Answers.Body: <p>I think the term 'intersect' is not really intuitive here.</p>
  <p>From c2, the white king can go to d2 and c3 in one move.</p>
  <p>That means the black king must find a square from which it can reach f3 and e3 in one move. That square is f4; this would be true for e2 and f2 as well, but then the black king is outside the pawn's square and it can run to promotion.</p>
  <p>It doesn't matter that it's not the same geometrical shape on the chessboard. (Actually, it <em>is</em> the same geometrical shape if you switch to the <a href="https://en.wikipedia.org/wiki/Chebyshev_distance" rel="noreferrer">Chebyshev distance metric</a>, but I digress.)</p>
  
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Answers.CreationDate: 2021-04-02T15:55:21.957Z
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Body: <p>I saw <a href="https://chess.stackexchange.com/questions/17596/identifying-corresponding-squares">this</a> question, but it does not answer my question. I was following a Dvoretsky endgame manual and came across this diagram.</p>
  <p><a href="https://i.sstatic.net/X38wJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/X38wJ.png" alt="enter image description here" /></a></p>
  <p>Now, I understand why c5 and c7 are corresponding squares and also the same for d6 and d8. I also understood how d6 and c5 intersect at d5 (going from those squares through files and ranks by straight lines and find intersection square) and how d8 and c7 intersect at c8, so the conclusion is that d5 and c8 are corresponding squares.</p>
  <p>A few pages later, Dvoretsky introduces this diagram:</p>
  <p><a href="https://i.sstatic.net/SPgUh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/SPgUh.png" alt="enter image description here" /></a></p>
  <p>I understand why d2 and f3 are corresponding squares and the same for e3 and c3. But Dvoretsky then concludes that based on those 2 pairs of corresponding squares we can conclude with an intersection that c2 and f4 are corresponding squares. How is that? d2 and c3 do intersect at c2. But how do e3 and f3 intersect at f4?</p>
  
Tags: <calculation><pawn-endgame>
Title: How to calculate corresponding squares
ThreadText: # How to calculate corresponding squares
  
  (Asked by: username_0 on 2021-04-02)
  
  I saw [this](https://chess.stackexchange.com/questions/17596/identifying-corresponding-squares) question, but it does not answer my question. I was following a Dvoretsky endgame manual and came across this diagram.
  
  [![enter image description here](https://i.sstatic.net/X38wJ.png)](https://i.sstatic.net/X38wJ.png)
  
  Now, I understand why c5 and c7 are corresponding squares and also the same for d6 and d8. I also understood how d6 and c5 intersect at d5 (going from those squares through files and ranks by straight lines and find intersection square) and how d8 and c7 intersect at c8, so the conclusion is that d5 and c8 are corresponding squares.
  
  A few pages later, Dvoretsky introduces this diagram:
  
  [![enter image description here](https://i.sstatic.net/SPgUh.png)](https://i.sstatic.net/SPgUh.png)
  
  I understand why d2 and f3 are corresponding squares and the same for e3 and c3. But Dvoretsky then concludes that based on those 2 pairs of corresponding squares we can conclude with an intersection that c2 and f4 are corresponding squares. How is that? d2 and c3 do intersect at c2. But how do e3 and f3 intersect at f4?
  
  **Tags:** calculation, pawn-endgame
  
  
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  ## 1 answers
  
  ### Answer by username_1 on 2021-04-02. Score: 6 | Accepted answer
  
  I think the term 'intersect' is not really intuitive here.
  
  From c2, the white king can go to d2 and c3 in one move.
  
  That means the black king must find a square from which it can reach f3 and e3 in one move. That square is f4; this would be true for e2 and f2 as well, but then the black king is outside the pawn's square and it can run to promotion.
  
  It doesn't matter that it's not the same geometrical shape on the chessboard. (Actually, it *is* the same geometrical shape if you switch to the [Chebyshev distance metric](https://en.wikipedia.org/wiki/Chebyshev_distance), but I digress.)
  
  
  ---
  
  