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 Page: [Previous] [Next] 1 2  (1) Posted by Mario Richter [Saturday, Apr 11, 2009 12:37]  Another popeye mystery popeye is quite fast in solving even long seriesmovers.
It has successfully been used in many longest seriesmover composition
contests as a correctnesschecking tool.
So I was surprised to hear that it could not solve a (what I thought
would be a relatively simple) seriesmover:
Die Schwalbe, Dec.2008 (Heft 234)
13940  Arno Tüngler
(= 11+16 )
(= 11+12 )
serA=>B 320
Since I couldn't believe it, I tested it myself  and yes, after 30 hours
of computation without any response from popeye, I canceled the popeye session.
(The input I used was:
forsyth ss1b4/p3P3/2PrRr2/ssrkBRP1/1Kb2q2/P2PrP2/1r6/1Bs3b1
stip serA=>B 320
forsyth ss6/4P3/2PrRr2/KBrk1RP1/P1bB1q2/3PrP2/1r6/2s3b1
option MoveNumber
)
So my questions are:
1. Is there something within the structure of this problem which makes it difficult for popeye to solve it?
2. (To the programmers of popeye) Where does popeye spend its time when trying to solve this problem?
3. I tested it under WindowsXP. Does the same behaviour occur under Linux?
(There is a small chance that WinXP's memory management, which has some influence
on popeye's hash mangement, may be 'suboptimal' ...)
4. Is there somebody out there, who has access to Alybadix or WinChloe (or any other tool,
which can handle the serA=>B 320 stipulation), and who can be so kind to test
this problem with his program?
Best regards,
mario   (2) Posted by Joost de Heer [Saturday, Apr 11, 2009 20:54]  The CVS version solves this without any problems:
Popeye Windows32Bit v20090400.52 (1024 MB)
Unknown
Unknown
+abcdefgh+
 
