Non-transitive dice
Let our dice select numbers on their faces with equal probability, i.e. fair dice. Dice may have more or less than six faces. (The possibility of there being a 3D physical shape that has that many "faces" that allow them to be fair dice, is ignored for this task - a die with 3 or 33 defined sides is defined by the number of faces and the numbers on each face).
Throwing dice will randomly select a face on each die with equal probability. To show which die of dice thrown multiple times is more likely to win over the others:
- calculate all possible combinations of different faces from each die
- Count how many times each die wins a combination
- Each combination is equally likely so the die with more winning face combinations is statistically more likely to win against the other dice.
If two dice X and Y are thrown against each other then X likely to: win, lose, or break-even against Y can be shown as: X > Y, X < Y, or X = Y respectively.
- Example 1
If X is the three sided die with 1, 3, 6 on its faces and Y has 2, 3, 4 on its faces then the equal possibility outcomes from throwing both, and the winners is:
X Y Winner = = ====== 1 2 Y 1 3 Y 1 4 Y 3 2 X 3 3 - 3 4 Y 6 2 X 6 3 X 6 4 X TOTAL WINS: X=4, Y=4
Both die will have the same statistical probability of winning, i.e.their comparison can be written as X = Y
- Transitivity
In mathematics transitivity are rules like:
if a op b and b op c then a op c
If, for example, the op, (for operator), is the familiar less than, <, and it's applied to integers
we get the familiar if a < b and b < c then a < c
- Non-transitive dice
These are an ordered list of dice where the '>' operation between successive
dice pairs applies but a comparison between the first and last of the list
yields the opposite result, '<'.
(Similarly '<' successive list comparisons with a final '>' between first and last is also non-transitive).
Three dice S, T, U with appropriate face values could satisfy
S < T, T < U and yet S > U
To be non-transitive.
- Notes
- The order of numbers on the faces of a die is not relevant. For example, three faced die described with face numbers of 1, 2, 3 or 2, 1, 3 or any other permutation are equivalent. For the purposes of the task show only the permutation in lowest-first sorted order i.e. 1, 2, 3 (and remove any of its perms).
- A die can have more than one instance of the same number on its faces, e.g.
2, 3, 3, 4
- Task
- ====
Find all the ordered lists of three non-transitive dice S, T, U of the form S < T, T < U and yet S > U; where the dice are selected from all four-faced die , (unique w.r.t the notes), possible by having selections from the integers one to four on any dies face.
Solution can be found by generating all possble individual die then testing all possible permutations, (permutations are ordered), of three dice for non-transitivity.
- Optional stretch goal
Find lists of four non-transitive dice selected from the same possible dice from the non-stretch goal.
Show the results here, on this page.
- References
- The Most Powerful Dice - Numberphile Video.
- Nontransitive dice - Wikipedia.
