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| (41) Posted by James Malcom [Saturday, Nov 14, 2020 17:17] |
Here's a mathematics question for you all of you smarties in the crowd.
A while ago I made a short Bosma in CSE.
#1
 (= 6+1 )
1. Se5#!
The point is that the checkmating piece REMOVES 2/4 of the attaks on the Black king to produce a legal checkmate.
Questions: Can removing 3 checks or higher be done in a single move, and if not, how can it be proven impossible? intuitively, I think 3+ cannot be done, but alas I lack the proof.
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| (42) Posted by Jan Hein Verduin [Saturday, Nov 14, 2020 23:47] |
Andrew:
QUOTE
This inspired Bosma in 1993 to define a fairy theme, (...)
The original article in which the theme was proposed appeared in 1992; it was the judgement of the theme tourney that was published in 1993.
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| (43) Posted by Andrew Buchanan [Sunday, Nov 15, 2020 03:52] |
Jan,
QUOTE
The original article in which the theme was proposed appeared in 1992; it was the judgement of the theme tourney that was published in 1993.
Thanks. I have updated the post here and also the PDB definition. Can you send me a scanned image of the judgement please? I will message you my email address. It's ok if it's Dutch.
I would really like to communicate with Dolf Wissmann. Do you have his contact details perhaps, please?
Thanks for everything,
Andrew
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| (44) Posted by Andrew Buchanan [Sunday, Nov 15, 2020 04:15] |
James,
QUOTE
Questions: Can removing 3 checks or higher be done in a single move, and if not, how can it be proven impossible? intuitively, I think 3+ cannot be done, but alas I lack the proof.
Think geometrically in terms of *changing* check status. Every check, even by knight, defines a straight line running through the centres of the squares containing king and checker. Most moves by a checker can only change the state of two line: the one the moving piece leaves and the one it joins. The only funny business might be castling or en passant, but castling stuck at the edge of the board doesn't offer any third line. For e.p. you can check easily that no third line can be changed with orthodox pieces. However if you have a nightrider (that mischievous thief of the letter "N") then an en passant can change the status of 3 lines. Try all possible squares for the king to see.
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| (45) Posted by Dolf Wissmann [Sunday, Nov 15, 2020 05:36] |
I should have contact details of this Dolf Wissmann here somewhere... Let's see ... Yes! Found it!
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| (46) Posted by Andrew Buchanan [Sunday, Nov 15, 2020 06:29] |
Hi Dolf,
OK, I will send a Note to this Dolf person, maybe you can relay it to him? :)
Thanks,
Andrew
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| (47) Posted by Hauke Reddmann [Sunday, Nov 15, 2020 10:18] |
In fairy chess, of course, everything is possible:
 (= 8+2 )
Madrasi #1
Replacing the Q by Kraken easily gives a 61-fold
(how do you say that in scientific Latin? :-) check.
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| (48) Posted by Joost de Heer [Sunday, Nov 15, 2020 10:50] |
Bernd Schwarzkopf
Problemkiste 2012
 (= 16+1 )
ser-+(15)19
Messigny
1. Sc2 2. Rd3 3. Sd5 4. Kb6 5. Kc5 6. Kd4 7.Ke3 8.Qd4 9.Bf4 10.Rf3 15.h8=S 16.Sg6 17.Sgh4 18.Sg2 19.Kg1«Ke3+++++++++++++++
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| (49) Posted by Dolf Wissmann [Sunday, Nov 15, 2020 16:32] |
In reply on post 39:
Thanks Andrew!, for looking at that #3 of mine, also for your comments about it in pdb.
It was sent for the Bosma-tournament in 1992, but Bosma rejected it: "is not thematical, during solution both Kings come shoulder to shoulder". Text still puzzles me.
Felt that it shouldn't be that easy to get rid of me and this resulted in publication in Probleemblad 1994, as an original fairyproblem, #3, with the condition "Bosma Rex Inclusief".
But this time I got into trouble with solver M.E.Nordlohne who just had ... tried all possible moves! Last move in his investigation was 1.Db8, the one that had seemed the least likely key to him. His thorough approach unmasked 1.Dc6! and 1.Nxb5+! as being cooks.
It was worth trying to correct I thought, so I did, but without satisfactory result.
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| (50) Posted by Dolf Wissmann [Sunday, Nov 15, 2020 16:33] |
About correcting the problem:
I remember possibilities for making that threat were limited and in the end my conclusion was that for a satisfactory correction a totally different scheme was needed.
I can't remember to have seen that with bQ on c8 one cook is gone or that transferring to key 1.Tg4-g6 does the whole trick, it is possible that the flight giving key pushed it outside my estimated limits for getting satisfaction.
You are right that 1.Db8! defends against threatening checks by black, and therefore is quite logic, still the downside of 1.Tg4-g6 is more obvious.
I will have another look at it but if there is nothing better than to settle for 1.Tg4-g6 I will accept your correction!
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| (51) Posted by Dolf Wissmann [Monday, Nov 16, 2020 00:12] |
I meant "flight taking" of course, not "flight giving".
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| (52) Posted by Andrew Buchanan [Monday, Nov 16, 2020 02:26] |
Hi Dolf,
Thanks for the entertaining info. It's a curious blind spot about the bumping kings - four of the other award winners are savagely and probably irretrievably cooked by WinChloe as a result. Christian's interpretation of the Bosma condition allows kings to be adjacent. I think this is a good call as (1) it's simplest & most consistent (2) it gives more design space for the composer to create novel arrangements. These 4 problems I will classify in PDB as Bosma(shy) - it’s not fair to declare them as cooked but they are not “canon”.
I will need to check WinChloe to make sure your problem is properly classified as not being in the tourney. I still haven't seen the Probleemblad issue with the results - someone said send they might send me a scan. When the dust settles I will update WinChloe & PDB with the status.
There are more Bosma compositions being made in 2020, as it’s somewhat cool and fun once defined properly. Once a solver "gets his eye in", a triple attack on wK in a d# or s# diagram stands out like a sore thumb as a triple unprovided check, which must be dealt with immediately. In your diagram with wQc7, Black immediately threatens 5 knight moves resulting in single or double checks, so the key is almost certain to have to address this. It becomes as obvious as flight-taking. I suspect the key for this problem is never going to be great, but it would be good if there is no initial triple check on White. This is one reason why wQb8 in the diagram appeals to me. Please let me know whether you find something.
All the best,
Andrew
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| (53) Posted by Jan Hein Verduin [Monday, Nov 16, 2020 17:43] |
QUOTE
I still haven't seen the Probleemblad issue with the results - someone said send they might send me a scan
You should have received it by now; the mail I sent hasn't been returned undeliverably (is that even a word?)
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| (54) Posted by Andrew Buchanan [Monday, Nov 16, 2020 19:26] |
Thank Jan just received it really appreciate. I’ve just sent a Note to Dolf with a new version of the #3.
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| (55) Posted by Joost de Heer [Tuesday, Nov 17, 2020 08:07] |
Just published on Julia's fairies:
Pierre Tritten
 (= 16+16 )
h+(7)#2.5, Make&Take, KobulKings
(helpmate in 2.5 with a 7-fold check)
https://juliasfairies.com/problems/no-1558/
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| (56) Posted by James Malcom [Sunday, Mar 5, 2023 19:23] |
Some time later, I've uploaded the scans of the relevant Probleemblad issues to the Internet Archive: https://archive.org/details/bosma-chess/
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