User:Clpippel/sandbox

R

Sistema Rutch-Berger.

modn <- 11
stopifnot(modn %% 2 == 1)
round <-  LETTERS[1:modn]
half  <- (modn + 1) / 2
step  <- 1
for (r in seq(modn)){
  cat(sprintf("Round %-2d: ", r))
  v0 <- c(round[1], "w")
  if (r %% 2 == 0) v0 <- rev(v0)
  cat(paste(c(v0[1], round[2:half]), c(v0[2], rev(tail(round, half - 1))), sep = "-"), "\n")
  # Mover una posición a la derecha.  
  round  <- c(tail(round, step), head(round, -step))
}

R

# Imprimir horario Berger.
# Jugadores desiguales, incluido uno mismo.
# Conteo de reloj basado en cero modulo n.
modn   <- 7
stopifnot(modn %%2 == 1)
venues <- seq(from=0, length.out=(modn + 1) / 2)
step   <- length(venues)
stopifnot((modn %% step) != 0)

# Crear primera ronda, diagonal == (0 %% n).
# Dos jugadores por sede (local, visitante).
round                 <- rep(0, length(venues) * 2)
round[c(TRUE, FALSE)] <- +venues
round[c(FALSE, TRUE)] <- -venues %% modn

for (r in seq_len(modn)){
  cat(sprintf("Round %2d:", r))

  # Imprimir sedes.
  for (v in venues) {
    # Reemplazar self por jugador añadido (n) en el primer sitio.
    # 1-indexación en la impresión.
    v0 <- round[v * 2 + 1:2]
    if (v == 0) v0[((r %% 2)) + 1] <- modn
    cat(sprintf("%2d - %2d  ", v0[1] + 1, v0[2] + 1))
  }
  cat('\n')
  
  # Ciclo a la siguiente diagonal.
  round <- (round + step) %% modn
}

Short Footnote

Landau.

Afbreken pre

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Table cell
Firstlinexxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Secondlineaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Mnemonic

To easily remember this method, the following mnemonic can be used. Starting from the first round,

                venue = 1
╭────────────────────────────────────────────────────┐

1—ω >>> 2—13 >>> 3—12 >>> 4—11 >>> 5—10 >>> 6—9 >>> 7—8

the next round is constructed:

ω—8 >>> 9—7 >>> 10—6 >>> 11—5 >>> 12—4 >>> 13—3 >>> 1—2

and then,

2—ω >>> 3—1 >>> 4—13 >>> 5—12 >>> 6—11 >>> 7—10 >>> 8—9
ω—9 >>> ...

If the number of players is odd, the player in the first venue gets a bye, or plays against the dummy player (ω).

---

To easily remember the pairing system, we can use the following mnemonic.

                      venue = 1 
  ╭────────────────────────────────────────────────┐
  |   7       6       5      4        3       2    |
  | ┌──┐    ┌──┐    ┌──┐    ┌──┐    ┌──┐    ┌──┐   |
1 — 1  2 — 13  3 — 12  4 — 11  5 — 10  6 —  9  7 — 8
8 — 8  9 —  7 10 —  6 11 —  5 12 —  4 13 —  3  1 — 2
2 — 2  3 —  1  4 — 13  5 — 12  6 — 11  7 — 10  8 — 9
9 — 9  ...

If the number of players is odd, the first field contains the player who has a bye, or plays against an extra player (14) if the number is even.

table

Percentage score p into rtg. difference

$p$	$d_{p}$	$d800_{p}$
1.00	+800	+400
0.99	+677	+392
0.9	+366	+320
0.8	+240	+240
0.7	+149	+160
0.6	+72	+80
0.5	0	0
0.4	−72	-80
0.3	−149	-160
0.2	−240	-240
0.1	−366	-320
0.01	−677	-392
0.00	−800	-400

The probability function D800 is given by D800(P) = 4C(P - 0.5), Class interval C = 200. (Elo, 1978, p.146). Rating performance using D800 produces the same result as "algorithm of 400".

quotations

(Elo, 1978, p. 25)

The coefficent reflects the ralative weights selected for the pre-event rating and the event performance rating. A high K gives high weight to the most recent performance. A low K gives more weight to earlier performances.

The coefficent K reflects the ralative weights selected for the pre-event rating and the event performance rating. A high K gives high weight to the most recent performance. A low K gives more weight to earlier performances (Elo, 1978, p. 25).