F#
<lang fsharp> // Find all 3(d1,d2,d3) four sided dice with faces valued 1,2,3,or 4 such that over a run d1 beats d2, d2 betas d3, but d3 beats d1. // Nigel Galloway: August 9th., 2020 let fN(n,g)=let n=seq{for n in n do for g in g->compare n g}|>Seq.countBy id|>Map.ofSeq
match Map.tryFind -1 n,Map.tryFind 1 n with (Some n,Some g)->n>g |(Some _,_)->true |_->false
let e=[for n0 in [1..4] do for n1 in [n0..4] do for n2 in [n1..4] do for n3 in [n2..4]->[n0;n1;n2;n3]] let i=[for n in e do for g in e->(n,g)]|>List.filter fN let l3=seq{for n in i do for g in i->(n,g)}|>Seq.filter(fun((a,n),(g,b))->n=g && List.contains (b,a) i) l3|>Seq.iter(fun((n,g),(_,g'))->printfn "%A<%A; %A<%A; %A>%A\n" n g g g' n g') let l4=seq{for n in i do for g in i do for i in i->(n,g,i)}|>Seq.filter(fun((a,n),(g,b),(x,y))->n=g && b=x && List.contains (y,a) i) l4|>Seq.iter(fun((n,g),(_,g'),(_,b))->printfn "%A<%A; %A<%A; %A<%A; %A>%A" n g g g' g' b n b) </lang>
- Output:
[1; 1; 4; 4]<[2; 2; 2; 4]; [2; 2; 2; 4]<[1; 3; 3; 3]; [1; 1; 4; 4]>[1; 3; 3; 3] [1; 3; 3; 3]<[1; 1; 4; 4]; [1; 1; 4; 4]<[2; 2; 2; 4]; [1; 3; 3; 3]>[2; 2; 2; 4] [2; 2; 2; 4]<[1; 3; 3; 3]; [1; 3; 3; 3]<[1; 1; 4; 4]; [2; 2; 2; 4]>[1; 1; 4; 4] [1; 1; 4; 4]<[2; 2; 2; 4]; [2; 2; 2; 4]<[2; 2; 3; 3]; [2; 2; 3; 3]<[1; 3; 3; 3]; [1; 1; 4; 4]>[1; 3; 3; 3] [1; 3; 3; 3]<[1; 1; 4; 4]; [1; 1; 4; 4]<[2; 2; 2; 4]; [2; 2; 2; 4]<[2; 2; 3; 3]; [1; 3; 3; 3]>[2; 2; 3; 3] [2; 2; 2; 4]<[2; 2; 3; 3]; [2; 2; 3; 3]<[1; 3; 3; 3]; [1; 3; 3; 3]<[1; 1; 4; 4]; [2; 2; 2; 4]>[1; 1; 4; 4] [2; 2; 3; 3]<[1; 3; 3; 3]; [1; 3; 3; 3]<[1; 1; 4; 4]; [1; 1; 4; 4]<[2; 2; 2; 4]; [2; 2; 3; 3]>[2; 2; 2; 4]
Factor
<lang factor>USING: grouping io kernel math math.combinatorics math.ranges prettyprint sequences ;
- possible-dice ( n -- seq )
[ [1,b] ] [ selections ] bi [ [ <= ] monotonic? ] filter ;
- cmp ( seq seq -- n ) [ - sgn ] cartesian-map concat sum ;
- non-transitive? ( seq -- ? )
[ 2 clump [ first2 cmp neg? ] all? ] [ [ last ] [ first ] bi cmp neg? and ] bi ;
- find-non-transitive ( #sides #dice -- seq )
[ possible-dice ] [ <k-permutations> ] bi* [ non-transitive? ] filter ;
! Task "Number of eligible 4-sided dice: " write 4 possible-dice length . nl
"All ordered lists of 3 non-transitive dice with 4 sides:" print 4 3 find-non-transitive . nl
"All ordered lists of 4 non-transitive dice with 4 sides:" print 4 4 find-non-transitive .</lang>
- Output:
Number of eligible 4-sided dice: 35
All ordered lists of 3 non-transitive dice with 4 sides:
V{
{ { 1 1 4 4 } { 2 2 2 4 } { 1 3 3 3 } }
{ { 1 3 3 3 } { 1 1 4 4 } { 2 2 2 4 } }
{ { 2 2 2 4 } { 1 3 3 3 } { 1 1 4 4 } }
}
All ordered lists of 4 non-transitive dice with 4 sides:
V{
{ { 1 1 4 4 } { 2 2 2 4 } { 2 2 3 3 } { 1 3 3 3 } }
{ { 1 3 3 3 } { 1 1 4 4 } { 2 2 2 4 } { 2 2 3 3 } }
