A doubt about SubsetCases

Questions
1

Hello all. Version 12.1 has several very useful functions. SubsetCases seems to be one of them. However, I have a doubt regarding its use. It comes from an example from the documentation:

SubsetCases[{1, 2, 3, a, a, b, c}, {a, _Integer}]

This returns {{a,1},{a,2}}. Why not {a,3}? Indeed it has an overlap with the previous two lists, but the chosen output also has the same overlap (that is, the symbol a}

If indeed overlaps are turned to true {a,3} and much more is included.
But if the overlaps are omitted, why does the code return {a,2} and not {a,3}?
Any clarification is welcomed.
Francisco

2

I think your interpretation of the documentation is reasonable. In fact, your understanding matched mine until I started experimenting.

Assuming that the SubsetCases[] works as designed, “overlap” is based on the elements of the first argument.

To illustrate, adding an additional a gives you the results we both expected:

SubsetCases[{1, 2, 3, a, a, a, b, c}, {a, _Integer}]

returns:

{{a, 1}, {a, 2}, {a, 3}}

So, “overlap” seems to be controlling whether elements of the first list can be reused.

Unfortunately, allowing overlaps probably does not give what you want. For example:

SubsetCases[{1, 2, 3, a, a, b, c}, {a, _Integer}, Overlaps -> True]

returns

{{a, 1}, {a, 2}, {a, 3}, {a, 1}, {a, 2}, {a, 3}}

Fortunately, I think there is a fix that might work for you:

Union[SubsetCases[{1, 2, 3, a, a, b, c}, {a, _Integer}, 
  Overlaps -> True]]

returns what we both originally expected to see:

{{a, 1}, {a, 2}, {a, 3}}

Knowing how it actually works, I think that the function might be working as designed. But, I think they could have done a better job explaining it in the documentation. And of course, some additional examples that highlights how overlap works would help.

Good luck.

3

Many thanks for this, Mike, now I fully understood.
All the best,
Fg

4

Francisco:

You are welcome. Happy to help.

5

Hello. I’m the developer of SubsetCases and related functions.

SubsetCases considers elements which appear multiple times as distinct, just like Subsets does:

In[1]:= Subsets[{a, a, b}, {2}]

Out[1]= {{a, a}, {a, b}, {a, b}}

To construct the solution of the given example, one can first generate all the subsets, then filter by an orderless version of the pattern:

In[2]:= Cases[Subsets[{1, 2, 3, a, a, b, c}, {2}], {OrderlessPatternSequence[a, _Integer]}]

Out[2]= {{1, a}, {1, a}, {2, a}, {2, a}, {3, a}, {3, a}}

Finally, to satisfy the Overlaps → False option value (the default for SubsetCases), one has to go through the list and drop each subset which contains any element already present in a previous subset.

One can see this more easily when looking at the result of SubsetPosition:

In[3]:= SubsetPosition[{1, 2, 3, a, a, b, c}, {a, _Integer}, Overlaps -> False]

Out[3]= {{4, 1}, {5, 2}}

Here one can see that no position element (integer) appears twice.

6

Toni:

Thanks for the clarification.

Is there any way to get this explanation and a few examples that illustrate how overlaps works into the documentation?

7

Many thanks Toni. I would suggest putting something of this unto the documentation, maybe with a couple of additional examples to prevent possible ambiguities.

8

Thanks for the question, we’ll improve the documentation.

9

I was having the same problem. This thread has been very helpful.