Round-robin tournament schedule
Round-robin tournament schedule is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
A round-robin tournament is also known as an all-play-all-tournament; each participant plays every other participant once.
For N participants the number of rounds is N-1 if N is an even number. When there are an odd number of participants then each round one contestor has no opponent (AKA as a "bye"). The number of rounds is N in that case.
- Task
Write a program that prints out a tournament schedule for 12 participants (represented by numbers 1 to 12).
- See also
AWK
# syntax: GAWK -f ROUND-ROBIN_TOURNAMENT_SCHEDULE.AWK
BEGIN {
main(1)
main(2)
main(5,"The Bizzaros")
main(12)
exit(0)
}
function main(n,description, arr,i,j,leng,tmp) {
if (n < 2) {
printf("\n%d is too few participants\n",n)
return
}
printf("\n%d players %s\n",n,description)
for (i=1; i<=n; i++) {
arr[i] = i
}
if (n % 2 == 1) {
arr[++n] = 0 # a "bye"
}
leng = length(n-1)
for (i=1; i<n; i++) {
printf("\nround %*d:",leng,i)
for (j=1; j<=n/2; j++) {
printf("%4s",arr[j]==0?"bye":arr[j])
}
printf("\n%*s",leng+7,"")
for (j=n; j>n/2; j--) {
printf("%4s",arr[j]==0?"bye":arr[j])
}
printf("\n")
tmp = arr[n]
for (j=n; j>2; j--) {
arr[j] = arr[j-1]
}
arr[2] = tmp
}
}
- Output:
1 is too few participants
2 players
round 1: 1
2
5 players The Bizzaros
round 1: 1 2 3
bye 5 4
round 2: 1 bye 2
5 4 3
round 3: 1 5 bye
4 3 2
round 4: 1 4 5
3 2 bye
round 5: 1 3 4
2 bye 5
12 players
round 1: 1 2 3 4 5 6
12 11 10 9 8 7
round 2: 1 12 2 3 4 5
11 10 9 8 7 6
round 3: 1 11 12 2 3 4
10 9 8 7 6 5
round 4: 1 10 11 12 2 3
9 8 7 6 5 4
round 5: 1 9 10 11 12 2
8 7 6 5 4 3
round 6: 1 8 9 10 11 12
7 6 5 4 3 2
round 7: 1 7 8 9 10 11
6 5 4 3 2 12
round 8: 1 6 7 8 9 10
5 4 3 2 12 11
round 9: 1 5 6 7 8 9
4 3 2 12 11 10
round 10: 1 4 5 6 7 8
3 2 12 11 10 9
round 11: 1 3 4 5 6 7
2 12 11 10 9 8
Common Lisp
Draft
Program
;; 221130 Draft
(defun planning-tournoi (joueurs &aux (stop ()))
(labels ((tour (sac &optional (lst ()) &aux (duo ()))
(unless (intersection lst stop :test #'equal)
(cond (sac
(dolist (i sac)
(dolist (j (rest (member i sac)))
(setf duo (list i j))
(tour (set-difference sac duo) (list* duo lst)))))
(t (print (reverse lst))
(setf stop (nconc stop lst)))))))
(tour joueurs)))
Execution
(planning-tournoi '(A B C D E *))
- Output:
((A B) (C D) (E *)) ((A C) (B E) (D *)) ((A D) (B *) (C E)) ((A E) (B D) (C *)) ((A *) (B C) (D E))
(planning-tournoi '(1 2 3 4 5 6 7 8 9 10 11 12))
- Output:
((1 2) (3 4) (5 6) (7 8) (9 10) (11 12)) ((1 3) (2 4) (5 7) (6 8) (9 11) (10 12)) ((1 4) (2 3) (5 8) (6 7) (9 12) (10 11)) ((1 5) (2 6) (3 9) (4 10) (7 11) (8 12)) ((1 6) (2 5) (3 10) (4 9) (7 12) (8 11)) ((1 7) (2 8) (3 11) (4 12) (5 9) (6 10)) ((1 8) (2 7) (3 12) (4 11) (5 10) (6 9)) ((1 9) (2 10) (3 7) (4 8) (5 11) (6 12)) ((1 10) (2 9) (3 8) (4 7) (5 12) (6 11)) ((1 11) (2 12) (3 5) (4 6) (7 9) (8 10)) ((1 12) (2 11) (3 6) (4 5) (7 10) (8 9))
cyril nocton (cyril.nocton@gmail.com) w/ google translate
FreeBASIC
function nob( n as uinteger, i as uinteger, bye as boolean ) as string
'helper function to allow byes to be printed intelligently
