I am facing the following research problem. We are given a matrix $M[1..p,1..p]$ of elements such that:

  • each element has value in the range $[0, \frac 1 j]$, $j <= p$, $j$ is given,
  • the sum of all elements in the matrix is 1.

The matrix is partitioned into $j$ axis-parallel non-overlapping rectangles. Each matrix cell with positive value is covered by exactly one rectangle. Whereas, each matrix cell with value 0 is covered by at most one rectangle. For each region, we sum the values inside it and obtain $region\_sum_i$, $i = 1..j$. We are given an optimal partitioning algorithm whose objective is to minimize the maximum $region\_sum_i$. The algorithm is Recursive Slice-and-Dice: Binary Space Partitions from Slice and Dice : A Simple, Improved Approximate Tiling Recipe. I need this algorithm due to the support for don't care regions and a general version of tiling problem. This is a DRTILE problem, but I translate this to RTILE using binary search.

The problem is to find an upper bound on $region\_sum_i$ for the optimal solution under worst-case input with given constraints. I need this as I am doing competitive analysis.

I appreciate any help! Many thanks!

  • $\begingroup$ I'm not sure I understand the problem. Do you mean that the partition into rectangles is given? (If so, why do we need an upper bound -- we can just compute subset_sum directly, right?) Or do you mean that the partition is not given and the goal is to find a partition with the specified constraints and minimizing the maximum subset_sum? Is $j$ given? What does it mean to say you already have an algorithm for partitioning? If your algorithm computes the optimal partition, doesn't that immediately give you a tight upper bound? $\endgroup$
    – D.W.
    Dec 6, 2013 at 1:17
  • $\begingroup$ I think he meant the upper bound for the optimal solution. $\endgroup$
    – Chao Xu
    Dec 6, 2013 at 1:58
  • $\begingroup$ I think it be nicer if you replace $m$ with $1$, as scaling doesn't matter. Do you want an upper bound in case of the $OPT$ or in case of $j$? $\endgroup$
    – Chao Xu
    Dec 6, 2013 at 8:11
  • $\begingroup$ @Chao Xu Thank you very much for your comments. I clarified the problem definition. I am not sure I understand your last question. I want an upper bound in case of OPT with given $j$. $\endgroup$ Dec 6, 2013 at 8:18

1 Answer 1


In the proof of theorem 2 in Improved Approximation Algorithms for Rectangle Tiling and Packing by Berman et al, they proved an upper bound of $\frac{11}{5} \max\{W/p,y\}$, where $W$ is the sum of the weight of all elements, $p$ is the number of rectangles and $y$ is the weight of the largest element.

This implies a upper bound of $\frac{11}{5j}$ for your problem.

  • $\begingroup$ Thank you very much. I update the problem definition to explain which algorithm I am using. $\endgroup$ Dec 6, 2013 at 12:03
  • $\begingroup$ This bound holds independent from the algorithm. $\endgroup$
    – Chao Xu
    Dec 6, 2013 at 12:16
  • $\begingroup$ Thanks. But this does not seem obvious to me. Can you please elaborate more? $\endgroup$ Dec 6, 2013 at 12:30
  • 1
    $\begingroup$ Every solution in this partition problem(P) is a solution to your partition(YP) problem. A solution of P is an upper bound for the optimal solution of YP. Note both P and YP have a lower bound of 1/j, and the proof of the theorem shows there exist a solution where each rectangle has weight at most $\frac{11}{5j}$ for P. Thus there exist a optimal solution where all rectangles have weight at most $\frac{11}{5j}$ for YP. $\endgroup$
    – Chao Xu
    Dec 6, 2013 at 12:43
  • 1
    $\begingroup$ I think I got it. The thing is that YP will not choose a worse solution than P, so $\frac{11}{5*j}$ is the upper bound. $\endgroup$ Dec 16, 2013 at 15:59

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