bipartite
bipartite is one function. Everything below came out of it during this
build, at parameters taken from the essays rather than invented for this page — so a figure
here is the same figure a reader meets in an essay, and if the generator changes, this page
changes with it.
At its defaults
show: "hall"
show: "regular"
show: "deficiency"
What it checks while it draws
Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.
- round 1 still has a matching covering every vertex ×3
- and every neighbour named is a vertex on the right ×1
- and no vertex is matched twice ×1
- and so does every vertex on the right ×1
- and the reason is a set with too few neighbours between them ×1
- and the rounds use up every edge exactly once ×1
- between two and four graphs the family knows are compared ×1
- every edge gets exactly one colour ×1
- every matched pair is an edge of the graph ×1
- every vertex on the left has a neighbour list ×1
- every vertex on the left has the same degree ×1
- the graph is one the family knows ×1
- the largest matching is exactly the left side less the worst deficiency ×1
- the table has both a graph that succeeds and a graph that fails ×1
- the view is one the family draws ×1
- this view is for a graph that has no complete matching ×1
- this view is for a graph whose matching covers the whole left side ×1
Where it is called
Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.
A lottery over whole assignments
A table of shares in which every person's shares add to one task and every task is exactly covered is never anything more than a mixture of whole assignments — and finding the mixture is a matter of taking one complete assignment out at a time.
DiscreteOne bottleneck and nothing else
A set of jobs can be filled by distinct people unless some group of jobs has too few candidates between them — and that single obstruction is the only one there is, which is what makes the theorem worth having.