8 S S . . . . . . 8
 
7 . . . . P . . . 7
 
6 . . P R R R . . 6
 
5 K B R K . R P . 5
 
4 P . B B . Q . . 4
 
3 . . . P R P . . 3
 
2 . R . . . . . . 2
 
1 . . S . . . B . 1
 
+abcdefgh+
sera=>b320 11 + 12
Initial (White >):
Unknown
Unknown
+abcdefgh+
 
8 S S . B . . . . 8
 
7 P . . . P . . . 7
 
6 . . P R R R . . 6
 
5 S S R K B R P . 5
 
4 . K B . . Q . . 4
 
3 P . . P R P . . 3
 
2 . R . . . . . . 2
 
1 . B S . . . B . 1
 
+abcdefgh+
sera=>b320 11 + 16
1 (e7e8=Q Time = 16.285 s)
2 (e7e8=S Time = 16.286 s)
3 (e7e8=R Time = 16.287 s)
4 (e7e8=B Time = 16.288 s)
5 (e7*d8=Q Time = 16.289 s)
6 (e7*d8=S Time = 16.291 s)
7 (e7*d8=R Time = 16.292 s)
8 (e7*d8=B Time = 16.294 s)
9 (Re6*d6 + Time = 16.295 s)
10 (Re6*f6 Time = 16.296 s)
11 (c6c7 Time = 16.297 s)
12 (g5g6 Time = 16.298 s)
13 (g5*f6 Time = 16.299 s)
14 (Rf5*f4 Time = 16.301 s)
15 (Rf5*f6 Time = 16.302 s)
16 (Be5*b2 + Time = 16.303 s)
17 (Be5c3 + Time = 16.304 s)
18 (Be5d4 + Time = 16.306 s)
19 (Be5*f4 + Time = 16.307 s)
20 (Be5*f6 + Time = 16.308 s)
21 (Be5*d6 + Time = 16.310 s)
22 (Kb4*c4 + Time = 16.311 s)
23 (Kb4*b5 Time = 16.312 s)
1.Kb4a4 2.Bb1c2 3.Bc2d1 4.Bd1e2 5.Be2f1 6.Bf1h3 7.Bh3g4 8.Bg4h5 9.Bh5e8 10.Be8d7 11.Bd7c8 12.Bc8a6 13.Ba6*b5 14.Bb5a6 15.Ba6c8 16.Bc8d7 17.Bd7e8 18.Be8h5 19.Bh5g4 20.Bg4h3 21.Bh3f1 22.Bf1e2 23.Be2d1 24.Bd1b3 25.Ka4b4 26.Kb4c3 27.Bb3c2 28.Kc3d2 29.Kd2d1 30.Bc2a4 31.Ba4b5 32.Bb5a6 33.Ba6c8 34.Bc8d7 35.Bd7e8 36.Be8h5 37.Bh5g4 38.Bg4h3 39.Bh3f1 40.Bf1e2 41.Kd1e1 42.Ke1f1 43.Kf1g2 44.Kg2h3 45.Be2d1 46.Bd1a4 47.Ba4b5 48.Bb5a6 49.Ba6c8 50.Bc8d7 51.Bd7e8 52.Be8h5 53.Bh5g4 54.Kh3h4 55.Kh4h5 56.Bg4h3 57.Bh3f1 58.Bf1e2 59.Be2d1 60.Bd1a4 61.Ba4b5 62.Bb5a6 63.Ba6c8 64.Bc8d7 65.Bd7e8 66.Be8g6 67.Kh5h6 68.Kh6g7 69.Bg6f7 70.Kg7f8 71.Kf8e8 72.Bf7h5 73.Bh5g4 74.Bg4h3 75.Bh3f1 76.Bf1e2 77.Be2d1 78.Bd1a4 79.Ba4b5 80.Bb5a6 81.Ba6c8 82.Bc8d7 83.Ke8*d8 84.Kd8e8 85.Bd7c8 86.Bc8a6 87.Ba6b5 88.Bb5a4 89.Ba4d1 90.Bd1e2 91.Be2f1 92.Bf1h3 93.Bh3g4 94.Bg4h5 95.Bh5f7 96.Ke8f8 97.Kf8g7 98.Bf7g6 99.Kg7h6 100.Kh6h5 101.Bg6e8 102.Be8d7 103.Bd7c8 104.Bc8a6 105.Ba6b5 106.Bb5a4 107.Ba4d1 108.Bd1e2 109.Be2f1 110.Bf1h3 111.Bh3g4 112.Kh5h4 113.Kh4h3 114.Bg4h5 115.Bh5e8 116.Be8d7 117.Bd7c8 118.Bc8a6 119.Ba6b5 120.Bb5a4 121.Ba4d1 122.Bd1e2 123.Kh3g2 124.Kg2f1 125.Kf1e1 126.Ke1d1 127.Be2f1 128.Bf1h3 129.Bh3g4 130.Bg4h5 131.Bh5e8 132.Be8d7 133.Bd7c8 134.Bc8a6 135.Ba6b5 136.Bb5a4 137.Ba4c2 138.Kd1d2 139.Kd2c3 140.Bc2b3 141.Kc3b4 142.Kb4a4 143.Bb3d1 144.Bd1e2 145.Be2f1 146.Bf1h3 147.Bh3g4 148.Bg4h5 149.Bh5e8 150.Be8d7 151.Bd7c8 152.Bc8a6 153.Ba6b5 154.Ka4*a5 155.Ka5a4 156.Bb5a6 157.Ba6c8 158.Bc8d7 159.Bd7e8 160.Be8h5 161.Bh5g4 162.Bg4h3 163.Bh3f1 164.Bf1e2 165.Be2d1 166.Bd1b3 167.Ka4b4 168.Kb4c3 169.Bb3c2 170.Kc3d2 171.Kd2d1 172.Bc2a4 173.Ba4b5 174.Bb5a6 175.Ba6c8 176.Bc8d7 177.Bd7e8 178.Be8h5 179.Bh5g4 180.Bg4h3 181.Bh3f1 182.Bf1e2 183.Kd1e1 184.Ke1f1 185.Kf1g2 186.Kg2h3 187.Be2d1 188.Bd1a4 189.Ba4b5 190.Bb5a6 191.Ba6c8 192.Bc8d7 193.Bd7e8 194.Be8h5 195.Bh5g4 196.Kh3h4 197.Kh4h5 198.Bg4h3 199.Bh3f1 200.Bf1e2 201.Be2d1 202.Bd1a4 203.Ba4b5 204.Bb5a6 205.Ba6c8 206.Bc8d7 207.Bd7e8 208.Be8g6 209.Kh5h6 210.Kh6g7 211.Bg6f7 212.Kg7f8 213.Kf8e8 214.Bf7h5 215.Bh5g4 216.Bg4h3 217.Bh3f1 218.Bf1e2 219.Be2d1 220.Bd1a4 221.Ba4b5 222.Bb5a6 223.Ba6c8 224.Bc8d7 225.Ke8d8 226.Kd8c8 227.Bd7e8 228.Be8h5 229.Bh5g4 230.Bg4h3 231.Bh3f1 232.Bf1e2 233.Be2d1 234.Bd1b3 235.Kc8b7 236.Kb7*a7 237.Ka7b7 238.Kb7c8 239.Bb3d1 240.Bd1e2 241.Be2f1 242.Bf1h3 243.Bh3g4 244.Bg4h5 245.Bh5e8 246.Be8d7 247.Kc8d8 248.Kd8e8 249.Bd7c8 250.Bc8a6 251.Ba6b5 252.Bb5a4 253.Ba4d1 254.Bd1e2 255.Be2f1 256.Bf1h3 257.Bh3g4 258.Bg4h5 259.Bh5f7 260.Ke8f8 261.Kf8g7 262.Bf7g6 263.Kg7h6 264.Kh6h5 265.Bg6e8 266.Be8d7 267.Bd7c8 268.Bc8a6 269.Ba6b5 270.Bb5a4 271.Ba4d1 272.Bd1e2 273.Be2f1 274.Bf1h3 275.Bh3g4 276.Kh5h4 277.Kh4h3 278.Bg4h5 279.Bh5e8 280.Be8d7 281.Bd7c8 282.Bc8a6 283.Ba6b5 284.Bb5a4 285.Ba4d1 286.Bd1e2 287.Kh3g2 288.Kg2f1 289.Kf1e1 290.Ke1d1 291.Be2f1 292.Bf1h3 293.Bh3g4 294.Bg4h5 295.Bh5e8 296.Be8d7 297.Bd7c8 298.Bc8a6 299.Ba6b5 300.Bb5a4 301.Ba4c2 302.Kd1d2 303.Kd2c3 304.Bc2b3 305.Kc3b4 306.Kb4a4 307.Bb3d1 308.Bd1e2 309.Be2f1 310.Bf1h3 311.Bh3g4 312.Bg4h5 313.Bh5e8 314.Be8d7 315.Bd7c8 316.Bc8a6 317.Ba6b5 318.Ka4a5 319.a3a4 320.Be5d4 a=>b
