WIP: Further improved stable marriage in python

[jdonszelmann]

In Discord we discussed the original implementation of gale shapley, and the improvement @Amaras made (#687 ). This is just a draft which improves it further, making it more similar to the rust implementation. It's very different to the original one so you'll just have to say what you think about it.

[jdonszelmann]

Not quite sure if this should be an error

[jdonszelmann]

nor if else is the right construct here, it's quite obscure

[jdonszelmann]

We determined that this algorithm doesn't show as clearly as previous implementations, how marriages are formed. Add to that the fact that it might not work, I don't think this PR should ever be merged, and I might close it soon.

[Amaras]

The first time I ran the algorithm, I got this output:

A: b, c, e, d, a
B: a, e, b, c, d
C: b, a, d, e, c
D: c, b, e, d, a
E: b, c, e, d, a

Preferences of the women:
a: E, C, A, D, B
b: D, B, A, E, C
c: C, D, E, A, B
d: E, D, C, A, B
e: E, D, A, B, C


The algorithm gave this solution:
A + b
B + a
C + d
D + c
E + e

Problem is: C prefers a over d, and a prefers C over B.
This pairing is thus unstable, so the algorithm is wrong.
As it's inspired by the Rust PR (#715), I recommend a careful analysis of the output code before considering merging it.

[lazyprop]

Let Alice and Bob both be engaged, but not to each other. Upon completion of the algorithm, it is not possible for both Alice and Bob to prefer each other over their current partners. If Bob prefers Alice to his current partner, he must have proposed to Alice before he proposed to his current partner. If Alice accepted his proposal, yet is not married to him at the end, she must have dumped him for someone she likes more, and therefore doesn't like Bob more than her current partner. If Alice rejected his proposal, she was already with someone she liked more than Bob.

This is from the Gale-Shapely WIkipedia page

Stability here means that there are no free men left. This is because engaged men do not propose and a change of partners only takes place if there is a free man who proposes to a woman and the woman likes him over her current partner.
The men propose in the order of their preference. And if they did get dumped, that must have been for a reason. So if the algorithm ends, the produced matching must be stable.

[Amaras]

Stability here means that there are no free men left.

No, that's not what stability means. It means that you can't find a pair like C - a where they prefer each other over their partner.

It is right that in the GS algorithm, the fact that there is no free man left means that the algorithm must stop, and thus that (as a consequence of the construction of the pairing) the pairing is stable. It doesn't mean that "all men have a partner" implies that "the pairing is stable".