- E contains (x, f(x)) for all x

And, for all y, and z != y, f(y), then either:

- E contains (f(y), z), or
- There is a path in G from f(y) to z such that each edge has larger weight than (z, f(y))

Now let x be such that (x, f(x)) is minimized. Must there be a path in G from f(x) to x where each edge has a larger weight than (x, f(x)) ?

Any insights, proofs, or counterexamples welcomed. I've been thinking about this problem for a while now.

]]>Using some of the ideas in your construction I found a slightly smaller one on 8 nodes

]]>I tweaked the example slightly as well but I don't think it was necessary.

]]>This is a great find, but remember all the edge weights should be unique! Nonetheless it's interesting to know that the conjecture does not hold in general.

Edit: Also, is there both an edge (l0, b1) and (b1, l0) ? I may be misinterpreting

Btw if you're curious for context, this is related to a question in https://arxiv.org/abs/2108.00542 that Simple Stable Voting will return a candidate in the Split Cycle set over uniquely weighted tournaments.

]]>From | c1 | c2 | c3 | l0 | l1 | l2 | b1 | b2 | d1 |
---|---|---|---|---|---|---|---|---|---|

c1 | - | -98 | - | - | - | - | - | - | - |

c2 | - | - | 4 | - | - | - | - | LO | - |

c3 | LO | - | - | 2 | - | - | 13 | 12 | - |

l0 | -99 | 1 | - | - | -10 | - | - | 9 | - |

l1 | LO | LO | LO | - | - | -11 | -8 | LO | - |

l2 | LO | LO | LO | -13 | - | - | - | 20 | - |

b1 | LO | 3 | - | -7 | - | -9 | - | - | - |

b2 | LO | - | - | - | - | - | 8 | - | -96 |

d1 | LO | LO | LO | -97 | LO | LO | LO | - | - |

f(c1)=c2 f(c2)=c3 f(c3)=l0

f(l0)=l1 f(l1)=l2 f(l2)=l0

f(b1)=l0 f(b2)=d1

f(d1)=l0

Edit: at the end of the row for l0, changed the last 3 values from 9/-7/- to -/9/-

Also changed the edge from l2 to b1 with weight -12 to an edge from b1 to l2 with weight -8.

The bottom graph shows the edges selected by f

]]>This is a good observation. You can construct such a graph where f creates a a single cycle by, for example, having a Hamiltonian cycle consisting of all the strongest edges in E.

Unfortunately, f may not be injective, so it will not necessarily be a permutation. I believe, however, that we can reduce to the case where f(x) (for specifically the x as chosen to minimize (x, f(x)) has a unique preimage

]]>Edit: If we take x such that (x,f(x)) is minimized, we can get a path from f(x) to x by repeatedly applying f and using the edge (f^i(x), f^i+1(x)). Since (x,f(x)) is minimal all of these edges will have weight higher than (x,f(x)). ((x, f(x)) will never have to be used because once x is reached, we're done.)

We are guaranteed to reach x because f is a permutation and permutations can always be decomposed into disjoint cycles.