dim as string pad
if n > 9 then pad = " " else pad = ""
if n = i and bye then
return pad+"B"
else
if i<10 then return pad + str(i) else return str(i)
end if
end function
sub roundrob( byval n as uinteger )
dim as boolean bye = false
if n mod 2 = 1 then 'if there is an odd number of competitors
bye = 1 'make note of this fact
n += 1 'and treat the tournament as having one more competitor
end if
dim as uinteger schd(1 to n), r, i, temp1, temp2
for i = 1 to n
schd(i) =i 'initial population of the array with numbers 1-n
next i
for r = 1 to n-1
print using "Round ##: ";r;
for i = 1 to n/2 'print the pairings according to the scheme
'1 2 3 4
'5 6 7 8
print using "(& - &) ";nob(n,schd(i),bye);nob(n,schd(i+n\2),bye);
next i
print
'now move positions 2-n around clockwise
temp1 = schd(n/2) 'need to track two temporary variables
temp2 = schd(n/2+1)
for i = n/2 to 3 step -1 'top row
schd(i) = schd(i-1)
next i
for i = n/2+1 to n-1 'bottom row
schd(i) = schd(i+1)
next i
schd(n) = temp1 'fill in the ones that "jumped" between rows
schd(2) = temp2
next r
end sub
print "Twelve teams"
roundrob(12)
print "Nine teams with byes"
roundrob(9)- Output:
Twelve teams Round 1: ( 1 - 7) ( 2 - 8) ( 3 - 9) ( 4 - 10) ( 5 - 11) ( 6 - 12) Round 2: ( 1 - 8) ( 7 - 9) ( 2 - 10) ( 3 - 11) ( 4 - 12) ( 5 - 6) Round 3: ( 1 - 9) ( 8 - 10) ( 7 - 11) ( 2 - 12) ( 3 - 6) ( 4 - 5) Round 4: ( 1 - 10) ( 9 - 11) ( 8 - 12) ( 7 - 6) ( 2 - 5) ( 3 - 4) Round 5: ( 1 - 11) (10 - 12) ( 9 - 6) ( 8 - 5) ( 7 - 4) ( 2 - 3) Round 6: ( 1 - 12) (11 - 6) (10 - 5) ( 9 - 4) ( 8 - 3) ( 7 - 2) Round 7: ( 1 - 6) (12 - 5) (11 - 4) (10 - 3) ( 9 - 2) ( 8 - 7) Round 8: ( 1 - 5) ( 6 - 4) (12 - 3) (11 - 2) (10 - 7) ( 9 - 8) Round 9: ( 1 - 4) ( 5 - 3) ( 6 - 2) (12 - 7) (11 - 8) (10 - 9) Round 10: ( 1 - 3) ( 4 - 2) ( 5 - 7) ( 6 - 8) (12 - 9) (11 - 10) Round 11: ( 1 - 2) ( 3 - 7) ( 4 - 8) ( 5 - 9) ( 6 - 10) (12 - 11) Nine teams with byes Round 1: ( 1 - 6) ( 2 - 7) ( 3 - 8) ( 4 - 9) ( 5 - B) Round 2: ( 1 - 7) ( 6 - 8) ( 2 - 9) ( 3 - B) ( 4 - 5) Round 3: ( 1 - 8) ( 7 - 9) ( 6 - B) ( 2 - 5) ( 3 - 4) Round 4: ( 1 - 9) ( 8 - B) ( 7 - 5) ( 6 - 4) ( 2 - 3) Round 5: ( 1 - B) ( 9 - 5) ( 8 - 4) ( 7 - 3) ( 6 - 2) Round 6: ( 1 - 5) ( B - 4) ( 9 - 3) ( 8 - 2) ( 7 - 6) Round 7: ( 1 - 4) ( 5 - 3) ( B - 2) ( 9 - 6) ( 8 - 7) Round 8: ( 1 - 3) ( 4 - 2) ( 5 - 6) ( B - 7) ( 9 - 8) Round 9: ( 1 - 2) ( 3 - 6) ( 4 - 7) ( 5 - 8) ( B - 9)
Go
package main
import "fmt"
func rotate(lst []int) {
len := len(lst)
last := lst[len-1]
for i := len - 1; i >= 1; i-- {
lst[i] = lst[i-1]
}
lst[0] = last
}
func roundRobin(n int) {
lst := make([]int, n-1)
for i := 0; i < len(lst); i++ {
lst[i] = i + 2
}
if n%2 == 1 {
lst = append(lst, 0) // 0 denotes a bye
n++
}
for r := 1; r < n; r++ {
fmt.Printf("Round %2d", r)
lst2 := append([]int{1}, lst...)
for i := 0; i < n/2; i++ {
fmt.Printf(" (%2d vs %-2d)", lst2[i], lst2[n-1-i])
}
fmt.Println()
rotate(lst)
}
}
func main() {
fmt.Println("Round robin for 12 players:\n")
roundRobin(12)
fmt.Println("\n\nRound robin for 5 players (0 denotes a bye) :\n")
roundRobin(5)
}
- Output:
Same as Wren example.
