Fair product distribution algorithm

Question

Here is my problem:

• There are n companies distributing products.
• All products should be distributed in k days
• Distributing products of company Ci should be consecutive - it means that it can be distributed on days 2,3,4,5 but not 2,3,6,7
• number of distributed products by company Ci on day j should be less than (or equal) on day j-1 (if there were any on day j-1)
• difference between distributed products between days i and j should not be greater than 1

Example:

We have 3 days to distribute products. Products of company A: a,a,a,a,a. Products of company B: b,b,b. Products of company C: c,c

Fair distribution: [aab,aabc,abc]

Invalid distribution: [aabc,aabc,ab] because on 1st day there are 4 products, on 3rd day 2 products (difference > 1)

Invalid distribution: [abc,aabc,aab] because on 1st day there is one product A, and on 2nd day there are 2 products A, so distribution of product A is not non-decreasing

EDIT if there is a case that makes fair distribution impossible please provide it with short description, I'll accept the answer

Answer 1

Gareth Rees's comment on djna's answer is right -- the following counterexample is unsolvable:

• 3 days, 7 items from company A and 5 items from company B

I tested this with the following dumbest-possible brute-force Perl program (which takes well under a second, despite being very inefficient):

my ($na, $nb) = (7, 5);
for (my $a1 = 0; $a1 <= $na; ++$a1) {
    for (my $a2 = 0; $a2 <= $na - $a1; ++$a2) {
        my $a3 = $na - $a1 - $a2;
        for (my $b1 = 0; $b1 <= $nb; ++$b1) {
            for (my $b2 = 0; $b2 <= $nb - $b1; ++$b2) {
                my $b3 = $nb - $b1 - $b2;
                if ($a1 >= $a2 && $a2 >= $a3 || $a1 == 0 && $a2 >= $a3 || $a1 == 0 && $a2 == 0) {
                    if ($b1 >= $b2 && $b2 >= $b3 || $b1 == 0 && $b2 >= $b3 || $b1 == 0 && $b2 == 0) {
                        if (max($a1 + $b1, $a2 + $b2, $a3 + $b3) - min($a1 + $b1, $a2 + $b2, $a3 + $b3) <= 1) {
                            print "Success! ($a1,$a2,$a3), ($b1,$b2,$b3)\n";
                        }
                    }
                }
            }
        }
    }
}

Please have a look and verify that I haven't made any stupid mistakes. (I've omitted max() and min() for brevity -- they just do what you'd expect.)

Answer 2

Since I thought the problem was fun, I did a model for finding solutions using MiniZinc. With the Gecode backend, the initial example is shown to have 20 solutions in about 1.6 ms.

include "globals.mzn";

%%% Data
% Number of companies
int: n = 3;
% Number of products per company
array[1..n] of int: np = [5, 3, 2];
% Number of days
int: k = 3;

%%% Computed values
% Total number of products
int: totalnp = sum(np);
% Offsets into products array to get single companys products 
% (shifted cumulative sum).
array[1..n] of int: offset = [sum([np[j] | j in 1..i-1]) 
                          | i in 1..n];

%%% Predicates
predicate fair(array[int] of var int: x) =
      let { var int: low,
            var int: high
      } in
        minimum(low, x) /\
        maximum(high, x) /\
        high-low <= 1;
predicate decreasing_except_0(array[int] of var int: x) =
        forall(i in 1..length(x)-1) (
                 (x[i] == 0) \/
                 (x[i] >= x[i+1])
        );
predicate consecutive(array[int] of var int: x) =
        forall(i in 1..length(x)-1) (
             (x[i] == x[i+1]) \/
             (x[i] == x[i+1]-1)
        );

%%% Variables
% Day of production for all products from all companies
array[1..totalnp] of var 1..k: products 
          :: is_output;
% total number of products per day
array[1..k] of var 1..totalnp: productsperday 
        :: is_output;

%%% Constraints 
constraint global_cardinality(products, productsperday);
constraint fair(productsperday);
constraint
    forall(i in 1..n) (
         let { 
             % Products produced by company i
             array[1..np[i]] of var int: pi
                = [products[j] | 
                 j in 1+offset[i]..1+offset[i]+np[i]-1],
             % Products per day by company i
             array[1..k] of var 0..np[i]: ppdi
         } in
           consecutive(pi) /\
           global_cardinality(pi, ppdi) /\
           decreasing_except_0(ppdi)
    );

%%% Find a solution, default search strategy
solve satisfy;

The predicates decreasing_except_0 and consecutive are both very naive, and have large decompositions. To solve larger instances, one should probably replace them with smarter variants (for example by using the regular constraint).