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MatPlus.Net Forum Retro/Math Moves that determine all the previous moves
 
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(41) Posted by Alex Levit [Sunday, Mar 11, 2012 18:31]

@Joost: Cool cook! I think that if we use trick with SAN notation than we should state it
in stipulation. So in 1st puzzle we should add: no tricks! and in the 2nd: tricks allowed :-)
By the way I composed 2nd just to show AUW together with 1st. But I think that such usage of
SAN notation itself may be interesting.

@Kostas: No! Firstly we should find all pairs of white 6th and black 7th moves that totaly
determines all the moves. So we have some set of games. Than we try to find in this set game with
white and black promotions on move 5. If only one such game exist than my puzzle is correct.
Your moves 6.Bfc4 and 7...Bg2+ doesn't determine all previous moves so it's not a cook.
I think that this type of problems is more logic puzzles than real chess compositions
but I like it a lot.
 
   
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(42) Posted by Kostas Prentos [Sunday, Mar 11, 2012 21:13]

@Alex. I was going for a position that would satisfy all conditions (two promotions and the said moves). I believe that I understand your explanation, though.
However, please try to explain one more thing. If all moves of the game are determined by white's 6th and black's 7th moves, then why do you need the extra condition that white and black promoted on the 5th move? Only to determine the move order? In your solution, this doesn't seem necessary.
 
   
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(43) Posted by Bojan Basic [Sunday, Mar 11, 2012 22:04]

 QUOTE 
If all moves of the game are determined by white's 6th and black's 7th moves, then why do you need the extra condition that white and black promoted on the 5th move?

I'm not Alex, but I believe that the purpose of this condition is to extract only the pair (Bcf4, Rxb1) among all the possible pairs of White's 6th and Black's 7th move that determine the game uniquely (there are many such pairs).
 
   
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(44) Posted by Alex Levit [Sunday, Mar 11, 2012 23:59]

 QUOTE 
I was going for a position that would satisfy all conditions (two promotions and the said moves). I believe that I understand your explanation, though.

White 6th and black 7th moves must determine all moves without any extra information. In your example, Kostas, 6th and 7th didn't determine the game itself, but only if we already knows that 5th move is white and black promotion. That's the difference.
To make things more clear I just compose as simple example as possible:

White 3rd move is checkmate. If I give you one move from that game than you can determine all the moves. How did the game go?
Answer:
=========================
I give you 2.exf5,
and the game is
1.e4 f5 2.exf5 g5 3.Qh5#

=========================
Note that this puzzle:
If I give you one move from that game than you can determine all the moves. How did the game go if white 3rd move is checkmate?
have no solution. But this one:
If I give you one move from that game than you can determine all the moves. How did the game go if black 3rd move is checkmate?
have unique solution.
 
   
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(45) Posted by Kostas Prentos [Monday, Mar 12, 2012 00:45]

Here is a new try: 1.a4 c5 2.a5 c4 3.a6 c3 4.axb7 cxb2 5.bxa8=R bxc1=Q 6.R1xa7 Qa5 7.Sa3 Qaxd2#
Now, the question is whether 5.Rxa7 Qa5 6.bxa8=R bxc1=Q is a refutation of this cook. The proof game is not unique as a whole, but if white's 6th and black's 7th moves are known, then the order is fixed. Furthermore, the fact that both promotions occurred on the 5th move, fixes the move order in a different way, as suggested in my previous post.
 
   
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(46) Posted by Alex Levit [Monday, Mar 12, 2012 19:00]

Yes, this is a real second solution! Thanks you very much, Kostas!
 
   
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(47) Posted by Per Olin [Wednesday, Mar 14, 2012 19:36]

Some comments:

- First problem by Alex: Very close to be correct in the original form. By adding more conditions it can perhaps survive: 'Find a game, where white promotes a knight in his 6th move and where this move and the 6th move by black uniquely determine all previous moves. Two solutions'. This keeps the knight promotions 6.exd8=N Qxh4# and 6.exf8=N Qhxh4# and excludes 6.exf8=B Qhxh4#. Too many conditions or acceptable? The addition of 'two solutions' might be confusing.

- Second problem by Alex: The 6th move by white and 7th move by black are to determine the previous moves and both parties are to promote in the 5th move. As I see it, such a condition for promotion in a certain move is there only to get a unique move order; valid also for the promotion in 6th move in previous problem. Anyhow, a problem with this type of stipulation will almost always have several solutions. In this thread we have seen two, by Alex and by Kostas. More will be found, here some own tries: A) 6.Bbf4 and 7. - Qcb6 give 1.d4 g5 5.cxb8B hxg1Q 6.Bbf4 Qc7 7.f3 Qcb6 B) 6.Bbf4 and 7. - Qbf2+ give 1.d4 g5 5.cxb8B hxg1Q 6.Bbf4 Qb6 7.f3 Qbf2+ C) 6.Bfa3 and 7. - Qbxd4 give 1.d4 c5 5.exf8B bxa1Q 6.Bfa3 Qb6 7.Qd4 Qbxd4.

- The prospects of getting things correct without giving the specific moves are small. What is then left? Well, in an earlier post in this thread we have the longest game with two given moves by long description. Why not try the longest one with two moves in short description?
 
