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forlano
17-05-2009, 05:29 AM
Hello,

I am an Italian arbiter and programmer (Vega's author). With a friend of mine (another arbiter) we are developing a pairing engine that implement the FIDE Dutch-Swiss. Currently we are performing several test of the program and comparing the results against those given by the certificated Swiss Master 5.5 (SM).
We obtained so far several discrepancies and we need to discuss them with expert arbiters to decide how to tune the program.
Googling internet I found only this site (in the past there was the FIDE forum now turned off) where many arbiters discuss the same problems at which we are interested too. So I hope to receive feedback from you. Moreover I hope that my questions may be of some interest for this forum.

My first post regards the color assignment between two players (in brackets the color hystori):

case 1)

A [WB]
B [--]

Player B entered the tournament after two forfeits so he has no color in the previous two games.
SM performar the pair

B - A

giving the white to player B and two consecutive black to player A. Is it correct? We would prefer

A - B

giving the alternation to A being the color preference of B practically 0.

case 2)

A [BWWBWB]
B [--WBWB]

it is related to the previous one. Who gets the white?

Thank you very much in advance for your answers.
Luigi Forlano

Oepty
17-05-2009, 11:43 AM
In case 1 the pairing should be A-B as you suggest.

In case 2 the pairing should be B-A

Scott

Edit: The answers above assume A is the higher ranked player.

Bill Gletsos
17-05-2009, 11:54 AM
In case 1 the pairing should be A-B as you suggest.

In case 2 the pairing should be B-A

ScottUnder what specific dutch pairing rules do you base your decisions on.

Oepty
17-05-2009, 11:58 AM
Under what specific dutch pairing rule do you base your decisions on.

Bill. The normal rules. Are my answers right or wrong?
Scott

Bill Gletsos
17-05-2009, 12:04 PM
Bill. The normal rules. Are my answers right or wrong?
ScottPossibly right and possibly wrong as there is insufficient information in the information provided to know which way to go.

In both cases you need to know if players A and B have the same score and secondly if they do have the same score which player is ranked higher.

Oepty
17-05-2009, 12:07 PM
Possibly right and possibly wrong as there is insufficient information in the information provided to know which way to go.

In both cases you need to know if players A and B have the same score and secondly if they do have the same score which player is ranked higher.

I assumed that A was higher ranked than B. I guess I should have said that.
Scott

Bill Gletsos
17-05-2009, 12:12 PM
I assumed that A was higher ranked than B. I guess I should have said that.
ScottI would assume that B is higher ranked than A given that is the pairing given by Swiss Master. ;)

Oepty
17-05-2009, 12:16 PM
I would assume that B is higher ranked than A given that is the pairing given by Swiss Master. ;)

Possibly although I would never put the scenario up in with the players listed in reverse ranking order. Perhaps though they are reversed in SM from the intended scenario and that is why is giving the other pairing.
Scott

Bill Gletsos
17-05-2009, 12:20 PM
When answering these sort of questions one should never assume anything that isnt stated by the person posing the question. ;)

forlano
17-05-2009, 04:15 PM
Possibly right and possibly wrong as there is insufficient information in the information provided to know which way to go.

In both cases you need to know if players A and B have the same score and secondly if they do have the same score which player is ranked higher.

Thanks all of you for the reply. Please forgive me for lack of info.
Both players have the same score and A is the higher ranked.

So, if I have understood, it depends by who is the higher ranked player while color alternation does not matter.

Many thanks,
Luigi

Kevin Bonham
17-05-2009, 05:09 PM
Case 1 is not adequately covered by the rules in my view. Applying E2 (Grant the stronger colour preference), neither player has an absolute or strong colour preference. Mild colour preferences are covered in A7c:

A mild colour preference occurs when a player's colour difference is zero, the preference being to alternate the colour with respect to the previous game.

Both A and B have colour difference zero. In A's case, there is a mild colour preference for white. In B's case, A7c seems to indicate that B has a mild colour preference, but as B has played no previous games, B has nothing specific to alternate to and there is no reason to consider that A-B grants B's colour preference any more or less than B-A does. But A-B clearly grants A's mild colour preference, while B-A clearly does not. I would allocate the colours A-B irrespective of the scores and the ranking of B. I cannot see any argument for the pairing B-A on any level as there is nothing to indicate that B has a colour preference for white.

