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Usually, in bin-packing, we have objects of sizes $a_1,...a_n$, and each bin has size 1, We need to minimize the number of bins, and for this, there are best fit/first-fit approximation algorithms.

What will happen if there are more parameters than just size, say 3 parameters? Each object $i$ has say parameters $(a_i,b_i,c_i)$ and there are bins with all the three parameters equal to 1. How can we extend the best-fit/first-fit approximation algorithms, so that in each bin, we have $\sum a_i \leq 1, \sum b_i \leq 1,\sum c_i \leq 1$ and we still have a constant approx factor?

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    $\begingroup$ The keywords are "multidimensional bin packing" and "vector bin packing". Multidimensional Bin Packing and Other Related Problems: A Survey, Section 4.1 says: "for fixed $d$ (In your example, $d=3$), vector bin packing can be approximated to within $O(\ln d)$ in polynomial time". You can check the survey for more detail. $\endgroup$
    – Dmitry
    Jan 25, 2023 at 7:52
  • $\begingroup$ O(ln d) is quite a lot. Like needing 50 bins for 100 items instead of 10. $\endgroup$
    – gnasher729
    Jan 25, 2023 at 13:26
  • $\begingroup$ @Dmitry, perhaps you would like to write that as an answer, so we can upvote it? I realize it is not a constant-factor approx. $\endgroup$
    – D.W.
    Jan 25, 2023 at 19:47
  • $\begingroup$ @D.W., it is a constant-factor approximation for constant $d$ (as in the post, with $d=3$). $\endgroup$
    – Dmitry
    Jan 25, 2023 at 20:11
  • $\begingroup$ I read the survey mentioned by Dmitry and it says that you cannot get a constant approximation factor for d-dimensional vector bin packing when $d>2$, where $d$ is an input parameter, not a constant. For the case $d=2$, some clever papers are there which produce very small approximation factors. link.springer.com/content/pdf/10.1007/BF02579456.pdf?pdf=button pg 354 (pg 6) gives an "easy-to-understand" approximation. $\endgroup$
    – Sandra
    Jan 25, 2023 at 20:15

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Here is the solution when the vector is two-dimensional. This process can be extended exactly in the same way to higher dimensions. As mentioned in the comments https://link.springer.com/content/pdf/10.1007/BF02579456.pdf?pdf=button shows the general solution.

Z is the set of vectors given to us. Let $L_i$ denote the set of indices of the vectors in $Z$ for each of which the largest component is the $ith$ one, breaking ties arbitrarily. Now consider all the vectors whose index has been put inside $L_i$ and then let's call $Z_i$ the list of the values of the $ith$ component of the vectors. An optimal packing of $Z_i$ will be an optimal packing for the vectors inside $L_i$. We already know that for 1-dimensional packing we can do a greedy best fit or first fit algorithm with an approximation factor of 2. We apply that greedy algorithm to this one-dimensional packing we got. Say we can optimally pack the $Z_i$ in $Z_i^*$ bins. Then for the two-dimensional case, we will not need any more than $2 \cdot (Z_1^* + Z_2^*)$ bins for packing all the vectors in $Z$.

Again observe that all those vectors that have indexed in $L_i$ cannot be packed in less than $Z_i^*$ bins. So if $Z^*$ is the optimal number of bins for all the vectors(i.e. OPT in this case), then $Z^* \geq max\{Z_1^*, Z_2^*\} \geq 1/2 \cdot (Z_1^* + Z_2^*)$. Combining both the inequalities, we see that our algorithm will obtain an approximation factor of 4, i.e. $2 \cdot (Z_1^* + Z_2^*) \leq 4 \cdot max\{Z_1^*, Z_2^*\} \leq 4 \cdot OPT$.

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