World Cup group stage

From Rosetta Code
Task
World Cup group stage
You are encouraged to solve this task according to the task description, using any language you may know.
It's World Cup season (or at least it was when this page was created)! The World Cup is an international football/soccer tournament that happens every 4 years. Countries put their international teams together in the years between tournaments and qualify for the tournament based on their performance in other international games. Once a team has qualified they are put into a group with 3 other teams. For the first part of the World Cup tournament the teams play in "group stage" games where each of the four teams in a group plays all three other teams once. The results of these games determine which teams will move on to the "knockout stage" which is a standard single-elimination tournament. The two teams from each group with the most standings points move on to the knockout stage. Each game can result in a win for one team and a loss for the other team or it can result in a draw/tie for each team. A win is worth three points in the standings. A draw/tie is worth one point. A loss is not worth any points.

Generate all possible outcome combinations for the six group stage games. With three possible outcomes for each game there should be 36 = 729 of them. Calculate the standings points for each team with each combination of outcomes. Show a histogram (graphical, ASCII art, or straight counts--whichever is easiest/most fun) of the standings points for all four teams over all possible outcomes.

Don't worry about tiebreakers as they can get complicated. We are basically looking to answer the question "if a team gets x standings points, where can they expect to end up in the group standings?".

Hint: there should be no possible way to end up in second place with less than two points as well as no way to end up in first with less than three. Oddly enough, there is no way to get 8 points at all.

C#[edit]

Translation of: Python
Works with: C# version 7+

Unlike the Python solution, this does not use a library for combinations and cartesian products but provides 4 1-liner Linq methods.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using static System.Console;
using static System.Linq.Enumerable;
 
namespace WorldCupGroupStage
{
public static class WorldCupGroupStage
{
static int[][] _histogram;
 
static WorldCupGroupStage()
{
int[] scoring = new[] { 0, 1, 3 };
 
_histogram = Repeat<Func<int[]>>(()=>new int[10], 4).Select(f=>f()).ToArray();
 
var teamCombos = Range(0, 4).Combinations(2).Select(t2=>t2.ToArray()).ToList();
 
foreach (var results in Range(0, 3).CartesianProduct(6))
{
var points = new int[4];
 
foreach (var (result, teams) in results.Zip(teamCombos, (r, t) => (r, t)))
{
points[teams[0]] += scoring[result];
points[teams[1]] += scoring[2 - result];
}
 
foreach(var (p,i) in points.OrderByDescending(a => a).Select((p,i)=>(p,i)))
_histogram[i][p]++;
}
}
 
// https://gist.github.com/martinfreedman/139dd0ec7df4737651482241e48b062f
 
static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> seqs) =>
seqs.Aggregate(Empty<T>().ToSingleton(), (acc, sq) => acc.SelectMany(a => sq.Select(s => a.Append(s))));
 
static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<T> seq, int repeat = 1) =>
Repeat(seq, repeat).CartesianProduct();
 
static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> seq) =>
seq.Aggregate(Empty<T>().ToSingleton(), (a, b) => a.Concat(a.Select(x => x.Append(b))));
 
static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> seq, int numItems) =>
seq.Combinations().Where(s => s.Count() == numItems);
 
private static IEnumerable<T> ToSingleton<T>(this T item) { yield return item; }
 
static new string ToString()
{
var sb = new StringBuilder();
 
var range = String.Concat(Range(0, 10).Select(i => $"{i,-3} "));
sb.AppendLine($"Points  : {range}");
 
var u = String.Concat(Repeat("─", 40+13));
sb.AppendLine($"{u}");
 
var places = new[] { "First", "Second", "Third", "Fourth" };
foreach (var row in _histogram.Select((r, i) => (r, i)))
{
sb.Append($"{places[row.i],-6} place: ");
foreach (var standing in row.r)
sb.Append($"{standing,-3} ");
sb.Append("\n");
}
 
return sb.ToString();
}
 
static void Main(string[] args)
{
Write(ToString());
Read();
}
}
}
 

Produces:

Points      : 0   1   2   3   4   5   6   7   8   9
─────────────────────────────────────────────────────
First  place: 0   0   0   1   14  148 152 306 0   108
Second place: 0   0   4   33  338 172 164 18  0   0
Third  place: 0   18  136 273 290 4   8   0   0   0
Fourth place: 108 306 184 125 6   0   0   0   0   0

Common Lisp[edit]

Works with: SBCL version 1.4.9
(defun histo ()    
(let ((scoring (vector 0 1 3))
(histo (list (vector 0 0 0 0 0 0 0 0 0 0) (vector 0 0 0 0 0 0 0 0 0 0)
(vector 0 0 0 0 0 0 0 0 0 0) (vector 0 0 0 0 0 0 0 0 0 0)))
(team-combs (vector '(0 1) '(0 2) '(0 3) '(1 2) '(1 3) '(2 3)))
(single-tupel)
(sum))
; six nested dotimes produces the tupels of the cartesian product of
; six lists like '(0 1 2), but without to store all tuples in a list
(dotimes (x0 3) (dotimes (x1 3) (dotimes (x2 3)
(dotimes (x3 3) (dotimes (x4 3) (dotimes (x5 3)
(setf single-tupel (vector x0 x1 x2 x3 x4 x5))
(setf sum (vector 0 0 0 0))
(dotimes (i (length single-tupel))
(setf (elt sum (first (elt team-combs i)))
(+ (elt sum (first (elt team-combs i)))
(elt scoring (elt single-tupel i))))
 
