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| (1) Posted by Hauke Reddmann [Friday, Jul 10, 2020 19:58]; edited by Hauke Reddmann [20-07-10] |
Triple Check (for Rewan)
Since in todays Zoom Chat (with Siegfried, Andrew, Shankar,
Rewan and some more I don't know) the theme came up:
This is based on a now fixed bug in the FIDE rules that
didn't account for the possibility of "parrying" a double
check by upgrading it to a triple check. (The FIDE would
be so in trouble without us problemists :-)
And since I was "accused" of slowing down (needing a
whopping minute to compose), this one was done in half
a minute. Which also means I didn't check it for correctness
(as far as that is possible with a joke problem).
 (= 8+5 )
#2 Triplechess
As you see, the wK is in a triple check, i.e., not in check
at all. Thus 1.S~ is illegal, as it would reduce it to a
double, i.e. selfcheck. The only way to progress is
to upgrade it to a quadruple check: 1. Rd2+! Bxd2
and now 2.Sd4# is possible. (Still has some flaws.)
One could also invert the idea: White is hindered from mate
because Black can upgrade his double to a triple check.
White must thus interference or whatnot. (Both versions
could also come with motive inversion, obviously.) EDIT:
 (= 11+3 )
#1.5 Triplechess
I leave it to the reader checking when after the move set
1...Bd8,f8,xd6,xf6 the reducing to double check with the move set
2.Sc8,g8,d7,b5 Black is mated or not :-)
Hauke
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| (2) Posted by seetharaman kalyan [Sunday, Jul 12, 2020 22:30] |
May I know what legal moves led to this triple check with white to play?. I see the only way was for WK to walk into it.
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| (3) Posted by Juraj Lörinc [Monday, Jul 13, 2020 00:27] |
If I am not mistaken, you are talking about fairy condition known as Bosma, I guess its the inventor's name. It is implemented in WinChloe. Probleemblad has run its 151st TT for problems featuring this fairy condition in 1993 (two sections, #2 and h#). Twomover sections was won by Henk le Grand with the following position:
Henk le Grand
1st Prize 151st TT Probleemblad 1993
 (= 8+5 )
#2
Bosma
1.Rc8! th. 2.Re8#
1…Q×d7 2.f8=S#
1…Sd6 2.f8=B#
1…Bd8 2.f8=R#
Selfblocks preventing triple checks involving white promotees.
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| (4) Posted by James Malcom [Monday, Jul 13, 2020 03:21]; edited by James Malcom [20-07-13] |
Nice problems, Hauke. Bet you can't beat my quintiple check in a directmate. :)
seetharam, that is correct-the idea is that it is legal for a king to walk into a triple check since it doesn't count as check anymore.
Juraj, Bosma Chess was a fairy vairant created based off the loophole in the laws. I added the problem that you showed into PBD awhile back. My source for it was here-https://www.chessvariants.com/problems.dir/bosma.html
This was (theortically) legal in the FIDE laws of chess from 1984, when the wording "the proposed in the 1983 Chess Congress and revoked in the 1993 rules. The earliest known problem that uses it is from 1988. Search K='triple' in Schwalbe for the old problems that I've entered in-more recent ones are going in soon.
I covered the history on it here in a straight timeline after the initial posting in the English Chess Forums-https://chess.stackexchange.com/questions/29990/when-was-it-possible-for-a-players-king-to-be-attacked-by-3-of-the-opponents-p
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| (5) Posted by Juraj Lörinc [Monday, Jul 13, 2020 12:48] |
I am glad that Bosma chess is known, thus it will be easier to explain. I have checked both Hauke's positions with WinChloe as Bosma #2 and set play of #2 respectively. The results are as follows:
#2 has no solution due to Bosma refutation: 1.Rd2+? Bxd2? 2.Sd4# is ok, however 1...Kd4+!! refutes.
#1.5 has two set plays:
1...Bxf6 2.Sc8#
1...Bxd6 2.Sg8#
Other two bishop moves allow no checkmate as bishop can always move away in the second moves, even both rooks on the first rank are not needed.