24 (Kb4a4 Time = 17.621 s)
25 (Kb4b3 Time = 17.622 s)
26 (Kb4*a5 Time = 17.624 s)
27 (Kb4*c5 + Time = 17.625 s)
28 (Kb4c3 Time = 17.626 s)
29 (d3d4 Time = 17.627 s)
30 (d3*c4 + Time = 17.628 s)
31 (a3a4 Time = 17.629 s)
32 (Bb1c2 Time = 17.631 s)
33 (Bb1a2 Time = 17.632 s)
solution finished. Time = 17.635 s   (3) Posted by Steven Dowd [Sunday, Apr 12, 2009 00:02]  Can someone explain this AB problem to me? Not that I don't understand the intent, but how long has it been around, etc. I must confess to knowing nothing about it. And it looks fascinating. Are such things solvable by humans (well, normal to below normal humans like me, anyway....)?
Could it be used to solve socalled double and triple seriesmovers, in which one side must get to a certain position, whereupon the other side begins his or her machinations to mate?
I must also admit I never used Popeye because I never could get comfortable with using it, the learning curve was very steep visavis Alybadix.   (4) Posted by Miodrag Mladenović [Sunday, Apr 12, 2009 07:40]  Steven wrote:
QUOTE Can someone explain this AB problem to me?
A>B means that white plays all the moves and has to achieve position B from the postion A. Black does not move at all. During the play white king cannot be in a check. It's OK if it's in check in the initial possition but on the first move white has to eliminate check. Either by moving king out of check or by placing some pice to protect king. A>B is usually not hard to solve. I solved this problem right away because there are no many options. Usually in this type of problems white plays a lot of moves by wK.   (5) Posted by Thomas Maeder [Sunday, Apr 12, 2009 07:48] 
QUOTE A>B means that white plays all the moves and has to achieve position B from the postion A. Black does not move at all.
I'm sure that this is just a typo in Miodrag's post, but to avoid confusion:
In regular (nonseries) A>B, both sides play.
In serA>B, White plays a series of moves that lead from position A to position B.   (6) Posted by Thomas Maeder [Sunday, Apr 12, 2009 07:57] 
QUOTE Can someone explain this AB problem to me? Not that I don't understand the intent, but how long has it been around, etc.
bernd ellinghoven wrote in feenschach that he "invented" A>B decades ago without publishing it. A>B was first made public at an Andernach meeting around 2000 IIRC, when it was used for the composing tourney.
QUOTE I must confess to knowing nothing about it. And it looks fascinating. Are such things solvable by humans (well, normal to below normal humans like me, anyway....)?
Yes. A>B is just a generalization of proof games.
QUOTE Could it be used to solve socalled double and triple seriesmovers, in which one side must get to a certain position, whereupon the other side begins his or her machinations to mate?
No. Not in is current definition, that is.
QUOTE I must also admit I never used Popeye because I never could get comfortable with using it, the learning curve was very steep visavis Alybadix.
Some people are more comfortable using a graphical frontend.
Cf. http://en.wikipedia.org/wiki/Popeye_(chess)#Windows_Popeye_Shell if you use Windows.
NB:
1. I have recently modified Popeye to support direct play in A>B, as was suggested by Friedrich Hariuc in a recent feenschach issue; this extension will be available in the next release (scheduled soon).
2. What you suggest above would be possible as well; not for the next release, though. Please submit a feature request (with example problems) on
http://sourceforge.net/tracker/?group_id=200122&atid=972240   (7) Posted by Joost de Heer [Sunday, Apr 12, 2009 08:00]; edited by Joost de Heer [090412]  A>B is a generalised type of proofgame, where the starting position is different from the regular starting position of a chess game. In other words: a normal proofgame is an A>B problem with position A = rsbqkbsr/pppppppp/8/8/8/8/PPPPPPPP/RSBQKBSR. SerA>B is a generalised form of serPG.
The first time I heard of it was in a review of the Andernach meeting in 2001.
