Answers.Body: <p>Although not a mathematician, my answer is the following.
  Consider the chessboard as having Cartesian coordinates, such as columns 0 to 7 and rows 0 to 7. Similar reasoning applies if 1 to 8 is chosen. The position of a king can be represented by a two item tuple, column first then row. Kings on opposite ends of a long diagonal would be W(0,0) and B(7,7) .
  In the above formulation the kings will be in opposition if the differences between their corresponding tuple items are both even. Consider zero as even and both differences being zero is prohibited as they cannot occupy the same square. This meets the definition A in the query. For example with W(0,0) &amp; B(2,4) and it's Black's move, White ( side not to move ) has the opposition.
  If the Black king moves he must change either item by one or both. White simple changes the corresponding item ( coordinate ) or both. If the White king is in a corner he can still change each coordinate, or both, by one. This satisfies B in the query. The other king can always walk back into opposition.</p>
  
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Answers.CreationDate: 2022-11-09T06:33:13.240Z
Answers.Text: Can you add some graphics.. Your answer is very interesting one.
Answers.UserDisplayName: ShadYantra
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Answers.CreationDate: 2022-11-09T09:50:33.033Z
Answers.Text: @dlemper: I was about to give the same answer today, but a) either more formal, or with a graphic as suggested, and b) you *must* also consider the case when the white king is restricted by board and/or enemy king for a *complete* proof. But this is easy subcase splitting, so feel free to edit it in.
Answers.UserDisplayName: Hauke Reddmann
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Answers.CreationDate: 2022-11-08T21:36:36.503Z
Answers.LastActivityDate: 2022-11-08T21:36:36.503Z
Answers.UserDisplayName: dlemper
Answers.Body: <p>Two kings are in <strong>direct opposition</strong> if they're on the same row or column separated by just one square. Two kings are in <strong>diagonal opposition</strong> if they're facing each other diagonally with one square in between. In both cases the player not having the turn has the opposition.</p>
  <p><strong>Indirect opposition</strong> is harder to define because different people will mean different things depending on context. The most practical definition is probably &quot;a player has the indirect opposition if he can force his opponent to concede the opposition.</p>
  <p>As you say in the question, the rectangle rule can give us a hint but squares defended by pawns can make a player lose this type of opposition.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2022-11-09T15:00:11.937Z
Answers.Text: See my edit; in case more stuff is on the board "opposition" may be meaningless and the theory of "corresponding squares" is needed anyway.
Answers.UserDisplayName: Hauke Reddmann
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Answers.CreationDate: 2022-11-09T11:20:05.637Z
Answers.LastActivityDate: 2022-11-09T11:20:05.637Z
Answers.UserDisplayName: David
Body: <p>This just came up in my youth chess training.</p>
  <p>A very <a href="https://chess.stackexchange.com/questions/35349/description-of-rectangular-opposition-in-silmans-complete-endgame-course">general</a> definition of &quot;having the opposition&quot;, including even rare cases, is</p>
  <ul>
  <li>A: All corners of the (axis-parallel) rectangle spanned by the two kings have the same color, the opponent has the move.</li>
  </ul>
  <p>A more &quot;operational&quot; description of &quot;having the opposition&quot; is</p>
  <ul>
  <li>B: Regardless how the opponent king walks out of opposition, you can walk in again.</li>
  </ul>
  <p>Prove A=&gt;B. (Danger, the board has edges, so you have to include in your proof that the mobility of your king may be limited!)</p>
  <p>EDIT: Of course, see comment below, it is assumed that the board is empty except the two kings. Otherwise &quot;opposition&quot; might be meaningless due to inaccessibility of squares to the kings, and the more general theory of &quot;corresponding squares&quot; sets in.</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T09:15:58.590Z
Comments.Text: Karsten Mueller (both a GM and a PhD in math) offers the same conclusion as your A statement: https://www.youtube.com/watch?v=QNYaY3lm-ds. However, he doesn't prove A=>B.
Comments.UserDisplayName: Inertial Ignorance
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T10:53:01.827Z
Comments.Text: "All **squares** of the rectangle ... have the same colour" is impossible. You mean "All **corners**, etc."
Comments.UserDisplayName: Evargalo
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T11:33:04.917Z
Comments.Text: @Evargalo: Argl. Edited. <facepalm>
Comments.UserDisplayName: Hauke Reddmann
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T14:27:32.043Z
Comments.Text: According to this definition, with kings on h8 and h6, does someone have the opposition?
Comments.UserDisplayName: NoseKnowsAll
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T14:40:04.873Z
Comments.Text: @NoseKnowsAll You could interpret the "four" corners as being h6, h8, h6, h8. In which case, yes someone has the opposition since all corners are dark. The rectangle here would have a length of 3 and width of 1.
Comments.UserDisplayName: Inertial Ignorance
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Comments.CreationDate: 2022-11-02T15:06:00.047Z
Comments.Text: @InertialIgnorance if that's true, then isn't than an example when A is true and B is not? If the king must go to g8, the other king is not "walking in" since there is no i7 square.
Comments.UserDisplayName: NoseKnowsAll
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T15:35:10.080Z
Comments.Text: @NoseKnowsAll Oh I interpreted B to mean that you can maintain the opposition. But for either interpretation of B, I'm sure you could find endless counterexamples where A is true but B is false. E.g., what if there are additional pieces that control key squares.
Comments.UserDisplayName: Inertial Ignorance
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-02T18:30:20.003Z
Comments.Text: No "walking in" required, and of course the board is assumed to be empty with only the two kings! And yes, h8/h6 is a 3:1 rectangle and counts.
Comments.UserDisplayName: Hauke Reddmann
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2022-11-09T01:48:11.947Z
Comments.Text: I think B->A is interesting, except it won’t apply on a 2xn board
Comments.UserDisplayName: Laska
Tags: <mathematics><opposition>
Title: Definition of opposition
ThreadText: # Definition of opposition
  
