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In a duel between naked kings

he who has a castle and steed shall prevail;

wrap the battlefield in and around itself, however,

and you'll find this is no longer the tale;

but before so doing, add a twist (actually, two!)

and you may find that as before,

our combatants no longer coexist.

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I think this is about

variants of the game of chess, played on boards where edges have been (as the mathematicians say) "identified".

In a duel between naked kings
he who has a castle and steed shall prevail;

In ordinary chess, KRN v K is an easy win for the player with the rook and knight. (Not sure why "naked", exactly.)

wrap the battlefield in and around itself, however,
and you'll find this is no longer the tale;

KRN v K is not a win in chess played on a torus, where the a and h files are "glued together" and so are the 1st and 8th ranks.

but before so doing, add a twist (actually, two!)
and you may find that as before,
our combatants no longer coexist.

I take it this is talking about chess on a projective plane where the sides are glued together as above, but flipped, so that e.g. when you go "north" from a8..h8 you get not a1..h1 but h1..a1, and likewise for the join between a and h files. (Hence two twists.)

However,

I haven't actually checked whether KRNvK is a win on a projective-plane board. It seems plausible that it might be because the R covers an awful lot more squares there than it does on a normal or torus board. (I did have a bit of a look on the internet to see if someone else has done the work, but didn't find anything conclusive.)

... Apparently I'm meant to prove this. OK then, let's have a go. The first thing is that

on a projective plane, the rook is rather a monster -- it covers two whole ranks and two whole files.

Now

suppose white has KRN against black's bare K. ("Bare" meaning "no other pieces", which doesn't seem to be the same as "naked" in the first line of the puzzle.) And suppose it's white's move with the BK not already in check.

First of all,

save the WR and/or WN if attacked by the BK. [HOLE: I think this is always possible because the WR controls two ranks and two files at once, but maybe there are some awkward situations where the two are forked by the BK and you can't save both.]

Then

get the WK protecting the WR, and then -- keeping it protected -- bring it to a corner (so that the BK is confined to the inner 6x6 region) and then gradually bring it inward, confining the BK to smaller and smaller areas, until the WR is on {c,f}{3,6} -- let's say c6 for definiteness -- and the BK is sitting in a 2x2 box.

Then

bring the WN to f3, so it's protected by the WR on c6 and controls two of the squares in that 2x2 box. This means that B's moves are now completely determined and by choosing the parity of W's moves we can bring the WK to e2 just before the BK moves to e4.

And now

Rc4 or Rc5 is mate.

Credit where due:

I am a twit and somehow got mixed up and described how to mate with KQR v K instead of KRN v K. Feryll pointed this out in comments and sketched how the KRN v K mate works, and my comments above are an elaboration of that sketch.

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  • $\begingroup$ Ah, yes, the full mystery revealed. Well done. But, before I give you the check mark, I should like a convincing argument that indeed, just as before, our combatants no longer coexist. That's right, this puzzle is a two-parter! $\endgroup$ – Feryll Jul 26 at 1:15
  • $\begingroup$ Uh, Gareth, you were meant to prove (rot13) XEA i X erfhygf va purpxzngr, abg XDA :c Ng fbzr cbvag lbh zhfg'ir tbggra zvkrq hc. Nygubhtu gur ebbx vf yrff zbafgebhfyl birecbjrerq ba gur cebwrpgvir purffobneq, gur nanylfvf bhtug gb or zber znantrnoyr—fvzcyl, jvgu fbzr unaqjnivat, pbeeny gur xvat vagb gur pragre sbhe fdhnerf, phg bss gjb bs gurz jvgu n xavtug, naq jvgu n fvzcyr gevnathyngvba, bccbfr gur xvat naq qryvire n "gjb-enax purpxzngr" jvgu bar ebbx. I'll go ahead and give you the check mark anyway, for your trouble. $\endgroup$ – Feryll Jul 27 at 22:33
  • $\begingroup$ Whoops, so I was. My apologies. I agree with your handwaving analysis and will make some appropriate comments in place of my handwaving analysis of something different. $\endgroup$ – Gareth McCaughan Jul 27 at 23:11
  • $\begingroup$ (Done now. Sorry about that.) $\endgroup$ – Gareth McCaughan Jul 27 at 23:23
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This sounds

like cutting a Mobius Strip with two twists along the center (of the battlefield, if you will), which will make two distinct, but interlocked loops (no longer coexisting).

you can

make a Mobius Strip with two twists by cutting a regular Mobius strip down the middle.

Not sure what the other things reference, so maybe this is off.

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  • $\begingroup$ On the right train of thought. But just to be clear: A "two-twist Mobius strip" is topologically indistinguishable from a simple cylinder :) $\endgroup$ – Feryll Jul 25 at 23:27

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