[Date Index]
[Thread Index]
[Author Index]
Re: generating submultisets with repeated elements
*To*: mathgroup at smc.vnet.net
*Subject*: [mg104028] Re: generating submultisets with repeated elements
*From*: Raffy <raffy at mac.com>
*Date*: Fri, 16 Oct 2009 07:19:04 -0400 (EDT)
*References*: <ha4r9k$d0h$1@smc.vnet.net>
On Oct 2, 5:22 am, David Bevan <david.be... at pb.com> wrote:
> I'm new to Mathematica, so if I've missed something obvious, my apologies.
>
> I want a function to generate a list of "submultisets" with up to k elements of a set s, allowing elements from s to be repeated.
>
> The following works, but is very inefficient since each multiset is generated multiple times and then sorted and then repeats deleted:
>
> coinSets[s_,k_]:=DeleteDuplicates[Sort/@Flatten[Tuples[s,#]&/@Range[k],1]]
>
> coinSets[{1,3,4},3]
>
> {{1},{3},{4},{1,1},{1,3},{1,4},{3,3},{3,4},{4,4},{1,1,1},{1,1,3},{1,1,4},{1 ,3,3},{1,3,4},{1,4,4},{3,3,3},{3,3,4},{3,4,4},{4,4,4}}
>
> I assumed there would be a suitable function in the Combinatorica package, but I can't see anything -- which would be a bit odd for a combinatorial package. What have I missed?
>
> Do I need to write my own (perhaps by looking at how KSubsets is implemented) or is there some easy way of generating these multisets?
>
> Thanks.
>
> David %^>
[I posted this yesterday but Google Groups is acting odd and placed my
reply in a different thread.]
Another way of looking at the "coin sets" is thinking about it in
terms of how each set combination is generated.
I'll start with a naive solution:
naive[k_, n_] := Union[Ceiling[#/n] & /@ Subsets[Range[k*n], {1, n}]];
naive[3, 3] = {{1}, {2}, {3}, {1, 1}, {1, 2}, {1, 3}, {2, 2}, {2, 3},
{3, 3}, {1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 2, 2}, {1, 2, 3}, {1, 3,
3}, {2, 2, 2}, {2, 2, 3}, {2, 3, 3}, {3, 3, 3}}
This set of sets can be analyzed by asking: for each set, how many 1's
are there, how many 2's, etc...
recipe[k_, n_] := Table[Count[v, i], {v, naive[k, n]}, {i, k}];
recipe[3, 3] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}, {2, 0, 0}, {1, 1, 0},
{1, 0, 1}, {0, 2, 0}, {0, 1, 1}, {0, 0, 2}, {3, 0, 0}, {2, 1, 0}, {2,
0, 1}, {1, 2, 0}, {1, 1, 1}, {1, 0, 2}, {0, 3, 0}, {0, 2, 1}, {0, 1,
2}, {0, 0, 3}}
For example:
set => recipe
{1} => {1, 0, 0} "a list with 1x1"
{1, 1} => {2, 0, 0} "a list with 2x1"
{2, 3, 3} => {0, 1, 2} "a list with 1x2 + 2x3"
If you take the element-sorted union of these recipes, you'll see a
pattern:
basis[k_, n_] := Union[Sort /@ recipe[k, n]]
basis[3, 3] = {{0, 0, 1}, {0, 0, 2}, {0, 0, 3}, {0, 1, 1}, {0, 1, 2},
{1, 1, 1}}
You can reverse this operation by performing:
recipe[3, 3] === Join @@ (Permutations /@ basis[3, 3])
(note they might have different sortings)
So the process to generate all the coin sets is the following:
1. We need a function that generates the "basis" from above, given k
and n:
basis[len_, sum_] :=
Reap[Do[If[Length[#1] == n, Sow[PadRight[#1, len]],
Do[#0[Append[#1, i], i, #3 - i], {i, #2, #3}]] &[{}, 1,
sum], {n, len}]][[2, 1]];
basis[3, 3] = {{1, 0, 0}, {2, 0, 0}, {3, 0, 0}, {1, 1, 0}, {1, 2, 0},
{1, 1, 1}}
2. We need to permutate each basis to generate all the recipes for
that combination.
recipes[k_, n_] := Join @@ Table[Permutations[Developer`ToPackedArray
[v]], {v, basis[k, n]}];
3. At this stage, we have our answer as a list of recipes, where each
recipe is a vector of length k, whose sum is between 1 and n, can be
converted into a coin set by using it as a recipe (see above).
{0, 1, 0} => 1x2 => {2}
{3, 0, 0} => 3x1 => {1, 1, 1}
{1, 1, 1} => 1x1 + 1x2 + 1x3 => {1, 2, 3}
Timing[Length[recipes[15, 12]]] => {4.62615, 17383859}
Timing[Length[recipes[15, 10]]] => {0.841442, 3268759}
Timing[Length[recipes[15, 7]]] => {0.052076, 170543}
Another advantage of the recipe form is it makes Intersect/Union/
Complement/MemberQ/FreeQ fast to implement.
Prev by Date:
**Re: Mathematica 7.01 and Mac OS 10.6 (Snow Leopard)**
Next by Date:
**Re: Mathematica 7.01 and Mac OS 10.6**
Previous by thread:
**Re: generating submultisets with repeated elements**
Next by thread:
**"Freezing" an Output cell?**
| |