J
Implementation (using the wikipedia circle method):
circ=: {{
if. 1=2|y do.
assert. 1<y
<:(#~ [: */"1 *)"2 circ y+1
else.
ids=. i.y
(-:y) ({.,.|.@}.)"_1] 0,.(}:ids)|."0 1}.ids
end.
}}
Task example:
rplc&'j:'"1":j./"1>:circ 12
1:12 2:11 3:10 4:9 5:8 6:7
1:2 3:12 4:11 5:10 6:9 7:8
1:3 4:2 5:12 6:11 7:10 8:9
1:4 5:3 6:2 7:12 8:11 9:10
1:5 6:4 7:3 8:2 9:12 10:11
1:6 7:5 8:4 9:3 10:2 11:12
1:7 8:6 9:5 10:4 11:3 12:2
1:8 9:7 10:6 11:5 12:4 2:3
1:9 10:8 11:7 12:6 2:5 3:4
1:10 11:9 12:8 2:7 3:6 4:5
1:11 12:10 2:9 3:8 4:7 5:6
(Here, circ uses index values which start at zero, so we need to add 1 to every index. Then we form the id pairs as complex numbers, replace the 'j' used to separate real from imaginary in their character representation with ':' for a hopefully compact and easy-to-read display.)
Julia
""" https://rosettacode.org/mw/index.php?title=Round-robin_tournament_schedule """
function schurig(N, verbose = true)
""" Taken from https://en.wikipedia.org/wiki/Round-robin_tournament
#Original_construction_of_pairing_tables_by_Richard_Schurig_(1886) """
nrows = isodd(N) ? N : N - 1
ncols = (N + 1) ÷ 2
players = mod1.(reshape(collect(1:nrows*ncols), ncols, nrows)', nrows)
opponents = zero(players)
table = [(0, 0) for _ in 1:nrows, _ in 1:ncols]
for i in 1:nrows
oldrow = i == nrows ? 1 : i + 1
verbose && print("\n", rpad("Round $i:", 10))
for j in 1:ncols
oldcol = ncols - j + 1
opponents[i, j] = players[oldrow, oldcol]
j == 1 && (opponents[i, j] = iseven(N) ? N : 0)
table[i, j] = (sort([players[i, j], opponents[i, j]])...,)
if verbose
s1, s2 = string.(table[i, j])
print(rpad("($(s1 == "0" ? "Bye" : s1) - $s2)", 10))
end
end
end
return table
end
print("Schurig table for round robin with 12 players:")
schurig(12)
print("\n\nSchurig table for round robin with 7 players:")
schurig(7)
- Output:
Schurig table for round robin with 12 players: Round 1: (1 - 12) (2 - 11) (3 - 10) (4 - 9) (5 - 8) (6 - 7) Round 2: (7 - 12) (6 - 8) (5 - 9) (4 - 10) (3 - 11) (1 - 2) Round 3: (2 - 12) (1 - 3) (4 - 11) (5 - 10) (6 - 9) (7 - 8) Round 4: (8 - 12) (7 - 9) (6 - 10) (5 - 11) (1 - 4) (2 - 3) Round 5: (3 - 12) (2 - 4) (1 - 5) (6 - 11) (7 - 10) (8 - 9) Round 6: (9 - 12) (8 - 10) (7 - 11) (1 - 6) (2 - 5) (3 - 4) Round 7: (4 - 12) (3 - 5) (2 - 6) (1 - 7) (8 - 11) (9 - 10) Round 8: (10 - 12) (9 - 11) (1 - 8) (2 - 7) (3 - 6) (4 - 5) Round 9: (5 - 12) (4 - 6) (3 - 7) (2 - 8) (1 - 9) (10 - 11) Round 10: (11 - 12) (1 - 10) (2 - 9) (3 - 8) (4 - 7) (5 - 6) Round 11: (6 - 12) (5 - 7) (4 - 8) (3 - 9) (2 - 10) (1 - 11) Schurig table for round robin with 7 players: Round 1: (Bye - 1) (2 - 7) (3 - 6) (4 - 5) Round 2: (Bye - 5) (4 - 6) (3 - 7) (1 - 2) Round 3: (Bye - 2) (1 - 3) (4 - 7) (5 - 6) Round 4: (Bye - 6) (5 - 7) (1 - 4) (2 - 3) Round 5: (Bye - 3) (2 - 4) (1 - 5) (6 - 7) Round 6: (Bye - 7) (1 - 6) (2 - 5) (3 - 4) Round 7: (Bye - 4) (3 - 5) (2 - 6) (1 - 7)
