Answers.Body: <p>The retro-logic is crystal clear:</p>
  <ol>
  <li>Either white or black castling rights are disrupted.</li>
  <li>If it's White to move then Black has lost castling rights.</li>
  </ol>
  <p>However, the forward play is messy. I approached Hans Gruber, retro editor of Feenschach, who also has detailed records of Die Schwalbe from this time.</p>
  <p>He says that the intention was to fool solvers into using point 1 above: so 1.Rd1? 0-0-0! and 1.0-0-0! 0-0-0? Solvers then would overlook that Black castling is illegal, so in reality both keys would work. But the composer didn’t spot that 1.Nc5/Ne5 also works.</p>
  <p>Therefore the problem is unsound: “cooked” in problemists’ terminology.</p>
  
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Answers.Text: There is also a problem with the assumption that the author preferred not to label AP tasks with the AP label (“in order not to betray anything”_V.Keym). There was another task of the author in the same issue and there is an AP mark there.
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Answers.Text: And here I didn't understand at all:
  
  1. 0-0-0? 0-0-0 (Illegally!)
  
  1. Rd1! (threat 2. h8=Q/R#) 0-0-0 2. Rb6# (Illegally!)
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Answers.Text: 0. ... h3 1. Rd1 0-0-0 2. Bb6# It is legal, but neither white nor black need it. Black has 0. ... 0-0-0, white has 1. 0-0-0.
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Answers.Text: An analysis of the initial position in the diagram shows that there was an interdependence of castling in the party. But at the same time, the analysis shows that black does not have a move that preserves castling. And it follows from this that there is neither PRA nor RS in the task. Nothing depends on anything and IN THE WHOLE GAME black ALWAYS has no right to castling, white has.
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Answers.Text: Yes, I also want to find an author's solution. I just have little idea how to do it.
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Answers.Text: There is an unconfirmed version that in the Ukrainian yearbook for 1997 the task was published as AP (indicating the publication in Schwalbe). I will give the solution given there in my answer. But that's all in words for now. There is no link or scan.
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Answers.Body: <p>To summarize.</p>
  <ol>
  <li><p>Obviously, the task cannot be correct in this form. There are 4 possible first moves by white. This is no good. 1. 0-0-0; 1. Rd1; 1. Nc5; 1. Ne5 и 2. h8=Q#</p>
  </li>
  <li><p>It is obvious that the error is either in the task (A) or in the diagram (B).</p>
  </li>
  </ol>
  <p>A1. This is the AP's task. A non-standard type of Anti-Keym (to prevent black from using AP to take a move in his position).
  Disadvantages of the version: the author is generally not seen in an effort to hide the use of AP in tasks (“in order not to betray anything”_V.Keym) and the idea itself is so innovative that it could not help but pop up in discussions of the task. It is impossible to say one hundred percent, but rather no than yes. The only solution is: 1. 0-0-0 ~~ 2. h8=(Q,R)#</p>
  <p><em>On the Internet with a link to (<strong>Щорічник 1997</strong>, 1998, page 92; Щорічник - in Ukrainian, the yearbook) it is claimed that the task was marked AP (it is indicated that the task was published in Schwalbe). The following author's solution is given:</em></p>
  <p><em>1.Rd1? 0-0-0 (требование ретропатом смены очередности хода; the retro stalemate requirement of changing the order of the move) 2. Са6#??</em></p>
  <p><em>1. 0-0-0! (отрицание смены очередности; denial of a change in the order of a move) ~ (0-0-0??) 2. h8=(Q,R)#.</em></p>
  <p>А2. An even more non-standard method: &quot;everything is possible in the past!&quot; (that is, we must take into account all the options when, for example, we lost the right to castling just like that - in fact, the method simply prohibits castling) and then the only solution will be 1. Rd1 (That's what they think here: <a href="https://www.janko.at/Retros/Schwalbe/Solutions.htm#9510" rel="nofollow noreferrer">https://www.janko.at/Retros/Schwalbe/Solutions.htm#9510</a>). But you and I remember about 1. Nc5/Ne5.
  Cons of the version: the method is so non-standard that it doesn't even have a name. Plus the duals. Absolutely not here.</p>
  <ol start="3">
  <li>It remains to check the error in the diagram. It may consist in the fact that &quot;black has the last move that preserves the castling.&quot; Then we have just the usual interdependence of black and white castling.