Double seriesmovers (i.e. black plays x moves, then white y moves) are supported for some stipulations (i.e. mate, stalemate, selfmate) in popeye: x>serSTIPy. But soubleseries proofgames don't work.   (8) Posted by Kevin Begley [Sunday, Apr 12, 2009 08:41]  Didn't René Millour compose some A>B problems (using Alice) way back in the late 80s?
Perhaps even earlier?
Perhaps he invented the stipulation, but, I could swear there was some earlier examples...
Since I first saw this idea, I have always kept it logged  because if ever proofgames show signs of slowing, this is a wonderful "get out of jail free card" to have in hand.
Too bad there is no such card to be found in some other genres!
Of course, there is always fairies  the ultimate get out of jail free card.   (9) Posted by René J. Millour [Sunday, Apr 12, 2009 12:10]  Having used it for the first time in Andernach 2001 [2nd Prize], I am not the inventor of the A→B stipulation.
A not so bad A→B problem using Alice is the following. Take care, A and B are already used in Alice, this explains the X→Y stipulation! In the X and Y positions, the Bpieces are upsidedown!
René J. MILLOUR
The Problemist, 2006
Position X
(= 7+8 )
X is A(6+5) and B(1+3)
Position Y
(= 1+2 )
Y is A(0+2) and B(1+0)
The stipulation is: h X→Y in 33 single moves, MONOCHROME ALICE!
Small RETRO reasoning on position Y.
1) In Monochrome Alice, 000 is forbidden and a K appearing in A state on the aceg files, or in B state on the bdfh files, cannot have played 00. On the other hand, a K appearing in B state on the aceg files, or in A state on the bdfh files, must have castled. In Y, White and Black played 00 !
2) Retraction of a move such as Bc3a1→A is a retrocheck, retraction of a wK move is illegal [in the forward play the K would have placed himself in check in A state before turning to B state]. Thus the Ba1Kg7 paralysis can be broken only by an interposition between a1 and g7. Now, there is no other blacksquare piece on the board and whitesquare pieces can only uncapture whitesquare pieces! An exception saves the day: Pf7 captured Pb2 e.p. !!! The paralysis is thus broken by restoring Pb2, which is legal because, in Alice forward play, the B could have crossed over, in B state, the Pb2 before turning to A state on a1.
Application to position X in FORWARDPLAY.
Pf7 must do f7xe6xd5xc4xb3e.p., the e.p. capture occurring with Ba1 and Kg7 already in place! On white squares, the pieces Qc4, Rh5, Be6 and Pg2 are necessarily concerned: 3 of them are captured by the bP, the 4th one captures the bP on b3. Already on e6, the wB must move, as Pf7 takes on e6 an Astate piece! For the same reason, Bb8xRa7 is not possible. Pg2 cannot g2xf3xe4xd5: a victim is missing. It can g2g4xf5xe6 but, in B state on e6, it cannot be captured by Pf7! The sole effective mobilization is the Excelsior g2g4xh5xg6xh7xg8, possible thanks to Ph7 and Sg8 as victims !
1.Re2→A!! Ra1→B!! 2.Ba7→A! Ba2→A!! 3.Re6→B Rd5→B 4.Bf5→A These strange first moves prepare the future and allow the Excelsior 4...Pg4→B 5.Ph5→B! This P cannot stay on h7, where the wP will capture a Bstate piece 5...Pxh5→A 6.Rg6→A ! Pxg6→B 7.Bh7→B Pxh7→A 8.Bd4→B Pxg8=B→B!! A Q in B state would not check, but wait ... 9.Bxa1→A Be6→A Here a Q would check and prevent the castling... 10.00→B! Ke7→B 11.Pxe6→B Kxf8→A [Not 10.Pxe6 Ke7 11.00 Kxf8, as 11.00 is in fact forbidden, the K in A state places himself in check from Qc4, before turning to B state.] 12.Pxd5→A Kg7→B!! This move is possible thanks to Pb2, still on board. 13.Pxc4→B Pb4→B 14.Pxb3e.p.→A! Bxb3→B+! 15.Kh7→A Be6→A 16.Kg6→B Bf5→B+ 17.Kxf5→A Position Y.
Castling! Excelsior! Precise promotion! Strange first moves! Unexpected e.p. capture! Pxg8=B + 00 + Pxb3e.p. = Valladao!   (10) Posted by Miodrag Mladenović [Sunday, Apr 12, 2009 15:27]  @Thomas,
Thank you for this correction. Yes, I even composed few problems of this type.