Perl
Even
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Round-robin_tournament_schedule
use warnings;
my $n = 12;
my @teams = 1 .. $n;
for (1 .. $n-1)
{
@teams[0,$n-1,1..$n-2] = @teams;
printf 'Round %2d:' . '%4d vs %2d'x($n/2) . "\n", $_, @teams[ map { $_, $n-1-$_} 0..($n/2)-1 ];
}
- Output:
Round 1: 1 vs 2 3 vs 12 4 vs 11 5 vs 10 6 vs 9 7 vs 8 Round 2: 1 vs 3 4 vs 2 5 vs 12 6 vs 11 7 vs 10 8 vs 9 Round 3: 1 vs 4 5 vs 3 6 vs 2 7 vs 12 8 vs 11 9 vs 10 Round 4: 1 vs 5 6 vs 4 7 vs 3 8 vs 2 9 vs 12 10 vs 11 Round 5: 1 vs 6 7 vs 5 8 vs 4 9 vs 3 10 vs 2 11 vs 12 Round 6: 1 vs 7 8 vs 6 9 vs 5 10 vs 4 11 vs 3 12 vs 2 Round 7: 1 vs 8 9 vs 7 10 vs 6 11 vs 5 12 vs 4 2 vs 3 Round 8: 1 vs 9 10 vs 8 11 vs 7 12 vs 6 2 vs 5 3 vs 4 Round 9: 1 vs 10 11 vs 9 12 vs 8 2 vs 7 3 vs 6 4 vs 5 Round 10: 1 vs 11 12 vs 10 2 vs 9 3 vs 8 4 vs 7 5 vs 6 Round 11: 1 vs 12 2 vs 11 3 vs 10 4 vs 9 5 vs 8 6 vs 7
Even and Odd
use strict;
use warnings;
use feature 'say';
use List::AllUtils <pairwise all>;
sub round_robin {
my($n) = @_;
my($round,@pairings);
my @players = (1,0)[$n%2] .. $n;
my $half = +@players / 2;
while () {
my @a = @players[ 0 .. $half-1];
my @b = reverse @players[$half .. $#players];
push @pairings, sprintf "Round %2d: %s\n", ++$round, join ' ', pairwise { sprintf "%3d vs %2d", $a, $b } @a, @b;
push @players, splice @players, 1, @players-2;
last if all { $players[$_-1] < $players[$_] } 1..$#players;
}
@pairings
}
say join '', round_robin 12;
say '';
say join '', map { s/0 vs /Bye: /r } round_robin 7;
- Output:
Round 1: 1 vs 12 2 vs 11 3 vs 10 4 vs 9 5 vs 8 6 vs 7 Round 2: 1 vs 11 12 vs 10 2 vs 9 3 vs 8 4 vs 7 5 vs 6 Round 3: 1 vs 10 11 vs 9 12 vs 8 2 vs 7 3 vs 6 4 vs 5 Round 4: 1 vs 9 10 vs 8 11 vs 7 12 vs 6 2 vs 5 3 vs 4 Round 5: 1 vs 8 9 vs 7 10 vs 6 11 vs 5 12 vs 4 2 vs 3 Round 6: 1 vs 7 8 vs 6 9 vs 5 10 vs 4 11 vs 3 12 vs 2 Round 7: 1 vs 6 7 vs 5 8 vs 4 9 vs 3 10 vs 2 11 vs 12 Round 8: 1 vs 5 6 vs 4 7 vs 3 8 vs 2 9 vs 12 10 vs 11 Round 9: 1 vs 4 5 vs 3 6 vs 2 7 vs 12 8 vs 11 9 vs 10 Round 10: 1 vs 3 4 vs 2 5 vs 12 6 vs 11 7 vs 10 8 vs 9 Round 11: 1 vs 2 3 vs 12 4 vs 11 5 vs 10 6 vs 9 7 vs 8 Round 1: Bye: 7 1 vs 6 2 vs 5 3 vs 4 Round 2: Bye: 6 7 vs 5 1 vs 4 2 vs 3 Round 3: Bye: 5 6 vs 4 7 vs 3 1 vs 2 Round 4: Bye: 4 5 vs 3 6 vs 2 7 vs 1 Round 5: Bye: 3 4 vs 2 5 vs 1 6 vs 7 Round 6: Bye: 2 3 vs 1 4 vs 7 5 vs 6 Round 7: Bye: 1 2 vs 7 3 vs 6 4 vs 5
Phix
Based on the circle with rotor diagrams on the wikipedia page, and implements home/away.
