[Retros] More ways in proof games

[per olin]

MORE WAYS IN PROOFGAMES


Introduction

It is with pleasure we notice the development in the proof game problem genre during the last two decades.  In some chess magazines there are even more proof games published than twomovers.  The splendid results in e.g. Donati 50 JT and latest problem congress in Greece show the immense possibilities there are when the retro lovers attack a certain theme.  It is evident that there is still much to come.

The intention here is to show some alternatives to the classic proof games. Here certainly is not shown all that has been done in this area; perhaps somebody can fill in to get a fuller picture. Anyhow, my wish is that this could sow the seed to some similar new approaches.


Several positions/solutions

The classic proof game has one solution. Gradually there are coming proof games with several positions/solutions.  Will we in the proof games see the same development as in helpmates, where a problem with only one phase starts to be a rarity. A fine example of a multi-solution proof game is the following:

1) P. Wong  1st Prize Problem Observer 1995 ts2k1st/ppp2ppp/8/8/8/2Pd2b1/PP1PPPPP/
TSBDKBST/ proof game in 6.5  three solutions. Solutions for all problems at the end of this article.



Repetition of position

In 1992 I wanted to confuse the solvers in the Finnish Solving Championship. 

2)   Per Olin,   Suomen Tehtäväniekat   4/1992

tsd2bst/pppkpppp/3p4/8/4P3/8/PP1P1PPP/TSLDKL1T

Stipulation:  The diagram is from a game, where white has made his 9th move.  What black move can lead to draw by repetition of the same position for the third time?  

All my problems with this theme were presented in an article in Springaren 73 / May 1998.  One special trick worth mentioning is combining repetition of position with either party giving piece(s) as odds/handicap; this can of course be done also in a normal proof game. 

Later I recall having seen proof games published with the stipulation ‘draw in x moves’. I have assumed that it must be shown that the same position can be reached three times in the stipulated number of moves. It is the choice of the composer does he want to show the end position or the position one move before it is repeated for the third time. - I have a suspicion that there is not so much to be done in this area anymore, but hope I am wrong.



Proof game with one or several moves given

In Problemkiste 91, February 1994 Bernt Scwarzkopf introduced a new type of question:  How did the game go up to 4.  -  Tg6xTh6 ?  As diagram given the initial position. He also gave an example with two given moves.  The longest with two given moves is probably from my article in Problemkiste 114, December 1997.  Here was used the stipulation ‘Ziel 5.Sc3xTd1 and 12.Df7xTc4#’ (solutions numbered 3 and 4 below).

There are unexplored areas with this problem type!



End position unknown

By using the 50 moves rule the end position of the asked proof game can be kept unknown.

5)  Per Olin,  Suomen Tehtavaniekat  5/1992

4kbst/ppp1L2p/5p1p/8/8/7P/P1PPPP1P/TS1DKL1d

Stipulation:  The game has ended draw after white’s 58th move according to the 50 moves rule. What was the beginning of the game up to white’s 8th move?  


Starting position and end position unknown

Yes, it is possible to have a proof game, where neither the starting position nor the end position is known.  This is the problem with which I have struggled most through my years of composing. It would be interesting if some mathematician would calculate all possibilities and estimate time needed for a computer to check everything, if there were a program able to perform the checking.

6)  Per Olin,  Suomen Tehtavaniekat 4-5/1996   dedicated to Hannu Harkola

2b1k3/2pp3p/1ppP2p1/P4ppt/P5P1/7P/P2PP3/1S1K1S1T 

Stipulation:  A game, where the order of the officers of white on the first rank and of black on the eighth rank was arranged by random (same officers on the same line), has ended draw after black’s 64th move (diagram) according to the 50 moves rule.  What was the order of the officers (a-h) and how did the game go up to black’s 14th move ?




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


In Andernach 2001 was asked can with any starting position a ´Babson-Frolkin’ be composed.  This was accomplished by the French grandmaster.

8)  Michael Caillaud,  1st Prize Andernach TT 2001

A)  8/P1PP2P1/1p2k2p/1s2p3/4P3/1K6/psPPp1pp/2b2b2
B)  s7/1b6/1pP4p/1P2p1k1/2s2b2/1K6/2P5/8

Stipulation:  Proof game in 17,5 from A to B

This excellent problem should inspire us to believe that the initial position does not have to be the starting point of a proof game. 

