Answers.Body: <p>In this situation, assuming the pairings are possible, #1 would play Black against #4 while #2 would play White against #3.</p>
  
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Answers.Text: Yes, but I think perhaps you should say why?
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Answers.Body: <p>The FIDE algorithm (I assume the standard is used) can be <a href="https://handbook.fide.com/chapter/C0403" rel="nofollow noreferrer">read online</a>. It is deterministic. I can think only of one worst case: No <a href="https://handbook.fide.com/chapter/C0402" rel="nofollow noreferrer">initial ranking</a> can be done as the players have same name and same rating :-)
  Considering the second part of your question, note that other Swiss variants exist.</p>
  
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Answers.Body: <blockquote>
  <p>There are four people currently tied with the most points, where #1 and #3 have just played White (for example), while #2 and #4 have just played Black.</p>
  </blockquote>
  <p>All else being equal, the pairings would likely be 1 vs 4 and 2 vs 3. This is, however, not guaranteed.</p>
  <p>The pairing rules for the commonly used Dutch system are <a href="https://handbook.fide.com/chapter/C0403" rel="noreferrer">here</a>. Let's walk through rule B (Pairing Process for a bracket) to see what happens.</p>
  <p>Rule B.2 says:</p>
  <blockquote>
  <p>To make the pairing, each bracket will be usually divided into two subgroups, called S1 and S2.<br />
  S1 initially contains the highest N1 players (sorted according to A.2), where N1 is either M1 (in a heterogeneous bracket) or MaxPairs (otherwise).<br />
  S2 initially contains all the remaining resident players.</p>
  </blockquote>
  <p>This is a fancy way of saying that the group is split into an upper and lower half. In your scenario, #1 and #2 would be in the first half while #3 and and #4 would be in the second half.</p>
  <p>Then, rule B.3 says:</p>
  <blockquote>
  <p>S1 players are tentatively paired with S2 players, the first one from S1 with the first one from S2, the second one from S1 with the second one from S2 and so on.</p>
  </blockquote>
  <p>So the tentative parings would be 1 vs 3 and 2 vs 4.</p>
  <p>Rule B.4 says we should then evaluate how this tentative pairing complies with the various criteria in rule C. When we do so, we find that this pairing violates the quality criteria C.10 &quot;minimize the number of players who do not get their colour preference.&quot; Since the candidate is not perfect, rule B.4 says we should apply rule B.5. Rule B.5 in turn says to apply rule B.6. (This is a homogeneous bracket, since everyone has the same score.)</p>
  <p>Rule B.6 says we should first try a transposition. A transposition means the order of the players in the second half are changed. Since there are only two players in the second half, only one transposition is possible and we can pretty much skip the rules saying which transpositions should be considered first. So the new tentative pairing is 1 vs 4 and 2 vs 3.</p>
  <p>As it turns out, this transposition results in a perfect candidate. So we stop there. The pairings would be 1 vs 4 and 2 vs 3.</p>
  <p>But there are several reasons why the pairings might not go this way:</p>
  <ul>
  <li>The players might have played each other in an earlier round. It is prohibited for players to play the same opponent twice in the same Swiss, so some other pairing would have to be generated.</li>
  <li>A player who played Black last round might actually be due Black again due to playing White twice in a row earlier (or vice versa.) Having a different color preference could result in different pairings.</li>
  <li>A problem with pairing the lower score groups could mean that the top group must be collapsed into the lower groups to be able to generate legal pairings. The group would then be entirely re-paired, which would almost certainly result in different pairings.</li>
  <li>Some other system besides the Dutch system is being used. In this case you'd obviously have to read the rules for that system instead.</li>
  </ul>
  