{ { 2 2 2 4 } { 2 2 3 3 } { 1 3 3 3 } { 1 1 4 4 } }
{ { 2 2 3 3 } { 1 3 3 3 } { 1 1 4 4 } { 2 2 2 4 } }
}
Go
<lang go>package main
import (
"fmt" "sort"
)
func fourFaceCombs() (res [][4]int) {
found := make([]bool, 256)
for i := 1; i <= 4; i++ {
for j := 1; j <= 4; j++ {
for k := 1; k <= 4; k++ {
for l := 1; l <= 4; l++ {
c := [4]int{i, j, k, l}
sort.Ints(c[:])
key := 64*(c[0]-1) + 16*(c[1]-1) + 4*(c[2]-1) + (c[3] - 1)
if !found[key] {
found[key] = true
res = append(res, c)
}
}
}
}
}
return
}
func cmp(x, y [4]int) int {
xw := 0
yw := 0
for i := 0; i < 4; i++ {
for j := 0; j < 4; j++ {
if x[i] > y[j] {
xw++
} else if y[j] > x[i] {
yw++
}
}
}
if xw < yw {
return -1
} else if xw > yw {
return 1
}
return 0
}
func findIntransitive3(cs [][4]int) (res [][3][4]int) {
var c = len(cs)
for i := 0; i < c; i++ {
for j := 0; j < c; j++ {
for k := 0; k < c; k++ {
first := cmp(cs[i], cs[j])
if first == -1 {
second := cmp(cs[j], cs[k])
if second == -1 {
third := cmp(cs[i], cs[k])
if third == 1 {
res = append(res, [3][4]int{cs[i], cs[j], cs[k]})
}
}
}
}
}
}
return
}
func findIntransitive4(cs [][4]int) (res [][4][4]int) {
c := len(cs)
for i := 0; i < c; i++ {
for j := 0; j < c; j++ {
for k := 0; k < c; k++ {
for l := 0; l < c; l++ {
first := cmp(cs[i], cs[j])
if first == -1 {
second := cmp(cs[j], cs[k])
if second == -1 {
third := cmp(cs[k], cs[l])
if third == -1 {
fourth := cmp(cs[i], cs[l])
if fourth == 1 {
res = append(res, [4][4]int{cs[i], cs[j], cs[k], cs[l]})
}
}
}
}
}
}
}
}
return
}
func main() {
combs := fourFaceCombs()
fmt.Println("Number of eligible 4-faced dice", len(combs))
it3 := findIntransitive3(combs)
fmt.Printf("\n%d ordered lists of 3 non-transitive dice found, namely:\n", len(it3))
for _, a := range it3 {
fmt.Println(a)
}
it4 := findIntransitive4(combs)
fmt.Printf("\n%d ordered lists of 4 non-transitive dice found, namely:\n", len(it4))
for _, a := range it4 {
fmt.Println(a)
}
}</lang>
- Output:
Number of eligible 4-faced dice 35 3 ordered lists of 3 non-transitive dice found, namely: [[1 1 4 4] [2 2 2 4] [1 3 3 3]] [[1 3 3 3] [1 1 4 4] [2 2 2 4]] [[2 2 2 4] [1 3 3 3] [1 1 4 4]] 4 ordered lists of 4 non-transitive dice found, namely: [[1 1 4 4] [2 2 2 4] [2 2 3 3] [1 3 3 3]] [[1 3 3 3] [1 1 4 4] [2 2 2 4] [2 2 3 3]] [[2 2 2 4] [2 2 3 3] [1 3 3 3] [1 1 4 4]] [[2 2 3 3] [1 3 3 3] [1 1 4 4] [2 2 2 4]]
Phix
Requires 0.8.2+ (added sq_cmp() builtin that returns nested -1/0/+1 compare() results, just like the existing sq_eq() does for equal().) <lang Phix>integer mx = 4, -- max number of a die side (later set to 6)
mn = 1 -- min number of a die side (later set to 0)
function possible_dice(integer sides)
--
-- construct all non-descending permutes of mn..mx,
-- ie/eg {1,1,1,1}..{4,4,4,4} with (say), amongst
-- others, {1,1,2,4} but not {1,1,4,2}.