However, it's an interesting question when condition is only A>B in some number of moves. Does black helps white to achieve position B or trying to prevent this goal? It looks to me like there should be direct A>B and help A>B stipulation to make differentiation between two different cases. So far I only composed few serA>B problems and then it does not matter if blacks helps white or not since black does not move at all.   (11) Posted by Steven Dowd [Sunday, Apr 12, 2009 16:44]  Thanks again for everyone's help. This is one of those things that intrigues me, but whether I will get around to trying it is another issue.
Some more questions:
1. Are these always expected to have great length, or can shorter versions be interesting (seriesmover type)?
2. Has anyone seen Albertston's book Chess Mazes, which strikes me as something very similar to, if not identical to this? He claims that solving such exercises are good for a chessplayer's visualization skills. It does seem that OTB players (in my experience) have the least resistance to seriesmovers as "unorthodox" problems go, the reason not quite clear to me.
3. Wouldn't a good direct A>B be notoriously difficult to compose? It's like trying to compose a correct and nontrivial =2 problem, which I find very difficult.
Again, thanks all.   (12) Posted by Steven Dowd [Sunday, Apr 12, 2009 16:52]  The Windows shell for Popeye does not appear to allow animation of the solution (in fact it looks like the kinds of diagrams one saw in the 1980s); this is not critical of the author of course, who offers this for free, but it would eliminate it for me, due to my sight problems, I need to play through solutions on the computer screen. I can't use a board and pieces.
It's just been that kind of a week. :(
Have fun, all.   (13) Posted by Thomas Maeder [Sunday, Apr 12, 2009 18:42] 
QUOTE However, it's an interesting question when condition is only A>B in some number of moves. Does black helps white to achieve position B or trying to prevent this goal? It looks to me like there should be direct A>B and help A>B stipulation to make differentiation between two different cases.
The next Popeye release will have both (and more: selfA>B, reflexA>B, semireflexA>B).
QUOTE So far I only composed few serA>B problems and then it does not matter if blacks helps white or not since black does not move at all.
In the tradition of seriesmoves, this sequence of moves is analoguous to that of a ser#.
And like there are serh#, sers#, serr#, we can think of serhA>B, sersA>B, serrA>B. And series play with introductory moves by the other side (e.g. black series followed by white series). And more.
The reason for help play being the default for A>B is because A>B was derived from proof games.   (14) Posted by Cornel Pacurar [Sunday, Apr 12, 2009 21:57]; edited by Cornel Pacurar [090413]  @ Steven
RE: Are these always expected to have great length, or can shorter versions be interesting (seriesmover type)?
Two 'longest series' tournaments using the sera=>b stipulation were held in recent years: 20062007, Nicolas Dupont (http://www.franceechecs.com/index.php?mode=showComment&art=20061119103001480) and 2008, Itamar Faybish (http://www.geocities.com/ifaybish/tournament2/results.html).
Although the chief aim was to find, within certain criteria, positions employing the largest number of moves, some of the problems with less pieces, and consequently with less moves, are also very interesting! Take a look, for instance, at this one:
James Willson, Cornel Pacurar, Pascal Wassong, Jacques Dupin, Miodrag Mladenovic, Arno Tüngler, Ralf Krätschmer, Nicolas Dupont, Guy Sobrecases
D/3/28, IF TT2
(= 2+3 )
sera=>b 28
(= 2+1 )
1.Ka1b1 2.Kb1c1 3.Kc1d1 4.Kd1e1 5.Ke1f1 6.Kf1g1 7.Kg1h2 8.Kh2g3 9.Kg3f4 10.Kf4e5 11.Ke5*d5 12.Kd5e4 13.Ke4f3 14.Kf3g2 15.Kg2f1 16.Kf1e1 17.Ke1d1 18.Kd1c1 19.Kc1b1 20.Kb1*a2 21.Ka2b1 22.Kb1c1 23.Kc1d1 24.Kd1e2 25.Ke2f3 26.Kf3g4 27.Kg4h5 28.Bf2e1
@ Thomas
RE: The next Popeye release will have both (and more: selfA>B, reflexA>B, semireflexA>B).
That is very good news, thank you!   (15) Posted by Joost de Heer [Monday, Apr 13, 2009 09:28]  A short A=>B problem:
Rene Millour
Die Schwalbe Heft 203, October 2003
(= 10+8 ) (= 9+7 )
