Knapsack problem

The unbounded knapsack problem (UKP) places no restriction on the number of copies of each kind of item. Besides, here we assume that ${\displaystyle x_{i}>0}$

${\displaystyle m[w']=\max \left(\sum _{i=1}^{n}v_{i}x_{i}\right)}$

subject to ${\displaystyle \sum _{i=1}^{n}w_{i}x_{i}\leq w'}$ and ${\displaystyle x_{i}>0}$

Observe that ${\displaystyle m[w]}$ has the following properties:

1. ${\displaystyle m[0]=0\,\!}$ (the sum of zero items, i.e., the summation of the empty set).

2. ${\displaystyle m[w]=\max(v_{1}+m[w-w_{1}],v_{2}+m[w-w_{2}],...,v_{n}+m[w-w_{n}])}$ , ${\displaystyle w_{i}\leq w}$, where ${\displaystyle v_{i}}$ is the value of the ${\displaystyle i}$-th kind of item.

The second property needs to be explained in detail. During the process of the running of this method, how do we get the weight ${\displaystyle w}$? There are only ${\displaystyle i}$ ways and the previous weights are ${\displaystyle w-w_{1},w-w_{2},...,w-w_{i}}$ where there are total ${\displaystyle i}$ kinds of different item (by saying different, we mean that the weight and the value are not completely the same). If we know each value of these ${\displaystyle i}$ items and the related maximum value previously, we just compare them to each other and get the maximum value ultimately and we are done.

Here the maximum of the empty set is taken to be zero. Tabulating the results from ${\displaystyle m[0]}$ up through ${\displaystyle m[W]}$ gives the solution. Since the calculation of each ${\displaystyle m[w]}$ involves examining at most ${\displaystyle n}$ items, and there are at most ${\displaystyle W}$ values of ${\displaystyle m[w]}$ to calculate, the running time of the dynamic programming solution is ${\displaystyle O(nW)}$. Dividing ${\displaystyle w_{1},\,w_{2},\,\ldots ,\,w_{n},\,W}$ by their greatest common divisor is a way to improve the running time.

Even if P≠NP, the ${\displaystyle O(nW)}$ complexity does not contradict the fact that the knapsack problem is NP-complete, since ${\displaystyle W}$, unlike ${\displaystyle n}$, is not polynomial in the length of the input to the problem. The length of the ${\displaystyle W}$ input to the problem is proportional to the number of bits in ${\displaystyle W}$, ${\displaystyle \log W}$, not to ${\displaystyle W}$ itself. However, since this runtime is pseudopolynomial, this makes the (decision version of the) knapsack problem a weakly NP-complete problem.

A demonstration of the dynamic programming approach.

A similar dynamic programming solution for the 0-1 knapsack problem also runs in pseudo-polynomial time. Assume ${\displaystyle w_{1},\,w_{2},\,\ldots ,\,w_{n},\,W}$ are strictly positive integers. Define ${\displaystyle m[i,w]}$ to be the maximum value that can be attained with weight less than or equal to ${\displaystyle w}$ using items up to ${\displaystyle i}$ (first ${\displaystyle i}$ items).

We can define ${\displaystyle m[i,w]}$ recursively as follows: (Definition A)

• ${\displaystyle m[0,\,w]=0}$
• ${\displaystyle m[i,\,w]=m[i-1,\,w]}$ if ${\displaystyle w_{i}>w\,\!}$ (the new item is more than the current weight limit)
• ${\displaystyle m[i,\,w]=\max(m[i-1,\,w],\,m[i-1,w-w_{i}]+v_{i})}$ if ${\displaystyle w_{i}\leqslant w}$.

The solution can then be found by calculating ${\displaystyle m[n,W]}$. To do this efficiently, we can use a table to store previous computations.

The following is pseudocode for the dynamic program:

// Input:
// Values (stored in array v)
// Weights (stored in array w)
// Number of distinct items (n)
// Knapsack capacity (W)
// NOTE: The array "v" and array "w" are assumed to store all relevant values starting at index 1.

array m[0..n, 0..W];
for j from 0 to W do:
    m[0, j] := 0
for i from 1 to n do:
    m[i, 0] := 0

for i from 1 to n do:
    for j from 0 to W do:
        if w[i] > j then:
            m[i, j] := m[i-1, j]
        else:
            m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])

This solution will therefore run in ${\displaystyle O(nW)}$ time and ${\displaystyle O(nW)}$ space. (If we only need the value m[n,W], we can modify the code so that the amount of memory required is O(W) which stores the recent two lines of the array "m".)