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Answers.Text: Well documented and clearly written! It's important to note however, that whilest it is to be prevented, it is not forbidden (and near impossible to prevent that) a player has the same colour twice in a row.  
  This is explained in `A.6`: the colour difference (e.g. 3 white plays and 4 black plays = +1) must vary between -1 and +1. 
  So you could imagine a 4B-3W=+1 : white paring is possible >which creates> 4B-4W=0: white pairing is possible >which creates> 4B-5W=-1 : an extra white paring would make -2, so a black pairing is now required.
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Answers.Text: Could you explain this E.2:
  "E.2 Grant the stronger colour preference. If both are absolute (topscorers, see A.7) grant the wider colour difference (see A.6)." I can understand A.7 and A.6, and colour difference but what is "grant the wider colour difference" exactly?
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Answers.Text: I am also thinking if this is the last round, it is better to pair #1B vs #2W, because they have the same score and color preferences are meet and the most important is to match the top players based from initial ranking or seeding.
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Answers.Text: @ferdy A player with XWBB has an absolute preference for White since he's just played Black twice in a row. A player with BBWB also has an absolute preference for White. But the first player has a difference of 1 while the second has a difference of 2, so player 2 would get their preference if they had to play each other.
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Answers.Text: @ferdy As for pairing #1 and #2 in the last round, that's an opinion the Dutch system doesn't share. If there's a winner in both games there will be a tie for first no matter how you pair them anyway; is it really better for that tie to be between #1 and #3 instead of #1 and #2?
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Answers.Text: @DM On "But the first player has a difference of 1 while the second has a difference of 2", that is technically wrong. The second player BBWB has a colour difference of -2, the formula to get the colour difference is `white - black`. `W=1, B=3`, `1 - 3 is -2`. What is `XWBB`?
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Answers.Text: OK, fine, -2. "X" is an unplayed game, such as a bye or forfeit win.
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Body: <p>Say we have a situation that's not super uncommon in swiss tournaments. There are four people currently tied with the most points, where #1 and #3 have just played White (for example), while #2 and #4 have just played Black. Everything else being equal, do FIDE rules say anything about what the pairings should be here? Information about any rules from individual federations is also welcome.</p>
  
Tags: <rules><fide><swiss-pairing>
Title: Correct swiss pairings for a group of 4 players
ThreadText: # Correct swiss pairings for a group of 4 players
  
  (Asked by: username_0 on 2023-06-30)
  
  Say we have a situation that's not super uncommon in swiss tournaments. There are four people currently tied with the most points, where #1 and #3 have just played White (for example), while #2 and #4 have just played Black. Everything else being equal, do FIDE rules say anything about what the pairings should be here? Information about any rules from individual federations is also welcome.
  
  **Tags:** rules, fide, swiss-pairing
  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_2 on 2023-07-20. Score: 5 | Accepted answer
  
  > There are four people currently tied with the most points, where #1 and #3 have just played White (for example), while #2 and #4 have just played Black.
  
  All else being equal, the pairings would likely be 1 vs 4 and 2 vs 3. This is, however, not guaranteed.
  
  The pairing rules for the commonly used Dutch system are [here](https://handbook.fide.com/chapter/C0403). Let's walk through rule B (Pairing Process for a bracket) to see what happens.
  
  Rule B.2 says:
  
  > To make the pairing, each bracket will be usually divided into two subgroups, called S1 and S2.  
  > S1 initially contains the highest N1 players (sorted according to A.2), where N1 is either M1 (in a heterogeneous bracket) or MaxPairs (otherwise).  
  > S2 initially contains all the remaining resident players.
  
  This is a fancy way of saying that the group is split into an upper and lower half. In your scenario, #1 and #2 would be in the first half while #3 and and #4 would be in the second half.
  
  Then, rule B.3 says:
  
  > S1 players are tentatively paired with S2 players, the first one from S1 with the first one from S2, the second one from S1 with the second one from S2 and so on.
  
  So the tentative parings would be 1 vs 3 and 2 vs 4.
  