The development of the Percentage Expectancy Table (table 2.11) is described in more detail by Elo (1978, p159) as follows:

The normal probabilities may be taken directly from the standard tables of the areas under the normal curve when the difference in rating is expressed as a z score. Since the standard deviation σ of individual performances is defined as 200 points, the standard deviation σ' of the differences in performances becomes √2 or 282.84. the z value of a difference then is D/282.84. This will then divide the area under the curve into two parts, the larger giving P for the higher rated player and the smaller giving P for the lower rated player.

For example, let D = 160. Then z = 160/282.84 = .566. The table gives .7143 and .2857 as the areas of the two portions under the curve. Thes probabilities are rounded to two figures in table 2.11. |author=Arad E. Elo |title=The Rating of Chessplayers

The table is actually build with 2000/7 as an approximation for √2.

Ceci n'est pas une pipe. This is Not a Pipe.

— René Magritte, The Treachery of Images

Foreign language

Zunächst sind ebenso viele Nummern zu verloosen, als Theilnehmer
vorhanden sein: bei 14 Theilnehmern z. B. die Nummern 1, 2, 3 bis 14.
 Die Paarung geschieht alsdann in folgender Weise: Man fertigt
ein Schema an, welches soviel senkrechte Reihen hat, als die Hälfte
der Theilnehmer beträgt. Bei einer ungeraden Anzähl ist die Anzahl
der senkrechten Reihen die Hälfte der um 1 vergrösserten Theilnemer-
zahl. Bei 14 Theilnehmern sind daher,  14/2 = 7, bei 13 Theilnehmern
(!3+1)/2= 14 / 2, also ebenfalls 7 senkrechte Reihen zu bilden.

Je suis English. Je suis English.

Math

${P(D)}={\frac {1}{1+10^{-D/4C}}}~.$

Base 10 or base ${\sqrt {10}}$

In The Rating of Chessplayers, 1978, page 141 we read at 8.43:

When the logarithms in equation (38) are taken to the base ${\sqrt {10}}$,

then the Verhulst and the logistic take the following forms:

${P(D)}={\frac {1}{1+10^{-D/2C}}}~$ (46)

$10^{1/2}$

And subsequently in chapter 8.46 Percentage Expectancy Table, page 143 the table does have the header:

Logistic Probabilities to base ${\sqrt {10}}$

Based on the above sources, it is ${\sqrt {10}}.$

LSM rating

Let Gm,m be bipartite with two disjoint parts A, B and m-regular. Let SA, SB be the sum of the scores si of Xi in A, B respectively.

Rating vector q equals:

   qi = (2.si + T / m) / γ,   where γ = n.m,  T = SA if Xi in A, and SB if Xi in B.

Cite

Cite direct journal.

Brouwer, Andries E.; Haemers, Willem H. (2012), "12.3.1 Hamming graphs" (PDF), Spectra of graphs, Universitext, New York: Springer, p. 178, doi:10.1007/978-1-4614-1939-6, ISBN 978-1-4614-1938-9, MR 2882891, retrieved 2022-08-08.

Elo, Arpad (1986) [1st pub. 1978]. The Rating of Chessplayers, Past and Present (Second ed.). Arco. ISBN 978-0-668-04721-0.

Jaan Kiusalaas (29 January 2010). "Hoofdstuk 2.7 Iterative Methods, 4.6 Systems of Equations". Numerical Methods in Engineering. With Python (2nd ed.). New York: Cambridge University Press. ISBN 978-0-521-19132-6.

Edmund Landau

[Dr. W. Ahrens]

[Citeer Arxiv]

Chance

Roblox

In practice, there is little difference between the shape of the logistic and normal curve. So it does not matter whether the logistic or normal distribution is used to calculate the expected scores.

Repeat pp
Repeat pp
Repeat at
Repeat loc

Diversen

Interwiki: pairwise comparison

(E1)          f(x) = x + 1

Deze iteratie is met pen en papier of pocket calculator uit te voeren. Het resultaat is een benadering. De relatie tussen ratingverschil (D) en winstkans (P) is niet lineair. Het rekenkundige gemiddelde van de tegenstanderratings is een benadering van van de rating waarvan de verwachting van de gespeelde partijen 50% is. Daarom zal deze procedure niet convergeren naar een oplossing, waarin de werkelijke en verwachte waarden identiek zijn.