However, if we take it a step or two further, we should know that the method will run in the time between ${\displaystyle O(nW)}$ and ${\displaystyle O(2^{n})}$. From Definition A, we know that there is no need to compute all the weights when the number of items and the items themselves that we chose are fixed. That is to say, the program above computes more than necessary because the weight changes from 0 to W often. From this perspective, we can program this method so that it runs recursively.

// Input:
// Values (stored in array v)
// Weights (stored in array w)
// Number of distinct items (n)
// Knapsack capacity (W)
// NOTE: The array "v" and array "w" are assumed to store all relevant values starting at index 1.

Define value[n, W]

Initialize all value[i, j] = -1

Define m:=(i,j)         // Define function m so that it represents the maximum value we can get under the condition: use first i items, total weight limit is j
{
    if i == 0 or j <= 0 then:
        value[i, j] = 0
        return

    if (value[i-1, j] == -1) then:     // m[i-1, j] has not been calculated, we have to call function m
        value[i-1, j] = m(i-1, j)

    if w[i] > j then:                      // item cannot fit in the bag
        value[i, j] = value[i-1, j]
    else: 
        if (value[i-1, j-w[i]] == -1) then:     // m[i-1,j-w[i]] has not been calculated, we have to call function m
            value[i-1, j-w[i]] = m(i-1, j-w[i])
        value[i, j] = max(value[i-1,j], value[i-1, j-w[i]] + v[i])
}

Run m(n, W)

For example, there are 10 different items and the weight limit is 67. So,

${\displaystyle {\begin{aligned}&w[1]=23,w[2]=26,w[3]=20,w[4]=18,w[5]=32,w[6]=27,w[7]=29,w[8]=26,w[9]=30,w[10]=27\\&v[1]=505,v[2]=352,v[3]=458,v[4]=220,v[5]=354,v[6]=414,v[7]=498,v[8]=545,v[9]=473,v[10]=543\\\end{aligned}}}$

If you use above method to compute for ${\displaystyle m(10,67)}$, you will get this, excluding calls that produce ${\displaystyle m(i,j)=0}$:

${\displaystyle {\begin{aligned}&m(10,67)=1270\\&m(9,67)=1270,m(9,40)=678\\&m(8,67)=1270,m(8,40)=678,m(8,37)=545\\&m(7,67)=1183,m(7,41)=725,m(7,40)=678,m(7,37)=505\\&m(6,67)=1183,m(6,41)=725,m(6,40)=678,m(6,38)=678,m(6,37)=505\\&m(5,67)=1183,m(5,41)=725,m(5,40)=678,m(5,38)=678,m(5,37)=505\\&m(4,67)=1183,m(4,41)=725,m(4,40)=678,m(4,38)=678,m(4,37)=505,m(4,35)=505\\&m(3,67)=963,m(3,49)=963,m(3,41)=505,m(3,40)=505,m(3,38)=505,m(3,37)=505,m(3,35)=505,m(3,23)=505,m(3,22)=458,m(3,20)=458\\&m(2,67)=857,m(2,49)=857,m(2,47)=505,m(2,41)=505,m(2,40)=505,m(2,38)=505,m(2,37)=505,m(2,35)=505,m(2,29)=505,m(2,23)=505\\&m(1,67)=505,m(1,49)=505,m(1,47)=505,m(1,41)=505,m(1,40)=505,m(1,38)=505,m(1,37)=505,m(1,35)=505,m(1,29)=505,m(1,23)=505\\\end{aligned}}}$

Besides, we can break the recursion and convert it into a tree. Then we can cut some leaves and use parallel computing to expedite the running of this method.

To find the actual subset of items, rather than just their total value, we can run this after running the function above:

/**
 * Returns the indices of the items of the optimal knapsack.
 * i: We can include items 1 through i in the knapsack
 * j: maximum weight of the knapsack
 */
function knapsack(i: int, j: int): Set<int> {
    if i == 0 then:
        return {}
    if m[i, j] > m[i-1, j] then:
        return {i} ∪ knapsack(i-1, j-w[i])
    else:
        return knapsack(i-1, j)
}

knapsack(n, W)

