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Dynamic Programming Application to the 0/1 Knapsack Problem, Study notes of Data Structures and Algorithms

University of Florida (UF)Data Structures and Algorithms

The concept of dynamic programming and its application to the 0/1 knapsack problem. The sequence of decisions, problem state, principle of optimality, and dynamic programming recurrence equations. It also provides a recursive code implementation and discusses time complexity and reducing run time.

Typology: Study notes

Pre 2010

Uploaded on 09/17/2009

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Dynamic Programming
•Sequence of decisions.
•Problem state.
•Principle of optimality.
•Dynamic Programming Recurrence
Equations.
•Solution of recurrence equations.
Sequence Of Decisions
•As in the greedy method, the solution to a
problem is viewed as the result of a
sequence of decisions.
•Unlike the greedy method, decisions are not
made in a greedy and binding manner.
0/1 Knapsack Problem
Let xi= 1 when item i is selected and let xi= 0
when item iis not selected.
i = 1
npixi
maximize
i = 1
nwixi<= c
subject to
and xi= 0 or 1 for all i
All profits and weights are positive.
Sequence Of Decisions
•Decide the xivalues in the order x1, x2, x3, …, xn.
•Decide the xivalues in the order xn,xn-1,xn-2, …,
x1.
•Decide the xivalues in the order x1,xn, x2,xn-1, …
•Or any other order.
pf3
pf4
pf5

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Dynamic Programming

  • Sequence of decisions.
  • Problem state.
  • Principle of optimality.
  • Dynamic Programming Recurrence Equations.
  • Solution of recurrence equations.

Sequence Of Decisions

  • As in the greedy method, the solution to a problem is viewed as the result of a sequence of decisions.
  • Unlike the greedy method, decisions are not made in a greedy and binding manner.

0/1 Knapsack Problem

Let xi = 1 when item i is selected and let xi = 0 when item i is not selected.

i = 1

n maximize pi^ xi

i = 1

n subject to wi^ xi^ <= c

and xi = 0 or 1 for all i All profits and weights are positive.

Sequence Of Decisions

  • Decide the xi values in the order x 1 , x 2 , x 3 , …, xn.
  • Decide the xi values in the order xn, xn-1, xn-2, …, x 1.
  • Decide the xi values in the order x 1 , xn, x 2 , xn-1, …
  • Or any other order.

Problem State

  • The state of the 0/1 knapsack problem is given by

ƒ the weights and profits of the available items ƒ the capacity of the knapsack

  • When a decision on one of the xi values is made, the problem state changes. ƒ item i is no longer available ƒ the remaining knapsack capacity may be less

Problem State

  • Suppose that decisions are made in the order x 1 , x 2 , x 3 , …, xn.
  • The initial state of the problem is described by the pair (1, c). ƒ Items 1 through n are available (the weights, profits and n are implicit). ƒ The available knapsack capacity is c.
  • Following the first decision the state becomes one of the following: ƒ (2, c) … when the decision is to set x 1 = 0. ƒ (2, c-w 1 ) … when the decision is to set x 1 = 1.

Problem State

  • Suppose that decisions are made in the order xn, xn-1, xn-2, …, x 1.
  • The initial state of the problem is described by the pair (n, c). ƒ Items 1 through n are available (the weights, profits and first item index are implicit). ƒ The available knapsack capacity is c.
  • Following the first decision the state becomes one of the following: ƒ (n-1, c) … when the decision is to set xn= 0. ƒ (n-1, c-wn) … when the decision is to set xn= 1.

