I was mistaken about there being a top down vs. lookahead option in SwissSys. I think that option existed in an earlier version of SwissSys but it’s either been removed or it’s been replaced by in-depth searches. You have the option enabled and it’s enabled by default, which is good (I think).
I do think there’s a problem with SwissSys’s pairing logic, and that you should report this to Thad Suits. The pairing that SwissSys produced in this section violates rule 29D1a.
They 80/200 limits are limits on transpositions and interchanges. I assumed that “turning them off” meant making it so that there are no limits. When there are no limits, any transposition or interchange within a score group which improves colors is allowed. Making the “limit” infinity (or 3000) is equivalent to disabling the limits. The opposite direction from turning them off is making them stricter, ultimately arriving at completely disallowing transpositions and interchanges for color, unless the two players involved have the same rating. (This is presumably what setting them to zero does).
Even when there are no 80/200 limits on transpositions and interchanges (as under FIDE rules), it is not as if anything goes. There are other inherent limits. Transpositions and interchanges still have to be within score groups. Between two pairings which have “equally good” color allocations, the one with fewer transpositions and interchanges is better. Transpositions are better than interchanges. It would be great if the USCF rules quantified some of these things. How many transpositions equals an interchange? Or is one interchange worse than any number of transpositions? etc.
I’ve just thought of something that might have caused the problem. Brian, did you have team pairings enabled?
That is what I meant: disable the arbitrary rating limits, as in set the limits to 3000 or infinity. I have found that makes it easier to pair small one-section Swisses, with players of a wide but not absurd ratings-spread.
In round four of a five-round Swiss with 12-14 players, let the TD decide if a 203-point switch to equalize colors at 2-2 each makes sense. For the kinds of events I (and perhaps you) direct, it often does.
For a larger tournament or section, or one with real money on the line, being chased by hungry chess lawyers…maybe less so. As in: For such events it is good to set some arbitrary standard; maybe.
However, that was not the issue in your tournament.
As for where the rulebook says you can ignore the 200/80-point limits: First see 1A. Then take a breath and think about the importance of Swiss pairing refinements vs. the hard rules that govern OTB play. Any algorithm that covers all potential Swiss quirks would likely make the software go tilt. Imagine the brain of a poor over-worked TD.
For example, in the case I cited above—round four of a small Swiss:
If you stick with the 200-point limit and give player X 3-1 colors, that player’s color is set in stone for the last round. (See if you find the place where the rulebook says score trumps 4-1 or 1-4 colors—with the ‘odd’ color in round three—or vice versa.)
That choice could happen in the last round of a small Swiss; very rarely, but it’s possible. Imagine trying to explain four Blacks out of five to a mild-mannered patzer/parent: “Cuz the rulebook says so.” Expanding the transposition rating limits to avoid such scenes is an example of pro-actively using the less quirky quirk.
Swiss pairings are messy more often than not. Try pairing by hand a few times, in small events with time between rounds. It’s eye-opening fun.
They were disabled.
Let me ask another one. I recall this was a 10 player event. Were the other four players withdrawn? I have seen strange board one pairings that didn’t make sense until I saw that the alternate created a problem on a much lower board.
All 10 players were paired for Round 4. Three of the other four players not shown had 1 point each. There was one player with 0 points. When I force the Alpha-Charlie pairing, and let SwissSys pair the rest, nothing bad happens. There is one player who does not equalize on colors, ending up with WBWW.
I went back and recreated the full tournament wallchart by hand, just as an exercise. (Slow work day.) I wanted to see exactly what the standings and colors (as close as possible, anyway) looked like after three rounds. (I’m assuming I found the right crosstable in MSA; if I didn’t, I just wasted about 45 minutes. 
)
It appears that SwissSys used top-down pairings for this, and simply did not examine the possibility of changing the pairing on board 1 once it was set. Here’s the tournament wallchart (colors included; some color assignments are deduced). Each round is listed by result, color, and pairing number of opponent. My advance apologies for lack of formatting.
It’s worth noting that the fourth round game on board 5 actually was recorded as a forfeit win. However, I don’t think these two players would have affected the higher boards.
Player 1: +b6, +w4, =b2, -b5
Player 2: +w7, +b3, =w1, +b4
Player 3: +b8, -w2, -w5, +b7
Player 4: +w9, -b1, -w6, -w2
Player 5: +b10, +w6, +b3, +w1
Player 6: -w1, -b5, +b4, -w8
Player 7: -b2, +w10, +b9, -w3
Player 8: -w3, -b9, +w10, -b6
Player 9: -b4, +w8, -w7, +b10
Player 10: -w5, -b7, -b8, -w9
It appears the “natural” (ignoring all possible restrictions) last round pairings, in descending board order, are: 5 - 1, 2 - 7, 6 - 3, 4 - 8, and 10 - 9. Of these pairings, the only one that doesn’t work is 2 - 7, as they’ve already played. So, my instinct is to switch players 1 and 2 (which also improves those two players’ color allocation), and leave the rest of the pairings alone. The two players I want to switch are close in rating, so that isn’t an issue.
Instead, the program cranked out the following pairings: 5 - 1, 4 - 2, 7 - 3, 6 - 8, 10 - 9.
