Answers.Body: <p>There is a crucial part missing from the definition. <em>To have the opposition you must have an odd number of squares between the kings when it is your opponent's turn to move.</em></p>
  <p>This is obvious when you think about it. Suppose you move your king to be diagonally opposite your opponent's king on a square of side 4. The two kings are on the same diagonal and so on the same colour. Then there are two squares between the kings and your opponent can simply move the king one square diagonally closer to your king giving your opponent the opposition.</p>
  <p>According to Silman's quoted rule each player had the opposition on their turn. This is nonsense. When you moved your king onto the same coloured square diagonally opposite your opponent's king but two squares away you lost the opposition.</p>
  <p>There is a good YouTube video by Hanging Pawns on the opposition <a href="https://youtu.be/SoZPZdnYYk8?t=474" rel="nofollow noreferrer">here</a>.</p>
  
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Answers.CreationDate: 2021-06-16T20:50:34.170Z
Answers.Text: Silman primarily gives the "odd number of squares" definition but offers this alternative that may be easier to see when the kings are far apart.  "...the rule is to make a square or a rectangle in which each corner is the same color with the other guy to move."  Silman's Complete Endgame Course - First Edition - page 58
Answers.UserDisplayName: Michael West
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Answers.Text: "According to Silman's quoted rule each player had the opposition on their turn." - No, I think you misunderstand (as I did when I first read the book.) You have to look at *all* the corners of the rectangle, not just the two the kings are on. If the kings are on, say, a1 and e5, the corners are a1, e1, e5, and a5, and all of those are black. After Kb2, the corners are b2, e2, b5, and e5, two of which are white.
Answers.UserDisplayName: D M
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Answers.LastActivityDate: 2021-06-16T17:08:29.240Z
Answers.UserDisplayName: Brian Towers
Answers.Body: <blockquote>
  <p>the rule is to make a square or rectangle in which each color is the
  same color with the other guy to move.</p>
  </blockquote>
  <p>I'm not quite sure how this sentence makes sense other than there is a typo, and it actually says &quot;...rectangle in which each <strong>corner</strong> is the same color with the other guy to move&quot;. If this is indeed the case, then obviously b4-d4-d8-b8 works and c4-d4-d7-c7 doesn't (two white and two black corners).</p>
  <p>An example to this would be if it's white to move, they play 1. Kd4 and black plays either Kd8 (direct long opposition) or Kb8 (indirect, rectangular opposition).</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-06-16T20:52:10.153Z
Answers.Text: That is the correct quote from my book.  So the example proves it rather than contradicts it.
Answers.UserDisplayName: Michael West
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Answers.CreationDate: 2021-06-16T23:46:15.367Z
Answers.Text: Ah okay, you clarified it for me. He meant all *four* corners need to be the same color, the wording somehow made me assume it was just the two the kings were sitting on.
Answers.UserDisplayName: James Ko
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Answers.UserDisplayName: sleepy
Body: <p>In <em>Silman's Complete Endgame Course</em>, the author gives this rule in order to gain the Opposition when there isn't a direct connection in between the kings:</p>
  <blockquote>
  <p>In this type of situation, the rule is to make a square or rectangle in which each color is the same color <em>with the other guy to move.</em></p>
  </blockquote>
  <p>But later in the book, this practice problem is given which seems to contradict this:</p>
  <pre><code>[FEN &quot;2k5/8/8/8/1PK5/8/8/8 b - - 0 1&quot;]
  
  1...Kb8 2.Kd4 Kc7?? 3.Kc5
  </code></pre>
  <blockquote>
  <p>2.Kd4 Kb7 with a draw (and not 2.Kc7?? 3.Kc5 when White gets the opposition and wins).</p>
  </blockquote>
  <p>Black moved his king to a square of the same color as White's king (forming a rectangle with corners c7-d7-d4-c4), but White still got the Opposition and won. Doesn't this invalidate the rule given above?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-06-16T18:07:11.140Z
Comments.Text: Looks like a mis-quote.  Should be "in which each corner"  I belive
Comments.UserDisplayName: Michael West
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-06-16T21:34:12.363Z
Comments.Text: Yes, I have the book right here, and it says "in which each corner is the same color".
Comments.UserDisplayName: D M
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-06-16T23:13:48.060Z
Comments.Text: He's just using bad wording. It should read "*all four corners* are the same color" which as discussed below is equivalent to a square or rect of odd dimensions (3x3, 5x5, 7x3 etc.)
Comments.UserDisplayName: BaseZen
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-06-17T10:00:01.200Z
Comments.Text: In addition to the answers already given, note that in the diagram above with Black to move, White *has* the opposition but can't *keep* it after Kb8 since the P is in the way. Opposition is merely a tool to penetrate with your king.
Comments.UserDisplayName: Hauke Reddmann
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-11-17T22:29:57.777Z
Comments.Text: There's an excellent book titled "The Final Countdown" (Hejenius & van Riemsdijk) which might help. It blends all the varieties of "opposition" into a set of corresponding squares to govern moves in a pawn endgame. In essence showing that opposition is merely a subset of the corresponding squares.
Comments.UserDisplayName: Arlen
Tags: <endgame><books><opposition>
Title: Description of rectangular opposition in Silman's Complete Endgame Course
ThreadText: # Description of rectangular opposition in Silman's Complete Endgame Course
  