Another algorithm for 0-1 knapsack, discovered in 1974[18] and sometimes called "meet-in-the-middle" due to parallels to a similarly named algorithm in cryptography, is exponential in the number of different items but may be preferable to the DP algorithm when ${\displaystyle W}$ is large compared to n. In particular, if the ${\displaystyle w_{i}}$ are nonnegative but not integers, we could still use the dynamic programming algorithm by scaling and rounding (i.e. using fixed-point arithmetic), but if the problem requires ${\displaystyle d}$ fractional digits of precision to arrive at the correct answer, ${\displaystyle W}$ will need to be scaled by ${\displaystyle 10^{d}}$, and the DP algorithm will require ${\displaystyle O(W10^{d})}$ space and ${\displaystyle O(nW10^{d})}$ time.

algorithm Meet-in-the-middle is
    input: A set of items with weights and values.
    output: The greatest combined value of a subset.

    partition the set {1...n} into two sets A and B of approximately equal size
    compute the weights and values of all subsets of each set

    for each subset of A do
        find the subset of B of greatest value such that the combined weight is less than W

    keep track of the greatest combined value seen so far

The algorithm takes ${\displaystyle O(2^{n/2})}$ space, and efficient implementations of step 3 (for instance, sorting the subsets of B by weight, discarding subsets of B which weigh more than other subsets of B of greater or equal value, and using binary search to find the best match) result in a runtime of ${\displaystyle O(n2^{n/2})}$. As with the meet in the middle attack in cryptography, this improves on the ${\displaystyle O(n2^{n})}$ runtime of a naive brute force approach (examining all subsets of ${\displaystyle \{1...n\}}$), at the cost of using exponential rather than constant space (see also baby-step giant-step).

As for most NP-complete problems, it may be enough to find workable solutions even if they are not optimal. Preferably, however, the approximation comes with a guarantee of the difference between the value of the solution found and the value of the optimal solution.

As with many useful but computationally complex algorithms, there has been substantial research on creating and analyzing algorithms that approximate a solution. The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. This means that the problem has a polynomial time approximation scheme. To be exact, the knapsack problem has a fully polynomial time approximation scheme (FPTAS).[19]

George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem.[20] His version sorts the items in decreasing order of value per unit of weight, ${\displaystyle v_{1}/w_{1}\geq \cdots \geq v_{n}/w_{n}}$. It then proceeds to insert them into the sack, starting with as many copies as possible of the first kind of item until there is no longer space in the sack for more. Provided that there is an unlimited supply of each kind of item, if ${\displaystyle m}$ is the maximum value of items that fit into the sack, then the greedy algorithm is guaranteed to achieve at least a value of ${\displaystyle m/2}$.

For the bounded problem, where the supply of each kind of item is limited, the above algorithm may be far from optimal. Nevertheless, a simple modification allows us to solve this case: Assume for simplicity that all items individually fit in the sack (${\displaystyle w_{i}\leq W}$ for all ${\displaystyle i}$). Construct a solution ${\displaystyle S_{1}}$ by packing items greedily as long as possible, i.e. ${\displaystyle S_{1}=\left\{1,\ldots ,k\right\}}$ where ${\displaystyle k=\textstyle \max _{1\leq k'\leq n}\textstyle \sum _{i=1}^{k'}w_{i}\leq W}$. Furthermore, construct a second solution ${\displaystyle S_{2}=\left\{k+1\right\}}$ containing the first item that did not fit. Since ${\displaystyle S_{1}\cup S_{2}}$ provides an upper bound for the LP relaxation of the problem, one of the sets must have value at least ${\displaystyle m/2}$; we thus return whichever of ${\displaystyle S_{1}}$ and ${\displaystyle S_{2}}$ has better value to obtain a ${\displaystyle 1/2}$-approximation.

It can be shown that the average performance converges to the optimal solution in distribution at the error rate ${\displaystyle n^{-1/2}}$ [21]

The fully polynomial time approximation scheme (FPTAS) for the knapsack problem takes advantage of the fact that the reason the problem has no known polynomial time solutions is because the profits associated with the items are not restricted. If one rounds off some of the least significant digits of the profit values then they will be bounded by a polynomial and 1/ε where ε is a bound on the correctness of the solution. This restriction then means that an algorithm can find a solution in polynomial time that is correct within a factor of (1-ε) of the optimal solution.[19]

algorithm FPTAS is 
    input: ε ∈ (0,1]
            a list A of n items, specified by their values, ${\displaystyle v_{i}}$, and weights
    output: S' the FPTAS solution

    P := max ${\displaystyle \{v_{i}\mid 1\leq i\leq n\}}$  // the highest item value
    K := ε ${\displaystyle {\frac {P}{n}}}$

    for i from 1 to n do
        ${\displaystyle v'_{i}}$ := ${\displaystyle \left\lfloor {\frac {v_{i}}{K}}\right\rfloor }$
    end for

    return the solution, S', using the ${\displaystyle v'_{i}}$ values in the dynamic program outlined above

Theorem: The set ${\displaystyle S'}$ computed by the algorithm above satisfies ${\displaystyle \mathrm {profit} (S')\geq (1-\varepsilon )\cdot \mathrm {profit} (S^{*})}$, where ${\displaystyle S^{*}}$ is an optimal solution.

