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Filling up a bin using balls with divisible weights


This post shows how the special case for this problem where there is exactly one bin, and each ball have weight a power of \(2\). It is one of the most popular unanswered problem on cs.stackexchange as of writing.

Problem1 We are interested in solving the following integer program, \begin{align*} \text{Minimize:} & \sum_{i=1}^n |x_i-a_i| \\ \text{subject to:} & \sum_{i=1}^n w_i x_i = c\\ & 0\leq x_i \leq b_i \text{ for all } 1\leq i \leq n\\ \end{align*}

where each \(w_i\) is a power of \(2\) for all \(1\leq i\leq n\). Assume \(w_i\leq w_{i+1}\).

In fact, we don't require power of \(2\)s, we can still establish polynomial time as long as \(w_{i+1}/w_i\) is bounded by a polynomial in terms of the input size for all \(i\). However, for simplicity of exposition, assume it's power of \(2\). If \(w_{i+1}/w_i\) it is not unbounded, then we are not sure.

Let's consider a more natural problem without the absolute values.

Problem2\(0\)-\(1\) exact knapsack problem with divisible weights We are interested in solving the following integer program, \begin{align*} \text{Minimize:} & \sum_{i=1}^n c_i x_i \\ \text{subject to:} & \sum_{i=1}^n w_i x_i = t\\ & x_i \in \{0,1\} \text{ for all } 1\leq i \leq n\\ \end{align*}

where \(w_i|w_{i+1}\) for all \(1\leq i\leq n\). \(w_i\) can be negative.

There are two things we have to show. We must show Problem 1 reduces to Problem 2 with polynomial blow up, and Problem 2 can be solved in polynomial time.

1 Reduction

The reduction goes through a few steps. We start with the integer program in Problem 1, and let \(y_i = a_i-x_i\), and we get

\begin{align*} \text{Minimize:} & \sum_{i=1}^n |y_i| \\ \text{subject to:} & \sum_{i=1}^n w_i y_i = \sum_{i=1}^n w_i a_i - c\\ & a_i-b_i\leq y_i \leq a_i \text{ for all } 1\leq i \leq n\\ \end{align*}

Hence, let \(c' = \sum_{i=1}^n w_ia_i -c\), and \(l_i = a_i-b_i\) and \(u_i = a_i\).

\begin{align*} \text{Minimize:} & \sum_{i=1}^n |y_i| \\ \text{subject to:} & \sum_{i=1}^n w_i y_i = c'\\ & l_i \leq y_i \leq u_i \text{ for all } 1\leq i \leq n\\ \end{align*}

Now, let \(y_i=y_i^+ - y_i^-\), where \(y_i^-,y_i^+\geq 0\), we can remove the absolute value.

\begin{align*} \text{Minimize:} & \sum_{i=1}^n y_i^+ + y_i^- \\ \text{subject to:} & \sum_{i=1}^n w_i y_i^+ + \sum_{i=1}^n -w_i y_i^- = c'\\ & l_i \leq y_i^+- y_i^- \leq u_i \text{ for all } 1\leq i \leq n\\ & y_i^-, y_i^+\geq 0 \text{ for all } 1\leq i \leq n\\ \end{align*}

Observe we can separate the inequalities involving \(y_i^+ - y_i^-\), because there is always an optimal solution where \(y_i^-\) or \(y_i^+\) is \(0\).

Remark

This observation fails when the number of bins is more than \(1\).

\begin{align*} \text{Minimize:} & \sum_{i=1}^n y_i^+ + y_i^- \\ \text{subject to:} & \sum_{i=1}^n w_i y_i^+ + \sum_{i=1}^n -w_i y_i^- = c'\\ & 0 \leq y_i^+ \leq u_i \text{ for all } 1\leq i \leq n\\ & 0 \leq y_i^- \leq -l_i \text{ for all } 1\leq i \leq n\\ \end{align*}

This is an integer program is in the following form as a bounded exact knapsack problem.

\begin{align*} \text{Minimize:} & \sum_{i=1}^n x_i \\ \text{subject to:} & \sum_{i=1}^n w_i x_i = c\\ & 0 \leq x_i \leq b_i \text{ for all } 1\leq i \leq n\\ \end{align*}

Finally, apply the standard technique that rewrites a bounded knapsack problem to \(0\)-\(1\)-knapsack problem (say, see Section 7.1.1 of [1]). Note the blow up is at most a factor of \(O(\log \max_i b_i)\). We can get the integer program in Problem 2, and also the weights are all powers of \(2\). The reduction runs in polynomial time with respect to input size.

2 Solving Problem 2

Yuzhou Gu noted that the integer program in Problem 2 could be solved by dynamic programming.

Let \(D[m,k]\) to be the optimal value to the following problem

\begin{align*} \text{Minimize:} & \sum_{j=1}^m c_j x_j \\ \text{subject to:} & \sum_{j=1}^m w_j x_j = k |w_m| + t \bmod |w_m|\\ & x_j \in \{0,1\} \text{ for all } 1\leq j \leq m\\ \end{align*}

The claim is that \(D[m,k]\) can be expressed by the following recurrence relation.

\[ D[m,k] = \min_{x_m\in \{0,1\}} D\left[m-1,\frac{|w_m|k- w_m x_m+(t\bmod |w_m| - t\bmod |w_{m-1}|)}{|w_{m-1}|}\right] \]

Note that \(|k|\) is at most \(m\). Therefore the table has at most \(O(n^2)\) entries. To obtain the solution to the original equation, we just have to find the minimum overall \(D[n, k]\). Clearly, this runs in polynomial time.

References

[1] H. Kellerer, U. Pferschy, D. Pisinger, Knapsack problems, Springer, 2004.

Posted by Chao Xu on 2017-03-09.
Tags: optimization, integer.