4 Help-Retractors

Arisktotle

Help-Retractor sounds as a fancy concept but it is not. You are playing an OTB game and someone walks by and whispers "you just missed a checkmate in 3". That is a help-retractor. If you missed it one move ago then you need to take back 1 move and forcefully checkmate your opponent in 3 moves.

All 4 diagrams have that precise assignment. You take back any white move you like and checkmate black in 3 moves instead. They are not just ordinary moves because you must take into account the retrograde analysis for each position related to the castling rights on both sides.

The 4 diagrams form a quadruplet. Each diagram only differs in 1 detail from the previous diagram, mostly by changing the position of just one piece or pawn. But as said, the similarity is deceptive when looking at the game history.

 

 

 

 

Rocky64

Here's my attempt at Diagram 1. Will try the others when time permits. 

White wants to retract a R to the d-file (e.g. Rd5-g5 or Rd1-a1) as that effectively immobilises the d7-B (in view of Rd8#), and then forces M3 with 1.a7 and 2.a8=Q+ (or 1...Kf8 2.Rxd7). Trouble is that 1...0-0! would refute as castling is currently legal. So the idea is that White must retract specifically to create a position where it can be proved by retro-analysis that castling is illegal.

Black castling is legal in the diagram because we can assume that the h8-R is the KR, but then how did the other R get to h7? If it reached h7 via the 8th rank, then it must have dislodged the BK previously from e8 and castling is actually illegal; the only alternative is that the R got there from the south side, meaning the g6- and h6-Ps have cross-captured to let it through. This cross-capture and the doubled c-Ps require 3 white units to be captured, and that's not a problem since White is missing 6 units.

However, White can start by retracting Rd5xPg5! The g5-P must have come from e7, implying the e6-P is from d7 and the c4-P is from a7. Such a P structure requires 5 captures, leaving only 1 missing white unit as a spare and that's not enough for the g6- and h6-Ps to have cross-captured. Therefore the h7-R must have gotten there via the 8th rank and had dislodged the BK previously. Black castling is thus illegal and White mates in 3 with 1.a7! 

Arisktotle
Rocky64 wrote:

Here's my attempt at Diagram 1. Will try the others when time permits. 

White wants to retract a R to the d-file (e.g. Rd5-g5 or Rd1-a1) as that effectively immobilises the d7-B (in view of Rd8#), and then forces M3 with 1.a7 and 2.a8=Q+ (or 1...Kf8 2.Rxd7). Trouble is that 1...0-0! would refute as castling is currently legal. So the idea is that White must retract specifically to create a position where it can be proved by retro-analysis that castling is illegal.

Black castling is legal in the diagram because we can assume that the h8-R is the KR, but then how did the other R get to h7? If it reached h7 via the 8th rank, then it must have dislodged the BK previously from e8 and castling is actually illegal; the only alternative is that the R got there from the south side, meaning the g6- and h6-Ps have cross-captured to let it through. This cross-capture and the doubled c-Ps require 3 white units to be captured, and that's not a problem since White is missing 6 units.

However, White can start by retracting Rd5xPg5! The g5-P must have come from e7, implying the e6-P is from d7 and the c4-P is from a7. Such a P structure requires 5 captures, leaving only 1 missing white unit as a spare and that's not enough for the g6- and h6-Ps to have cross-captured. Therefore the h7-R must have gotten there via the 8th rank and had dislodged the BK previously. Black castling is thus illegal and White mates in 3 with 1.a7! 

Exquisite analysis + a correct solution!

I am glad I annotated the solutions myself years ago. I would have trouble reproducing them now on an uninhabited island. I am a bad archivist. Lost many problems or don't remember the solutions. Occasionally other composers bring them back to me happy.png

Rocky64

So you're saying this is an old problem of yours and not an original? That's a relief cause I was thinking you're mad to be publishing such a high-quality work in a forum post, rather than sending it to a proper magazine! wink.png

Anyway, Diagram 2 was very tough and took me ages to work out...

With a WR already on the d-file, a possible mating plan is to bring the other R to d1, followed by Rxd7 and Rd8#. White can even try the retraction Rd1-a1 and aim for M2 with 1.Rxd7, but once again 1...0-0! refutes. How does White retract to preclude black castling as a legal move? 

Looking ahead at the forward M3, it's a big hint that White 0-0-0 is possible, but why wouldn't Rad1 work just as well to threaten Rxd7? Consider the 3 WPs on f2, g2, and h2, which combined with an unmoved WK would have trapped the original h1-R in the corner. By playing 0-0-0 rather than Rad1, that proves that the WK hadn't moved before and thus the h1-R had never escaped (before it got captured), meaning the WR on d4 is a promoted piece. 

