Wythoff's Nim - Going for the Gold

[@wilderuhl3450]

The presence of the golden ratio here is phenomenal. Good video

[@mostly_mental]

Glad you liked it. Thanks for watching!

[@shaunokane9600]

I appreciated the series. Thank you kindly.

[@mostly_mental]

Glad you liked it. Thanks for watching!

[@enbyd]

I enjoy how you approach topics as well as the topics you approach, especially alongside carefully hand-illustrated diagrams (this video included). I recently drew a geometric diagram to help someone answer a numerical question, and had another person respond by noting their appreciation of hand-drawn math. It got me thinking... Maybe it could be seen as a type of nostalgia for well-made hand-drawn math (since CAD is widespread), but my main reason for appreciation is that the form factor can't help but showcase the effort that went into its construction right alongside the information it is designed to convey, and I appreciate that. It's certainly possible to see effort behind computer animations (see: 3b1b), but usually the clear showcase I see there is in deciding *what* the animation depicts, rather than the substance of the animation itself (even though it no doubt takes skill to use Manim as a tool, that effort is less immediately visible). Whereas here, I feel like I can see both the planning and the execution in action and can appreciate that synthesis more tangibly. Or maybe I'm reading between the marker strokes/pixels :) Thanks for making and sharing stuff!

[@mostly_mental]

I'm glad you like it. Thanks for watching!

[@oddlyspecificmath]

Nice playthrough and description; I feel like I actually understand this better instead of just learning an algorithm. Is a game like Othello compatible with analyses like this, or does the way the game works make it a lot more complicated?

[@mostly_mental]

The analysis we did here relies on the fact that Wythoff's Nim is an "impartial" game. That is, both players have the same moves available to them from any given position. But Othello is a "partisan" game; one player can only place white pieces and one can only place black. Partisan games tend to be a bit harder to analyze. You can use a lot of the same logic. For instance, the value of a position still depends on the values of the positions you can move to. But now in addition to P and N positions, there are also L positions, where the left player wins regardless of who moves next, and R positions, where the right player wins. I talked a bit about partisan games in my video on the surreal numbers (kzbin.info/www/bejne/b17RlqWhnNeskJo ). And I'd definitely like to revisit them in a future series.

[@oddlyspecificmath]

@@mostly_mental Wow, thank you -- more than I knew to ask :) I suspected "impartial" but "partisan" wasn't there with me, and it's a richer distiction to have those words. I appreciate your related reference too; I had skimmed your list and not realized / I'll plan to watch that tomorrow!

[@Skyb0rg]

Excellent video! I wonder how this strategy generalizes to more than 2 piles, or if there’s a rule set for every Beatty sequence.

[@mostly_mental]

Glad you liked it, and those are both great questions. I'm pretty sure there's a known strategy for 3 piles, since there's a Project Euler problem about it (projecteuler.net/problem=260 ), but I haven't solved it and I try to avoid spoilers, so I don't know what that strategy is. At the very least none of the plots I made for this video have led me in the right direction. There's are some trivial rule sets for any sequence (from a red point you can move down or left, and from any other point you can move to (0, 0)), but I assume you want something a bit more natural. If there is one, I'd guess you need to take x from one pile and ax from the other for some a based on your sequence (with appropriate rounding). Basically, you're trying to find a diagonal direction that will always hit exactly one of the lines. Play around with it a bit, and if you find something, let me know.

[@mostly_mental]

I was curious, so I tried out various values of a, and it doesn't seem to generate Beatty sequences (or lines at all). I'm not sure what kind of rule you would need to make it work out.

[@ra1u]

Having different difference n for every winning pair is necessary condition for strategy to work. If there would be two pairs of winning positions with same difference one could made a single move to get from one winning position to another winning position, but that would be a contradition, because every move from winning position needs to be to loosing position. In same way we can show that all numbers in all winning pairs must be different.

[@mostly_mental]

Sounds like you've got the idea of the proof.