with javascript_semantics function round_robin(integer n) -- -- As per the wikipedia page, we do something like this: -- -- even(n), say 6: in round 1 we have 6 & 1,2,3,4,5 -> {{6,1},{2,5},{3,4}}, -- 2 2,3,4,5,1 -> {{6,2},{3,1},{4,5}}, -- 3 3,4,5,1,2 -> {{6,3},{4,2},{5,1}}, -- 4 4,5,1,2,3 -> {{6,4},{5,3},{1,2}}, -- 5 5,1,2,3,4 -> {{6,5},{1,4},{2,3}} -- -- for an odd(n), say 5, simply replace all the 6 above with 0 (a bye). -- -- As per the wikipedia diagram, we pick a rotor (6/0) and arrange the rest -- in a circle, and as it rotates around the circle we play it against that -- one then pick off second/last, third/last-but-one, and so forth from the -- perspective of the rotor. There must obviously be an odd number of teams -- in the circle itself, otherwise pairing-offs won't meet in the middle. -- -- However, rather than physically rotate the {1,2,3,4,5}, we'll just say -- that anything past 5 starts from 1 again (the -= n below), and use the -- shorthand of l [===length(result)] as our starting position/offset. -- -- Not shown above, but we'll also use even/odd rules for home/away matches. -- integer rotor = iff(even(n)?n:0), l = 0 -- length(result), shorthand n -= even(n) -- (circle must be odd) sequence result = {} for rownd=1 to n do -- (since "round" is a builtin) sequence games = {iff(even(rownd) or rotor=0?{rownd,rotor}:{rotor,rownd})} integer opponent = n -- pair rest off from last inwards, for m=2 to (n+1)/2 do -- such that m plays current opponent integer rom = m+l, -- all shifted by rop = opponent+l -- l as an offset if rom>n then rom -= n end if if rop>n then rop -= n end if games &= iff(odd(m)?{{rom,rop}}:{{rop,rom}}) opponent -= 1 end for result = append(result,games) l += 1 -- (obviously "l = length(result)" works fine here too) end for return result end function function vs(sequence pair) -- (display helper) return sprintf(iff(pair[2]=0?"%2d bye ":"%2d vs %-2d"),pair) end function for test=12 to 3 by -9 do sequence res = round_robin(test) printf(1,"\nFor %d teams:\n",test) for r=1 to length(res) do printf(1,"Round %2d: %s\n",{r,join(apply(res[r],vs))}) end for end for
- Output:
For 12 teams: Round 1: 12 vs 1 11 vs 2 3 vs 10 9 vs 4 5 vs 8 7 vs 6 Round 2: 2 vs 12 1 vs 3 4 vs 11 10 vs 5 6 vs 9 8 vs 7 Round 3: 12 vs 3 2 vs 4 5 vs 1 11 vs 6 7 vs 10 9 vs 8 Round 4: 4 vs 12 3 vs 5 6 vs 2 1 vs 7 8 vs 11 10 vs 9 Round 5: 12 vs 5 4 vs 6 7 vs 3 2 vs 8 9 vs 1 11 vs 10 Round 6: 6 vs 12 5 vs 7 8 vs 4 3 vs 9 10 vs 2 1 vs 11 Round 7: 12 vs 7 6 vs 8 9 vs 5 4 vs 10 11 vs 3 2 vs 1 Round 8: 8 vs 12 7 vs 9 10 vs 6 5 vs 11 1 vs 4 3 vs 2 Round 9: 12 vs 9 8 vs 10 11 vs 7 6 vs 1 2 vs 5 4 vs 3 Round 10: 10 vs 12 9 vs 11 1 vs 8 7 vs 2 3 vs 6 5 vs 4 Round 11: 12 vs 11 10 vs 1 2 vs 9 8 vs 3 4 vs 7 6 vs 5 For 3 teams: Round 1: 1 bye 3 vs 2 Round 2: 2 bye 1 vs 3 Round 3: 3 bye 2 vs 1
While I "optimised away" the need for a physical rotate, obviously not because I was concerned with performance but more in the hope of creating shorter and more elegant code, in the end it made little difference. Should it be more to your taste, you can remove "l" and replace the inner loop above with:
sequence circle = tagset(n) for rownd=1 to n do -- (since "round" is a bultin) integer r = circle[1] sequence games = {iff(even(rownd) or rotor=0?{r,rotor}:{rotor,r})} integer ldx = 2, rdx = n while ldx<rdx do integer teama = circle[ldx], teamb = circle[rdx] games &= {iff(odd(ldx)?{teama,teamb},{teamb,teama})} ldx += 1 rdx -= 1 end while result = append(result,games) circle = circle[2..$]&circle[1] -- (physically rotate it) end for
Picat
Constraint modelling
import sat.
main =>
nolog,
N = 12,
tournament_cp(N, NumRounds,NumPairs,_,X,Bye),
print_tournament(X,NumRounds,NumPairs,Bye),
nl.
tournament_cp(N, NumRounds,NumPairs,Extras, X,Bye) =>
% Adjust for odd number of players.