  (Asked by: username_0 on 2021-06-16)
  
  In *Silman's Complete Endgame Course*, the author gives this rule in order to gain the Opposition when there isn't a direct connection in between the kings:
  
  > In this type of situation, the rule is to make a square or rectangle in which each color is the same color *with the other guy to move.*
  
  But later in the book, this practice problem is given which seems to contradict this:
  
  ```
  [FEN "2k5/8/8/8/1PK5/8/8/8 b - - 0 1"]
  
  1...Kb8 2.Kd4 Kc7?? 3.Kc5
  
  ```
  
  > 2.Kd4 Kb7 with a draw (and not 2.Kc7?? 3.Kc5 when White gets the opposition and wins).
  
  Black moved his king to a square of the same color as White's king (forming a rectangle with corners c7-d7-d4-c4), but White still got the Opposition and won. Doesn't this invalidate the rule given above?
  
  **Tags:** endgame, books, opposition
  
  > **Comments:**
  > **username_1**: Looks like a mis-quote.  Should be "in which each corner"  I belive  
  > **username_2**: Yes, I have the book right here, and it says "in which each corner is the same color".  
  > **username_3**: He's just using bad wording. It should read "*all four corners* are the same color" which as discussed below is equivalent to a square or rect of odd dimensions (3x3, 5x5, 7x3 etc.)  
  > **username_4**: In addition to the answers already given, note that in the diagram above with Black to move, White *has* the opposition but can't *keep* it after Kb8 since the P is in the way. Opposition is merely a tool to penetrate with your king.  
  > **username_5**: There's an excellent book titled "The Final Countdown" (Hejenius & van Riemsdijk) which might help. It blends all the varieties of "opposition" into a set of corresponding squares to govern moves in a pawn endgame. In essence showing that opposition is merely a subset of the corresponding squares.  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_7 on 2021-06-16. Score: 7 | Accepted answer
  
  > the rule is to make a square or rectangle in which each color is the
  > same color with the other guy to move.
  
  I'm not quite sure how this sentence makes sense other than there is a typo, and it actually says "...rectangle in which each **corner** is the same color with the other guy to move". If this is indeed the case, then obviously b4-d4-d8-b8 works and c4-d4-d7-c7 doesn't (two white and two black corners).
  
  An example to this would be if it's white to move, they play 1. Kd4 and black plays either Kd8 (direct long opposition) or Kb8 (indirect, rectangular opposition).
  
  > **Comments:**
  > **username_1**: That is the correct quote from my book.  So the example proves it rather than contradicts it.  
  > **username_0**: Ah okay, you clarified it for me. He meant all *four* corners need to be the same color, the wording somehow made me assume it was just the two the kings were sitting on.  
  
  ---
  
  ### Answer by username_6 on 2021-06-16. Score: 4
  
  There is a crucial part missing from the definition. *To have the opposition you must have an odd number of squares between the kings when it is your opponent's turn to move.*
  
  This is obvious when you think about it. Suppose you move your king to be diagonally opposite your opponent's king on a square of side 4. The two kings are on the same diagonal and so on the same colour. Then there are two squares between the kings and your opponent can simply move the king one square diagonally closer to your king giving your opponent the opposition.
  
  According to Silman's quoted rule each player had the opposition on their turn. This is nonsense. When you moved your king onto the same coloured square diagonally opposite your opponent's king but two squares away you lost the opposition.
  
  There is a good YouTube video by Hanging Pawns on the opposition [here](https://youtu.be/SoZPZdnYYk8?t=474).
  
  > **Comments:**
  > **username_1**: Silman primarily gives the "odd number of squares" definition but offers this alternative that may be easier to see when the kings are far apart.  "...the rule is to make a square or a rectangle in which each corner is the same color with the other guy to move."  Silman's Complete Endgame Course - First Edition - page 58  
  > **username_2**: "According to Silman's quoted rule each player had the opposition on their turn." - No, I think you misunderstand (as I did when I first read the book.) You have to look at *all* the corners of the rectangle, not just the two the kings are on. If the kings are on, say, a1 and e5, the corners are a1, e1, e5, and a5, and all of those are black. After Kb2, the corners are b2, e2, b5, and e5, two of which are white.  
  
  ---
  
  