In case 2, both players have the same colour preference (a weak colour preference for white) so E2 is useless. E3 is also useless since there is no previous round in which the two players played with different colours (there are four rounds where they played with the same colours, and two where one played and one did not, which is not the same thing). Therefore apply E4, "Grant the colour preference of the higher ranked player." It is impossible to determine colours in this case without this information.

Bill Gletsos
17-05-2009, 05:57 PM
I would like to see the ranking list and the full parings for rounds 1 & 2 for case 1 and the ranking list and the full parings for rounds 1 - 6 for case 2.

Bill Gletsos
17-05-2009, 06:24 PM
Case 1 is not adequately covered by the rules in my view. Applying E2 (Grant the stronger colour preference), neither player has an absolute or strong colour preference. Mild colour preferences are covered in A7c:

A mild colour preference occurs when a player's colour difference is zero, the preference being to alternate the colour with respect to the previous game.

Both A and B have colour difference zero. In A's case, there is a mild colour preference for white. In B's case, A7c seems to indicate that B has a mild colour preference, but as B has played no previous games, B has nothing specific to alternate to and there is no reason to consider that A-B grants B's colour preference any more or less than B-A does. But A-B clearly grants A's mild colour preference, while B-A clearly does not. I would allocate the colours A-B irrespective of the scores and the ranking of B. I cannot see any argument for the pairing B-A on any level as there is nothing to indicate that B has a colour preference for white.One way one could possibly get the B-A pairing would be firstly if B outranked A and secondly using the argument that since E1 (grant both colour prefs) does not apply then the colour allocated to B requires that A does not get their colour pref.


In case 2, both players have the same colour preference (a weak colour preference for white) so E2 is useless. E3 is also useless since there is no previous round in which the two players played with different colours (there are four rounds where they played with the same colours, and two where one played and one did not, which is not the same thing). Therefore apply E4, "Grant the colour preference of the higher ranked player." It is impossible to determine colours in this case without this information.Agree.

Kevin Bonham
17-05-2009, 06:59 PM
One way one could possibly get the B-A pairing would be firstly if B outranked A and secondly using the argument that since E1 (grant both colour prefs) does not apply then the colour allocated to B requires that A does not get their colour pref.

I considered whether it makes sense to consider the colour allocated to B at the start of the tournament, or the colour B would have received at the start based on odd/even ranking. But the colour received at the start has no stated bearing because it only dictates what colour a player receives in the first round (E5); the rules don't say anything about that player still having a preference for or against that colour in later rounds if they do not play round 1.

Beyond that, I don't see how any colour can be allocated to B such that A doesn't get preference. There is no basis for doing so unless a decision is made to give B a specific weak colour preference arbitrarily or randomly.

I did a trial with SP to see what it did in this situation (note: SP doing something is not evidence that it is right!)


Place No Opponents Colours Float Score


1-2 7 : 3,2,10 WWB 2
10 : -,4,7 -WW 2

3-8 2 : 6,7,5 WBB 1
3 : 7,9,8 BWB u 1
4 : 8,10,9 WBW 1
5 : 9,6,2 BBW 1
8 : 4,-,3 B-W d 1
9 : 5,3,4 WBB d 1

9-10 1 : -,-,6 --W d 0
6 : 2,5,1 BWB 0

In this instance player 1 is rated 2000 and the remaining players have no rating. SP pairs 1-6 in the third round giving player 6 alternating colour.

Not sure why SP says player 1 has had a downfloat though.

Kevin Bonham
17-05-2009, 07:05 PM
Crosstable for above:


No Name Feder Rtg 1 2 3

1. j, 2000 : : 6:
2. a, 6:W 7:L 5:
3. b, 7:L 9:W 8:
4. c, 8:W 10:L 9:
5. d, 9:L 6:W 2:
6. e, 2:L 5:L 1:
7. f, 3:W 2:W 10:
8. g, 4:L :W 3:
9. h, 5:W 3:L 4:
10. i, :W 4:W 7:

I manually changed the pairing 5 vs 6 in round 2 to 6 vs 5 so I could make sure the player playing vs 1 had had two different colours.

losboba
17-05-2009, 09:33 PM
Hello everybody.

I am the other Italian arbiter who is working with Luigi (who posed the original question).

I made a bunch of tests (with an older version of SwissMaster, though) and came up with the following results.

Player A is always the higher ranked, both players have the same score.

Case 1
-------
A(WB) B() white to A
A(BW) B() white to B
A() B(WB or BW) depends on PTC of A

I guess you are asking what the hell is the "PTC of A"...

Well, it seems that SM keeps track of the original color that the player should have had in the first round (PTC = Prefixed Theoretical Color - the acronym is mine).