Solving the unbounded knapsack problem can be made easier by throwing away items which will never be needed. For a given item ${\displaystyle i}$, suppose we could find a set of items ${\displaystyle J}$ such that their total weight is less than the weight of ${\displaystyle i}$, and their total value is greater than the value of ${\displaystyle i}$. Then ${\displaystyle i}$ cannot appear in the optimal solution, because we could always improve any potential solution containing ${\displaystyle i}$ by replacing ${\displaystyle i}$ with the set ${\displaystyle J}$. Therefore, we can disregard the ${\displaystyle i}$-th item altogether. In such cases, ${\displaystyle J}$ is said to dominate ${\displaystyle i}$. (Note that this does not apply to bounded knapsack problems, since we may have already used up the items in ${\displaystyle J}$.)

Finding dominance relations allows us to significantly reduce the size of the search space. There are several different types of dominance relations,[11] which all satisfy an inequality of the form:

${\displaystyle \qquad \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}}$, and ${\displaystyle \sum _{j\in J}v_{j}\,x_{j}\ \geq \alpha \,v_{i}\,}$ for some ${\displaystyle x\in Z_{+}^{n}}$

where ${\displaystyle \alpha \in Z_{+}\,,J\subsetneq N}$ and ${\displaystyle i\not \in J}$. The vector ${\displaystyle x}$ denotes the number of copies of each member of ${\displaystyle J}$.

Collective dominance

The ${\displaystyle i}$-th item is collectively dominated by ${\displaystyle J}$, written as ${\displaystyle i\ll J}$, if the total weight of some combination of items in ${\displaystyle J}$ is less than w_i and their total value is greater than v_i. Formally, ${\displaystyle \sum _{j\in J}w_{j}\,x_{j}\ \leq w_{i}}$ and ${\displaystyle \sum _{j\in J}v_{j}\,x_{j}\ \geq v_{i}}$ for some ${\displaystyle x\in Z_{+}^{n}}$, i.e. ${\displaystyle \alpha =1}$. Verifying this dominance is computationally hard, so it can only be used with a dynamic programming approach. In fact, this is equivalent to solving a smaller knapsack decision problem where ${\displaystyle V=v_{i}}$, ${\displaystyle W=w_{i}}$, and the items are restricted to ${\displaystyle J}$.

Threshold dominance

The ${\displaystyle i}$-th item is threshold dominated by ${\displaystyle J}$, written as ${\displaystyle i\prec \prec J}$, if some number of copies of ${\displaystyle i}$ are dominated by ${\displaystyle J}$. Formally, ${\displaystyle \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}}$, and ${\displaystyle \sum _{j\in J}v_{j}\,x_{j}\ \geq \alpha \,v_{i}\,}$ for some ${\displaystyle x\in Z_{+}^{n}}$ and ${\displaystyle \alpha \geq 1}$. This is a generalization of collective dominance, first introduced in[13] and used in the EDUK algorithm. The smallest such ${\displaystyle \alpha }$ defines the threshold of the item ${\displaystyle i}$, written ${\displaystyle t_{i}=(\alpha -1)w_{i}}$. In this case, the optimal solution could contain at most ${\displaystyle \alpha -1}$ copies of ${\displaystyle i}$.

Multiple dominance

The ${\displaystyle i}$-th item is multiply dominated by a single item ${\displaystyle j}$, written as ${\displaystyle i\ll _{m}j}$, if ${\displaystyle i}$ is dominated by some number of copies of ${\displaystyle j}$. Formally, ${\displaystyle w_{j}\,x_{j}\ \leq w_{i}}$, and ${\displaystyle v_{j}\,x_{j}\ \geq v_{i}}$ for some ${\displaystyle x_{j}\in Z_{+}}$ i.e. ${\displaystyle J=\{j\},\alpha =1,x_{j}=\lfloor {\frac {w_{i}}{w_{j}}}\rfloor }$. This dominance could be efficiently used during preprocessing because it can be detected relatively easily.

