Timeline for A different equivalence relation on partizan combinatorial games
Current License: CC BY-SA 4.0
7 events
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| Sep 22, 2018 at 13:02 | comment | added | Paolo Lipparini | You're right, I have taken things the other way round (it sometimes happens ;). Yes, additive inverses do work, and the whole stuff becomes a Ring, I have edited the answer. I suggest you to write a paper about the subject! | |
| Sep 22, 2018 at 12:56 | history | edited | Paolo Lipparini | CC BY-SA 4.0 |
added further considerations: the last possibility E seems viable
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| Sep 19, 2018 at 11:35 | history | edited | Paolo Lipparini | CC BY-SA 4.0 |
clarifications
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| Sep 17, 2018 at 22:01 | comment | added | Gro-Tsen | I want $≜$ to imply $≐$ (i.e., the sought-after equivalence relation $≜$ is stronger, i.e., has smaller classes, and I call this "finer", than Conway's $≐$). So in fact I want (E) but apparently what you call "coarser" is what I (and Wikipedia) call "finer". Sorry about the confusion. I believe additive inverses should still work if I define $G≜G'$ as something like "$G⊗H≐G'⊗H$ for all $H$", which is what I proposed. | |
| Sep 17, 2018 at 20:28 | comment | added | Paolo Lipparini | I have understood your question in the sense that $\triangleq$ should be such that if $G \doteq H$, then $G \triangleq H$. So if $1/2 + 1/2 \doteq 1$, then $1/2 + 1/2 \triangleq 1$. Of course, we can consider instead a relation coarser than $\doteq $, as I mentioned in (E), but then it is hard to get a group, that is, to have additive inverses. | |
| Sep 16, 2018 at 11:03 | comment | added | Gro-Tsen | Of course, there is a game $x$ that corresponds to (say) the Hackenbush realization of Conway's $½$, which satisfies $x+x≐1$, but it will not satisfy $x+x≜1$. So my question is about renouncing $½$ if you want, but not renouncing it in the sense that it ceases to be a game, but that it does not have a representative satisfying $x+x≜1$. | |
| Sep 15, 2018 at 20:05 | history | answered | Paolo Lipparini | CC BY-SA 4.0 |