3
2 0
3 0
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3
2 0
3 0
4 0
i = 1
vi = 3
wi = 2
w = 3
w-wi = 1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3
2 0
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3
2 0
3 0
4 0
i = 1
vi = 3
wi = 2
w = 4
w-wi = 2
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3
2 0
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3
2 0
3 0
4 0
i = 1
vi = 3
wi = 2
w = 5
w-wi = 3
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0
3 0
4 0
i = 2
vi = 4
wi = 3
w = 1
w-wi = -2
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0
3 0
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0
3 0
4 0
i = 2
vi = 4
wi = 3
w = 2
w-wi = -1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3
3 0
4 0
i = 2
vi = 4
wi = 3
w = 3
w-wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4
3 0
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4
3 0
4 0
i = 2
vi = 4
wi = 3
w = 4
w-wi = 1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4
3 0
4 0
i = 2
vi = 4
wi = 3
w = 5
w-wi = 2
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0
4 0

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0
4 0
i = 3
vi = 5
wi = 4
w = 1..3
w-wi = -3..-1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4
4 0
i = 3
vi = 5
wi = 4
w = 4
w-wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5
4 0
i = 3
vi = 5
wi = 4
w = 5
w-wi = 1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0
i = 4
vi = 6
wi = 5
w = 1..4
w-wi = -4..-1
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5
i = 4
vi = 6
wi = 5
w = 5
w-wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
B[i,w] = vi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]

We’re DONE!!
The max possible value that can be carried in this knapsack is $7
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7

• This algorithm only finds the max possible
value that can be carried in the knapsack
– The value in B[n,W]
• To know the items that make this maximum
value, we need to trace back through the
table.

Finding the Items
• Let i = n and k = W
if B[i, k] ≠ B[i-1, k] then
mark the ith item as in the knapsack
i = i-1, k = k-wi
else
i = i-1 // Assume the ith item is not in the knapsack
// Could it be in the optimally packed knapsack?

Finding the Items
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i = 4
k = 5
vi = 6
wi = 5
B[i,k] = 7
B[i-1,k] = 7
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
i = n , k = W
while i, k > 0
i = i-1, k = k-wi
else
i = i-1
Knapsack:

Finding the Items
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i = 3
k = 5
vi = 5
wi = 4
B[i,k] = 7
B[i-1,k] = 7
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
i = n , k = W
while i, k > 0
i = i-1, k = k-wi
else
i = i-1
Knapsack:

Finding the Items
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i = 2
k = 5
vi = 4
wi = 3
B[i,k] = 7
B[i-1,k] = 3
k – wi = 2
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
i = n , k = W
while i, k > 0
i = i-1, k = k-wi
else
i = i-1
Knapsack:
Item 2

Finding the Items
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i = 1
k = 2
vi = 3
wi = 2
B[i,k] = 3
B[i-1,k] = 0
k – wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
i = n , k = W
while i, k > 0
i = i-1, k = k-wi
else
i = i-1
Knapsack:
Item 1
Item 2

Finding the Items
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
i = 1
k = 2
vi = 3
wi = 2
B[i,k] = 3
B[i-1,k] = 0
k – wi = 0
i / w 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
k = 0, so we’re DONE!
The optimal knapsack should contain:
Item 1 and Item 2
Knapsack:
Item 1
Item 2

Knapsack 0-1 Problem – Run Time
for w = 0 to W
B[0,w] = 0
for i = 1 to n
B[i,0] = 0
for i = 1 to n
for w = 0 to W
< the rest of the code >
What is the running time of this algorithm?
O(n*W)
Remember that the brute-force algorithm takes: O(2n)
O(W)
O(W)
Repeat n times
O(n)

Knapsack Problem
1) Fill out the
dynamic
programming
table for the
knapsack
problem to the
right.
2) Trace back
through the
table to find the
items in the Dr. AMIT KUMAR @JUET

References
• Slides adapted from Arup Guha’s Computer
Science II Lecture notes:
http://www.cs.ucf.edu/~dmarino/ucf/cop3503/
lectures/
• Additional material from the textbook:
Data Structures and Algorithm Analysis in Java (Second
Edition) by Mark Allen Weiss
• Additional images:
www.wikipedia.com
xkcd.com

Dynamic Programming knapsack 0 1

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