(setf (elt sum (second (elt team-combs i)))
(+ (elt sum (second (elt team-combs i)))
(elt scoring (- 2 (elt single-tupel i))))))
 
(dotimes (i (length (sort sum #'<)))
(setf (elt (nth i histo) (elt sum i))
(1+ (elt (nth i histo) (elt sum i)))))
))))))
(reverse histo)))
 
; friendly output
(dolist (el (histo))
(dotimes (i (length el))
(format t "~3D " (aref el i)))
(format t "~%"))
Output:
  0   0   0   1  14 148 152 306   0 108 
  0   0   4  33 338 172 164  18   0   0 
  0  18 136 273 290   4   8   0   0   0 
108 306 184 125   6   0   0   0   0   0 

D[edit]

This imports the module of the third D solution of the Combinations Task.

Translation of: Python
void main() {
import std.stdio, std.range, std.array, std.algorithm, combinations3;
 
immutable scoring = [0, 1, 3];
/*immutable*/ auto r3 = [0, 1, 2];
immutable combs = 4.iota.array.combinations(2).array;
uint[10][4] histo;
 
foreach (immutable results; cartesianProduct(r3, r3, r3, r3, r3, r3)) {
int[4] s;
foreach (immutable r, const g; [results[]].zip(combs)) {
s[g[0]] += scoring[r];
s[g[1]] += scoring[2 - r];
}
 
foreach (immutable i, immutable v; s[].sort().release)
histo[i][v]++;
}
writefln("%(%s\n%)", histo[].retro);
}
Output:
[0, 0, 0, 1, 14, 148, 152, 306, 0, 108]
[0, 0, 4, 33, 338, 172, 164, 18, 0, 0]
[0, 18, 136, 273, 290, 4, 8, 0, 0, 0]
[108, 306, 184, 125, 6, 0, 0, 0, 0, 0]

This alternative version is not fully idiomatic D, it shows what to currently do to tag the main function of the precedent version as @nogc.

import core.stdc.stdio, std.range, std.array, std.algorithm, combinations3;
 
immutable uint[2][6] combs = 4u.iota.array.combinations(2).array;
 
void main() nothrow @nogc {
immutable uint[3] scoring = [0, 1, 3];
uint[10][4] histo;
 
foreach (immutable r0; 0 .. 3)
foreach (immutable r1; 0 .. 3)
foreach (immutable r2; 0 .. 3)
foreach (immutable r3; 0 .. 3)
foreach (immutable r4; 0 .. 3)
foreach (immutable r5; 0 .. 3) {
uint[4] s;
foreach (immutable i, immutable r; [r0, r1, r2, r3, r4, r5]) {
s[combs[i][0]] += scoring[r];
s[combs[i][1]] += scoring[2 - r];
}
 
foreach (immutable i, immutable v; s[].sort().release)
histo[i][v]++;
}
 
foreach_reverse (const ref h; histo) {
foreach (immutable x; h)
printf("%u ", x);
printf("\n");
}
}
Output:
0 0 0 1 14 148 152 306 0 108 
0 0 4 33 338 172 164 18 0 0 
0 18 136 273 290 4 8 0 0 0 
108 306 184 125 6 0 0 0 0 0 

Elena[edit]

Translation of: Java

ELENA 3.2.1 :

import system'routines.
import extensions.
 
program =
{
static games := ("12", "13", "14", "23", "24", "34").
 
static results := "000000".
 
nextResult
[
if (results=="222222") [ ^ false ].
 
results := (results toInt(3) + 1) toLiteral(3); padLeft($48, 6).
 
^ true
]
 
// program entry point
eval
[
var points := IntMatrix new(4, 10).
 
[
var r := results.
var records := IntArray new(4).
 
0 till:6 do(:i)
[
(results[i]) =>
"2" [ records[games[i][0] toInt - 49] += 3 ];
"1" [
records[games[i][0] toInt - 49] += 1.
records[games[i][1] toInt - 49] += 1
];
"0" [ records[games[i][1] toInt - 49] += 3 ].
].
 
records := records ascendant.
 
0 to:3 do(:i)[ points[i][records[i]] += 1 ].
 
] repeatUntil:$($self nextResult).
 