  Indirectly, this is indicated by the fact that both in the discussions and in the Turevsky database, black's defense in the form of 0-0-0 appears widely (which is impossible in the original diagram).</li>
  </ol>
  <p>&quot;Проверим, проверим&quot; (с)</p>
  <pre><code>r3k3/2pN3P/5PPP/6pp/8/6Pp/N5P1/R3KB2 w - - 1 13
  </code></pre>
  <p>And, by the way, this is a great reason to finally understand PRA and RS.</p>
  <p><em>In the examples nos. 3–7 one can see typical common features of PRA and RS: identical retrograde analysis and partially identical solution. If two castlings exclude each other, the PRA problem has a (partial) solution for each of the two cases (both partial solutions together form the complete solution), the RS problem, however, has a complete solution for one of the two cases.</em>(V.Keym,2010) <a href="https://www.janko.at/Retros/Glossary/Castling-and-En-passant.htm" rel="nofollow noreferrer">https://www.janko.at/Retros/Glossary/Castling-and-En-passant.htm</a></p>
  <p>If we continue further, we can simply say that PRA and RS are twin brothers (like &quot;the party and Lenin&quot;). They can either be together, or there is no one at all. Even more importantly, both methods use the same approach (strict compliance with 16.1 and 16.2 - that is, neither method uses the options of depriving an optimistic right without grounds or obtaining a pessimistic right without grounds - this is already to RV, LM and similar quack methods).</p>
  <p>And even more importantly, understand the Meta-Agreement here.</p>
  <ol>
  <li>First, a simple method is applied (without any methods at all)</li>
  <li>If it doesn't work out, use PRA or RV (according to the rules, PRA has priority, all other things being equal).</li>
  </ol>
  <p>The method is just a tool. No one can force us to use a tool unless we need it to solve a problem.
  The simple method always takes precedence over any method.</p>
  <p>===</p>
  <p>Finally, to the point.</p>
  <p>There is no simple solution. It is necessary to use methods. First on the PRA list.</p>
  <p><em><strong>(we remember that both methods divided the past into two realities: A - white has castling, black does not; B - black has castling, white does not. Theoretically, the possible reality of C (castling is not available to anyone) is not considered by any of the methods).</strong></em></p>
  <p>The PRA method will easily solve B. But in A we get the same thing (4 possible first moves of the solution!). There's a dead end here! The problem can be solved using the PRA method, but the solution will not satisfy other requirements for the solution (uniqueness of the solution).</p>
  <p>Let's try through RS.</p>
  <ol>
  <li><p>0-0-0 (stripped black of castling) 2. h8=(Q,R)#. Everything is fine here!</p>
  </li>
  <li><p>Rd1 (c threat h8=Q#) 0-0-0 (the only defense!) 2. Ba6# Everything is fine here too!</p>
  </li>
  </ol>
  <h2>But there are two solutions! But are there two?</h2>
  <p>In the first case 1.0-0-0 by legalizing reality A, we have destroyed the legality of reality B (a classic of the genre!).
  And what happened in the second case? Yes, the same thing!</p>
  <p>With the move 1. Rd1, we destroyed the legality of reality A and the only legal reality remained reality B. <strong>By breaking our castling, we proved the right to castling black!</strong></p>
  <p>It's beautiful! <strong>It can be proved not only by performing castling, but also by refusing to perform castling.</strong></p>
  <p>And we end up with two partial solutions. One for reality A, the other for reality B. They are impossible at the same time! And, most importantly, we dispensed with the RRA so that he would be healthy!  Hocus pocus!</p>
  <p>P.S. It's a shame that the task with two such brilliant ideas has remained misunderstood for incomprehensible technical reasons...</p>
  
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Body: <pre><code>[Title &quot;#2 - Leonid M. Borodatow. Die Schwalbe 163 02/1997&quot;]
  [FEN &quot;r3k3/2pN3P/5PPP/6pp/7p/6P1/N5P1/R3KB2 w - - 0 1&quot;]
  </code></pre>
  <p>Different sources give different answers.</p>
  
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Comments.CreationDate: 2024-07-24T06:29:43.040Z
Comments.Text: I have fixed the question, but the rest of the text is confusing. I suggest that you tidy it up and remove the mention of changing the diagram, maybe shift it to an answer
Comments.UserDisplayName: Laska
Tags: <problems><castling><retrograde-analysis>
Title: What's the answer to this castling rights problem?
ThreadText: # What's the answer to this castling rights problem?
  