So if PTC of A is white, A gets white. If PTC of A is black, A gets black. The sequence of B doesn't matter. Very strange behaviour (that I have no intentions to duplicate, by the way).

Case 2
-------
This is easy (but I got it only after reading what Kevin Bonham wrote - thanks!): the higher ranked player gets his/her preference.

Many thanks and cheers,
Roberto Ricca

Bill Gletsos
17-05-2009, 10:13 PM
I just duplicated Kevin's tournament in Swiss Master 5.5 (build 15) with the difference that I gave players a-i descending ratings from 1990 down to 1910.

I then did two tests.

I added player j with a rating of 2000.

The pairing for round 3 paired player j with e with white being allocated to player a. Note in this case we have j() and e(BW) and the PTC of j would have been white.

I then cleared the pairings for the third round, changed player j to have a rating of 1985. The computer still allocated player j with the white pieces. Note in this case the PTC of j would have been black.

The above results seemingly contradict Roberto's observation above regarding PTC.

in the above even if j is ranked below e, j always gets white.

Bill Gletsos
17-05-2009, 10:20 PM
Following on from the above, I changed the colour allocation in round 1 of a-e to e-a and in round 2 from e-d to d-e.

Thus we still have j() but now have e(WB)

Irrespective of j's rank, e always gets white in round 3.

This disagree's with Luigi's case 1 situation with Swiss Master 5.5 (unknown build)

Bill Gletsos
17-05-2009, 10:26 PM
Player B entered the tournament after two forfeits so he has no color in the previous two games.What exactly do you mean by two forfeits?

Do you mean he was paired in rounds 1 & 2 with opponents who won on forfeit, do you mean he was given a zero point bye in rounds 1 & 2 or do you mean he was simply added to the tournament starting in round 3 with no recorded results whatsoever in rounds 1 & 2?

Basil
17-05-2009, 10:28 PM
May I have 'U' in colour in the thread headline please? Please? Oh go on! Pretty please for Gunner. Kiss.

Thought not.

Oh go on.

losboba
18-05-2009, 05:19 AM
I just duplicated Kevin's tournament in Swiss Master 5.5 (build 15) with the difference that I gave players a-i descending ratings from 1990 down to 1910.

I then did two tests.

I added player j with a rating of 2000.

The pairing for round 3 paired player j with e with white being allocated to player a.

Note in this case we have j() and e(BW) and the PTC of j would have been white.

I then cleared the pairings for the third round, changed player j to have a rating of 1985. The computer still allocated player j with the white pieces. Note in this case the PTC of j would have been black.

The above results seemingly contradict Roberto's observation above regarding PTC.

Interesting. Maybe it is the different version of SwissMaster (I used 4.9).

But, just to be sure that we are on the same page: how did you retrieve the PTC for player j in your tests?

I probably didn't make myself clear, but the PTC is a very concrete datum for SwissMaster, as it is kept in column 64 of the .smw file (in the line dedicated to any player).

If you look at it, you will probably find a 'W' for the player j (it's a guess, but I'm pretty sure). That datum won't change when you move j around. Therefore, if j is ranked above e, j will get always white.
When j is ranked below e, PTC doesn't matter, player e gets his/her preference (which is black) and j gets white again.

You can change manually the PTC in the .smw file. Put a 'B' in place of 'W' (in column 64, I mean) and you will probably get different results.

Unless I am completely wrong, which is always a possibility :-)

Cheers,
Roberto

losboba
18-05-2009, 05:52 AM
Following on from the above, I changed the colour allocation in round 1 of a-e to e-a and in round 2 from e-d to d-e.

Thus we still have j() but now have e(WB)

Irrespective of j's rank, e always gets white in round 3.

This disagree's with Luigi's case 1 situation with Swiss Master 5.5 (unknown build)
I am the culprit. I gave Luigi wrong information.

The player who was playing his first game was #85. The other player (with the sequence WB) was #121. So the higher ranked played was actually "B" (in Luigi's example).

As we discovered later, the decisive factor was the PTC of 85, which was 'W'.

But, coming back to your observation. Is 'j' the same player as in the previous example? If so, I assume his PTC is White. Therefore when he is ranked above e, he should get White.

Unless pairing considerations don't take into account the players' rating but just the pairing numbers (j has the lowest, doesn't he?). In version 4.9, that seems to be the case. What about 5.5?

Cheers,
Roberto

forlano
18-05-2009, 06:10 AM
What exactly do you mean by two forfeits?