A=>B3 (6 solutions, black starts, =Nightrider, =Grasshopper)   (16) Posted by Mario Richter [Thursday, Apr 16, 2009 11:43]  @joost Thanks for testing it with the current CVS version of popeye!
Since Joost provides an executable at http://sanguis.xs4all.nl/popeye/ , I should have had the idea to use this executable myself, but I simply didn't think of it ...
Instead I used the last official version 4.47.
Since other defects in that version have shown up in the past, my question to the popeye team: Wouldn't it be good to release a new official version?
Btw., even if popeye now solves the given problem relatively fast, there is still room for improvements:
QUOTE 1 (e7e8=Q Time = 16.285 s)
...
solution finished. Time = 17.635 s
so popeye spends more than 90% of the solving time with the investigation of a subvariant, which definitively cannot lead to a solution (rule: in an orthodox A>B problem, if number_of_white_pawns(A)==number_of_white_pawns(B), then no wP may be captured or promote) ...
QUOTE bernd ellinghoven wrote in feenschach that he "invented" A>B decades ago without publishing it. A>B was first made public at an Andernach meeting around 2000 IIRC, when it was used for the composing tourney.
Those interested in the history of A>B should please also look at Per Ingvar Olin's comment on that issue, which he made in a posting to the Retros Mailing List:
http://www.pairlist.net/pipermail/retros/2006December/001644.html
@René J. Millour
I was surprised to see that e.p. capture is allowed in AliceChess. Are there precise rules if and when such a capture is allowed?
Best,
mario   (17) Posted by Joost de Heer [Thursday, Apr 16, 2009 12:24]  If I remember correctly, Popeye will do a breadthfirst search. So it will generate all positions in 1 move, in 2, in 3, etc etc. So the first 16 seconds are used to generate all 319moveandless positions. Because of optimalisations, this only takes a short while.   (18) Posted by René J. Millour [Thursday, Apr 16, 2009 16:20]  Answer to (16): “@René J. Millour. I was surprised to see that e.p. capture is allowed in AliceChess. Are there precise rules if and when such a capture is allowed?”.
Your surprise surprises me!
It is logic and very simple. The P candidate to be captured e.p. necessarily just made a doublestep, therefore it is necessarily in B state at the end of its move (in B state on the 5th rank if it is a BP). This means the P capturing e.p. must itself be in B state before the capturing move (in B state on the 5th rank in case of WPxBPe.p.)! This is all the “rule”!
There are several problems using that. For example the following not really short Retro with only 4 pieces!
René J. Millour
StrateGems, 20012, 3rd Prize
Wenigsteiner, 2002, 1st Place
(= 3+1 )
Monochrome Alice A(2+0) B(1+1)
3 questions are asked: a) Last 2 single moves? b) By which piece was Sb8 captured? c) By which piece and where was Bc1 captured?
In Monochrome, a move is legal only when both starting and arrival squares are of the same color. Therefore, 00 is possible, 000 is not, Ss don't move, e.p. captures are legal. In Monochrome Alice, 000 is forbidden and a K appearing in A state on the aceg files, or in B state on the bdfh files, cannot haved played 00, a K appearing in B state on the aceg files, or in A state on the bdfh files, must have castled. Here, White played 00, Black did not. Since it could not come from a1 in Monochrome, Rd4 is a promotee. Only Rd4 and Kf4 are retromobile because, in Alice, retraction of a move such as Bb7a8 is a retrocheck, retraction of Kg2h1 is illegal [in the forward play the K would have placed itself in check in A state before turning to B state] Therefore, to avoid Black retrostalemate, White must immediately uncapture. Getting a BP to 5th rank in A state excludes the possibility of a doublestep and implies 2 captures, in B state on 4th rank 3 captures, etc... and promoting a P requires 4 or 6 captures depending whether it began with a doublestep.