In de beschouwing wordt uitgegaan van een binomiale verdeling, uitsluitend winst (1) en verlies (0) komen voor. In de schaakpraktijk is remise geen uitzondering. Daardoor wordt de verdeling vlakker. De binomiale variantie van een partij tussen even sterke spelers is per definitie gelijk aan: μ = (0 + 1)/2, en σ² = (1 - μ)²/2 + (0 - μ)²/2 = 1/4. Inclusief remise (½) wordt dit: μ = (0 + ½ + 1)/3, en σ² = (1 - μ)²/3 + (½ - μ)²/3 + (0 - μ)²/3 = 1/6. Dit is een factor 2/3 kleiner.

Calculating of the Rating Difference table

The standard normal distribution, with mean value μ = 0 and standard deviation σ = 1, in x = 2.17 equals to 98.500%. This is the first value in the 99% category. The corresponding rating is 620. Therefore the standard deviation employed in the table equals to 620 / 2.17 = 2000 / 7. The Elo table contains a few irregularities. A more accurate expectation of 620 equals to 98.4997%, which falls into the 98% range. The table assigns 344 to expectation 88%. However the expectation due to the normal distribution is 88.5705%, which is clearly in the 89%. Needed are references to the construction of this table in order to explain the underlying calculation.

If Player A has a rating of $R_{A}$ and Player B a rating of $R_{B}$, the exact formula (using the logistic curve). for the expected score of Player A is

Next

Elo made references to the papers of Good 1955, David 1959, Trawinski and David 1963 and Buhlman and Huber 1963.

The above schedule can also be represented by a graph, as shown below:

Round Robin Schedule Span Diagram

This image was created with Inkscape.

Designing Schedules for Leagues and Tournaments.

(http://www.emba.uvm.edu/~jdinitz/preprints/design_tourney_talk.pdf)

Brewer's Dictionary of Phrase and Fable

Ramon Llull Ars Magna (cyclic pattterns), Martin Gardner, Last Recreations.

Round robin references

nl:IJmuiden IJmuiden (nnle)

"Damclub IJmuiden" (in Dutch). Retrieved 2014-01-01.

Édouard Lucas, Récréations Mathématiques, four volumes, Gauthier-Villars, Paris, 1882-94 (https://archive.org/stream/rcrationsmathma01lucagoog#page/n0/mode/2up)

M. Walecki , professeur de Mathémntiques spéciales au lycée Condorcet

http://www.les-mathematiques.net/phorum/read.php?17,569123,569771

Pour Walecki: Walecki Felix Charles Louis

Né le 12/01/1844 à Moselle - Metz. Agrégé de mathématiques. A eu la légion d'honneur. A enseigné au lycée de Nancy avant de sévir à Paris, au lycée Condorcet. » C’est donc son anniversaire aujourd’hui : 166 ans aujourd’hui. Mais il faut aller plus loin : qui est Walecki ? Qu’a-t-il publié ? démontré ? qu’est-ce ce qui pousse à s’intéresser aujourd’hui à lui ? Comment procéder méthodologiquement pour en savoir plus sur lui ?

Schurig does not provide a proof nor a motivation for his algorithm. For more historical details, see Ahrens.

This method was discovered by Felix Walecki and reported by Lucas (1883) as a solution to the Walking schoolgirl problem (Les Promenades du Pensionnat).

Lucas, who describes the method as simple and ingenious, attributes the solution to Felix Walecki. Lucas also shows an alternative solution by means of a sliding puzzle

Other articlea are: and .

Teeds method .

Alternatively Berger tables,

In France this is called the Carousel-Berger system (Système Rutch-Berger).

See also Combinatorial design

Schurig tables.

According to the Oxford Companion, the pairing tables were first published by Richard Schurig in 1886 in Deutsche Schachzeitung v41. "With scant regard for chess history FIDE calls Schurig's creation 'Berger tables' because Johann Berger gave them, duly acknowledged, in his two Schachjahrbucher (1892-3 and 1899-1900) "How old are the Berger round-robin tables?".,

Try to do something in Beta.

A quad is a Round-robin tournament.

Berger published the pairing tables in his two Schachjahrbucher, with due reference to its inventor Richard Schurig.