Do you mean he was paired in rounds 1 & 2 with opponents who won on forfeit, do you mean he was given a zero point bye in rounds 1 & 2 or do you mean he was simply added to the tournament starting in round 3 with no recorded results whatsoever in rounds 1 & 2?

Yes, I mean he was simply added to the tournament starting in round 3 with no recorded results whatsoever in rounds 1 & 2.

Luigi

forlano
18-05-2009, 04:59 PM
Unless pairing considerations don't take into account the players' rating but just the pairing numbers (j has the lowest, doesn't he?). In version 4.9, that seems to be the case. What about 5.5?


Very interesting.
Because the PTC, if exists, should be determined since first round (please confirm) I wonder if it may change in case the pairing number is even or odd as happen to the other players at first round.
So far the players was added at round 3. But what about if he is already in the list of registered players at the very beginning but simply not available for the first two rounds? Have you noticed something? Otherwise I will perform these tests.

EDIT: One more question: what if we enter two players at round 3? I am curios to know if the PTC make the difference. I'll do some test tonight.

Luigi

Bill Gletsos
18-05-2009, 08:26 PM
If you create a new tournament in SM 5.5 and then add 9 players and save the tournament then if you check the .smw file you will find that all 9 players have a W in column 64.

Now after you do the first round pairings the colour shown in column 64 is as follows:

a - W
b - B
c - W
d - B
e - B
f - W
g - B
h - W
i - B

Player i gets the bye in the first round.

Scenario 1
In round 2 I swapped the pairing d-e to e-d as per Kevin's test.

Player j when added after round two always gets an W in column 64 irrespective of whether it is ranked above or below player e.

Pairing round 3 you always get j-e.

Scenario 2
Now I unpaired round 3, deleted player j, went and swapped the round 1 pairing to be e-a from a-e and set round 2's e-d back to d-e.

Again when Player j is added after round two they always gets a W in column 64 irrespective of whether it is ranked above or below player e and the pairing is always e-j.

I also tried adding another player k at the same time as adding player j.
Irrespective of ranking k always gets a W in column 64.

The pairing of j-e in Scenario 1 remains as j-e if j is ranked higher than k and k-e if k is ranked higher.

The pairing of e-j in Scenario 2 remains as e-j if j is ranked higher than k and e-k if k is ranked higher.

So it appears that PTC of white has no bearing the colour allocation in round 3.

losboba
18-05-2009, 11:54 PM
If you create a new tournament in SM 5.5 and then add 9 players and save the tournament then if you check the .smw file you will find that all 9 players have a W in column 64.

Very interesting find. So the rule seems: a new player's PTC is always 'W'.



Now after you do the first round pairings the colour shown in column 64 is as follows:

a - W
b - B
c - W
d - B
e - B
f - W
g - B
h - W
i - B


And it won't change (unless it's done manually).



[...]
So it appears that PTC of white has no bearing the colour allocation in round 3.

Now I am under the impression that, also in SwissMaster 5.5, only the pairing number counts. 'j' has the lowest one. Therefore his PTC will never enter into the equation.

Can you try with a player in the middle of the field (as it was in the original question)? I know that the version 4.9 allows for changing and rearranging pairing numbers, but a simple way to do it is preparing the whole field in the beginning and then let the player #3 (for instance) and then #4 forfeit their first two games.

Cheers,
Roberto

forlano
19-05-2009, 05:03 AM
Can you try with a player in the middle of the field (as it was in the original question)? I know that the version 4.9 allows for changing and rearranging pairing numbers, but a simple way to do it is preparing the whole field in the beginning and then let the player #3 (for instance) and then #4 forfeit their first two games.


As noted by Bill the PTC of a late entry is always W (SwissMaster 5.5 build 3).

I've done a test removing from first round the higher rated player, A. Since first round his PTC turned to "-", neither W nor B. Perhaps it is initialized to W and so remain only for late entries.
At round 3 player A entered the tournement at 0 point and I was able to obtain

G(WB) - A (--) (both at 0 points),

and clearing the pairing and modifying the result of round 2,

A (--) - H (BW) (both at 0 points)

It seems that the pairing number it is not important. It look like it prefers to alternate the colour as we wanted. But for late entry this is not always done.

Luigi

Bill Gletsos
20-05-2009, 12:18 AM
I've done a test removing from first round the higher rated player, A. Since first round his PTC turned to "-", neither W nor B. Perhaps it is initialized to W and so remain only for late entries.When you say removed him from the first round what do you mean exactly as just changing his first round result to a 0 point bye does not seem to reset his PTC to "-" and removing him from the player list altogether deletes his whole entry from the .smw file.