% The bye (Dummy) player is N+1.
if N mod 2 == 1 then
N := N + 1,
Bye = N
end,
NumRounds = N-1,
NumPairs = N div 2,
X = new_array(NumRounds,NumPairs,2),
X :: 1..N,
% ensure that all players play each other
foreach(P1 in 1..N, P2 in P1+1..N)
sum([X[Round,P,1] #= P1 #/\ X[Round,P,2] #= P2 : Round in 1..NumRounds, P in 1..NumPairs]) #= 1
end,
foreach(Round in 1..NumRounds)
all_different([X[Round,I,J] : I in 1..NumPairs, J in 1..2]),
% symmetry breaking
% - all first players in increasing order
increasing_strict([X[Round,I,1] : I in 1..NumPairs]),
% - player 1 < player 2
foreach(P in 1..NumPairs)
X[Round,P,1] #< X[Round,P,2]
end
end,
if Extras != [] then
foreach([P1,P2,Round] in Extras)
sum([X[Round,P,1] #= P1 #/\ X[Round,P,2] #= P2 : P in 1..NumPairs]) #= 1
end
end,
solve($[ff,split],X).
print_tournament(X,NumRounds,NumPairs,Bye) =>
N = X[1].len,
foreach(Round in 1..NumRounds)
printf("Round %2d: ", Round),
if N > 10 then nl end,
foreach(P in 1..NumPairs)
P2Val = X[Round,P,2],
if var(Bye) ; P2Val != Bye then
printf("(%2w vs %2w) ",X[Round,P,1],P2Val),
if N > 10 then nl end
end
end,
nl
end,
nl.- Output:
Round 1: ( 1 vs 11) ( 2 vs 5) ( 3 vs 6) ( 4 vs 12) ( 7 vs 9) ( 8 vs 10) Round 2: ( 1 vs 5) ( 2 vs 4) ( 3 vs 10) ( 6 vs 7) ( 8 vs 9) (11 vs 12) Round 3: ( 1 vs 6) ( 2 vs 8) ( 3 vs 5) ( 4 vs 11) ( 7 vs 10) ( 9 vs 12) Round 4: ( 1 vs 12) ( 2 vs 11) ( 3 vs 7) ( 4 vs 6) ( 5 vs 8) ( 9 vs 10) Round 5: ( 1 vs 9) ( 2 vs 6) ( 3 vs 12) ( 4 vs 5) ( 7 vs 8) (10 vs 11) Round 6: ( 1 vs 4) ( 2 vs 3) ( 5 vs 7) ( 6 vs 10) ( 8 vs 12) ( 9 vs 11) Round 7: ( 1 vs 2) ( 3 vs 4) ( 5 vs 10) ( 6 vs 9) ( 7 vs 12) ( 8 vs 11) Round 8: ( 1 vs 3) ( 2 vs 12) ( 4 vs 10) ( 5 vs 9) ( 6 vs 8) ( 7 vs 11) Round 9: ( 1 vs 10) ( 2 vs 7) ( 3 vs 8) ( 4 vs 9) ( 5 vs 11) ( 6 vs 12) Round 10: ( 1 vs 8) ( 2 vs 10) ( 3 vs 9) ( 4 vs 7) ( 5 vs 12) ( 6 vs 11) Round 11: ( 1 vs 7) ( 2 vs 9) ( 3 vs 11) ( 4 vs 8) ( 5 vs 6) (10 vs 12)
Constraint model with extra constraints
The constraint model is slower than the algorithmic approach for larger number of players. The advantage of a constraint model is that it is quite easy to add extra constraint, such that some players must play in a certain round (e.g. for availability reasons etc).
Here are some extra constraints:
- 1 vs 2 must be played the third round
- 5 vs 9 must be played in the 7th round
- 2 vs 3 must be played in the last round
- 7 vs 12 must be played in the last round
main =>
nolog,
N = 12,
Extras = [[1,2,3],
[5,9,7],
[2,3,N-1],
[7,12,N-1]],
tournament_cp(N, NumRounds,NumPairs,Extras,X,Bye),
print_tournament(X,NumRounds,NumPairs,Bye).- Output:
Round 1: ( 1 vs 11) ( 2 vs 4) ( 3 vs 12) ( 5 vs 8) ( 6 vs 9) ( 7 vs 10) Round 2: ( 1 vs 12) ( 2 vs 11) ( 3 vs 9) ( 4 vs 7) ( 5 vs 10) ( 6 vs 8) Round 3: ( 1 vs 2) ( 3 vs 10) ( 4 vs 12) ( 5 vs 11) ( 6 vs 7) ( 8 vs 9) Round 4: ( 1 vs 4) ( 2 vs 6) ( 3 vs 11) ( 5 vs 12) ( 7 vs 8) ( 9 vs 10) Round 5: ( 1 vs 10) ( 2 vs 7) ( 3 vs 5) ( 4 vs 6) ( 8 vs 12) ( 9 vs 11) Round 6: ( 1 vs 6) ( 2 vs 5) ( 3 vs 4) ( 7 vs 9) ( 8 vs 11) (10 vs 12) Round 7: ( 1 vs 3) ( 2 vs 12) ( 4 vs 8) ( 5 vs 9) ( 6 vs 10) ( 7 vs 11) Round 8: ( 1 vs 7) ( 2 vs 8) ( 3 vs 6) ( 4 vs 5) ( 9 vs 12) (10 vs 11) Round 9: ( 1 vs 8) ( 2 vs 9) ( 3 vs 7) ( 4 vs 10) ( 5 vs 6) (11 vs 12) Round 10: ( 1 vs 9) ( 2 vs 10) ( 3 vs 8) ( 4 vs 11) ( 5 vs 7) ( 6 vs 12) Round 11: ( 1 vs 5) ( 2 vs 3) ( 4 vs 9) ( 6 vs 11) ( 7 vs 12) ( 8 vs 10)
For this small tournament it took about the same time with and without these extra constraints (0.08s).