0 repeatTill:4; zip:("1st", "2nd", "3rd", "4th") forEach(:i:l)
[
console printLine(l,": ", points[3 - i])
].
]
}.
Output:
1st: 0,0,0,1,14,148,152,306,0,108
2nd: 0,0,4,33,338,172,164,18,0,0
3rd: 0,18,136,273,290,4,8,0,0,0
4th: 108,306,184,125,6,0,0,0,0,0

Elixir[edit]

Translation of: Erlang
defmodule World_Cup do
def group_stage do
results = [[3,0],[1,1],[0,3]]
teams = [0,1,2,3]
allresults = combos(2,teams) |> combinations(results)
allpoints = for list <- allresults, do: (for {l1,l2} <- list, do: Enum.zip(l1,l2)) |> List.flatten
totalpoints = for list <- allpoints, do: (for t <- teams, do: {t, Enum.sum(for {t_,points} <- list, t_==t, do: points)} )
sortedtotalpoints = for list <- totalpoints, do: Enum.sort(list,fn({_,a},{_,b}) -> a > b end)
pointsposition = for n <- teams, do: (for list <- sortedtotalpoints, do: elem(Enum.at(list,n),1))
for n <- teams do
for points <- 0..9 do
Enum.at(pointsposition,n) |> Enum.filter(&(&1 == points)) |> length
end
end
end
 
defp combos(1, list), do: (for x <- list, do: [x])
defp combos(k, list) when k == length(list), do: [list]
defp combos(k, [h|t]) do
(for subcombos <- combos(k-1, t), do: [h | subcombos]) ++ (combos(k, t))
end
 
defp combinations([h],list2), do: (for item <- list2, do: [{h,item}])
defp combinations([h|t],list2) do
for item <- list2, comb <- combinations(t,list2), do: [{h,item} | comb]
end
end
 
format = String.duplicate("~4w", 10) <> "~n"
:io.format(format, Enum.to_list(0..9))
IO.puts String.duplicate(" ---", 10)
Enum.each(World_Cup.group_stage, fn x -> :io.format(format, x) end)
Output:
   0   1   2   3   4   5   6   7   8   9
 --- --- --- --- --- --- --- --- --- ---
   0   0   0   1  14 148 152 306   0 108
   0   0   4  33 338 172 164  18   0   0
   0  18 136 273 290   4   8   0   0   0
 108 306 184 125   6   0   0   0   0   0

Erlang[edit]

This solution take advantage of the expressiveness power of the list comprehensions expressions. Function combos is copied from panduwana blog.

 
-module(world_cup).
 
-export([group_stage/0]).
 
group_stage() ->
Results = [[3,0],[1,1],[0,3]],
Teams = [1,2,3,4],
Matches = combos(2,Teams),
AllResults =
combinations(Matches,Results),
AllPoints =
[lists:flatten([lists:zip(L1,L2) || {L1,L2} <- L]) || L <- AllResults],
TotalPoints =
[ [ {T,lists:sum([Points || {T_,Points} <- L, T_ == T])} || T <- Teams] || L <- AllPoints],
SortedTotalPoints =
[ lists:sort(fun({_,A},{_,B}) -> A > B end,L) || L <- TotalPoints],
PointsPosition =
[ [element(2,lists:nth(N, L))|| L <- SortedTotalPoints ] || N <- Teams],
[ [length(lists:filter(fun(Points_) -> Points_ == Points end,lists:nth(N, PointsPosition) ))
|| Points <- lists:seq(0,9)] || N <- Teams].
 
combos(1, L) ->
[[X] || X <- L];
combos(K, L) when K == length(L) ->
[L];
combos(K, [H|T]) ->
[[H | Subcombos] || Subcombos <- combos(K-1, T)]
++ (combos(K, T)).
 
combinations([H],List2) ->
[[{H,Item}] || Item <- List2];
combinations([H|T],List2) ->
[ [{H,Item} | Comb] || Item <- List2, Comb <- combinations(T,List2)].
 

Output:

[[0,0,0,1,14,148,152,306,0,108],
 [0,0,4,33,338,172,164,18,0,0],
 [0,18,136,273,290,4,8,0,0,0],
 [108,306,184,125,6,0,0,0,0,0]]

Go[edit]

Translation of: Kotlin
package main
 
import (
"fmt"
"sort"
"strconv"
)
 
var games = [6]string{"12", "13", "14", "23", "24", "34"}
var results = "000000"
 
func nextResult() bool {
if results == "222222" {
return false
}
res, _ := strconv.ParseUint(results, 3, 32)
results = fmt.Sprintf("%06s", strconv.FormatUint(res+1, 3))
return true
}
 
func main() {
var points [4][10]int
for {
var records [4]int
for i := 0; i < len(games); i++ {
switch results[i] {
case '2':
records[games[i][0]-'1'] += 3
case '1':
records[games[i][0]-'1']++
records[games[i][1]-'1']++
case '0':
records[games[i][1]-'1'] += 3
}
}
sort.Ints(records[:])
for i := 0; i < 4; i++ {
points[i][records[i]]++
}
if !nextResult() {
break
}
}
fmt.Println("POINTS 0 1 2 3 4 5 6 7 8 9")
fmt.Println("-------------------------------------------------------------")
places := [4]string{"1st", "2nd", "3rd", "4th"}
for i := 0; i < 4; i++ {
fmt.Print(places[i], " place ")
for j := 0; j < 10; j++ {
fmt.Printf("%-5d", points[3-i][j])
}
fmt.Println()
}
}
Output:
POINTS       0    1    2    3    4    5    6    7    8    9
-------------------------------------------------------------
1st place    0    0    0    1    14   148  152  306  0    108  
2nd place    0    0    4    33   338  172  164  18   0    0    
3rd place    0    18   136  273  290  4    8    0    0    0    
4th place    108  306  184  125  6    0    0    0    0    0 