Thus we find that the last move cannot have been b7xa8=B. This B in A state implies, in Alice, a 6move excelsior, which in Monochrome must be 6 captures, which may include an e.p. If a2xBb3xRc4xPd5xP(e.p.)c6xPb7xXa8=B, the BP captured in A state on d5 would itself have captured d/f7xBc/e6xRd5 and the piece X captured at a8 would be resulting from f/d7f/d5xPe4xQd3x??c2xSb1=X. Now, this last sequence is impossible: a capturee is lacking for d3 or c2, the unmatched states of the Ps making impossible, in a 5move excelsior, both an e.p. capture (here of WPd) and the capture of a P at home (here of WPc).
Thus the Ba8Kh1 paralysis can be broken only by an interposition between a8 and h1. Now, there is no other whitesquare piece on the board and blacksquare pieces can only uncapture blacksquare pieces! An exception saves the day: the WP promoted to R captured e.p. BPb on white square! The paralysis is thus broken by restoring Pb7, which is legal because, in Alice forward play, the B (which did not result from a7a8=B, illegal in Monochrome) could have crossed over, in B state, Pb7 before turning to A state on a8!
A WP capturing e.p. must be in B state on the 5th rank! Such an Alice Monochrome excelsior [example: b2xQc3xBb4xPc5xP(e.p.)b6xPc7xRd8=R] implies 6 moves and 6 captures. Note that the P therefore is in B state before promoting and a S (here Sb8), in A state because it never moves in Monochrome, cannot be taken by the P!
In addition to the BPb e.p. capture, the excelsior captures 5 blacksquare pieces: Q, R, B on squares inaccessible to the Ps, and 2 Ps on squares requiring the fewest captures on their part, thus one of the 2 Ps on its home square, and the other on the 5th rank, which itself captured 2 times because it is required to be in A state in order to be captured by a WP, itself in A state on the 4th rank!
One more P, also a blacksquare one, must be uncaptured on the first retromove to avoid Black retrostalemate. The WK would produce a BPf4 in B state, which would have made 3 captures, in fact impossible (see below the number of captures by Black on black squares).
The WR restores a BPd4 in A state, which came from the 7th rank by a doublestep and a single capture. Black therefore has only two retromoves and, without delay, White must uncapture a piece by unpromotion!
The last black move, was it c5xd4 or e5xd4? If c5xd4, the P came from c7 and in that case before it promotes to R on b8 the WP must have captured BPe on c5 (after e7xd6xc5), BPb e.p., and BPa on a7. Now, the retroplay Rd8xPd4 c5xYd4 (the White piece Y being undetermined for the time being) Rb8d8 c7c5 a7xb8=R... does not work because the R is on b8 in B state, signifying an excelsior in 5 moves, which is inconsistent with the e.p. capture needed on b6. To have the R in A state on b8, requires an extra White retromove, which is impossible due to Black retrostalemate! However, with promotion on d8, the retroplay Rd8xPd4 e5xYd4 Y_d4 e7e5 c7xd8=R..., or simply Rd8xPd4 e5xYd4 c7xd8=R, is legal!
The last move Rd8xPd4 is thus known with precision. The White unit Y involved in the move e5xYd4 will be determined later (note that on e5 the P retroparalyzes the WK without retrocheck). If c7x[Q/R/B]d8=R replaces a unit to give Black an infinity of retromoves, sooner or later White must restore BPc to c7. Then comes the uncapture by e.p. and the return to b7 of BPb, making possible the retraction Kg2h1 and the retroplay of Ba8. Thus there are restored a BP on c/a5 and 2 other units on the 4th and 3rd ranks.
The BP restored on c/a5 in A state is neither BPe captured by the WR on d4, nor BPc captured by the WP on c7. Is it BPa after a7xb6xc/a5? The 2 victims would have been pieces, not Ps, because Ps would have required too many blacksquare captures by White (6 blacksquare captures were already made by White: 5 in the excelsior, 1 in the last move). Thus BPa would have captured 1) WBc on b6 because WRa cannot reach b6 in Monochrome 2) WRa on c/a5. Thus, the capture of the B on b6 won’t work because in this case Sb8 would still have to be present!!! In fact, it could not be captured by the WP upon promoting, as shown above, nor by any of the following pieces:
 the WK because of their differing states (capturing the S would imply a K in A state on a7 or c7, which is not possible given that it has castled),