Number of solutions
Here are the number of different solutions for N = [2,4,6,8] with the symmetry constraints (but without the extra round constraints). The number of odd N players is the same as the number of N-1 players.
Here the cp solver is used since it's faster than the sat solver for generating all solutions.
import cp.
main =>
foreach(N in 2..2..8)
Count = count_all(tournament_cp(N, _NumRounds,_NumPairs,_Extras,_X,_Bye)),
println(N=Count)
end.- Output:
2 = 1 4 = 6 6 = 720 8 = 31449600
This seems to be related to the OEIS sequence "A036981: (2n+1) X (2n+1) symmetric matrices each of whose rows is a permutation of 1..(2n+1)". The next term (for N=10) would be 444733651353600 which takes too long to check.
Raku
my @players = (1,0)[$_%2] .. $_ given 12;
my $half = +@players div 2;
my $round = 0;
loop {
printf "Round %2d: %s\n", ++$round, "{ zip( @players[^$half], @players[$half..*].reverse ).map: { sprintf "(%2d vs %-2d)", |$_ } }";
@players[1..*].=rotate(-1);
last if [<] @players;
}
- Output:
Round 1: ( 1 vs 12) ( 2 vs 11) ( 3 vs 10) ( 4 vs 9 ) ( 5 vs 8 ) ( 6 vs 7 ) Round 2: ( 1 vs 11) (12 vs 10) ( 2 vs 9 ) ( 3 vs 8 ) ( 4 vs 7 ) ( 5 vs 6 ) Round 3: ( 1 vs 10) (11 vs 9 ) (12 vs 8 ) ( 2 vs 7 ) ( 3 vs 6 ) ( 4 vs 5 ) Round 4: ( 1 vs 9 ) (10 vs 8 ) (11 vs 7 ) (12 vs 6 ) ( 2 vs 5 ) ( 3 vs 4 ) Round 5: ( 1 vs 8 ) ( 9 vs 7 ) (10 vs 6 ) (11 vs 5 ) (12 vs 4 ) ( 2 vs 3 ) Round 6: ( 1 vs 7 ) ( 8 vs 6 ) ( 9 vs 5 ) (10 vs 4 ) (11 vs 3 ) (12 vs 2 ) Round 7: ( 1 vs 6 ) ( 7 vs 5 ) ( 8 vs 4 ) ( 9 vs 3 ) (10 vs 2 ) (11 vs 12) Round 8: ( 1 vs 5 ) ( 6 vs 4 ) ( 7 vs 3 ) ( 8 vs 2 ) ( 9 vs 12) (10 vs 11) Round 9: ( 1 vs 4 ) ( 5 vs 3 ) ( 6 vs 2 ) ( 7 vs 12) ( 8 vs 11) ( 9 vs 10) Round 10: ( 1 vs 3 ) ( 4 vs 2 ) ( 5 vs 12) ( 6 vs 11) ( 7 vs 10) ( 8 vs 9 ) Round 11: ( 1 vs 2 ) ( 3 vs 12) ( 4 vs 11) ( 5 vs 10) ( 6 vs 9 ) ( 7 vs 8 )
Ruby
def round_robin( n )
rotating_players = (2..n).map(&:to_s) #player 1 to be added later
rotating_players << "bye" if n.odd?