At round 3 player A entered the tournement at 0 point and I was able to obtain

G(WB) - A (--) (both at 0 points),

and clearing the pairing and modifying the result of round 2,

A (--) - H (BW) (both at 0 points)

It seems that the pairing number it is not important. It look like it prefers to alternate the colour as we wanted.That is what I would expect.

But for late entry this is not always done.
I wonder what specifically you are doing to sometimes cause it not to happen as I have so far been unable to create a situation where it does not always happen.

forlano
20-05-2009, 05:02 AM
When you say removed him from the first round what do you mean exactly as just changing his first round result to a 0 point bye does not seem to reset his PTC to "-" and removing him from the player list altogether deletes his whole entry from the .smw file.

I mean make him not available for the first two rounds. That is when I run "pairings/make new pairing" I simply remove player A from the pairing. His record in the .smw file is the following:

1 1 A 2000 - d -0-- d -0-- 05B1#7



I wonder what specifically you are doing to sometimes cause it not to happen as I have so far been unable to create a situation where it does not always happen.

At this point I believe that this problem, produced by Roberto using swiss49, is speciic of this old DOS version. Roberto should be able to replicate it with swiss49.

Luigi

losboba
20-05-2009, 08:20 AM
I wonder what specifically you are doing to sometimes cause it not to happen as I have so far been unable to create a situation where it does not always happen.
Maybe you could let #4 (which usually has the PTC 'B') forfeit his first two games, then let him play in the third round with someone with a lower pairing number who has a sequence of "BW". #4 will get black and his opponent will get a second white in a row.

Then start all over and repeat the double-forfeit trick with player #3 meeting in the third round someone lower numbered with a sequence of "WB". #3 will get white.

Cheers,
Roberto

forlano
20-05-2009, 08:01 PM
Maybe you could let #4 (which usually has the PTC 'B') forfeit his first two games, then let him play in the third round with someone with a lower pairing number who has a sequence of "BW". #4 will get black and his opponent will get a second white in a row.

Then start all over and repeat the double-forfeit trick with player #3 meeting in the third round someone lower numbered with a sequence of "WB". #3 will get white.


I have done several tests. I obtained with C

G[WB] - C[--]
C[--] - E[BW]

and then with D

G[WB] - D[--]
D[--] - H[BW]

I was waiting these results. I do not see any influence of PTC in color allocation... at least for SM5.5 build 3. The PTC in all these tests was "-".
Luigi

forlano
21-05-2009, 08:58 AM
I performed new tests and I can confirm what Roberto said about the influence of the PTC. To replicate the case one need to perform the normal pairing and then assign the forfeited result. In this case the PTC remain W in case of player C, and B in case of players D for the first two forfeited games.

Case 1) player C get two forfeited games, PTC=W

at round 3 I got
C(--) - F(WB)
and changing the result
C(--) - D(BW)

Case 2) player D get two forfeited games, PTC=B

at round 3 I got
F(WB) - D(--)
and changing the result
E(BW) - D(--)

In the test of the above post I simply gave player C and D a forfeit loss by using the option "Default Result" in the "Make Pairing" window. In that case the PTC assume the value "-" as reported.
So the PTC make the difference in the color allocation. Perhaps after the forfeited game it is not set to "-" how it should be.

Luigi

Bill Gletsos
21-05-2009, 10:34 AM
n the test of the above post I simply gave player C and D a forfeit loss by using the option "Default Result" in the "Make Pairing" window.Use of the "Default Result" function is not to assign forfeits but to assign full point, half point and zero point byes.

A forfeit occurs when a player is paired against another player and fails to turn up and the result entered via the normal results function not by using "Default Result".

Bill Gletsos
21-05-2009, 10:38 AM
I have done several tests. I obtained with C

G[WB] - C[--]
C[--] - E[BW]

and then with D

G[WB] - D[--]
D[--] - H[BW]

I was waiting these results. I do not see any influence of PTC in color allocation... at least for SM5.5 build 3. The PTC in all these tests was "-".
LuigiSo in these tests how did you actaully set C & D's results.
Did you use "Default Resul" here?

ER
21-05-2009, 10:50 AM
Hey Bill,
Don't provide all our technological expertees and intellectual knowledge to the Europeans for free! Unless they are prepared to accept NSWCA in the EU of course! :lol:

forlano
21-05-2009, 03:53 PM
So in these tests how did you actaully set C & D's results.
Did you use "Default Resul" here?