J[edit]

There might be a more elegant way of expressing this.

require'stats'
outcome=: 3 0,1 1,:0 3
pairs=: (i.4) e."1(2 comb 4)
standings=: +/@:>&>,{<"1<"1((i.4) e."1 pairs)#inv"1/ outcome

Here, standings represents all the possible outcomes:

   $standings
729 4

Of course, not all of them are distinct:

   $~.standings
556 4

With only 556 distinct outcomes, there must be some repeats. Looking at this more closely (gathering the outcomes in identical groups, counting how many members are in each group, and then considering the unique list of group sizes):

   ~.#/.~standings
1 2 3 4 6

Some standings can be attained two different ways, some three different ways, some four different ways, and some six different ways. Let's look at the one with six different possibilities:

   (I.6=#/.~standings){{./.~standings
4 4 4 4

That's where every team gets 4 standing.

How about outcomes which can be achieved four different ways?

   (I.4=#/.~standings){{./.~standings
6 6 3 3
6 3 3 6
6 3 6 3
3 6 6 3
3 6 3 6
3 3 6 6

Ok, that's simple. So how about a histogram. Actually, it's not clear what a histogram should represent. There are four different teams, and possible standings range from 0 through 9, so we could show the outcomes for each team:

   +/standings =/ i.10
27 81 81 108 162 81 81 81 0 27
27 81 81 108 162 81 81 81 0 27
27 81 81 108 162 81 81 81 0 27
27 81 81 108 162 81 81 81 0 27

Each team has the same possible outcomes. So instead, let's order the results so that instead of seeing each team as a distinct entity we are seeing the first place, second place, third place and fourth place team (and where there's a tie, such as 4 4 4 4, we just arbitrarily say that one of the tied teams gets in each of the tied places...):

   +/(\:"1~ standings)=/"1 i.10
0 0 0 1 14 148 152 306 0 108
0 0 4 33 338 172 164 18 0 0
0 18 136 273 290 4 8 0 0 0
108 306 184 125 6 0 0 0 0 0

Here, the first row represents whichever team came in first in each outcome, the second row represents the second place team and so on.

Meanwhile, the leftmost column represents the number of outcomes where that team would have zero standing, the second column represents the number of outcomes where that team would have one standing and so on. (The rightmost column represents the number of outcomes where that team would have 9 standing.)

Java[edit]

Works with: Java version 1.5+

This example codes results as a 6-digit number in base 3. Each digit is a game. A 2 is a win for the team on the left, a 1 is a draw, and a 0 is a loss for the team on the left.

import java.util.Arrays;
 
public class GroupStage{
//team left digit vs team right digit
static String[] games = {"12", "13", "14", "23", "24", "34"};
static String results = "000000";//start with left teams all losing
 
private static boolean nextResult(){
if(results.equals("222222")) return false;
int res = Integer.parseInt(results, 3) + 1;
results = Integer.toString(res, 3);
while(results.length() < 6) results = "0" + results; //left pad with 0s
return true;
}
 
public static void main(String[] args){
int[][] points = new int[4][10]; //playing 3 games, points range from 0 to 9
do{
int[] records = {0,0,0,0};
for(int i = 0; i < 6; i++){
switch(results.charAt(i)){
case '2': records[games[i].charAt(0) - '1'] += 3; break; //win for left team
case '1': //draw
records[games[i].charAt(0) - '1']++;
records[games[i].charAt(1) - '1']++;
break;
case '0': records[games[i].charAt(1) - '1'] += 3; break; //win for right team
}
}
Arrays.sort(records); //sort ascending, first place team on the right
points[0][records[0]]++;
points[1][records[1]]++;
points[2][records[2]]++;
points[3][records[3]]++;
}while(nextResult());
System.out.println("First place: " + Arrays.toString(points[3]));
System.out.println("Second place: " + Arrays.toString(points[2]));
System.out.println("Third place: " + Arrays.toString(points[1]));
System.out.println("Fourth place: " + Arrays.toString(points[0]));
}
}
Output:
First place: [0, 0, 0, 1, 14, 148, 152, 306, 0, 108]
Second place: [0, 0, 4, 33, 338, 172, 164, 18, 0, 0]
Third place: [0, 18, 136, 273, 290, 4, 8, 0, 0, 0]
Fourth place: [108, 306, 184, 125, 6, 0, 0, 0, 0, 0]

Kotlin[edit]

Translation of: Java
// version 1.1.2
 
val games = arrayOf("12", "13", "14", "23", "24", "34")
var results = "000000"
 
fun nextResult(): Boolean {
if (results == "222222") return false
val res = results.toInt(3) + 1
results = res.toString(3).padStart(6, '0')
return true
}
 
fun main(args: Array<String>) {
val points = Array(4) { IntArray(10) }
do {
val records = IntArray(4)
for (i in 0..5) {
when (results[i]) {
'2' -> records[games[i][0] - '1'] += 3
'1' -> { records[games[i][0] - '1']++ ; records[games[i][1] - '1']++ }
'0' -> records[games[i][1] - '1'] += 3
}
}
records.sort()
for (i in 0..3) points[i][records[i]]++
}
while(nextResult())
println("POINTS 0 1 2 3 4 5 6 7 8 9")
println("-------------------------------------------------------------")
val places = arrayOf("1st", "2nd", "3rd", "4th")
for (i in 0..3) {
print("${places[i]} place ")
points[3 - i].forEach { print("%-5d".format(it)) }
println()
}
}
Output:
POINTS       0    1    2    3    4    5    6    7    8    9
-------------------------------------------------------------
1st place    0    0    0    1    14   148  152  306  0    108  
2nd place    0    0    4    33   338  172  164  18   0    0    
3rd place    0    18   136  273  290  4    8    0    0    0    
4th place    108  306  184  125  6    0    0    0    0    0    