Array.new(rotating_players.size) do |r|
all = ["1"] + rotating_players.rotate(-r)
[all[0, all.size/2], all[all.size/2..].reverse]
end
end
round_robin(12).each.with_index(1) do |round, i|
puts "Round #{i}"
round.each do |players|
puts players.map{|player| player.ljust(4)}.join
end
puts
end
- Output:
Round 1 1 2 3 4 5 6 12 11 10 9 8 7 Round 2 1 12 2 3 4 5 11 10 9 8 7 6 Round 3 1 11 12 2 3 4 10 9 8 7 6 5 Round 4 1 10 11 12 2 3 9 8 7 6 5 4 Round 5 1 9 10 11 12 2 8 7 6 5 4 3 Round 6 1 8 9 10 11 12 7 6 5 4 3 2 Round 7 1 7 8 9 10 11 6 5 4 3 2 12 Round 8 1 6 7 8 9 10 5 4 3 2 12 11 Round 9 1 5 6 7 8 9 4 3 2 12 11 10 Round 10 1 4 5 6 7 8 3 2 12 11 10 9 Round 11 1 3 4 5 6 7 2 12 11 10 9 8
Wren
import "./fmt" for Fmt
var rotate = Fn.new { |lst|
var last = lst[-1]
for (i in lst.count-1..1) lst[i] = lst[i-1]
lst[0] = last
}
var roundRobin = Fn.new { |n|
var lst = (2..n).toList
if (n % 2 == 1) {
lst.add(0) // 0 denotes a bye
n = n + 1
}
for (r in 1...n) {
Fmt.write("Round $2d", r)
var lst2 = [1] + lst
for (i in 0...n/2) Fmt.write(" ($2d vs $-2d)", lst2[i], lst2[n - 1 - i])
System.print()
rotate.call(lst)
}
}
System.print("Round robin for 12 players:\n")
roundRobin.call(12)
System.print("\n\nRound robin for 5 players (0 denotes a bye) :\n")
roundRobin.call(5)- Output:
Round robin for 12 players: Round 1 ( 1 vs 12) ( 2 vs 11) ( 3 vs 10) ( 4 vs 9 ) ( 5 vs 8 ) ( 6 vs 7 ) Round 2 ( 1 vs 11) (12 vs 10) ( 2 vs 9 ) ( 3 vs 8 ) ( 4 vs 7 ) ( 5 vs 6 ) Round 3 ( 1 vs 10) (11 vs 9 ) (12 vs 8 ) ( 2 vs 7 ) ( 3 vs 6 ) ( 4 vs 5 ) Round 4 ( 1 vs 9 ) (10 vs 8 ) (11 vs 7 ) (12 vs 6 ) ( 2 vs 5 ) ( 3 vs 4 ) Round 5 ( 1 vs 8 ) ( 9 vs 7 ) (10 vs 6 ) (11 vs 5 ) (12 vs 4 ) ( 2 vs 3 ) Round 6 ( 1 vs 7 ) ( 8 vs 6 ) ( 9 vs 5 ) (10 vs 4 ) (11 vs 3 ) (12 vs 2 ) Round 7 ( 1 vs 6 ) ( 7 vs 5 ) ( 8 vs 4 ) ( 9 vs 3 ) (10 vs 2 ) (11 vs 12) Round 8 ( 1 vs 5 ) ( 6 vs 4 ) ( 7 vs 3 ) ( 8 vs 2 ) ( 9 vs 12) (10 vs 11) Round 9 ( 1 vs 4 ) ( 5 vs 3 ) ( 6 vs 2 ) ( 7 vs 12) ( 8 vs 11) ( 9 vs 10) Round 10 ( 1 vs 3 ) ( 4 vs 2 ) ( 5 vs 12) ( 6 vs 11) ( 7 vs 10) ( 8 vs 9 ) Round 11 ( 1 vs 2 ) ( 3 vs 12) ( 4 vs 11) ( 5 vs 10) ( 6 vs 9 ) ( 7 vs 8 ) Round robin for 5 players (0 denotes a bye) : Round 1 ( 1 vs 0 ) ( 2 vs 5 ) ( 3 vs 4 ) Round 2 ( 1 vs 5 ) ( 0 vs 4 ) ( 2 vs 3 ) Round 3 ( 1 vs 4 ) ( 5 vs 3 ) ( 0 vs 2 ) Round 4 ( 1 vs 3 ) ( 4 vs 2 ) ( 5 vs 0 ) Round 5 ( 1 vs 2 ) ( 3 vs 0 ) ( 4 vs 5 )
XPL0
def N = 12; \number of players (must be even)
int I, Player(N+1), Round, Temp;
[for I:= 1 to N do Player(I):= I;
for Round:= 1 to N-1 do
[IntOut(0, Round); ChOut(0, ^:);
for I:= 1 to N/2 do
[ChOut(0, 9\tab\); IntOut(0, Player(I))];
CrLf(0);
for I:= N downto N/2+1 do
[ChOut(0, 9\tab\); IntOut(0, Player(I))];
CrLf(0); CrLf(0);
Temp:= Player(N); \rotate
for I:= N-1 downto 2 do
Player(I+1):= Player(I);
Player(2):= Temp;
];
]- Output:
1: 1 2 3 4 5 6
12 11 10 9 8 7
2: 1 12 2 3 4 5
11 10 9 8 7 6
3: 1 11 12 2 3 4
10 9 8 7 6 5
4: 1 10 11 12 2 3
9 8 7 6 5 4
5: 1 9 10 11 12 2
8 7 6 5 4 3
6: 1 8 9 10 11 12
7 6 5 4 3 2
7: 1 7 8 9 10 11
6 5 4 3 2 12
8: 1 6 7 8 9 10
5 4 3 2 12 11
9: 1 5 6 7 8 9
4 3 2 12 11 10
10: 1 4 5 6 7 8
3 2 12 11 10 9
11: 1 3 4 5 6 7
2 12 11 10 9 8