Erlang Thurday – ordsets:is_disjoint/2

Today’s Erlang Thurday covers ordsets:is_disjoint/2.

There are times in your in your coding day where you have problems where you need to know if given a list of items that none of those items appear in a secondary list.

Your first intuition might be to write out the code as described, like such:

not( lists:any(fun(Item) -> lists:member(Item, [1, 3, 5, 7]) end, [2, 4, 6])).
% true

And while that is accurate, if you redefine your problem in more mathematical terms, you can start to think in sets. When you start to think in terms of sets, you realize that you can check to see if the intersection of the two sets is the empty set.

ordsets:intersection([1, 3, 5, 7], [2, 4, 6]) =:= [].
% true

This is becoming not only more concise, but also more explicit about what you are trying to check.

We can do better still, by checking if the lists are disjoint sets. Enter ordsets:is_disjoint/2.

ordsets:is_disjoint/2 takes two lists, and returns true if no elements are in common.

ordsets:is_disjoint([1, 3, 5, 7], [2, 4, 6]).
% true
ordsets:is_disjoint([1, 2, 3, 5, 7], [2, 4, 6]).
% false

Because ordsets:is_disjoint/2 operates against two lists, we do not have to make sure the elements are unique prior to calling ordsets:disjoint/2.

ordsets:is_disjoint([1, 1, 3, 5, 7, 5, 3], [2, 4, 2, 2, 6]).
% true
€ordsets:is_disjoint([1, 2, 3, 5, 7], [2, 4, 2, 2, 6]).
% false

And if either list passed to ordsets:is_disjoint/2 is an empty list, the result is that the lists are disjoint.

ordsets:is_disjoint([1, 2, 3, 5, 7], []).
% true
ordsets:is_disjoint([], [2, 4, 6]).
% true
ordsets:is_disjoint([], []).
% true

And if you are curious, by running ordsets:is_disjoint/2 through timer:tc/3, we can see that as soon as Erlang knows that the sets are not disjoint, it returns false. And if you remember from the previous Erlang Thrusday on timer:tc/3, the return value is a tuple with the first element being the number of microseconds it took to complete.

timer:tc(ordsets, is_disjoint, [lists:seq(1, 1000000), lists:seq(2000000, 3000000)]).
% {19032,true}
€timer:tc(ordsets, is_disjoint, [lists:seq(1, 1000000), lists:seq(1, 3000000)]).
% {2,false}


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