1 problem(s) found in 2547 milliseconds (displaying 1 problem(s)). [PROBID='P1004030'] [download as LaTeX]

1. Kc8+ Tc6+ 2. Kb7 Tc5+ Remis (Dead Position, da 3. Kb6+ Tc6+ 4. Kb7 dreimalige Stellungswiederholung ergibt)

WPs have made 11 pawn captures, Black none. So [BaP] was captured on the a file by a piece. [BbP] did promote on b1, so White's last move cannot have been bxc4. Other WPs which captured are currently retro-blocked. Conclusion: White's last move was not a capture. So Black's last move was a K move.

Since 0. ... Kc5-d5? would imply an illegal check, White's last move was not with K. To avoid Black's prior move being an illegal check, White must have just played 0. ... Rc6-a6. What were the prior moves? Possible chains are:

-1. Kb6-b7 Rc5-c6++ -2. Kb7-b6+ Rc6-c5 (loop)

-1. Kc8-b7 Ra/b6-c6+ -2. Kb7-c8+ Rc6-a/b6+ (loop)

-1. Kc8-b7 Bc7-d6+ which is the way that we must have originally entered this, and then followed up with zero or more loops.

Since neither loop may have occurred, we can ignore them. There are just 4 prior positions which must have recently occurred, including the current position: 0. ... Rc6-a6 -(2n+1). Kc8-b7 Bc7-d6+.

Now let's look forward. There is no helpstalemate in 2.0, so we must look for a position which must forcibly lead to a position repeated for the third time. Up to the diagram, no position has necessarily been repeated, and we cannot repeat a loop. Therefore, in our 2.0 moves, we *must* repeat one of the 4 proven past positions:

- If we repeat the current diagram it takes 2.0 moves, with 1. Kc8+ Rc6+* 2. Kb7* Ra6*, (* denoting repeat) but then there is no way to force a third repeat: e.g. 3. Kc8 Kc5.

- If we repeat the position prior to 0 ... Rc6-a6, it takes 1.5 moves 1. Kc8+ Rc6+* 2. Kb7*. Then 2. ... Rc5+ forces 3. Kb6+ Rc6+ Kb7* repeating the key position. This is the solution!

- If we repeat the position prior to -(2n+1). Kc8-b7, it takes 1.0 moves 1. Kc8+ Rc6+*. Then there is no way to get a forced repeatable position except by 2. Kb7 which we have already explored.

- We can't repeat the position prior to -(2n+1). ... Bc7-d6+ in less than 2.5 moves.

This beautiful problem relies on the interpretation of the interaction of Laws and Codex whereby the DP calculator "knows" that claiming draw by repetition is mandatory, and so a future forced third occurrence of a position implies that the game ends now. At one stage I was against this interpretation but currently I am agnostic. At the very least, it leads to a rich and self-consistent problem here.

WPs have made 11 pawn captures, Black none. So [BaP] was captured on the a file by a piece. [BbP] did promote on b1, so White's last move cannot have been bxc4. Other WPs which captured are currently retro-blocked. Conclusion: White's last move was not a capture. So Black's last move was a K move.

Since 0. ... Kc5-d5? would imply an illegal check, White's last move was not with K. To avoid Black's prior move being an illegal check, White must have just played 0. ... Rc6-a6. What were the prior moves? Possible chains are:

-1. Kb6-b7 Rc5-c6++ -2. Kb7-b6+ Rc6-c5 (loop)

-1. Kc8-b7 Ra/b6-c6+ -2. Kb7-c8+ Rc6-a/b6+ (loop)

-1. Kc8-b7 Bc7-d6+ which is the way that we must have originally entered this, and then followed up with zero or more loops.

Since neither loop may have occurred, we can ignore them. There are just 4 prior positions which must have recently occurred, including the current position: 0. ... Rc6-a6 -(2n+1). Kc8-b7 Bc7-d6+.

Now let's look forward. There is no helpstalemate in 2.0, so we must look for a position which must forcibly lead to a position repeated for the third time. Up to the diagram, no position has necessarily been repeated, and we cannot repeat a loop. Therefore, in our 2.0 moves, we *must* repeat one of the 4 proven past positions:

- If we repeat the current diagram it takes 2.0 moves, with 1. Kc8+ Rc6+* 2. Kb7* Ra6*, (* denoting repeat) but then there is no way to force a third repeat: e.g. 3. Kc8 Kc5.

- If we repeat the position prior to 0 ... Rc6-a6, it takes 1.5 moves 1. Kc8+ Rc6+* 2. Kb7*. Then 2. ... Rc5+ forces 3. Kb6+ Rc6+ Kb7* repeating the key position. This is the solution!

- If we repeat the position prior to -(2n+1). Kc8-b7, it takes 1.0 moves 1. Kc8+ Rc6+*. Then there is no way to get a forced repeatable position except by 2. Kb7 which we have already explored.

- We can't repeat the position prior to -(2n+1). ... Bc7-d6+ in less than 2.5 moves.

This beautiful problem relies on the interpretation of the interaction of Laws and Codex whereby the DP calculator "knows" that claiming draw by repetition is mandatory, and so a future forced third occurrence of a position implies that the game ends now. At one stage I was against this interpretation but currently I am agnostic. At the very least, it leads to a rich and self-consistent problem here.

**Keywords:**Help Dead Position, Joke, Draw by repetition, Dead Position

**Genre:**Retro, Fairies

**FEN:**b7/Pk1p4/R2B4/3KP1PP/2PQBNpP/6N1/6PP/R7

**Input:**Gerd Wilts, 2002-04-05

**Last update:**Alfred Pfeiffer, 2019-01-16 more...

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The problems of this query have been registered by the following contributors:

Gerd Wilts (1)
Brett um 180 Grad drehen, dann 1. e1=S axb5 2. Sf3 Txf3= Die Diagrammstellung kann wie folgt aufgelöst werden: R: 1. Tc6-a6 Kc8-b7 2. Lc7-d6+), aber es gibt keine Lösung für das Hilfspatt.

But a much simpler position would result in the same solution, so there is clearly something else going on.

Note:

(1) "fairy" because any help problem except for h# is automatically fairy.

(2) "joke" because of playful interpretation of "h=2" as draw not stalemate as is standard.

Henrik Juel: The = in the stipulation should be interpreted as draw, not stalemate. With retroplay -1.Tc6 Kc8-2.Ta6-c6 Kb7, the forward play 0... Kc8+ 1.Tc6+ Kb7 2.cTa6 yields draw by triple repetition! (2004-04-26)

Henrik Juel: Here is a better explanation: The joke is not turning the board, but interpreting the stipulation as 'help-draw in 2' rather than the customary 'help-stalemate in 2'. White captured [Pa7] above wPa and the other missing black men with P's, so last move must have been Tc6-a6. After 0... Kc8+ 1.Tc6+ Kb7 2.Tc5+, where Black and White help each other for 2.0 moves, the play must continue 2... Kb6+ 3.Tc6+ Kb7, and the game is drawn by triple repetition. The problem won 3rd Commended. (2004-12-10)comment