Original construction of pairing tables by Richard Schurig (1886)

For 7 or 8 players, Schurig builds a table with $n/2$ vertical rows and $n-1$ horizontal rows, as follows:

1.	Round	1	2	3	4
2.	,,	5	6	7	1
3.	,,	2	3	4	5
4.	,,	6	7	1	2
5.	,,	3	4	5	6
6.	,,	7	1	2	3
7.	,,	4	5	6	7

Then a second table is constructed as shown below:

1.	Round	. 1	. 7	. 6	. 5
2.	,,	. 5	. 4	. 3	. 2
3.	,,	. 2	. 1	. 7	. 6
4.	,,	. 6	. 5	. 4	. 3
5.	,,	. 3	. 2	. 1	. 7
6.	,,	. 7	. 6	. 5	. 4
7.	,,	. 4	. 3	. 2	. 1

By merging above tables we arrive at:

1.	Round	1, 1	2, 7	3, 6	4, 5
2.	,,	5, 5	6, 4	7, 3	1, 2
3.	,,	2, 2	3, 1	4, 7	5, 6
4.	,,	6, 6	7, 5	1, 4	2, 3
5.	,,	3, 3	4, 2	5, 1	6, 7
6.	,,	7, 7	1, 6	2, 5	3, 4
7.	,,	4, 4	5, 3	6, 2	7, 1

Then the first column is updated. The first or second position are alternating substituted by a bye, if $n$ is not even, or by $n$.

The pairing tables were published as an annex concerning the arrangements for the holding of master tournaments.

Relatieve ratings uit de periode Morphy

Als er toernooiresultaten bekend zijn over een langere periode, dan kunnen de relatieve ratings worden vastgesteld, ook als spelers niet tegen elkaar hebben gespeeld. Elo werkt dit uit, op basis van de onderstaande kruistabel.

												W	P
Rating		Player	A	Ha	Ho	K	L	M	P	S	W	Wins	Pct.
552		Anderssen		10½	1½	10½	5	4	5	4		40½	.513
518		Harrwitz	7½		14½	1½	16	3½			21	64	.542
406		Horwitz	½	11½		1	1			11	7½	32½	.378
516		Kolisch	9½	2½	3				17			32	.500
505		Lowenthal	3	11	5			4½		2	11	36½	.474
695		Morphy	13	5½			10½		9½			38½	.726
502		Paulsen	4			19		2½				25½	.447
508		Staunton	1	7	20						11	39	.591
425		Williams		6	9½		8			10		33½	.399
514		Losses:L	38½	54	53½	32	40½	14½	31½	27	50½	342

De relatieve rating van een speler wordt berekend op basis de formule

• R_p = R_c + d(p) (E1)

Hierin is R_p de eigen rating en R_c de gemiddelde rating van de tegenstanders, gewogen per gespeelde partij.

De relatieve rating wordt nu door successieve benaderingen berekend:

1. Wijs aan alle spelers één initiële rating R_i toe, groot genoeg om tijdens de iteratie positief te blijven
2. Vind voor iedere speler de d(p) op basis van de werkelijke score en de relatie tussen winstkans en rating verschil.
3. Bereken vervolgens voor iedere speler de eerste correctie R1 op basis van regel (E1), met R_c = R_i
4. Bepaal vervolgens voor iedere speler het gewogen gemiddelde van de tegenstanderratings R_c1.
5. Bepaal de tweede benadering op basis van formule (E1), met R_c = R_c1
6. Vervolg de berekening totdat de berekende ratings weinig veranderen.

Deze methode convergeert niet bijzonder snel.

De oplossing kan ook beschouwd worden als het nulpunt van de vergelijking:

• We(x) = W,

waarbij We(x) de verwachte score is, afhankelijk van de relatieve ratings. Het nulpunt kan met moderne iteratieve methodes efficiënt worden bepaald. Zie ook voor een moderne benadering van dit probleem.

Irregularities in The Percentage Expectancy Table

Tables of Normal Probability Functions U. S. Department of Commerce National Bureau of Standards Applied Mathematics Series #23 Issued June 5, 1953

Fide Table Handbook Fide, https://handbook.fide.com/chapter/B022022 The Rating of Chessplayers.

