# Filling up a bin using balls with divisible weights

This post shows how to solve the special case for this problem. The special case has 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.

We are interested in solving the following integer program, \displaystyle \begin{aligned} \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{aligned}

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 do not require the w_is are powers of 2. We can 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 w_is are powers of 2. We do not know the case when w_{i+1}/w_i is unbounded.

Consider a more natural problem without the absolute values.

We are interested in solving the following integer program, \displaystyle \begin{aligned} \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{aligned}

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

We 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

\displaystyle \begin{aligned} \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{aligned}

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

\displaystyle \begin{aligned} \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{aligned}

Let y_i=y_i^+ - y_i^-, where y_i^-,y_i^+\geq 0, we can remove the absolute value.

\displaystyle \begin{aligned} \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{aligned}

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

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

\displaystyle \begin{aligned} \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{aligned}

This is an integer program as a bounded exact knapsack problem.

\displaystyle \begin{aligned} \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{aligned}

Finally, apply the standard technique that rewrites a bounded knapsack problem to 0-1-knapsack problem (see Section 7.1.1 of [1]). The blow up in problem size 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 has a dynamic programming solution.

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

\displaystyle \begin{aligned} \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{aligned}

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

\displaystyle 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 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.