Perl[edit]

Translation of: Perl 6
use Math::Cartesian::Product;
 
@scoring = (0, 1, 3);
push @histo, [(0) x 10] for 1..4;
push @aoa, [(0,1,2)] for 1..6;
 
for $results (cartesian {@_} @aoa) {
my @s;
my @g = ([0,1],[0,2],[0,3],[1,2],[1,3],[2,3]);
for (0..$#g) {
$r = $results->[$_];
$s[$g[$_][0]] += $scoring[$r];
$s[$g[$_][1]] += $scoring[2 - $r];
}
 
my @ss = sort @s;
$histo[$_][$ss[$_]]++ for 0..$#s;
}
 
$fmt = ('%3d ') x 10 . "\n";
printf $fmt, @$_ for reverse @histo;
Output:
  0   0   0   1  14 148 152 306   0 108
  0   0   4  33 338 172 164  18   0   0
  0  18 136 273 290   4   8   0   0   0
108 306 184 125   6   0   0   0   0   0

Perl 6[edit]

Works with: rakudo version 2016.12
Translation of: Python
constant scoring = 0, 1, 3;
my @histo = [0 xx 10] xx 4;
 
for [X] ^3 xx 6 -> @results {
my @s;
 
for @results Z (^4).combinations(2) -> ($r, @g) {
@s[@g[0]] += scoring[$r];
@s[@g[1]] += scoring[2 - $r];
}
 
for @histo Z @s.sort -> (@h, $v) {
++@h[$v];
}
}
 
say .fmt('%3d',' ') for @histo.reverse;
Output:
  0   0   0   1  14 148 152 306   0 108
  0   0   4  33 338 172 164  18   0   0
  0  18 136 273 290   4   8   0   0   0
108 306 184 125   6   0   0   0   0   0

Python[edit]

from itertools import product, combinations, izip
 
scoring = [0, 1, 3]
histo = [[0] * 10 for _ in xrange(4)]
 
for results in product(range(3), repeat=6):
s = [0] * 4
for r, g in izip(results, combinations(range(4), 2)):
s[g[0]] += scoring[r]
s[g[1]] += scoring[2 - r]
 
for h, v in izip(histo, sorted(s)):
h[v] += 1
 
for x in reversed(histo):
print x
Output:
[0, 0, 0, 1, 14, 148, 152, 306, 0, 108]
[0, 0, 4, 33, 338, 172, 164, 18, 0, 0]
[0, 18, 136, 273, 290, 4, 8, 0, 0, 0]
[108, 306, 184, 125, 6, 0, 0, 0, 0, 0]

Racket[edit]

#lang racket
;; Tim Brown 2014-09-15
(define (sort-standing stndg#)
(sort (hash->list stndg#) > #:key cdr))
 
(define (hash-update^2 hsh key key2 updater2 dflt2)
(hash-update hsh key (λ (hsh2) (hash-update hsh2 key2 updater2 dflt2)) hash))
 
(define all-standings
(let ((G '((a b) (a c) (a d) (b c) (b d) (c d)))
(R '((3 0) (1 1) (0 3))))
(map
sort-standing
(for*/list ((r1 R) (r2 R) (r3 R) (r4 R) (r5 R) (r6 R))
(foldr (λ (gm rslt h)
(hash-update
(hash-update h (second gm) (λ (n) (+ n (second rslt))) 0)
(first gm) (curry + (first rslt)) 0))
(hash) G (list r1 r2 r3 r4 r5 r6))))))
 
(define histogram
(for*/fold ((rv (hash)))
((stndng (in-list all-standings)) (psn (in-range 0 4)))
(hash-update^2 rv (add1 psn) (cdr (list-ref stndng psn)) add1 0)))
 
;; Generalised histogram printing functions...
(define (show-histogram hstgrm# captions)
(define (min* a b)
(if (and a b) (min a b) (or a b)))
(define-values (position-mn position-mx points-mn points-mx)
(for*/fold ((mn-psn #f) (mx-psn 0) (mn-pts #f) (mx-pts 0))
(((psn rw) (in-hash hstgrm#)))
(define-values (min-pts max-pts)
(for*/fold ((mn mn-pts) (mx mx-pts)) ((pts (in-hash-keys rw)))
(values (min* pts mn) (max pts mx))))
(values (min* mn-psn psn) (max mx-psn psn) min-pts max-pts)))
 
(define H
(let ((lbls-row# (for/hash ((i (in-range points-mn (add1 points-mx)))) (values i i))))
(hash-set hstgrm# 'thead lbls-row#)))
 