  (Asked by: username_0 on 2022-11-02)
  
  This just came up in my youth chess training.
  
  A very [general](https://chess.stackexchange.com/questions/35349/description-of-rectangular-opposition-in-silmans-complete-endgame-course) definition of "having the opposition", including even rare cases, is
  
  * A: All corners of the (axis-parallel) rectangle spanned by the two kings have the same color, the opponent has the move.
  
  A more "operational" description of "having the opposition" is
  
  * B: Regardless how the opponent king walks out of opposition, you can walk in again.
  
  Prove A=>B. (Danger, the board has edges, so you have to include in your proof that the mobility of your king may be limited!)
  
  EDIT: Of course, see comment below, it is assumed that the board is empty except the two kings. Otherwise "opposition" might be meaningless due to inaccessibility of squares to the kings, and the more general theory of "corresponding squares" sets in.
  
  **Tags:** mathematics, opposition
  
  > **Comments:**
  > **username_1**: Karsten Mueller (both a GM and a PhD in math) offers the same conclusion as your A statement: https://www.youtube.com/watch?v=QNYaY3lm-ds. However, he doesn't prove A=>B.  
  > **username_2**: "All **squares** of the rectangle ... have the same colour" is impossible. You mean "All **corners**, etc."  
  > **username_0**: @username_2: Argl. Edited. <facepalm>  
  > **username_3**: According to this definition, with kings on h8 and h6, does someone have the opposition?  
  > **username_1**: @username_3 You could interpret the "four" corners as being h6, h8, h6, h8. In which case, yes someone has the opposition since all corners are dark. The rectangle here would have a length of 3 and width of 1.  
  > **username_3**: @InertialIgnorance if that's true, then isn't than an example when A is true and B is not? If the king must go to g8, the other king is not "walking in" since there is no i7 square.  
  > **username_1**: @username_3 Oh I interpreted B to mean that you can maintain the opposition. But for either interpretation of B, I'm sure you could find endless counterexamples where A is true but B is false. E.g., what if there are additional pieces that control key squares.  
  > **username_0**: No "walking in" required, and of course the board is assumed to be empty with only the two kings! And yes, h8/h6 is a 3:1 rectangle and counts.  
  > **username_4**: I think B->A is interesting, except it won’t apply on a 2xn board  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_5 on 2022-11-08. Score: 2
  
  Although not a mathematician, my answer is the following.
  Consider the chessboard as having Cartesian coordinates, such as columns 0 to 7 and rows 0 to 7. Similar reasoning applies if 1 to 8 is chosen. The position of a king can be represented by a two item tuple, column first then row. Kings on opposite ends of a long diagonal would be W(0,0) and B(7,7) .
  In the above formulation the kings will be in opposition if the differences between their corresponding tuple items are both even. Consider zero as even and both differences being zero is prohibited as they cannot occupy the same square. This meets the definition A in the query. For example with W(0,0) & B(2,4) and it's Black's move, White ( side not to move ) has the opposition.
  If the Black king moves he must change either item by one or both. White simple changes the corresponding item ( coordinate ) or both. If the White king is in a corner he can still change each coordinate, or both, by one. This satisfies B in the query. The other king can always walk back into opposition.
  
  > **Comments:**
  > **username_6**: Can you add some graphics.. Your answer is very interesting one.  
  > **username_0**: @username_5: I was about to give the same answer today, but a) either more formal, or with a graphic as suggested, and b) you *must* also consider the case when the white king is restricted by board and/or enemy king for a *complete* proof. But this is easy subcase splitting, so feel free to edit it in.  
  
  ---
  
  ### Answer by username_7 on 2022-11-09. Score: 1
  
  Two kings are in **direct opposition** if they're on the same row or column separated by just one square. Two kings are in **diagonal opposition** if they're facing each other diagonally with one square in between. In both cases the player not having the turn has the opposition.
  
  **Indirect opposition** is harder to define because different people will mean different things depending on context. The most practical definition is probably "a player has the indirect opposition if he can force his opponent to concede the opposition.
  
  As you say in the question, the rectangle rule can give us a hint but squares defended by pawns can make a player lose this type of opposition.
  
  > **Comments:**
  > **username_0**: See my edit; in case more stuff is on the board "opposition" may be meaningless and the theory of "corresponding squares" is needed anyway.  
  
  ---
  
  