The number of diagrams does not have to be two. 9)  Per Olin & Unto Heinonen,  Prize   Suomen Tehtavaniekat  Summer Tourney 2003, final award in number 4-5 2004

A)  8/1p6/1p2s3/1p1kb2d/sP6/1K6/1PPt4/T6T
B)  st6/1p7/k7/pp2b2d/1P6/1K6/1P2s3/T6T
C)  st2d3/8/8/pp1K4/k7/p2s4/1P6/T6T
D)  4d3/8/s7/3K4/p7/pk6/p2t4/T6T


Stipulation:  Proof game in 9,5 moves a) from A to B (black begins)    b) from B to C (white begins)  c) from C to D (black begins)   


Technical remark concerning multi-diagram proof games

In proof games from A to B it has been suggested that the black move should be first, as in e.g. helpmates.  However, if the first part of the proof game starts from initial position the convention is that white starts and white’s move is mentioned first.  If we have a problem a) from initial position to A  and b ) from A to B, we would only have confusions if white’s move is not mentioned first.  Therefore I think there can be only one way:  white’s move has to be mentioned first.   



Proof games with fairy conditions

There are published proof games with fairy conditions such as Andernach, Circe and Randzuger.  Personally I am not very fond of these.  It is true that some spectacular paradoxes can be shown by changing the rules. But I am very much more impressed when a paradox is a paradox and not a consequence of changing the rules. 


Summary

Start experimenting!

Espoo,  Finland in November 2004

Per Olin


Solutions:

1) 1.Sc3 d5 2.Sd5  Le6 3.Sxe7 Ld5 4.Sxd5 Ld6 5.Sc3 Lg3 6.Sb1 Dd3 7.c3;  1.Sf3 e5 2.Sxe5 Ld6 3.Sxd7 Lg3 4.Se5 Lg4 5.c3 Lf3 6.Sxf3 Dd3 7.Sg1;  1.c3 e6 2.Da4 Ld6 3.Dxd7 Kf8 4.Dxc8 Lg3 5.De6 Dd3 6.Db3 Ke8 7.Dd1

2)  9. – Dd8.  Play 1.e3 d6 2.Se2 Lf5 3.Sec3 Lxc2 4.e4 Lxb1 5.Sxb1 Kd7 and so forth.  The try is to move 5. – Dc8, whereby 9.    -  Ke8 could return to the same position for the third time. However, black would then have lost his castling rights, which means that the optically similar positions are not the same.

3) 1.a4 a5 2.Ta3 Ta6 3.Th3 Tg6 4.Th6 Tg6xTh6

4). 1.d4 a5 2.d5 Ta6 3.d6 Txd6 4.Sc3 Txd1 5.Sc3xTd1 d6 6.g4 Kd7 7.g5 Kc6 8.g6 Kb5 9.gxh7 Ka4 10.hxg8D Th4 11.Dxf7 Tc4 12.Df7xTc4#

5) 1.Sc3 d5  2.Sxd5 Lh3  3.Sxe7 Dd4  4.Sxg8 Dxb2  5.Sh6  Dxa1  6.Lb2  gxh6  7.Lxh8 f6 8.gxh3 and 50 moves without captures or pawn moves leading to the diagram position.

6) Piece order: DTSSTKLL  and 1.g4 g6  2.Lc6 bxc6  3.f4 Tb3  4.Lb6 axb6  5.cxb3 Da4  6.bxa4  Sb7  7.b4 Sa5  8.Df6 exf6  9.bxa5  Te5  10.Tb5 Th5  11.Tg5 fxg5  12.h3 Le5  13.fxe5 Sd6  14.exd6 f5;  after this there are 50 moves without pawn moves or captures leading to the diagram position.

7) 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. 

8)  1 – c8D  2.Kf6 Dc5  3.Kg5 Dg1  4.hxg1D a8S  5.Dc5 Sc7  6.Dc6 Sd5  7.Sc7 Sc3  8.Sa8 Sb1 9.axb1S d8T  10.Sc3 dxc3  11.Lf4 Td1  12.exd1T g8L  13.La6 Lc4  14.Td5 Lf1  15.gxf1L exd5  16.Lfb5 c4  17.Lb7 cxb5  18.Sc4 dxc6  (see technical remark above!)

9)    a)  1.  -  Kc6  2.c4 Sc3   3.Ta5 bxa5  4.Ta1 Kb6  5.c5 Ka6  6.c6 Td8  7.c7 Tb8  8.c8T Se2  9.T8c1 Sc7  10.Th1 Sa8      b)  1.Ta4 bxa4  2.Kc4 a3  3.Kd5 Kb5  4.Ta1  Ka4  5.b5   De8  6.b6 Lc7  7.bxc7 Sc1  8.c8T Sd3  9.T8c1 b5 10.Th1    c)  1.  -  b4    2.Ta2 b3  3.Tha1 bxa2  4.b4 Sc5  5.b5 Sa6  6.b6 S8c7 7.bxc7 Tb2  8.c8T Kb3  9.T8c1 a4 10. Th1 Td2#


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