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  1. Assume all balls with the same color are indistinguishable.
  2. The order in which balls are put in a bin does not matter.
  3. No bins are allowed to have the same distribution of balls!

For example, this configuration

{RRGGGB} {RRRRGBBBB} {RGB} {RGB}

is not ok, because the two last bins contains the same distribution of balls: one Red, one Green, one Blue.

Actualy I'm most intressted in the procedure for solving this problem where the number of balls, colors and bins are much larger numbers.

Penlect
  • 73

2 Answers2

4

This problem is a straightforward application of the Polya Enumeration Theorem. Suppose we treat the case of $r$ red balls, $g$ green balls and $b$ blue balls and $n$ indistinguishable bins where no bins are left empty.

Recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of the set operator $\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=n}$ on $n$ slots, which is $$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$$

We need to employ the set operator because the elements of a distribution are supposed to be unique.

We have for example, $$Z(P_3) = 1/6\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}$$ and $$Z(P_4) = 1/24\,{a_{{1}}}^{4}-1/4\,a_{{2}}{a_{{1}}}^{2} +1/3\,a_{{3}}a_{{1}}+1/8\,{a_{{2}}}^{2}-1/4\,a_{{4}}.$$

Applying PET it now follows almost by inspection that the desired count is given by using the repertoire

$$-1 + \sum_{q=0}^r R^q \sum_{q=0}^g G^q \sum_{q=0}^b B^q$$

with $Z(P_n)$ to get

$$[R^r G^g B^b] Z\left(P_n; -1 + \sum_{q=0}^r R^q \sum_{q=0}^g G^q \sum_{q=0}^b B^q \right).$$

The minus one term cancels empty bins.

The answer for eight red, six green and seven blue balls in four indistinguishable bins turns out to be $$60040.$$

The following Maple code implements two routines, q1 and q2. The first of these computes the value for $(r,g,b)$ and $n$ by brute force (enumerate all configurations) and can be used to verify the correctness of the PET formula for small values of the parameters. The second one uses the PET for instant computation of the desired coefficient.

Observe that we can compute values that are utterly out of reach of brute force enumeration, e.g. with ten balls of each color and five bins we obtain $$7098688.$$

With twenty balls of each color and six bins we get $$194589338219.$$

with(combinat);

pet_cycleind_set :=
proc(n)
option remember;


if n=0 then return 1; fi;

expand(1/n*
       add((-1)^(l-1)*a[l]*pet_cycleind_set(n-l), l=1..n));



end;


pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;


res := ind;

polyvars := indets(poly);
indvars := indets(ind);

for v in indvars do
    pot := op(1, v);

    subs1 :=
    [seq(polyvars[k]=polyvars[k]^pot,
         k=1..nops(polyvars))];

    subs2 := [v=subs(subs1, poly)];

    res := subs(subs2, res);
od;

res;



end;


allparts :=
proc(val, size)
    option remember;
    local res, els, p, pp, q;


res := [];

for p in partition(val) do
    if nops(p) <= size then
        pp := [op(p), 0$(size-nops(p))];
        res := [op(res), pp];
    fi;
od;

res;



end;


q1 :=
proc(RC, GC, BC, n)
    option remember;
    local p, q, pr, pg, pb, res,
    sr, sg, sb, dist;


pr := [];
for p in allparts(RC, n) do
    pr := [op(pr), op(permute(p))];
od;

pg := [];
for p in allparts(GC, n) do
    pg := [op(pg), op(permute(p))];
od;

pb := [];
for p in allparts(BC, n) do
    pb := [op(pb), op(permute(p))];
od;

res := {};

for sr in pr do
    for sg in pg do
        for sb in pb do
            dist :=
            {seq(R^sr[pos]*G^sg[pos]*B^sb[pos],
                 pos=1..n)};

            if nops(dist) = n and
            not member(1, dist) then
                res := res union {dist};
            fi;
        od;
    od;
od;

nops(res);



end;


q2 :=
proc(RC, GC, BC, n)
    option remember;
    local q, rep, sind;


rep := add(R^q, q=0..RC)
* add(G^q, q=0..GC)
* add(B^q, q=0..BC);

sind := pet_varinto_cind(rep-1,
                         pet_cycleind_set(n));

coeff(coeff(coeff(expand(sind), R, RC),
            G, GC), B, BC);



end;
Marko Riedel
  • 61,317
  • Thank you Marko for the great and detailed answer. I have not studied the Polya Enumeration Theorem, so I have some noobish follow-up questions :). I'm a bit uncertain about the square-bracket notation here: [R^rG^gB^b], and what are the difference between captial R and and lower case r? Also, what are the definition/values of the a_n coefficients? Finally, I don't fully understand the Newton binomial series expression, what does the square-bracket notation and Mu mean here? Thanks again. – Penlect Jan 11 '15 at 10:27
  • Thanks. I don't have the time for a detailed reply at the moment but I suggest you study the Maple code, which should make it clear what the variables mean. Not only does it compute values it is also a commentary on the text. There are many more PET examples including introductory ones at MSE Meta. – Marko Riedel Jan 11 '15 at 11:14
  • What's the complexity of this algorithm? – Elliot Gorokhovsky Oct 30 '15 at 17:42
0

Here is a simpler example to explain the mechanism above.

Suppose we have a rectangle marked in every corner with several points, such that the sum of points is 7 and each distribution fixes the rectangle, otherwise being acted by the Klein group.

The asymmetry cycle index is

$Z(rectangle) = 1/4 \ a_1^3 - 3/4 \,a_2^2 + 1/2 \ a_4$

The asymmetry cycle index goes back to Rota 1964 and, totally counterintuitive, it involves negative terms as well as 4-cycles.

We can securely use the pet_varinto_cind procedure delivered above.

pet_varinto_cind (poly, Z(rectangle))

The polynomial would be $poly = (p+p^2+p^3+...p^{10} + ..+p^m)$. In fact, power 10 is more than enough to follow the example, and there is no need to sum too many terms.

We then extract the desired coefficient representing the sum of points over the four corners.

sum of points 5 -----> 1 fixed rectangle

sum of point 6 -----> 1 fixed rectangle

sum of points 7 -----> 5 fixed rectangles

sum of points 8 -----> 7 fixed rectangles

for 7 points, the fixed rectangles are

4111, 3211, 3121, 3112 and 2221

Counting partitions without equal parts follows the same mechanism.

Note the asymmetry series/polynomial is not a function of cycle index only, because it depends on some lattice of subgroups, not uniquely characterized by the cycle index.

Boyku
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