That the d4-R is a promoted piece can help us because the path of the promoting P could have somehow disabled black castling. How to make this happen with the correct white retraction is the hardest part of this problem! Note that the h5-P must have taken 3 of Black's 5 missing units. With 2 spare black units to capture, it seems easy for a WP to promote on the Q-side without disturbing the BK. Let's try to create a barrier against the promoting P by retracting b5xPa6. It turns out that White's original a-P could still have promoted on a8 because the BPs on a6 and b6 could have cross-captured to let it through (White's 5 missing units are enough for (1) this cross-capture, (2) the K-side cross-capture to let in the h7-R, plus (3) the doubled c-Ps). And the promoted R could also have escaped via the a-file without disturbing the BK.

The actual solution is to retract b5xPa6 e,p,!! Such an en passant uncapture means that Black's previous move was ...a7-a5. With the BP on a7 rather than a6, no cross-capture is possible and White's a-P couldn't have marched down to a8 and promote. White must have promoted the d-P instead, with only a dark-squared BB as a possible captive along the way (because the h5-P had captured 3 black pieces on light squares, namely the Q and 2 Ns). Whether the d-P had promoted on d8 (passing through d7) or e8 (after capturing the BB on e7), either would have dislodged the BK from e8. Therefore after 1.0-0-0!, black castling is illegal and White mates with 2.Rxd7 and 3.Rd8.

Awesome variation of the mutually exclusive castling idea. If White can legally castle, then Black can't, and vice versa. White, who has the turn, can take advantage of this situation by castling first, i.e. White "proves" that castling is legal by doing it (!), and thereby renders black castling illegal. The problem also shows the Valladao theme - combination of castling, en passant, and promotion.

Arisktotle
Rocky64 wrote:

... The problem also shows the Valladao theme - combination of castling, en passant, and promotion.

Yes it does, but I don't count it as such since there is no underpromotion and the queen promotions are not all that exciting in themselves. I might count it after an excelsior!

Be back on this post later.

Arisktotle

Well done, again, Rocky64! There is one point worth noting: It is possible to retract: wPb4xPc5 and replace it with w0-0-0.The retro strategy is OK as black loses castling right. But of course he now has the simple defense bPc5xd4!

I can tell you that diagram 3 is not as retro-filled as the other ones but it operates as a bridge to diagram 4 which fits in with dia 1 and 2! Hope this will give you some relief from your torment.

Rocky64

Great to hear that I got it right. When you wrote that the problem doesn't have an underpromotion, I thought my solution was off the mark since its retro-play does feature a R promotion!

Yes, the b4xPc5 retraction is a nice try which I considered before seeing the correct play.

Arisktotle
Rocky64 wrote:

Great to hear that I got it right. When you wrote that the problem doesn't have an underpromotion, I thought my solution was off the mark since its retro-play does feature a R promotion!

Oops, you are right! I never considered it that way because it comes in one package with the white castling right - but it is nice to have a Valadao in my name! Have to check my other problems too; might have overlooked another one in the same fashion wink.png

Rocky64

Here's the solution to Diagram 3, I think...

It's tempting to retract 0-0-0 because that implies the d4-R is promoted, as in the previous solution. But without a BP on a7 here, White's original a-P could have promoted without disturbing the BK, so uncastling followed by 1.0-0-0 or 1.Rad1 to threaten 2.Rxd7 fails to 1...0-0! again. There seems no way to prevent black castling by means of retro-analysis like that for the first two diagrams. Rather desperately, we can try to prevent it directly by retracting the WB to e7 or g7. Such a retraction loses control of the BK's flight-square on e7, but to compensate, here both WRs are already on the d-file and so after the retraction, White can play 1.Rxd7 immediately. 

Whether the WB retracts to e7 or g7, it should uncapture something on f6 to block that potential flight-square, e.g. retract Be7xPf6 and play 1.Rxd7, which threatens both 2.Rd8+ Kxe7 3.R1d7 and 2.a7 and 3.a8=Q. But this try is defeated by 1...f5! unblocking the f6-flight to neutralise 2.Rd8+, and if 2.a7 then 2...f6! lets the BK escape to f7, since the WB has cut off the WR's control of that square. 

The solution is to retract Bg7xPf6! for 1.Rxd7! with two threats as before. 1...Rxg7 forces 2.Rd8+ Ke7 3.R1d7 since 2.a7? is too slow against 2...0-0!, while 1...e5 necessitates 2.a7 and 3.a8=Q since 2.Rd8+? Ke7 and the BK escapes to e6. The WB must uncapture a P on f6 as a BB or BN would easily prevent the M3, while a BR would allow 1.Rxd7 e5! 2.a7 Rd6!