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| (6) Posted by Hauke Reddmann [Monday, Jul 13, 2020 16:07] |
And that's why I stay orthodox :-)
<weasel mode> It was just to illustrate the idea :-O
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| (7) Posted by Andrew Buchanan [Sunday, Oct 25, 2020 17:12] |
A.Buchanan & J.Malcom
 (= 5+7 )
#3 Bosma
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| (8) Posted by edderiofer [Tuesday, Oct 27, 2020 17:35] |
First time learning about this chess variant.
Does this therefore allow this mate in two (at least, I think it's a mate in two; I don't have a copy of WinChloe to properly verify this)?
 (= 10+9 )
(The intended solution, encoded in ROT13 because I don't know how to make spoilers on this forum, is: bar xvat gb qrygn guerr ebbx gnxrf puneyvr sbhe gjb xvat gb rpub sbhe naq guvf vf zngr orpnhfr juvgr vf va purpx guerr gvzrf naq guhf pnag or pncgherq juvyr rirel fdhner nebhaq oynpx vf nggnpxrq rvgure bapr be gjvpr.)
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| (9) Posted by Andrew Buchanan [Wednesday, Oct 28, 2020 18:07] |
WinChloe says that the #2 by edderiofer is sound. (Also the #3 by me and James, btw.)
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| (10) Posted by Joost de Heer [Wednesday, Oct 28, 2020 21:09] |
In edderiofer's composition, Sd2 isn't necessary, because capturing a king isn't allowed anyway.
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| (11) Posted by edderiofer [Thursday, Oct 29, 2020 08:58] |
Sure, it isn't, and I could easily have switched out the bPa4 for a bB. Ah well, oversights abound, I'm sure I'll get better at constructing with practice.
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| (12) Posted by Hauke Reddmann [Thursday, Oct 29, 2020 09:30] |
@Joost: So this means something *like* this is possible?
 (= 1+26 )
"Like", as some of the Q must be "downgraded" or the wK can take
a shortcut on the edge (6.Kxb7#). I'm too lazy to get it correct :-)
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| (13) Posted by James Malcom [Thursday, Oct 29, 2020 14:40] |
BLANK
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| (14) Posted by Andrew Buchanan [Thursday, Oct 29, 2020 14:41] |
 (= 8+6 )
#2
Removing wSd2 means that the intended key robs a flight. So let's have a different key, and remove some pawns. The new position even has four different defences to tries!
Hauke's spiral is very entertaining but I find I am too tired to repair it: can't count up to three tonight!
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| (15) Posted by James Malcom [Thursday, Oct 29, 2020 14:47] |
Inspired by Hauke's "stalemate," here's a handy dandy little seriesmover.
ser-!=10, Bosma
 (= 2+6 )
Also, if we're sharing, here's one from Chess Stack Exchange (linked above) that realizes one of the implications of the triple check rule/Bosma.
Me, CSE 8/12/2020
#2
 (= 6+5 )
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| (16) Posted by Joost de Heer [Thursday, Oct 29, 2020 16:59] |
The ser-!= has no solution.
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| (17) Posted by edderiofer [Thursday, Oct 29, 2020 17:26] |
@Andrew Buchanan Oh wow, this uses quite a bit more of the Bosma condition than in mine; I really like how getting the Ra5 to the f-file isn't actually a threat because Black now has Kf4. I hope I get this good at composition someday. :D
(However, I'm evidently not good enough at Bosma Chess to tell what exactly the tries and defenses are. And if the solution is 1.Kf3 zugzwang, don't both 2.Ke4 and 2.Kf4 mate, giving a dual, which is a flaw?)
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| (18) Posted by Andrew Buchanan [Thursday, Oct 29, 2020 18:30] |
Haha you are too kind I am full of errors. For example, it's indeed a dual! The main difficulty with Bosma is that the kings can get stuck to one another, like boxers. I must go to bed now, but I am sure someone else more competent will fix it :D
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| (19) Posted by Joost de Heer [Thursday, Oct 29, 2020 18:55] |
 (= 9+7 )
If Bosma can help white, it can help black too (Ke4 is no longer mate because f5 is guarded thrice).
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| (20) Posted by Andrew Buchanan [Friday, Oct 30, 2020 01:35] |
Hurray and this allows even more simplification:
 (= 8+5 )
#2 Bosma
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