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### Re: Bin packing problem variation repost (see[834245])

by jethro (Monsignor)
 on Apr 14, 2010 at 16:22 UTC ( #834722=note: print w/replies, xml ) Need Help??

I assume a solution should have no leftovers, i.e. every number should be in exactly one group and only valid groups (with totals in the range 840 to 900) should exist.

One question is whether it can be assumed that such a solution always exists? In any case there should be a time limit after which the program stops and declares failure. My guess is that the more numbers are given the less likely it is that no solution exists. But you can always construct a not working data sample of arbitrary length, for example the array (101)x \$n for any value of \$n

It sounds to me like a genetic algorithm or at least a stochastic algorithm (or whatever it is called in CS) might be good at this:

Basically you create an initial packing (without looking really for a correct solution), then adapt that packing randomly but sensibly until you find a solution. A real genetic algorithm would work with many packings and cross the "best" of them. A simpler stochastic algorithm would only work on one packing but switch between incremental optimization (to get to a local "best" packing) and random destruction (to get out of local paths without a solution).

I began with the programming but now I have to do something else, maybe I can finish the script tonight. There are "only" a few subs to fill out if anyone wants to do it. I post it here to show how it can be done, read the comments in the main loop

```#!/usr/bin/perl

use strict;
use warnings;

use Data::Dumper;

my @a = qw(
121 182 111 160 105 113 121 97 123 157 133 161 141 135 137 145
133 137 151 118 126 141 174 181 154 109 198 114 122 162 91 99
116 122 195 199 150 192 163 88 112 157 182 210 124 105 144 166
144 257 164 156 173 154 193 142 143 126 118 130 107 86 131 154
131 147 134 118 115 135 141 158 129 143 126 128 134 129 167 130
135 117 127 146 96 117 99 99 139 152 149 136 105 124 136 160
160 139 177 115 123 103 150 183 132 171 121 114 111 113 131
144 122 141 111 139 145 109 114 122 103 160 153 147 172 155 122
296 124 112 161 124 311 99 157 122 120 198 152 140 162 177 98
138 156 177 103 180 187 173 150 135 168 132 196 112 195 126 113
116 105 116 151 216 188 158 121 166 148 132 89 197 92 115 98
130 103 120 261 143 126 167 203 95 165 129
);

# count the sum of all numbers in a and derive from that the minimum a
+nd maximum number of packs

my \$sum;
\$sum+= \$_ for @a;

my \$minimum= int((\$sum-1)/900)+1;
my \$maximum= int(\$sum/840);

print "Sum is \$sum, minimal \$minimum packs, maximal \$maximum packs\n";
die "No solution possible\n" if (\$maximum<1);

# put numbers into packs (an array of arrays) without looking much at
+correctness

my @packs=[];
my @packsum;
while (@a) {
\$_= shift @a;
push @{\$packs[\$#packs]}, \$_;
\$packsum[\$#packs]+= \$_;
if (\$packsum[\$#packs]>840 and @a) {
push @packs, [];
}
}

# print Dumper(\@packs,\@packsum);

# now the loop with optimization and destruction

my \$steps;
my \$wanttogrow=0;

while (1) {
# if not enough packs, remove values from other packs to create a
+new one
while (@packs<\$minimum) { grow_another_pack(\@packs,\@packsum); }
# too much packs can't happen, we avoid this

#switch two numbers, if possible between an overfilled pack and an
+ underfilled pack
optimized_switching(\@packs,\@packsum);
if (finished(\@packsum)) { last };

# if we did lots of incremental optimization without success, turn
+ to destruction
\$steps++;
if (\$wanttogrow) {
if (@packs==\$maximum) {
\$wanttogrow= 0;
}
else {
grow_another_pack(\@packs,\@packsum);
}
}
else {
if (@packs==\$minimum) {
\$wanttogrow= 1;
}
else {
remove_a_pack(\@packs,\@packsum);
}
}
#just randomly switch some numbers, check for success even though
+unlikely
if (random_switching(\@packs,\@packsum)) { last };
if (finished(\@packsum)) { last };
}

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Re^2: Bin packing problem variation repost (see[834245])
by gskoli (Novice) on Apr 15, 2010 at 11:56 UTC
Thanks
Also try to show index of picked number
And what will be left for you? :) You can even parse the solution and compute the indexes back.

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