  (Asked by: username_0 on 2024-07-23)
  
  ```
  [Title "#2 - Leonid M. Borodatow. Die Schwalbe 163 02/1997"]
  [FEN "r3k3/2pN3P/5PPP/6pp/7p/6P1/N5P1/R3KB2 w - - 0 1"]
  
  ```
  
  Different sources give different answers.
  
  **Tags:** problems, castling, retrograde-analysis
  
  > **Comments:**
  > **username_1**: I have fixed the question, but the rest of the text is confusing. I suggest that you tidy it up and remove the mention of changing the diagram, maybe shift it to an answer  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_1 on 2024-07-23. Score: 1
  
  The retro-logic is crystal clear:
  
  1. Either white or black castling rights are disrupted.
  2. If it's White to move then Black has lost castling rights.
  
  However, the forward play is messy. I approached Hans Gruber, retro editor of Feenschach, who also has detailed records of Die Schwalbe from this time.
  
  He says that the intention was to fool solvers into using point 1 above: so 1.Rd1? 0-0-0! and 1.0-0-0! 0-0-0? Solvers then would overlook that Black castling is illegal, so in reality both keys would work. But the composer didn’t spot that 1.Nc5/Ne5 also works.
  
  Therefore the problem is unsound: “cooked” in problemists’ terminology.
  
  > **Comments:**
  > **username_0**: There is also a problem with the assumption that the author preferred not to label AP tasks with the AP label (“in order not to betray anything”_V.Keym). There was another task of the author in the same issue and there is an AP mark there.  
  > **username_0**: And here I didn't understand at all:  1. 0-0-0? 0-0-0 (Illegally!)  1. Rd1! (threat 2. h8=Q/R#) 0-0-0 2. Rb6# (Illegally!)  
  > **username_0**: 0. ... h3 1. Rd1 0-0-0 2. Bb6# It is legal, but neither white nor black need it. Black has 0. ... 0-0-0, white has 1. 0-0-0.  
  > **username_0**: An analysis of the initial position in the diagram shows that there was an interdependence of castling in the party. But at the same time, the analysis shows that black does not have a move that preserves castling. And it follows from this that there is neither PRA nor RS in the task. Nothing depends on anything and IN THE WHOLE GAME black ALWAYS has no right to castling, white has.  
  > **username_0**: Yes, I also want to find an author's solution. I just have little idea how to do it.  
  > **username_0**: There is an unconfirmed version that in the Ukrainian yearbook for 1997 the task was published as AP (indicating the publication in Schwalbe). I will give the solution given there in my answer. But that's all in words for now. There is no link or scan.  
  
  ---
  
  ### Answer by username_0 on 2024-07-23. Score: 1
  
  To summarize.
  
  1. Obviously, the task cannot be correct in this form. There are 4 possible first moves by white. This is no good. 1. 0-0-0; 1. Rd1; 1. Nc5; 1. Ne5 и 2. h8=Q#
  2. It is obvious that the error is either in the task (A) or in the diagram (B).
  
  A1. This is the AP's task. A non-standard type of Anti-Keym (to prevent black from using AP to take a move in his position).
  Disadvantages of the version: the author is generally not seen in an effort to hide the use of AP in tasks (“in order not to betray anything”\_V.Keym) and the idea itself is so innovative that it could not help but pop up in discussions of the task. It is impossible to say one hundred percent, but rather no than yes. The only solution is: 1. 0-0-0 ~~ 2. h8=(Q,R)#
  
  *On the Internet with a link to (**Щорічник 1997**, 1998, page 92; Щорічник - in Ukrainian, the yearbook) it is claimed that the task was marked AP (it is indicated that the task was published in Schwalbe). The following author's solution is given:*
  