   
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(48) Posted by Alex Levit [Thursday, Mar 15, 2012 14:44]; edited by Alex Levit [12-03-15]

@Per I think that if we use moves which depend on SAN disambiguation then we should stay it directly in stipulation.
So in first puzzle we can add that 6th black move is independent on SAN disambiguation. I think this is better than
two solution form. In second puzzle only in my solution black 7th move is independent on disambiguation.
May be if we add this condition my puzzle would be sound.
But I think you are right: without given the specific moves it's hard to compose something deep.

Yet another type of puzzle:
Black play 7...Qcd7+ and checkmate white on next move. If I give you one more move from the game than you can determine
all the moves. How did the game go?


Answer:
=========================
I can give e.g. 2.Ke2
1.e3 d5 2.Ke2 d4
3.Kd3 dxe3+ 4.Kc4 exd2
5.Qg4 d1Q 6.Qxc8 Qxc8
7.Kb5 Qcd7+ 8.Kc5 Q1d4#

=========================
 
   
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(49) Posted by Per Olin [Thursday, Mar 15, 2012 22:49]

Stipulation: Find a game with 8.Qac7+ and 10. - Q4b3+. - Fairly easy to solve, more pleasure would come from cooking it. Solution later.
 
   
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(50) Posted by Kostas Prentos [Thursday, Mar 15, 2012 23:55]

Per, this looks like a cook: 1.d4 b5 2.d5 b4 3.d6 b3 4.dxc7 bxc2 5.cxb8=Q (or 5.Qd6) cxb1=Q (or Qb6) 6.Qxa7 Qb6 7.Qd6 Kd8 8.Qac7+ Ke8 9.b4 Q6xb4+ 10.Kd1 Q4b3+
 
   
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(51) Posted by Per Olin [Friday, Mar 16, 2012 20:20]

Yes, Kostas, you're right. Normal optimism from my side putting one or two halfmoves too much. My guess is that along these lines the longest game with two given moves (standard algebraic notation) will be found: promotion by both parties and play with the promoted and/or original pieces. Whenever we find this, we will have to wait a considerable time before the computer verifies the unbeatable record.
 
   
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(52) Posted by Alex Levit [Friday, Mar 16, 2012 20:42]

My try: 6...Q8d7+ and 11.Ke1+
 
   
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(53) Posted by Per Olin [Friday, Mar 16, 2012 21:30]

Alex, recalling one of your previous puzzles makes solving easier. However, there seems to be cooks: 1.e3 d5 2.Ke2 d4 3.Kd3 dxe3+ 4.Kc4 exd2 5.Q - d1Q 6.Kb5 Q8d7+ 7.Kc4 Qb5+ 8.Kc3 Kd8 9.Qxd1+ Bd7 10.Kd2 Bh3 11.Ke1+.
 
   
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(54) Posted by Alex Levit [Friday, Mar 16, 2012 22:19]; edited by Alex Levit [12-03-17]

Thank you Per! And what about 6...Q1d7+ and 9...Kc5+ ?

Upd. 6...Q1d7+ and 10...Ne8+
 
   
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(55) Posted by Per Olin [Saturday, Mar 17, 2012 21:45]

Both look good, Alex! If the latter one holds, as I suppose it will, then we know that the longest one so far is up to 10th move by black. Can this be improved, remains to be seen. I will not make a guess...!
 
   
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(56) Posted by Kostas Prentos [Sunday, Mar 18, 2012 08:54]

Alex, the first version looks OK, but the longer one is cooked: 1.e3 d5 2.Ke2 d4 3.Kd3 dxe3+ 4.Kc4 exd2 5.Qg4 d1=Q 6.Kb5 Q1d7+, and now white is not obliged to capture on d7: 7.Kc4/Kb4 Sf6 8.Qxg7 Bxg7 9.Kc3 0-0/Kf8 10.~ Se8+
 
   
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(57) Posted by Per Olin [Sunday, Mar 18, 2012 12:14]

As we can see, these are very difficult ones! Here an attempt with two promotions: Find a game with 7. - Bfc2+ and 10.D8xe5#. Solution: 1.b4 d5 4.bxc7 dxc2 5.cxb8Q cxb1B 6.Qb3 Bcf5 7.Kd1 Bfc2+ 8.Ke1 e5 9.Qb5+ Ke7 10.Q8xe5#
 
   
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(58) Posted by Alex Levit [Sunday, Mar 18, 2012 19:39]; edited by Alex Levit [12-03-18]

Thanks, Kostas! Another try:
1)6...Q8d7+ and 10.Rxd1#
2)6...Q8d7+ and 11.Rxd4#
3)6...Q8d7+ and 13.Rexd5#
 
   
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(59) Posted by Alex Levit [Tuesday, Mar 20, 2012 15:15]

I'm sorry 1) and 2) have second solution:
1.e3 d5 2.Ke2 d4 3.Kd3 dxe3+ 4.Kc4 exd2 5.Qg4 d1Q
6.Kb5 Q8d7+ 7.Qxd7 Kxd7 8.Bc4 Kd6 9.Nf3 Qd4 10.Rd1 ~ 11.Rxd4#
But this cook don't work in 3)
 
   
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(60) Posted by Kostas Prentos [Tuesday, Mar 20, 2012 18:28]

Alex, the 13-mover is very ambitious! So far, I only have the intent: 8.Bc4 Qxg1 9.Bxf7 Qxc1xb1xa2-d5.
I will keep you updated.
 
   
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MatPlus.Net Forum Retro/Math Moves that determine all the previous moves