Modular dominance

Let ${\displaystyle b}$ be the best item, i.e. ${\displaystyle {\frac {v_{b}}{w_{b}}}\geq {\frac {v_{i}}{w_{i}}}\,}$ for all ${\displaystyle i}$. This is the item with the greatest density of value. The ${\displaystyle i}$-th item is modularly dominated by a single item ${\displaystyle j}$, written as ${\displaystyle i\ll _{\equiv }j}$, if ${\displaystyle i}$ is dominated by ${\displaystyle j}$ plus several copies of ${\displaystyle b}$. Formally, ${\displaystyle w_{j}+tw_{b}\leq w_{i}}$, and ${\displaystyle v_{j}+tv_{b}\geq v_{i}}$ i.e. ${\displaystyle J=\{b,j\},\alpha =1,x_{b}=t,x_{j}=1}$.

This variation changes the goal of the individual filling the knapsack. Instead of one objective, such as maximizing the monetary profit, the objective could have several dimensions. For example, there could be environmental or social concerns as well as economic goals. Problems frequently addressed include portfolio and transportation logistics optimizations.[22][23]

As an example, suppose you ran a cruise ship. You have to decide how many famous comedians to hire. This boat can handle no more than one ton of passengers and the entertainers must weigh less than 1000 lbs. Each comedian has a weight, brings in business based on their popularity and asks for a specific salary. In this example, you have multiple objectives. You want, of course, to maximize the popularity of your entertainers while minimizing their salaries. Also, you want to have as many entertainers as possible.

In this variation, the weight of knapsack item ${\displaystyle i}$ is given by a D-dimensional vector ${\displaystyle {\overline {w_{i}}}=(w_{i1},\ldots ,w_{iD})}$ and the knapsack has a D-dimensional capacity vector ${\displaystyle (W_{1},\ldots ,W_{D})}$. The target is to maximize the sum of the values of the items in the knapsack so that the sum of weights in each dimension ${\displaystyle d}$ does not exceed ${\displaystyle W_{d}}$.

Multi-dimensional knapsack is computationally harder than knapsack; even for ${\displaystyle D=2}$, the problem does not have EPTAS unless P${\displaystyle =}$NP.[24] However, the algorithm in[25] is shown to solve sparse instances efficiently. An instance of multi-dimensional knapsack is sparse if there is a set ${\displaystyle J=\{1,2,\ldots ,m\}}$ for ${\displaystyle m<D}$ such that for every knapsack item ${\displaystyle i}$, ${\displaystyle \exists z>m}$ such that ${\displaystyle \forall j\in J\cup \{z\},\ w_{ij}\geq 0}$ and ${\displaystyle \forall y\notin J\cup \{z\},w_{iy}=0}$. Such instances occur, for example, when scheduling packets in a wireless network with relay nodes.[25] The algorithm from[25] also solves sparse instances of the multiple choice variant, multiple-choice multi-dimensional knapsack.

The IHS (Increasing Height Shelf) algorithm is optimal for 2D knapsack (packing squares into a two-dimensional unit size square): when there are at most five square in an optimal packing.[26]

This variation is similar to the Bin Packing Problem. It differs from the Bin Packing Problem in that a subset of items can be selected, whereas, in the Bin Packing Problem, all items have to be packed to certain bins. The concept is that there are multiple knapsacks. This may seem like a trivial change, but it is not equivalent to adding to the capacity of the initial knapsack. This variation is used in many loading and scheduling problems in Operations Research and has a Polynomial-time approximation scheme.[27]

The quadratic knapsack problem maximizes a quadratic objective function subject to binary and linear capacity constraints.[28] The problem was introduced by Gallo, Hammer, and Simeone in 1980,[29] however the first treatment of the problem dates back to Witzgall in 1975.[30]

The subset sum problem is a special case of the decision and 0-1 problems where each kind of item, the weight equals the value: ${\displaystyle w_{i}=v_{i}}$. In the field of cryptography, the term knapsack problem is often used to refer specifically to the subset sum problem and is commonly known as one of Karp's 21 NP-complete problems.[31]

The generalization of subset sum problem is called multiple subset-sum problem, in which multiple bins exist with the same capacity. It has been shown that the generalization does not have an FPTAS.[32]

In the geometric knapsack problem, there is a set of rectangles with different values, and a rectangular knapsack. The goal is to pack the largest possible value into the knapsack.[33]