(define cap-col-width (for/fold ((m 0)) ((v (in-hash-values captions))) (max m (string-length v))))
 
(for ((plc (in-sequences
(in-value 'thead)
(in-range position-mn (add1 position-mx)))))
(define cnts (for/list ((pts (in-range points-mn (add1 points-mx))))
(~a #:align 'center #:width 3 (hash-ref (hash-ref H plc) pts 0))))
(printf "~a ~a~%"
(~a (hash-ref captions plc (curry format "#~a:")) #:width cap-col-width)
(string-join cnts " "))))
 
(define captions
(hash 'thead "POINTS:"
1 "1st Place:"
2 "2nd Place:"
3 "Sack the manager:"
4 "Sack the team!"))
 
(show-histogram histogram captions)
Output:
POINTS:            0    1    2    3    4    5    6    7    8    9 
1st Place:         0    0    0    1   14   148  152  306   0   108
2nd Place:         0    0    4   33   338  172  164  18    0    0 
Sack the manager:  0   18   136  273  290   4    8    0    0    0 
Sack the team!    108  306  184  125   6    0    0    0    0    0 

REXX[edit]

version 1, static game sets[edit]

Translation of: Java
/* REXX -------------------------------------------------------------------*/
results = '000000' /*start with left teams all losing */
games = '12 13 14 23 24 34'
points.=0
records.=0
Do Until nextResult(results)=0
records.=0
Do i=1 To 6
r=substr(results,i,1)
g=word(games,i); Parse Var g g1 +1 g2
Select
When r='2' Then /* win for left team */
records.g1=records.g1+3
When r='1' Then Do /* draw */
records.g1=records.g1+1
records.g2=records.g2+1
End
When r='0' Then /* win for right team */
records.g2=records.g2+3
End
End
Call sort_records /* sort ascending, */
/* first place team on the right */
r1=records.1
r2=records.2
r3=records.3
r4=records.4
points.0.r1=points.0.r1+1
points.1.r2=points.1.r2+1
points.2.r3=points.2.r3+1
points.3.r4=points.3.r4+1
End
ol.='['
sep=', '
Do i=0 To 9
If i=9 Then sep=']'
ol.0=ol.0||points.0.i||sep
ol.1=ol.1||points.1.i||sep
ol.2=ol.2||points.2.i||sep
ol.3=ol.3||points.3.i||sep
End
Say ol.3
Say ol.2
Say ol.1
Say ol.0
Exit
 
nextResult: Procedure Expose results
/* results is a string of 6 base 3 digits to which we add 1 */
/* e.g., '000212 +1 -> 000220 */
If results="222222" Then Return 0
res=0
do i=1 To 6
res=res*3+substr(results,i,1)
End
res=res+1
s=''
Do i=1 To 6
b=res//3
res=res%3
s=b||s
End
results=s
Return 1
 
sort_records: Procedure Expose records.
Do i=1 To 3
Do j=i+1 To 4
If records.j<records.i Then
Parse Value records.i records.j With records.j records.i
End
End
Return
Output:
[0, 0, 0, 1, 14, 148, 152, 306, 0, 108]
[0, 0, 4, 33, 338, 172, 164, 18, 0, 0]
[0, 18, 136, 273, 290, 4, 8, 0, 0, 0]
[108, 306, 184, 125, 6, 0, 0, 0, 0, 0]

version 2, generated game sets[edit]

Translation of: Java
Translation of: REXX


This REXX version allows the number of teams to be specified to be used in the calculations,   and
also generates the character string used for the   list of games   to be played   (game sets).

This REXX version can also simulate a Cricket World Cup   (by specifying   2   for the   win   variable).

