Presumably you are using this third party module since Python doesn't have a built-in orderedset. A quick look through the source code on github shows that the issuperset function is implemented as def issuperset(self, other): return other <= self Taking a look at how the less than or equal operator...

A terminal object in a category by definition is an object T with the property that for any object C in the category, there is exactly one arrow C -> T. In the category of sets, a singleton object, for example {1} has this property: For any set S there...

algorithm,computer-science,combinatorics,set-theory

We already know the answer for these cases (as you wrote in your examples): n=2 n=3 n=4 For n=5: You can partition from 2: 1 2 - 3 4 5. This is like dividing the 5 member set into two sets, first one n=2, and second one n=3. We can...

java,set,set-intersection,set-theory

What would be the best way to do this in Java? The first part you are describing is a powerset (as I edited your question to include last week). You would then get the intersection for each set of sets in the powerset. Because you are doing a powerset...

python,django,orm,django-queryset,set-theory

The answer in the link you provided should be correct. Basically what you need to do is exclude set A from set B. setA = MyModel.objects.filter(…) # Query for items belonging in set A setB = MyModel.objects.filter(…) # Query for items belonging in set B # setA and setB are...

javascript,arrays,union,intersection,set-theory

filter() can at least hide the loops for you: A = [1,2,3,4,5]; B = [1,2,3,6,7]; C = intersection(A, B); D = arrayDiff(A, C); E = arrayDiff(B, C); console.log(JSON.stringify(C)); console.log(JSON.stringify(D)); console.log(JSON.stringify(E)); function intersection(a, b) { return a.filter( function(el) { return b.indexOf(el) >= 0; } ); } function arrayDiff(a, b) { return...

select distinct least(x,y), greatest(x,y) from the_table; least() and greatest() put the values into an order so that 1,2 and 2,1 are returned as 1,2. The distinct then removes the duplicates...

mapping,combinatorics,computation-theory,np-complete,set-theory

You could create a bipartite graph in the following manner: For each element in the set X create a node in the U disjoint set of the graph For each subset in the set S create a node in the V disjoint set of the graph If element of X...

javascript,math,underscore.js,set-theory

I don't think this would perform faster with real data, but you can do it like this function reverseCartesian(cp, arr) { return _.chain(cp) .map(_.partial(_.difference, _, arr)) .uniq(function(currentItem) { return currentItem.join("|"); }) .value(); } console.log(reverseCartesian(cp, [3, 4])); // [ [ 1, 'a' ], [ 1, 'b' ], [ 2, 'a' ],...