By the "Edit Result" window and inserting the results 7->1-0 forf or 8->0-1 forf.

I think that from point of view of next pairing should make no difference if the point is awarded by forfeit or bye. In fact the unplayed game should be considered as no color... at least for the losing player. I remind that in a Aeroflot tournament the player that got a point without to play was considered to play with white color. Perhaps to compensate this advantage with black piece in the next round.

Luigi

EDIT: Just now I've recognised that you were referring to the previous test in post #31. Unfortunatly the forum does not quote the whole post but only the last reply and I missed the link.

In the post 31 I used "Default Result" to assign a bye loss as you correctly reported.

The "Edit Result" window and forfeited result was used for test of post #32.

I'm sorry for the confusion.

Luigi

Bill Gletsos
21-05-2009, 11:21 PM
If I use the "Default Result" feature and set any player to a zero point bye in round 1 and then check the .smw the PTC field is never set to "-".

Similarly if I use the "Edit Result" feature and set a result using 7->1-0 forf or 8->0-1 forf and then check the .smw the PTC field is never set to "-".

forlano
22-05-2009, 01:07 AM
If I use the "Default Result" feature and set any player to a zero point bye in round 1 and then check the .smw the PTC field is never set to "-".

I am not an expert of SwissMaster (v 5.5 build 3). Perhaps I missed something. I have set in "default result" a 0 point bye to player C and then made the pairing. I saved the file .smw and got the following


8 4 0 1 0
1 1 a 2000 W 01W1#5 v -1-- v -1-- v -1--
2 2 b 1950 B 02B1#6 v -1-- v -1-- v -1--
3 3 c 1900 - d -0-- v -1-- v -1-- v -1--
4 4 d 1850 W 03W1#7 v -1-- v -1-- v -1--
5 5 e 1800 B 01B0#1 v -1-- v -1-- v -1--
6 6 f 1750 W 02W0#2 v -1-- v -1-- v -1--
7 7 g 1700 B 03B0#4 v -1-- v -1-- v -1--
8 8 h 1650 W v -1-- v -1-- v -1-- v -1--


where C has "-".



Similarly if I use the "Edit Result" feature and set a result using 7->1-0 forf or 8->0-1 forf and then check the .smw the PTC field is never set to "-".

I agree.
Luigi

Bill Gletsos
22-05-2009, 01:15 AM
I am not an expert of SwissMaster (v 5.5 build 3). Perhaps I missed something. I have set in "default result" a 0 point bye to player C and then made the pairing. I saved the file .smw and got the following


8 4 0 1 0
1 1 a 2000 W 01W1#5 v -1-- v -1-- v -1--
2 2 b 1950 B 02B1#6 v -1-- v -1-- v -1--
3 3 c 1900 - d -0-- v -1-- v -1-- v -1--
4 4 d 1850 W 03W1#7 v -1-- v -1-- v -1--
5 5 e 1800 B 01B0#1 v -1-- v -1-- v -1--
6 6 f 1750 W 02W0#2 v -1-- v -1-- v -1--
7 7 g 1700 B 03B0#4 v -1-- v -1-- v -1--
8 8 h 1650 W v -1-- v -1-- v -1-- v -1--


where C has "-".I did the same thing but it does not become a "-".

I am using SwissMaster 5.5 build 15. Maybe that is the difference.

losboba
22-05-2009, 08:31 AM
Don't provide all our technological expertees and intellectual knowledge to the Europeans for free!

It's not for free!

We gave you back the PTC!

I bet you never heard of PTC before - and will probably never hear again:), as the acronym is mine and it is not a very good one. But please, let's continue with PTC, for the sake of clarity.

All jokes aside, I think that we followed your exhortation:) and, working together, we discover that:
(1) there is an interesting (weird?) concept in SwissMaster, called PTC, a character kept in the column 64 of the .smw file
(2) a PTC is assigned to each player *paired* in the first round; it is coincident with the colour the player gets ('W' or 'B'); also the bye gets 'W' or 'B'
(3) an absent player gets a PTC equal to '-' (neutral?)
(4) a later round entry has PTC 'W'
(5) ratings don't matter; only pairing numbers count for SM (well, this was an inference more than a fact, but I also verified it in a sample tournament).

We just checked the PTC regarding the colour allocation, but its use is more widespread than that.