Below we find the irregularities in the table:

         σ=2000/7
 Rtg H  |z-score |table x           |Tabel y 50% + 2y |Calc             | P
 Dif    |        |                  |                 |                 |
 54  58%| 0,1890 |0,149907183209998 |0,574953591604999|0,574953591604999| 57%
     88%| 1,2000 |0,769860659556583 |0,884930329778292|0,884930329778292|
 343    | 1,2005 |                  |                 |0,885027364557095| 89%
     88%| 1,2010 |0,770248798671795 |0,885124399335897|0,885124399335898|
 344 88%| 1,2040 |0,771410421454696 |0,885705210727348|0,885705210727348| 89% *
        |        |                  |                 |                 |
 358 90%| 1,253  |0,789794293303286 |0,894897146651643|0,894897146651643| 89% *
 392 92%| 1,372  |0,829936562186873 |0,914968281093437|0,914968281093436| 91% *
 620 99%| 2,170  |0,969993154052536 |0,984996577026268|0,984996577026268| 98% *
 

Inflation in FIDE ratings

There is no relationship between strength and rating increase. Elo rating is differential. So if FIDE decides to lower the Elo rating of all its members in the pool by 100, the scoring probabilities between players remain the same. As more rating points are added, the average rating per player in the pool will increase.

Rating development chess world champions (Chess Fide Ratings)

The graph shows the expectation of the champion's rating from the FIDE rating list, compared to the ratings of the following 19 players. Robert Fischer's superiority is undeniable. The line represents the average rating development. The rating inflation since 1985 is clear.

Magnus Carlsen's rating as of February 2022 is 2865, and the average rating difference with the next 19 players is 102 (64%). Rating inflation have stabilized.

The meaning of the K factor

The K-factor determines how we weigh past and present. A low K gives more weight to past performances. (Elo, 1978, p. 25). In the following example we show the effect of the K-factor on the rating development in more detail. The K factor in this example is set to 32.

The new rating (Rn) is approximately equal to the rating performance (Rp) calculated over the games played in the rating period (N = 5) supplemented by (800/K - N) = 20 fictitious draws against own rating.

Let N = 5, Ne = 20, N + Ne = 25, K = 4C / 25, R0 = 1613.

The class interval C = 200 is rooted in tradition (Elo, 1978, p. 19). In this example, the past weighs 4 times as much as the present. If K is set to 10 then the ratio between present and past becomes 1 to 15.

K-factor example: Rn ≈≈ Rp

	R0 =1613				Elo	Elo	800	Lin	A400
Pl	Rc	N	W	D	P(D)	We	P(D)	We	Rp
S0	1613	20	10	0	50,0%	10,00	50,0%	10,00	0
S1	1609	1	½	4	50,6%	0,51	50,5%	0,51	0
S2	1477	1	½	136	68,3%	0,68	67,0%	0,67	0
S3	1388	1	1	225	78,5%	0,78	78,1%	0,78	400
S4	1586	1	1	27	53,8%	0,54	53,4%	0,53	400
S5	1720	1	0	-107	35,4%	0,35	36,6%	0,37	-400
	1601,600	25	13,0			12,865		12,856	400/25
		P =	52,00%					P =	52,00%
		D(P) =	14,330					D800(P) =	16

We note the following:

Rn_Elo   = 1617,33 = 1613 + 32(13,000 – 12,865)
Rn_P800 = 1617,60 = 1613 + 32(13,000 – 12,856)
Rp_D800 = 1617,60 = 1601,600 + 16
Rp_Elo  = 1615,93 = 1601,600 + 14,330

As the expectation curve between -300 and +300 is almost linear we have:

Rn_Elo ≈≈ Rn_D800 == Rp_D800 ≈≈ Rp_Elo, calculated over the above 25 games.

Percentage Expectancy Curve

P(D) = norm.dist(D, 0, 2000 / 7, cumulative), (Elo, 1978, p.28)

Linear Approximation Formulae ( "algorithm of 400")

P800(D) = D / 4C + 50%, slope = 1 / 4C, intercept = 50%.

D800(P) = (P - 50%)4C.

kolommen

${\begin{array}{lll}Rn_{Elo}&=&1601.33=&1613+32(12.500-12.865)\\Rn_{D800}&=&1601.60=&1613+32(12.500-12.856)\\Rp_{DP800}&=&1601.60=&1601.600+0\\Rp_{Elo}&=&1601.60=&1601.600+0\end{array}}$