/*REXX pgm calculates world cup standings based on the number of games won by the teams.*/
parse arg teams win . /*obtain optional argument from the CL.*/
if teams=='' | teams=="," then teams= 4 /*Not specified? Then use the default.*/
if win=='' | win=="," then win= 3 /* " " " " " " */
sets=0; gs= /*the number of sets (so far). */
do j=1 for teams
do k=j+1 to teams; sets= sets+1 /*bump the number of game sets. */
games.sets= j || k; gs= gs j || k /*generate the game combinations. */
end /*j*/
end /*k*/
z= 1; setLimit= copies(2, sets) /*Z: max length of any number shown. */
say teams ' teams, ' sets " game sets: " gs /*display what's being used for calcs. */
results = copies(0, sets); say /*start with left-most teams all losing*/
points. = 0 /*zero all the team's point. */
do until \nextResult(results); @.= 0
do j=1 for sets; r= substr(results, j, 1)
parse var games.j A +1 B /*get the A and B teams*/
if r==0 then @.B= @.B + win /*win for right─most team.*/
if r==1 then do; @.A= @.A + 1; @.B= @.B + 1; end /*draw for both teams*/
if r==2 then @.A= @.A + win /*win for left─most team. */
end /*j*/
call sort teams
do t=1 for teams; tm= t - 1; _= @.t
points.tm._ = points.tm._ + 1; z= max(z, length(points.tm._) )
end /*t*/
end /*until*/
$.=
do j=0 for teams+6
do k=0 for teams; $.k= $.k || right(points.k.j, z)'│ '; end /*k*/
end /*j*/
say; /* [↓] build grid line for the box*/
L=length($.1) -2; $$= translate( translate( left($.1, L), , 0123456789), '─', " ")
say left('', 15) center('points', L) /*display the boxed title. */
say left('', 15) "╔"translate($$, '═╤', "─│")'╗' /*display the bottom sep for title.*/
p=0
do m=teams-1 by -1 for teams; p = p+1
say right('('th(p) "place)", 14) " ║"left($.m, L)'║'
if m>0 then say right(' ', 14) " ╟"translate($$, '┼', "│")'╢'
end /*m*/
say left('', 15) "╚"translate($$, '═╧', "─│")'╝' /*display the bottom sep for title.*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
nextResult: if results==setLimit then return 0
res= 0 /* [↓] do arithmetic in base three.*/
do k=1 for sets; res= res * 3 + substr(results, k, 1)
end /*j*/
results=; res= res + 1
do sets; results= res // 3 || results; res= res % 3
end /*sets*/
return 1
/*──────────────────────────────────────────────────────────────────────────────────────*/
sort: procedure expose @.; arg #; do j=1 for #-1 /*a bubble sort, ascending order.*/
do k=j+1 to # /*swap two elements out of order.*/
if @.k<@.j then parse value @.j @.k with @.k @.j
end /*k*/
end /*j*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
th: arg th; return (th/1) || word('th st nd rd', 1 +(th//10) *(th//100%10\==1)*(th//10<4))
output   when using the default input of:     4
4  teams,  6  game sets:   12 13 14 23 24 34


                                    points
               ╔═══╤════╤════╤════╤════╤════╤════╤════╤════╤════╗
  (1st place)  ║  0│   0│   0│   1│  14│ 148│ 152│ 306│   0│ 108║
               ╟───┼────┼────┼────┼────┼────┼────┼────┼────┼────╢
  (2nd place)  ║  0│   0│   4│  33│ 338│ 172│ 164│  18│   0│   0║
               ╟───┼────┼────┼────┼────┼────┼────┼────┼────┼────╢
  (3rd place)  ║  0│  18│ 136│ 273│ 290│   4│   8│   0│   0│   0║
               ╟───┼────┼────┼────┼────┼────┼────┼────┼────┼────╢
  (4th place)  ║108│ 306│ 184│ 125│   6│   0│   0│   0│   0│   0║
               ╚═══╧════╧════╧════╧════╧════╧════╧════╧════╧════╝
output   when using the input of:     5
5  teams,  10  game sets:   12 13 14 15 23 24 25 34 35 45


                                                 points
               ╔═════╤══════╤══════╤══════╤══════╤══════╤══════╤══════╤══════╤══════╤══════╗
  (1st place)  ║    0│     0│     0│     0│     1│    74│  2409│ 11520│ 16230│ 10860│ 14310║
               ╟─────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────╢
  (2nd place)  ║    0│     0│     0│     5│   181│  7314│ 15609│ 26400│  5610│  3660│   270║
               ╟─────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────╢
  (3rd place)  ║    0│     0│    30│   825│ 10401│ 25794│ 16119│  5790│    30│    60│     0║
               ╟─────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────╢
  (4th place)  ║    0│   270│  3990│ 13185│ 28871│ 10414│  2289│    30│     0│     0│     0║
               ╟─────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────╢
  (5th place)  ║ 3645│ 14310│ 17850│ 15145│  7931│   144│    24│     0│     0│     0│     0║
               ╚═════╧══════╧══════╧══════╧══════╧══════╧══════╧══════╧══════╧══════╧══════╝
output   when using the input of:     6
6  teams,  15  game sets:   12 13 14 15 16 23 24 25 26 34 35 36 45 46 56


                                                                 points
               ╔═══════╤════════╤════════╤════════╤════════╤════════╤════════╤════════╤════════╤════════╤════════╤════════╗
  (1st place)  ║      0│       0│       0│       0│       0│       1│     434│   68910│ 1049904│ 2333079│ 4056210│ 3149820║
               ╟───────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────╢
  (2nd place)  ║      0│       0│       0│       0│       6│    1165│  207308│ 1803570│ 5266944│ 3648879│ 2822850│  392580║
               ╟───────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────╢
  (3rd place)  ║      0│       0│       0│      45│    4926│  366445│ 2580578│ 6232110│ 3900744│ 1055889│  206550│     540║
               ╟───────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────╢
  (4th place)  ║      0│       0│     540│   34965│  623466│ 3793085│ 5521498│ 3916830│  410364│   47889│     270│       0║
               ╟───────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────╢
  (5th place)  ║      0│   10935│  261360│ 1395225│ 4446336│ 5645615│ 2210128│  378300│     864│     144│       0│       0║
               ╟───────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────┼────────╢
  (6th place)  ║ 354294│ 1760535│ 3281040│ 3884175│ 3782616│ 1176803│  108874│     570│       0│       0│       0│       0║
               ╚═══════╧════════╧════════╧════════╧════════╧════════╧════════╧════════╧════════╧════════╧════════╧════════╝