The forward M3 play isn't easy either for this part!

 

Arisktotle

Excellent once more! You didn't fall for any of the traps in this one either! Interesting try too is: retract Bg7xRf6? replaced by 1. Rxd7 e5! 2. a7 Rd6!

I am pretty confident the 4th diagram won't be a problem to you either since these variants must start feeling like home to you now! After you solved it, I'll explain why I made the quadruplet. There is a bit of theory behind it. 

Rocky64

Yes, Diagram 4 seems relatively straightforward...

With the BP on a7 stopping the a6-P from queening, the solution of Diagram 3 becomes a try here: retract Bg7xPf6? and 1.Rxd7 e5! 2.Rd8+ Ke7.

As indicated previously, the natural retraction of 0-0-0 ensures that the d4-R is promoted, and thanks to the a7-P, White couldn't have simply promoted on a8 and then moved the R out via the a-file. Since Black is missing 5 units and 3 were taken by the h5-P, that leaves 2 spare units for a WP to capture on its way to promote. Any promotion on the Q-side would require the R to escape the 8th rank via d8, dislodging the BK. The d-P could have promoted on f8, attacking e8 again, but not on g8 as that needed 3 captures. So in this position after White's retraction, the BK must have moved previously and cannot castle.

Now White has potential M3 with 1.Rad1 or 1.0-0-0, but 1.Rad1? fails to 1...Rxh5! 2.Rxd7 Re5+/Rd5! and 1.0-0-0? fails to 1...Rxh5! 2.Rxd7 Rd5! - I almost fell for the latter. The key is actually 1.c6! Rxh5 2.cxd7+ Kf8 3.d8=Q.  

Excellent quadruplet in which the four parts not only juggle different retro-analysis, but they have distinct and precise key-moves for the M3.

Arisktotle

Very well done for 4 flawless solutions! Interestingly, your understanding of the retractor-logic appears to be greater than that of the retro-community at large. In 1958, R. Kofman published a famous #3 which you would see is flawed right away. Only you and me and 3 more retro-experts wink.png understand that the proof game for the solution consists of (a) the history before the diagram (b) the retraction move, and (c) the solution with the substitute move. That is the implied understanding of the term retraction. You do something, take it back and correct it in the same time line. Kofman doesn't think so:

The author would of course retract 0-0-0! and solve it with 1. 0-0-0, but considering you solved my 4th diagram, you would be equally happy to play 1. dxc3!. Black didn't lose his castling right by the forward move 1. 0-0-0 but by the retraction 0-0-0 because they are in the same time line / proof game.

The quadruplet was set up as a trap. If you disagree with this interpretation of contiguous time, You cannot solve the 2nd diagram either. After all, there is no reason to assume the captured pawn came from a7 if you forgot the retraction move!

Peculiarly, the same Kofman made another composition where he retracted and reexecuted an e.p. move with retraction memory - the same logic as my 2nd diagram, Anyone using inconsistent logics like Kofman would stumble on either my diagram 2 or diagram 4!

Solution: self evident.

Note: I have come to understand why the retro-community think as it thinks. It doesn't think at all - strange as it may sound for a bunch of highly intelligent people. "Thinking" requires the manipulation of some sort of dynamic tool leading you from A to B. What they have, is the magic "mutually exclusive castling" potion which you shouldn't endeavour to understand. They see it happening in Kofmans problem before their eyes and their brains light up - they don't see it in diagram 4 and their brains shut down. Well, I'm working on it, and things will be better one day.

 

Rocky64

That first Kofman diagram looks familiar and I found it in my copy of Nunn's Solving in Style, which I read a long time ago when I would've been too "green" as a problemist to notice this retro isn't really sound. I'm surprised to hear that this is controversial. The intended solution of retracting and replaying 0-0-0 is so thematic that it's easy to assume it's the only solution, but when solvers - and indeed the composer - are made aware that 1.dxc3 also works, do they really dispute it and insist the M3 is solved by 1.0-0-0 only!? Or is it possible they view it as a dual rather than a more serious cook, since the initial 0-0-0 retraction is forced?

I haven't seen the second Kofman diagram before and it's such a gem!

Arisktotle
Rocky64 wrote:

That first Kofman diagram looks familiar and I found it in my copy of Nunn's Solving in Style, which I read a long time ago when I would've been too "green" as a problemist to notice this retro isn't really sound. I'm surprised to hear that this is controversial. The intended solution of retracting and replaying 0-0-0 is so thematic that it's easy to assume it's the only solution, but when solvers - and indeed the composer - are made aware that 1.dxc3 also works, do they really dispute it and insist the M3 is solved by 1.0-0-0 only!? Or is it possible they view it as a dual rather than a more serious cook, since the initial 0-0-0 retraction is forced?