  *1.Rd1? 0-0-0 (требование ретропатом смены очередности хода; the retro stalemate requirement of changing the order of the move) 2. Са6#??*
  
  *1. 0-0-0! (отрицание смены очередности; denial of a change in the order of a move) ~ (0-0-0??) 2. h8=(Q,R)#.*
  
  А2. An even more non-standard method: "everything is possible in the past!" (that is, we must take into account all the options when, for example, we lost the right to castling just like that - in fact, the method simply prohibits castling) and then the only solution will be 1. Rd1 (That's what they think here: <https://www.janko.at/Retros/Schwalbe/Solutions.htm#9510>). But you and I remember about 1. Nc5/Ne5.
  Cons of the version: the method is so non-standard that it doesn't even have a name. Plus the duals. Absolutely not here.
  
  3. It remains to check the error in the diagram. It may consist in the fact that "black has the last move that preserves the castling." Then we have just the usual interdependence of black and white castling.
     Indirectly, this is indicated by the fact that both in the discussions and in the Turevsky database, black's defense in the form of 0-0-0 appears widely (which is impossible in the original diagram).
  
  "Проверим, проверим" (с)
  
  ```
  r3k3/2pN3P/5PPP/6pp/8/6Pp/N5P1/R3KB2 w - - 1 13
  
  ```
  
  And, by the way, this is a great reason to finally understand PRA and RS.
  
  *In the examples nos. 3–7 one can see typical common features of PRA and RS: identical retrograde analysis and partially identical solution. If two castlings exclude each other, the PRA problem has a (partial) solution for each of the two cases (both partial solutions together form the complete solution), the RS problem, however, has a complete solution for one of the two cases.*(V.Keym,2010) <https://www.janko.at/Retros/Glossary/Castling-and-En-passant.htm>
  
  If we continue further, we can simply say that PRA and RS are twin brothers (like "the party and Lenin"). They can either be together, or there is no one at all. Even more importantly, both methods use the same approach (strict compliance with 16.1 and 16.2 - that is, neither method uses the options of depriving an optimistic right without grounds or obtaining a pessimistic right without grounds - this is already to RV, LM and similar quack methods).
  
  And even more importantly, understand the Meta-Agreement here.
  
  1. First, a simple method is applied (without any methods at all)
  2. If it doesn't work out, use PRA or RV (according to the rules, PRA has priority, all other things being equal).
  
  The method is just a tool. No one can force us to use a tool unless we need it to solve a problem.
  The simple method always takes precedence over any method.
  
  ===
  
  Finally, to the point.
  
  There is no simple solution. It is necessary to use methods. First on the PRA list.
  
  ***(we remember that both methods divided the past into two realities: A - white has castling, black does not; B - black has castling, white does not. Theoretically, the possible reality of C (castling is not available to anyone) is not considered by any of the methods).***
  
  The PRA method will easily solve B. But in A we get the same thing (4 possible first moves of the solution!). There's a dead end here! The problem can be solved using the PRA method, but the solution will not satisfy other requirements for the solution (uniqueness of the solution).
  
  Let's try through RS.
  
  1. 0-0-0 (stripped black of castling) 2. h8=(Q,R)#. Everything is fine here!
  2. Rd1 (c threat h8=Q#) 0-0-0 (the only defense!) 2. Ba6# Everything is fine here too!
  
  ## But there are two solutions! But are there two?
  
  In the first case 1.0-0-0 by legalizing reality A, we have destroyed the legality of reality B (a classic of the genre!).
  And what happened in the second case? Yes, the same thing!
  
  With the move 1. Rd1, we destroyed the legality of reality A and the only legal reality remained reality B. **By breaking our castling, we proved the right to castling black!**
  
  It's beautiful! **It can be proved not only by performing castling, but also by refusing to perform castling.**
  
  And we end up with two partial solutions. One for reality A, the other for reality B. They are impossible at the same time! And, most importantly, we dispensed with the RRA so that he would be healthy! Hocus pocus!
  
  P.S. It's a shame that the task with two such brilliant ideas has remained misunderstood for incomprehensible technical reasons...
  
  
  ---
  
  