Principle Of Optimality

  • An optimal solution satisfies the following property: ƒ No matter what the first decision, the remaining decisions are optimal with respect to the state that results from this decision.
  • Dynamic programming may be used only when the principle of optimality holds.

x 1 = a 1 = 1

  • If not, this instance has a better solution b 2 , b 3 , …, bn.

i = 2

n maximize pi^ xi

i = 2

n subject to wi^ xi^ <= c- w^1

and xi = 0 or 1 for all i

i = 2

n pi bi > i = 2

n pi ai

x 1 = a 1 = 1

  • x 1 = a 1 , x 2 = b 2 , x 3 = b 3 , …, xn = bn is a better solution to the original instance than is x 1 = a 1 , x 2 = a 2 , x 3 = a 3 , …, xn = an.
  • So x 1 = a 1 , x 2 = a 2 , x 3 = a 3 , …, xn = an cannot be an optimal solution … a contradiction with the assumption that it is optimal.

0/1 Knapsack Problem

  • Therefore, no matter what the first decision, the remaining decisions are optimal with respect to the state that results from this decision.
  • The principle of optimality holds and dynamic programming may be applied.

Dynamic Programming Recurrence

  • Let f(i,y) be the profit value of the optimal solution to the knapsack instance defined by the state (i,y). ƒ Items i through n are available. ƒ Available capacity is y.
  • For the time being assume that we wish to determine only the value of the best solution. ƒ Later we will worry about determining the xis that yield this maximum value.
  • Under this assumption, our task is to determine f(1,c).

Dynamic Programming Recurrence

  • f(n,y) is the value of the optimal solution to the knapsack instance defined by the state (n,y). ƒ Only item n is available. ƒ Available capacity is y.
  • If wn <= y, f(n,y) = pn.
  • If wn > y, f(n,y) = 0.

Dynamic Programming Recurrence

  • Suppose that i < n.
  • f(i,y) is the value of the optimal solution to the knapsack instance defined by the state (i,y). ƒ Items i through n are available. ƒ Available capacity is y.
  • Suppose that in the optimal solution for the state (i,y), the first decision is to set xi= 0.
  • From the principle of optimality (we have shown that this principle holds for the knapsack problem), it follows that f(i,y) = f(i+1,y).

Dynamic Programming Recurrence

  • The only other possibility for the first decision is xi= 1.
  • The case xi= 1 can arise only when y >= wi.
  • From the principle of optimality, it follows that f(i,y) = f(i+1,y-wi) + pi.
  • Combining the two cases, we get ƒ f(i,y) = f(i+1,y) whenever y < wi. ƒ f(i,y) = max{f(i+1,y), f(i+1,y-wi) + pi}, y >= wi.

Recursive Code

/** @return f(i,y) */ private static int f(int i, int y) { if (i == n) return (y < w[n])? 0 : p[n]; if (y < w[i]) return f(i + 1, y); return Math.max(f(i + 1, y), f(i + 1, y - w[i]) + p[i]); }

Integer Weights

  • Assume that each weight is an integer.
  • The knapsack capacity c may also be assumed to be an integer.
  • Only f(i,y)s with 1 <= i <= n and 0 <= y <= c are of interest.
  • Even though level i of the recursion tree has up to 2 i-1^ nodes, at most c+1 represent different f(i,y)s.

Integer Weights Dictionary

  • Use an array fArray[][] as the dictionary.
  • fArray[1:n][0:c]
  • fArray[i][y] = -1 iff f(i,y) not yet computed.
  • This initialization is done before the recursive method is invoked.
  • The initialization takes O(cn) time.

No Recomputation Code

private static int f(int i, int y)

{

if (fArray[i][y] >= 0) return fArray[i][y]; if (i == n) {fArray[i][y] = (y < w[n])? 0 : p[n]; return fArray[i][y];} if (y < w[i]) fArray[i][y] = f(i + 1, y); else fArray[i][y] = Math.max(f(i + 1, y), f(i + 1, y - w[i]) + p[i]); return fArray[i][y];

}

Time Complexity

  • t(n) = O(cn).
  • Analysis done in text.
  • Good when cn is small relative to 2 n.
  • n = 3, c = 1010101 w = [100102, 1000321 , 6327] p = [102, 505 , 5]
  • 2 n^ = 8
  • cn = 3030303