This looks like a case where SwissSys didn’t do an optimal job. If I had to guess, I’d say the program simply pulled players 1 and 5 out of the refinement process entirely, and then proceeded to re-pair the rest of the tournament. (If you do that here, the actual last round pairings on boards 2 through 5 appear to make sense. For example, if you simply delete players 1 and 5, the next pairing is 7 - 2, which is illegal. So, 3 - 2 then…which is also illegal. Soooo…4 - 2, which is what happened.)
When I use any pairing program, I manually review the top several pairings before posting them during the last few rounds. I’ve actually switched late-round pairings at large tournaments using both WinTD and SwissSys because I felt the pairing I wanted to make was more logical. In a small tournament, this is even more necessary.
While we’ve been focused on the top two boards for these pairings, it turns out that the bottom three boards also present interesting challenges. Once we pair the top four players (one 3.0, two 2.5, and one 2.0), there are five players with 1.0 and one with 0.0 remaining. Of the five players with 1.0, one is unrated. Both WinTD and SwissSys end up dropping the unrated player to be the opponent of the (unrated) player with 0.0, seemingly in contradiction with rule 29D1a. There are also four players who are due for a particular color to avoid three of the same color in a row. (Luckily, two of those are due white and two are due black.) Pairing these six players so that there are no matches, no one has three of the same color in a row (29E5f), and the unrated player with 1.0 is not dropped from the score group (29D1a) is tricky, but I believe I found a pairing that meets all the constraints.
As one STD put it: Sometimes NTD’s make up the pairing rules as they go along, consistently-wise.
FWIW, for scholastics, I set the 80/200 settings to 101/221 to try to minimize bad color allocations toward the last rounds. Scholastic ratings are less predictve than the 2200+ pairing situations for cash prizes that led to the 80/200 rule.
And in the end, everyone has to play someone…you win, good…you lose, bad…
I’d probably avoid using the full abbreviation for Senior Tournament Director. Just sayin’. 
Consider it motivation to move up to the next level. 
What’s wrong with SrTD?
Cheers!
Guys,
I think some of the problem is that we’re trying to create pairing programs that pair using the same methods as humans. This was the early idea with chess playing programs and it didn’t work well. In the end brute force created chess playing problems that are excellent players. I think we should look at the same idea for pairings.
The problem the computer has with starting with the natural pairings and then trying to modify them to find the optimum pairings is the local minimum problem. That’s when all small changes lead to worse pairings. So, if the optimum pairing requires a major change in the natural pairings, it is unlikely that we can find that searching from the natural pairings.
Instead, what if we just look at all possible pairings. This isn’t as big of a solution space as you might think. If n is the number of players and n is even (to make thing simple), then the total number of pairings is just n!/(2^(n/2)*(n/2)!). So, the solution space grows slower than n!. In tabular form it looks like:
n number of parings
2 1
4 3
6 15
8 105
10 945
20 654,729,075
So, all we need is what Brian has asked for, an objective way to determine if one set of pairings is better than other set. The we just find the set with the best score and we’re done.
Since the tournaments with the biggest problems are ones with a low number of players relative to the number of rounds, small tournaments are the ones with the most problems so method should not take very long for them. For a 100 player tournament, it might be a problem but unless it has a high number of rounds, it won’t be hard to pair using the normal methods.
Overall, It will avoid the local minimum problem and might even do better than an NTD 
Mike Regan
I believe that there was a change made in Swissys that prevents the #1 pair number to be selected as “oddman”. If the TD wants to allow this, he/she must make that change. There are pros and cons for doing this and I have previously brought this up to the author of Swissys’s attention.
This is an interesting approach, and I think it has merit. One potential complication: I believe the solution space is bigger than you think. This is due to color allocation.
For ease of illustration, let’s say we’re pairing a two-player event, with players A and B. There are two possible sets of pairings: A-B, or B-A. That’s two solutions.
Now, we move on to the slightly more complex case of four players. You have three possible opponent pairings in any given round: A-B and C-D, A-C and B-D, and A-D and B-C. However, this doesn’t account for possible color allocation differences in each set of pairings. I count 12 possible pairings each round, when you take all possible color assignments under consideration.
If you take six players, it gets even more complex when you account for all the possible color swaps in each set of opponent pairings.
Why? Once the initial toss for White on the top board has been made, aren’t the color allocation rules completely deterministic for any given pairing? Well, there is one exception to that. If both players have no due color (for instance, if both are late entries who were given a half point bye for the first round, then both players have no due color, and the color assignment is random.) But if at least one player has a due color, then the color allocation rules are completely deterministic (unless I’m missing something).
I’m not really talking about first-round pairings, which is what I assume you’re referring to when you mention an “initial toss”. The suggestion, if I read Mr. Regan’s proposal correctly, was to use brute-force calculation to do pairings for all rounds. I was just pointing out that the number of permutations is increased by the possibility of switching colors between one, some, most or all of the pairings in each tree.
I may not have expressed myself clearly. What I meant to write was that, after the first round, if you show me any two players on the wallchart, I can tell you completely deterministically what the color allocation must be for those two players under the USCF pairing rules. Therefore, I do not understand why a pairing program that examines random pairings must also examine all possible combinations of color allocation; there is exactly one combination that is correct under the pairing rules.
There are two exceptions I can think of to my “you show me two players in round two or later, I can tell you what the color allocation must be” statement. First, if both players have no due color (because both have not yet played a game – maybe late entrants), the color allocation is random. Second, if the variant of tossing for color in the last round when both players have the same color history is used, then (obviously) the color allocation is random. Otherwise, it seems to me that determinism reigns supreme.