# Set consolidation

**Set consolidation**

You are encouraged to solve this task according to the task description, using any language you may know.

See also:

Given two sets of items then if any item is common to any set then the result of applying consolidation to those sets is:

- The two input sets if no common item exists between those two sets of items.
- Or one set of all the items from both sets if they share a common item.

Given N sets of items then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible.

**Example 1:**

Given the two sets {A,B} and {C,D} then there is no common element between the sets and the result is the same as the input.

**Example 2:**

Given the two sets {A,B} and {B,D} then there is a common element B between the sets and the result is the set {B,D,A}. (Note that order of items in a set is immaterial - {A,B,D} is the same as {B,D,A} and {D,A,B} etc).

**Example 3:**

Given the three sets {A,B} and {C,D} and {D,B} then there is no common element between the sets {A,B} and {C,D} but the sets {A,B} and and {D,B} do share a common element that consolidates to produce the result {B,D,A}. On examining this result with the remaining set {C,D} they share a common element and so consolidate to the final output of the single set {A,B,C,D}

**Example 4:**

The consolidation of the five sets: {H,I,K}, {A,B}, {C,D}, {D,B}, and {F,G,H} Is the two sets: {'A', 'C', 'B', 'D'}, and {'G', 'F', 'I', 'H', 'K'}

## Python

The docstring contains solutions to all the examples as well as a check to show the order-independence of the sets given to the consolidate function. <lang python>def consolidate(sets):

>>> # Define some variables >>> A,B,C,D,E,F,G,H,I,J,K = 'A,B,C,D,E,F,G,H,I,J,K'.split(',') >>> # Consolidate some lists of sets >>> consolidate([{A,B}, {C,D}]) [{'A', 'B'}, {'C', 'D'}] >>> consolidate([{A,B}, {B,D}]) [{'A', 'B', 'D'}] >>> consolidate([{A,B}, {C,D}, {D,B}]) [{'A', 'C', 'B', 'D'}] >>> consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}]) [{'A', 'C', 'B', 'D'}, {'G', 'F', 'I', 'H', 'K'}] >>> consolidate([{A,H}, {H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}]) [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}] >>> consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}]) [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}] >>> # Confirm order-independence >>> import itertools >>> sets = [{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}] >>> answer = consolidate(sets) >>> for perm in itertools.permutations(sets): assert consolidate(perm) == answer

>>> len(list(itertools.permutations(sets))) 720 >>> setlist = [s for s in sets if s] for i, s1 in enumerate(setlist): if s1: for j, s2 in enumerate(setlist[i+1:], i+1): intersection = s1.intersection(s2) if intersection: s2.update(s1) s1.clear() s1 = s2 return [s for s in setlist if s]

</lang>