Analyzing other .smw files generated by SM, I had some interesting finds:

(a) a tournament -which I directed in February, by the way- where in the second round there was a score bracket with 16 people expecting white, 14 expecting black and 3 who were to play their first game. SM uses the PTC as a preference in the pairings generation phase. That is, who had PTC='W' played with white; who had PTC='B' played with black. It didn't happen by chance (although it could have). It was wanted (or maybe a side-effect; I actually don't know).
I can write down the full composition of the score bracket, if it is of any interest.

(b) another tournament where in the third round there was a group with 8 players expecting W, 7 expecting B and one, whose PTC was 'W', playing his first game.
This is the list of the sixteen players, with their preference, divided in S1/S2:

S1 = 076w 077b 087w 101w 103w 104b 105A 107b
S2 = 110b 116b 123w 125w 126b 130b 131w 132w

To my astonishment, SM considered 105 as expecting White and therefore put x=1. The pairing generated by SM was 76-110 123-77 87-116 101-125(!) 103-126 131-104 105-130 132-107.

I would have generated: 76-110 123-77 87-116 101-126 103-130 125-104 131-105 132-107. Anything wrong with that?

======================
Trying to synthesize things, it seems to me that the PTC works like a (weak) preference. In the examples that we beat to death, the higher "ranked" player (i.e. the player with the lower pairing number in SM world) got the colour indicated by his PTC because of E.4 (that's the simplest explanation!)

To me, it (i.e. The PTC existence and its use/abuse) doesn't sound right.

As Luigi said, we are developing a pairing engine. But I see no basis to use the PTC in it.

Does anybody see one?

Thanks for all the contributions.

Cheers,
Roberto

drbean
01-07-2009, 04:45 PM
Analyzing other .smw files generated by SM, I had some interesting finds:

...

(b) another tournament where in the third round there was a group with
8 players expecting W, 7 expecting B and one, whose PTC was 'W',
playing his first game.

This is the list of the sixteen players, with their preference, divided in S1/S2:

S1 = 076w 077b 087w 101w 103w 104b 105A 107b
S2 = 110b 116b 123w 125w 126b 130b 131w 132w


To my astonishment, SM considered 105 as expecting White and therefore
put x=1. The pairing generated by SM was

76-110 123-77 87-116 101-125(!) 103-126 131-104 105-130 132-107.

I would have generated:

76-110 123-77 87-116 101-126 103-130 125-104 131-105 132-107.

Anything wrong with that?



That looks correct to me, too.

Isn't this exactly the same problem as the one which started this
thread? It appears to be a BUG that results from SM's questionable
design decision to give players a default White preference.

The behavior seems inconsistent with F2, where it says byes and
unplayed pairings are not taken into account in determining color. F3,
which says invalid games are pushed to the back and the colors of
games previous to them used to determine pairings, also suggests that
respecting a color not actually played is not in the spirit of the
FIDE Rules.




======================
Trying to synthesize things, it seems to me that the PTC works like a
(weak) preference. In the examples that we beat to death, the higher
"ranked" player (i.e. the player with the lower pairing number in SM
world) got the colour indicated by his PTC because of E.4 (that's the
simplest explanation!)



I can't see any reason to grant players any preference, until after E5
and A7c, about how the (top) player's preference is determined by lot,
are carried out.

And then only the players in S1 should receive a preference. The
pairing mechanism needs to be able to handle players with no
preference.





To me, it (i.e. The PTC existence and its use/abuse) doesn't sound right.

As Luigi said, we are developing a pairing engine. But I see no basis to use the PTC in it.



I think this all means the preferences defined by the Rules need to be
augmented with a null preference.

That's what I did with
http://cpansearch.perl.org/dist/Games-Tournament-Swiss>Games::Tournament::Swiss




Does anybody see one?



The program needs some way of handling the problem of what to do with
preferences, whether it is a default preference or code that copes
with players with no preferences.

But there is another problem here that will continue to exist, even if
there is no PTC. Even if there is no PTC, players are going to get
colors awarded in rounds and these have to be recorded by the program.

Now, whether or not these games are actually played, is another
question. Whether or not allocations need to be respected in
pairing the next round is something that the program has to check up
on. That is the real problem.

It has to go back and look at the results of the matches and edit the
preference history, depending on whether the game was played or not.

Which is a lot of work, pulling all the data together. I don't think I did that
with my own program. I don't think it is able to handle inconsistent lineups
from round to round.




Thanks for all the contributions.

Cheers,
Roberto

Can you post the entire crosstable/pairing tables for all brackets in
that round?