 the Ra1, which cannot reach the 8th rank in Monochrome,
 the Rd4 which, having only recently been promoted, played only the move Rd8xd4 (see above),
 the Bc1 which, before dying by a7xBb6, was blocked from b8 by Ps at a7 and c7 (capturing the S requires a B in A state, which could not cross over a Pc7 in A state)!
One comes to the paradox that BPa could not be captured as a P nor as a promotee, by the White excelsior which required an extra BP. In fact, the Bc1 captured the Sb8 via a7, before being captured by the BPg, which itself was then captured by the White excelsior, as we shall see.
g7xBg6xRe5xPd4 allows theoretically the promoting WP to capture Pg7 on d4, but the other WP, here captured in A state on d4, must itself capture 2 times. This would make too many captures. On the other hand, the excelsior of BPg, in 5 moves and 4 captures [e.p. not possible / g7g5xPf/h4xRg3xBh2xSg1=X], does not involve any capture by White.
The unmoved S is captured on g1. A WP (f or h), uncapturable at home, in taken in B state on the 4th rank. Ra1 ends up on the 3rd rank because the 2nd is unavailable to it in Monochrome. Finally, Bc1 cannot be captured at f2 where the capturing P would give check the WK before it played 00, which could not yet have occurred since Sg1 has not yet been captured ! Pg7 captured Bc1 on h2. [Now that the promotion of BPg has been shown, one discovers that the piece captured on c/a5 is not necessarily the promoted BPg: the capture squares of the BPg promotee and the 3 other original pieces (Q, R and B) can be interchanged!]
From the captured White pieces, one finds that the piece Y which was captured at d4 was necessarily WPd, which came straight from d2 without capturing. The last Black move therefore was e5xPd4.
Finally, the answers are a) Rd8xPd4 e5xPd4, b) Sb8 captured by Bc1, c) Bc1 captured by Pg7 on h2.
Henryk Juel: The Monochrome condition allows economic positions, but it is exciting that adding the Alice condition has enable the author to construct a Wenigsteiner with interesting content. A delight for the connoisseur.   (19) Posted by Mario Richter [Friday, Apr 17, 2009 09:00]  @René J. Millour, Post(18)
Dear René, thank you very much for this instructive example!
Just one remark:
QUOTE Your surprise surprises me!
It is logic and very simple.
It's not logic, it's an axiom which must be explicitly stated.
I'm only an amateur chess problem admirer. I do not have tons of old issues of 'diagrammes', 'rex multiplex', 'feenschach', 'FCR' or other chess problem literature on my bookshelf, which I could consult for an exact definition of 'Alice Chess'.
So my only source is the internet, and the website with seemed to me most informative (and complete) with respect to 'Alice Chess' is
http://www.bcvs.ukf.net/alice.htm
where one can read:
QUOTE (c) En passant capture is abolished. The inventor of Alice Chess, V. R. Parton, said nothing about the rule for en passant capture, but since the rule for orthodox chess can be interpreted in at least two different ways (does the capturing pawn have to be on the first or second board?) and is subverted by the fact that the square passed over may be occupied on the other board, it is usual to forgo it.
Best regards,
mario   (20) Posted by Per Olin [Tuesday, Apr 28, 2009 11:37]  In this chain was earlier mentioned the history of A to B proofgames. Here copied what I wrote in Retrocorner in 2004:
Proofgame starting from an other position than the initial position
The earliest example of a proof game starting from an other position than the initial position is probably the following:
7) Per Olin, Suomen Tehtavaniekat 1/1993
A) 8/PPP2P2/8/8/8/p3pkp1/1T4p1/T3K3
B) B1S2T1D/8/8/8/8/8/6k1/s1K2btd
Stipulation: Shortest proof game from A to B with either white or black to begin
Solution: 1.a2 000 2.a1S Tf2+ 3.exf2 Th1 4.f1L b8D 5.gxh1D Dh8 6.g2 c8S 7.g1T f8T+ 8.Kg2 a8L+. The try where white starts by castling 1.000 ? fails due to the impossibility of the last moves by black.
Sincerely yours
Per Olin   Read more...  Page: [Previous] [Next] 1 2
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