  Rule B.4 says we should then evaluate how this tentative pairing complies with the various criteria in rule C. When we do so, we find that this pairing violates the quality criteria C.10 "minimize the number of players who do not get their colour preference." Since the candidate is not perfect, rule B.4 says we should apply rule B.5. Rule B.5 in turn says to apply rule B.6. (This is a homogeneous bracket, since everyone has the same score.)
  
  Rule B.6 says we should first try a transposition. A transposition means the order of the players in the second half are changed. Since there are only two players in the second half, only one transposition is possible and we can pretty much skip the rules saying which transpositions should be considered first. So the new tentative pairing is 1 vs 4 and 2 vs 3.
  
  As it turns out, this transposition results in a perfect candidate. So we stop there. The pairings would be 1 vs 4 and 2 vs 3.
  
  But there are several reasons why the pairings might not go this way:
  
  * The players might have played each other in an earlier round. It is prohibited for players to play the same opponent twice in the same Swiss, so some other pairing would have to be generated.
  * A player who played Black last round might actually be due Black again due to playing White twice in a row earlier (or vice versa.) Having a different color preference could result in different pairings.
  * A problem with pairing the lower score groups could mean that the top group must be collapsed into the lower groups to be able to generate legal pairings. The group would then be entirely re-paired, which would almost certainly result in different pairings.
  * Some other system besides the Dutch system is being used. In this case you'd obviously have to read the rules for that system instead.
  
  > **Comments:**
  > **username_4**: Well documented and clearly written! It's important to note however, that whilest it is to be prevented, it is not forbidden (and near impossible to prevent that) a player has the same colour twice in a row.   This is explained in `A.6`: the colour difference (e.g. 3 white plays and 4 black plays = +1) must vary between -1 and +1.  So you could imagine a 4B-3W=+1 : white paring is possible >which creates> 4B-4W=0: white pairing is possible >which creates> 4B-5W=-1 : an extra white paring would make -2, so a black pairing is now required.  
  > **username_5**: Could you explain this E.2: "E.2 Grant the stronger colour preference. If both are absolute (topscorers, see A.7) grant the wider colour difference (see A.6)." I can understand A.7 and A.6, and colour difference but what is "grant the wider colour difference" exactly?  
  > **username_5**: I am also thinking if this is the last round, it is better to pair #1B vs #2W, because they have the same score and color preferences are meet and the most important is to match the top players based from initial ranking or seeding.  
  > **username_2**: @username_5 A player with XWBB has an absolute preference for White since he's just played Black twice in a row. A player with BBWB also has an absolute preference for White. But the first player has a difference of 1 while the second has a difference of 2, so player 2 would get their preference if they had to play each other.  
  > **username_2**: @username_5 As for pairing #1 and #2 in the last round, that's an opinion the Dutch system doesn't share. If there's a winner in both games there will be a tie for first no matter how you pair them anyway; is it really better for that tie to be between #1 and #3 instead of #1 and #2?  
  > **username_5**: @DM On "But the first player has a difference of 1 while the second has a difference of 2", that is technically wrong. The second player BBWB has a colour difference of -2, the formula to get the colour difference is `white - black`. `W=1, B=3`, `1 - 3 is -2`. What is `XWBB`?  
  > **username_2**: OK, fine, -2. "X" is an unplayed game, such as a bye or forfeit win.  
  
  ---
  
  ### Answer by username_1 on 2023-06-30. Score: 1
  
  In this situation, assuming the pairings are possible, #1 would play Black against #4 while #2 would play White against #3.
  
  > **Comments:**
  > **username_2**: Yes, but I think perhaps you should say why?  
  
  ---
  
  ### Answer by username_3 on 2023-06-30. Score: 1
  
  The FIDE algorithm (I assume the standard is used) can be [read online](https://handbook.fide.com/chapter/C0403). It is deterministic. I can think only of one worst case: No [initial ranking](https://handbook.fide.com/chapter/C0402) can be done as the players have same name and same rating :-)
  Considering the second part of your question, note that other Swiss variants exist.
  
  
  ---
  
  