I think this would be valuable, because it is the pairing of a real
tournament (as opposed to a theoretical thought experiment.) The
complex mixture of variables involved in tournaments makes it
difficult to tell whether the Rules are followed or not. It's easier
if all the data (real data) is there.

I think a public list of tests that should be passed also helps the
development of better pairing programs.

One of the discussions here, eg http://www.chesschat.org/showthread.php?t=6619,

I put in the tests for Games::Tournament::Swiss.

http://cpansearch.perl.org/src/DRBEAN/Games-Tournament-Swiss-0.16/t/cc6619.t

However, wrapped up in a perl test, it is not easy to interpret. A pairing
table would be easier to interpret.

drbean
22-07-2009, 07:50 AM
At http://chesschat.org/showpost.php?p=241948&postcount=38,
losboba said:


(b) another tournament where in the third round there was a group with 8 players expecting W, 7 expecting B and one, whose PTC was 'W', playing his first game.
This is the list of the sixteen players, with their preference, divided in S1/ S2:

S1 = 076w 077b 087w 101w 103w 104b 105A 107b
S2 = 110b 116b 123w 125w 126b 130b 131w 132w

To my astonishment, SM considered 105 as expecting White and therefore put x=1. The pairing generated by SM was 76-110 123-77 87-116 101-125(!) 103-126 131-104 105-130 132-107.

I would have generated: 76-110 123-77 87-116 101-126 103-130 125-104 131-105 132-107. Anything wrong with that?


Another easy way to generate the same mistaken behavior, and one that Games::Tournament::Swiss was guilty of, is to take a shortcut and calculate the number of players with one preference by subtracting the number with the other preference from the total number of players in the bracket.

If every player has a preference, no error is generated. But in the case above, if we calculate b as 7 and then assume w is 16-7, or 9, we conclude x=1.

The FIDE calculation actually allows the possibility of negative x.



A.8 Definition of "x"
The number of pairings which can be made in a score bracket, either
homogeneous or heterogeneous, not fulfilling all colour preferences,
is represented by the symbol x.
x can be calculated as follows:
w = number of players having a colour preference white.
b = number of players having a colour preference black.
q = number of players in the score bracket divided by 2, rounded
upwards.
If b >> w then x = b-q, else x = w-q.


If there are 5 players in the bracket, and w=2, b=2 and q=3, ie, there is one late entry, then x=-1.

So, we have to 'round up' to zero.

Pepechuy
03-09-2011, 11:41 AM
This raises another interesting question. Lets have the following scenario. We have a 5 round Swiss system tournament with 12 players. Two of them have clearly stated that they will play, but will be late and unable to attend the first round. So they both remain unpaired and get 0 points in the first round. All the other ten players end up drawing their games. So, in the second round it is unavoidable to pair the two late comers, who have no colour preference.
Which one plays white?

Kevin Bonham
04-09-2011, 01:48 AM
This raises another interesting question. Lets have the following scenario. We have a 5 round Swiss system tournament with 12 players. Two of them have clearly stated that they will play, but will be late and unable to attend the first round. So they both remain unpaired and get 0 points in the first round. All the other ten players end up drawing their games. So, in the second round it is unavoidable to pair the two late comers, who have no colour preference.
Which one plays white?

Random draw would seem just fine to me. This is a very rare situation so there is no need to regulate it.

Another idea would be to work out which colour the highest seeded of the two players would have had in that round based on seeding if that player had been present for round one, and go with that.

Denis_Jessop
04-09-2011, 01:01 PM
This raises another interesting question. Lets have the following scenario. We have a 5 round Swiss system tournament with 12 players. Two of them have clearly stated that they will play, but will be late and unable to attend the first round. So they both remain unpaired and get 0 points in the first round. All the other ten players end up drawing their games. So, in the second round it is unavoidable to pair the two late comers, who have no colour preference.
Which one plays white?

In Canberra club tournaments the players who could not attend would receive a half-point bye, thus avoiding your problem though I doubt that half-point byes were introduced for that reason.

DJ

Garvinator
04-09-2011, 07:27 PM
In Canberra club tournaments the players who could not attend would receive a half-point bye, thus avoiding your problem though I doubt that half-point byes were introduced for that reason.Wouldn't having players receiving half point byes in the first round increase the odds of this occurring? The only players the newcomers could be paired against are either the floaters, or those who drew in round one.

I have come across this issue quite a few times of who is supposed to get the colour when a player has not played round one.