Ruby[edit]

teams = [:a, :b, :c, :d]
matches = teams.combination(2).to_a
outcomes = [:win, :draw, :loss]
gains = {win:[3,0], draw:[1,1], loss:[0,3]}
places_histogram = Array.new(4) {Array.new(10,0)}
 
# The Array#repeated_permutation method generates the 3^6 different
# possible outcomes
outcomes.repeated_permutation(6).each do |outcome|
results = Hash.new(0)
 
# combine this outcomes with the matches, and generate the points table
outcome.zip(matches).each do |decision, (team1, team2)|
results[team1] += gains[decision][0]
results[team2] += gains[decision][1]
end
 
# accumulate the results
results.values.sort.reverse.each_with_index do |points, place|
places_histogram[place][points] += 1
end
end
 
fmt = "%s :" + "%4s"*10
puts fmt % [" ", *0..9]
puts fmt % ["-", *["---"]*10]
places_histogram.each.with_index(1) {|hist,place| puts fmt % [place, *hist]}
Output:
  :   0   1   2   3   4   5   6   7   8   9
- : --- --- --- --- --- --- --- --- --- ---
1 :   0   0   0   1  14 148 152 306   0 108
2 :   0   0   4  33 338 172 164  18   0   0
3 :   0  18 136 273 290   4   8   0   0   0
4 : 108 306 184 125   6   0   0   0   0   0

Scala[edit]

object GroupStage extends App { //team left digit vs team right digit
val games = Array("12", "13", "14", "23", "24", "34")
val points = Array.ofDim[Int](4, 10) //playing 3 games, points range from 0 to 9
var results = "000000" //start with left teams all losing
 
private def nextResult: Boolean = {
if (results == "222222") false
else {
results = Integer.toString(Integer.parseInt(results, 3) + 1, 3)
while (results.length < 6) results = "0" + results //left pad with 0s
true
}
}
 
do {
val records = Array(0, 0, 0, 0)
for (i <- results.indices.reverse by -1) {
results(i) match {
case '2' => records(games(i)(0) - '1') += 3
case '1' => //draw
records(games(i)(0) - '1') += 1
records(games(i)(1) - '1') += 1
case '0' => records(games(i)(1) - '1') += 3
}
}
java.util.Arrays.sort(records) //sort ascending, first place team on the right
 
points(0)(records(0)) += 1
points(1)(records(1)) += 1
points(2)(records(2)) += 1
points(3)(records(3)) += 1
} while (nextResult)
 
println("First place: " + points(3).mkString("[",", ","]"))
println("Second place: " + points(2).mkString("[",", ","]"))
println("Third place: " + points(1).mkString("[",", ","]"))
println("Fourth place: " + points(0).mkString("[",", ","]"))
 
}

Tcl[edit]

Translation of: Java
Works with: Tcl version 8.6
package require Tcl 8.6
proc groupStage {} {
foreach n {0 1 2 3} {
set points($n) {0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0}
}
set results 0
set games {0 1 0 2 0 3 1 2 1 3 2 3}
while true {
set R {0 0 1 0 2 0 3 0}
foreach r [split [format %06d $results] ""] {A B} $games {
switch $r {
2 {dict incr R $A 3}
1 {dict incr R $A; dict incr R $B}
0 {dict incr R $B 3}
}
}
foreach n {0 1 2 3} r [lsort -integer [dict values $R]] {
dict incr points($n) $r
}
 
if {$results eq "222222"} break
while {[regexp {[^012]} [incr results]]} continue
}
return [lmap n {3 2 1 0} {dict values $points($n)}]
}
 
foreach nth {First Second Third Fourth} nums [groupStage] {
puts "$nth place:\t[join [lmap n $nums {format %3s $n}] {, }]"
}
Output:
First place:	  0,   0,   0,   1,  14, 148, 152, 306,   0, 108
Second place:	  0,   0,   4,  33, 338, 172, 164,  18,   0,   0
Third place:	  0,  18, 136, 273, 290,   4,   8,   0,   0,   0
Fourth place:	108, 306, 184, 125,   6,   0,   0,   0,   0,   0

zkl[edit]

Translation of: D
Translation of: Python
combos :=Utils.Helpers.pickNFrom(2,T(0,1,2,3)); // ( (0,1),(0,2) ... )
scoring:=T(0,1,3);
histo  :=(0).pump(4,List().write,(0).pump(10,List().write,0).copy); //[4][10] of zeros
 
foreach r0,r1,r2,r3,r4,r5 in ([0..2],[0..2],[0..2],[0..2],[0..2],[0..2]){
s:=L(0,0,0,0);
foreach i,r in (T(r0,r1,r2,r3,r4,r5).enumerate()){
g:=combos[i];
s[g[0]]+=scoring[r];
s[g[1]]+=scoring[2 - r];
}
foreach h,v in (histo.zip(s.sort())){ h[v]+=1; }
}
foreach h in (histo.reverse()){ println(h.apply("%3d ".fmt).concat()) }
Output:
  0   0   0   1  14 148 152 306   0 108 
  0   0   4  33 338 172 164  18   0   0 
  0  18 136 273 290   4   8   0   0   0 
108 306 184 125   6   0   0   0   0   0