Combinations with repetition

问题:

I'm using Mathematica 7 and with a combinatorica package function I can get all combinations of a certain number from a list of elements where the order doesn't matter and there is no repetition.e.g:

in: KSubsets[{a, b, c, d}, 3]
out: {{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}}

I cannot find a function that will give me all combinations of a certain number from a list of elements where the order doesn't matter and there is repetition. i.e. the above example would include elements like {a,a,b},{a,a,a},{b,b,b}...etc in the output.

It may require a custom function. If I can come up with one I will post an answer but for now I don't see an obvious solution.

Edit: Ideally the output will not contain duplication of a combination e.g. Tuples[{a, b, c, d}, 3] will return a list that contains two elements like {a,a,b} and {b,a,a} which from a combinations point of view are the same.

回答1:

DeleteDuplicates[Map[Sort, Tuples[{a, b, c, d}, 3]]]

回答2:

You can encode each combination as {na,nb,nc,nd} where na gives the number of times a appears. The task is then to find all possible combinations of 4 non-negative integers that add up to 3. IntegerPartition gives a fast way to generate all such such combinations where order doesn't matter, and you follow it with Permutations to account for different orders

vars = {a, b, c, d};
len = 3;
coef2vars[lst_] := 
 Join @@ (MapIndexed[Table[vars[[#2[[1]]]], {#1}] &, lst])
coefs = Permutations /@ 
   IntegerPartitions[len, {Length[vars]}, Range[0, len]];
coef2vars /@ Flatten[coefs, 1]

Just for fun, here's timing comparison between IntegerPartitions and Tuples for this task, in log-seconds

approach1[numTypes_, len_] := 
  Union[Sort /@ Tuples[Range[numTypes], len]];
approach2[numTypes_, len_] := 
  Flatten[Permutations /@ 
    IntegerPartitions[len, {numTypes}, Range[0, len]], 1];

plot1 = ListLinePlot[(AbsoluteTiming[approach1[3, #];] // First // 
       Log) & /@ Range[13], PlotStyle -> Red];
plot2 = ListLinePlot[(AbsoluteTiming[approach2[3, #];] // First // 
       Log) & /@ Range[13]];
Show[plot1, plot2]

http://yaroslavvb.com/upload/save/tuples-vs-partitions.png

回答3:

A slight variant on the elegant method given by High Performance Mark:

Select[Tuples[{a, b, c, d}, 3], OrderedQ]

Permutations is slightly more versatile (but not what you are looking for?)

For example:

Select[Permutations[
  Sort@Flatten@ConstantArray[{a, b, c, d}, {3}], {2, 3}], OrderedQ]

gives the following

Edit:

Select[Tuples[Sort@{a, b, d, c}, 3], OrderedQ]

is probably better

Edit-2

Of course, Cases may also be used. For example

Cases[Permutations[
  Sort@Flatten@ConstantArray[{a, b, d, c}, {3}], {2, 3}], _?OrderedQ]

Edit-3.

The two approaches will differ if the list contains a repeated element. The output from the following (approach 2), for example, will contain duplicates (which may or may not be desired):

Select[Tuples[{a, b, c, d, a}, 3], OrderedQ]

They may easily be got rid of:

Union@Select[Tuples[{a, b, c, d, a}, 3], OrderedQ]

The following evaluates to 'True' (remove duplicate elements from the list presented to approach 2, and Sort the list produced by approach 1 (High Performance Mark method):

lst = RandomInteger[9, 50]; 
Select[Union@Sort@Tuples[lst, 3], OrderedQ] == 
 Sort@DeleteDuplicates[Map[Sort, Tuples[lst, 3]]]

as does the following (remove duplicates from output of approach 2, Sort output of approach 1):

lst = RandomInteger[9, 50]; 
Union@Select[Sort@Tuples[lst, 3], OrderedQ] == 
 Sort@DeleteDuplicates[Map[Sort, Tuples[lst, 3]]]

Sorry about that!

回答4:

Here's a simple solution that takes advantage of Mathetmatica's built-in function Subsets and thus is a nice balance between simplicity and performance. There is a simple bijection between k-subsets of [n+k-1] and k-combinations of [n] with repetition. This function changes subsets into combinations with repetition.

CombWithRep[n_, k_] := #-(Range[k]-1)&/@Subsets[Range[n+k-1],{k}]

So

CombWithRep[4,2]

yields

{{1,1},{1,2},{1,3},{1,4},{2,2},{2,3},{2,4},{3,3},{3,4},{4,4}}