I haven't seen the second Kofman diagram before and it's such a gem!

Do you think John Nunn would have published Kofmans 1st problem without mentioning the dual had he known about it? Do you think that solving World Champion John Nunn would fail to find the dual in Kofmans 1st problem himself - unless he believed it was invalid?

I've seen it many times. The deepest truth is this one. Problemists do not really believe that rational retro logic exists. They fear it will all crumble when you zoom in too closely. And they solve the unesase for themselves with a lighthearted disrespect for reason, replacing it with the collegial smile of mutual 'understanding'.

The exception is PRA. They know partial retro analysis as it runs parallel to the scientific method. But they cannot identify with retro-strategy and a posteriori logic (basically the same thing) and keep it at arm's length.

R. Turnbull has written a long article about 'abominable retro problems' - fairy retro's - which was endorsed by a list of high ranking philosophers and mathematicians. It showed he had no understanding whatsoever of the existing orthodox retro logics which were much better suited to handle his problems. And they would have made some of the flawed problems correct as well. The same happened to MDR (minimum deviation from the rules) developed by another composer. Added his own logics which made his own pivotal problems fail. Common retro-logics would have saved them. It will all be in the book, I will probably never write wink.png

Arisktotle

By the way, there was a dual solution to the original of Kofmans e.p. problem as well - related to my 2nd quadruplet diagram. That's why I posted my corrected Kofman version here. As with the castling problem, I never saw anyone comment on the error though it was even more obvious than there. Only collegial smiles!

Rocky64

Hmm, I took another look at the first Kofman diagram and, in defence of the composer, it seems that he was aware that white uncastling was sufficient to disable black castling. The evidence is the BP on g4! wink.png

I found the original Kofman showing e.p. and indeed it's badly unsound. Good job to fix it so neatly! 

Arisktotle
Rocky64 wrote:

Hmm, I took another look at the first Kofman diagram and, in defence of the composer, it seems that he was aware that white uncastling was sufficient to disable black castling. The evidence is the BP on g4! 

Don't think so - even though I did consider that possibility as well! After (retract 0-0-0) 1. dxc3 g3 2. 0-0-0 white still mates with the help of the castling move. The point is that he need not castle on his first forward move which is what the composer made us believe! I agree that the presence of Pg4 throws some confusion on the composer's thoughts but in the end it is clear that he would never have allowed the dual 1.dxc3 if he had understood that it works. Note that Pg4 is effective against 1.Rd1? g3! but such is irrelevant with the dxc3 dual in force.

My theory is confirmed by the error in the e.p. problem 10 years later which is basically the same. Kofmans stern belief is that the e.p. move must be reexecuted in order to take away the black castling right as did white castling in the 1958 problem. And again, it is untrue.

His mind - and the minds of almost all problemists - see these problems as instances of the proof it by doing it principle which is somewhat associated with retro-strategy. Actually, neither of the Kofman problems is a retro-strategy problem! Only their intended solutions made them look like one! The simple point is this: in a retro-strategy solution, an uncertainty is first turned into a certainty; another phase of the solution then takes advantage of an unusual consequence of that certainty! In the Kofman problems, the castling and e.p. rights are already 100% certain when the forward solution starts. Instead of retracting, one could publish the diagrams after retractions with the stipulation texts like "white has castling right" and "white has e.p. right". No retro-strategy.

Note that retractions have nothing to do with retro-strategy (except in a natural language understanding) , because they are free. There are no conventions stating preferred assumptions for retracting e.p. or castling moves or anything. Even in a diagram with apparent white castling 'right', white can retract a king or rook move without retribution. In the end there is a massive gap between absolute rights which cannot be undone and conventional permissions (licenses).

Rocky64
Arisktotle wrote:
Rocky64 wrote:

Hmm, I took another look at the first Kofman diagram and, in defence of the composer, it seems that he was aware that white uncastling was sufficient to disable black castling. The evidence is the BP on g4! 

Don't think so - even though I did consider that possibility as well! After (retract 0-0-0) 1. dxc3 g3 2. 0-0-0 white still mates with the help of the castling move. The point is that he need not castle on his first forward move which is what the composer made us believe! I agree that the presence of Pg4 throws some confusion on the composer's thoughts but in the end it is clear that he would never have allowed the dual 1.dxc3 if he had understood that it works. Note that Pg4 is effective against 1.Rd1? g3! but such is irrelevant with the dxc3 dual in force.

The g4-P serves a specific purpose - to stop the ruinous dual, Rd1, i.e. uncastle & 1.Rd1 g3! 2.dxc3 gxf2+! If (as you suggest) the composer had thought that black castling is disabled only after the forward 1.0-0-0 is played ("prove by doing") and not before, then he would've been very happy to see 1.Rd1 as a thematic try that's defeated by 1...0-0! (or 1...cxd2+ 2.Rxd2 0-0!), which would allow the g4-P to be placed elsewhere, e.g. g5. But no, he specifically placed the P on g4 to stop 1.Rd1 in another (more mundane) way because he realises it's necessary, i.e. he knows that after White's uncastling, black castling is already illegal and thus incapable of defeating 1.Rd1.

The difference between dxc3 and Rd1 is that the latter dual would completely spoil the problem, since the main idea is to show retracting and replaying 0-0-0. The actual dxc3 dual is bad but doesn't kill the problem, so my theory is that the composer either overlooked it or thought it was tolerable. Your theory that he didn't think it was a dual at all is less plausible because he went to the trouble of stopping the other dual, 1.Rd1.

Arisktotle wrote:

My theory is confirmed by the error in the e.p. problem 10 years later which is basically the same. Kofmans stern belief is that the e.p. move must be reexecuted in order to take away the black castling right as did white castling in the 1958 problem. And again, it is untrue.

This retro is unsound but to confirm your theory, it needs a dual that's analogous to dxc3 in the first problem. That is, in the uncorrected position, after retracting the e.p., White should have an alternative move to capturing e.p. that also solves the M3. Such an alternative move would back your view that Kofman mistakenly thought only the forward e.p. capture would disable black castling and hence the dual is invalid. However, I can't see such a dual in the e.p. problem, and why it's unsound seems unrelated to the "prove by doing" issue.

Rocky64

Mystery solved... I've just opened Solving in Style again to check Nunn's analysis of the problem, and that's when I noticed that he in fact quoted a different version of it, with the g4-P on g3 instead!! This is a correction that eliminates the 1.dxc3 dual (and 1.Rd1 too, obviously), because with Black threatening 1...gxf2+ after White's uncastling, 1.0-0-0! is forced. So the problem is now perfectly sound and Nunn didn't miss anything. Both the original and corrected versions can be found on the Schwalbe problem database. The correction isn't attributed to anyone besides Kofman, so it shows that he's aware that the original 1.dxc3 was a dual that needs to be fixed.

Arisktotle

Final edit:

Well, I couldn't know in my comment that a different version was in Nunns book (Sorry, John). My original article was published in 2007 in Probleemblad and I don't know what changed in the databases since. I also left comments in one of the databases (don't know if it was die Schwalbe) and these posts might have triggered changes to them. Some people might be waking up to rationality. I can see from your comments you are uncomfortable with the asserted irrationality and incompetence in the retro-logic-field but it is there and it is almost total. I would have won about 3 more 1e prizes had it not been so. Happy I didn't because it proved my point! The incompetence is caused by the confusion stemming from not getting the fundamentals. I am working on it with problems such as the quadruplet which forces solvers to ignore the fixed patterns. Before solving the quadruplet, did you ever ask yourself about the Kofman problems if the castling right was lost after the retraction or after the 1st solution move? I bet not because they didn't require you to ask that question. Now the answer is so obvious to you that you can't even imagine others missed it. But they did, just as with countless other retro-logic issues.

The wrong Kofman version was quoted in Tim Krabbés "Schaakcuriosa" booklet with the incorrect mutual castling argument and without any inquisitive comments. The pivotal issue is not that a correct version exists but that the wrong version is explicitly annotated as wrong in the databases and publications including the reasons of why it is wrong. You do not leave a cooked original uncommented in a database when you know you have a correction. It's not just a matter of "may be you like this" or "may be you like that". Do they know that one is wrong and one is the correction?

Kofman didn't add Pg4 to stop Rd1 but to turn 1.dxc3? into a decent try which he hoped to refute not by ..g3? 2. 0-0-0! but only by ...0-0!. That he corrected the whole problem later with a heavy heart was certainly not because he knew at the time there was an error but likely because someone told him there was! In the meantime, 1.dxc3 was a fatal flaw no author would ever find 'acceptable for the time being'.

I lost the wrong version of Kofmans e.p. problem (can you post it?) but I recall that white did indeed have a dual M3 solution by not reexecuting the e.p. move. That would make it similar to the dxc3 dual. If something else was wrong, my mind played tricks on me and I withdraw my 'repeat offender' argument. My 2007 article was not about the error in the e.p. retraction problem and that is why I only used my corrected version in it.