--
sequence die = repeat(mn,sides), -- (main work area)
res = {die}
while true do
-- find rightmost incrementable side
-- ie/eg {1,2,4,4} -> set rdx to 2 (if 1-based indexing)
integer rdx = rfind(true,sq_lt(die,mx))
if rdx=0 then exit end if
-- set that and all later to that incremented
-- ie/eg {1,2,4,4} -> {1,3,3,3}
die[rdx..$] = die[rdx]+1
res = append(res,die)
end while
printf(1,"There are %d possible %d..%d %d-sided dice\n",{length(res),mn,mx,sides})
return res
end function
function Dnn(sequence die)
-- reconstruct the python die numbering (string)
-- essentially just treat it as a base-N number.
integer l = length(die), -- (sides)
N = mx-mn+1, -- (base)
n = 0 -- (result)
for k=1 to l do
n += (die[k]-mn)*power(N,l-k)
end for
return sprintf("D%d",n+1)
end function
function cmpd(sequence die1, die2)
-- compares two die returning -1, 0, or +1 for <, =, >
integer res = 0
for i=1 to length(die1) do
res += sum(sq_cmp(die1[i],die2))
end for
return sign(res)
end function
integer lp = 0 -- (last progress length, to be wiped out) procedure progress(string msg, sequence args = {})
if length(args) then msg = sprintf(msg,args) end if
integer lm = length(msg)
if lm=0 then
puts(1,repeat(' ',lp))
lp = 0
else
if lm<lp then msg[$..$] = repeat(' ',lp-lm)&msg[$] end if
puts(1,msg)
lp = iff(msg[$]='\r'?lm:0)
end if
end procedure
integer rotations function find_non_trans(sequence dice, integer n=3)
atom t1 = time()+1
integer l = length(dice), sk, sk1, c
sequence set = repeat(1,n), -- (indexes to dice)
cache = repeat(repeat(-2,l),l),
res = {}
while true do
bool valid = true
for k=1 to n-1 do
sk = set[k]
sk1 = set[k+1]
c = cache[sk][sk1]
if c=-2 then
c = cmpd(dice[sk],dice[sk1])
cache[sk][sk1] = c
end if
if c!=-1 then
valid = false
-- force k+1 to be incremented next:
set[k+2..$] = l
exit
end if
end for
if valid then
sk = set[1]
sk1 = set[$]
c = cache[sk][sk1]
if c=-2 then
c = cmpd(dice[sk],dice[sk1])
cache[sk][sk1] = c
end if
if c=+1 then
res = append(res,sort(set))
end if
end if
-- find rightmost incrementable die index
-- ie/eg if l is 35 and set is {1,2,35,35}
-- -> set rdx to 2 (if 1-based indexing)
integer rdx = rfind(true,sq_lt(set,l))
if rdx=0 then exit end if
-- increment that and reset all later
-- ie/eg {1,2,35,35} -> {1,3,1,1}
set[rdx] += 1
set[rdx+1..$] = 1
if time()>t1 then
progress("working... (%d/%d)\r",{set[1],l})
t1 = time()+1
end if
end while
progress("")
rotations = length(res)
res = unique(res)
rotations -= length(res)
return res
end function
function verbose_cmp(sequence die1, die2)
-- compares two die returning their relationship of their names as a string
integer c = cmpd(die1,die2)
string op = {"<","=",">"}[c+2],
n1 = Dnn(die1),
n2 = Dnn(die2)
return sprintf("%s %s %s",{n1,op,n2})
end function
function verbose_dice_cmp(sequence dice, set)
sequence c = {}, d1, d2
for j=1 to length(set)-1 do
d1 = dice[set[j]]
d2 = dice[set[j+1]]
c = append(c,verbose_cmp(d1,d2))
end for
d1 = dice[set[1]]
d2 = dice[set[$]]
c = append(c,verbose_cmp(d1,d2))
return join(c,", ")
end function
procedure show_dice(sequence dice, non_trans, integer N)
integer l = length(non_trans),
omissions = 0,
last = 0
if N then
printf(1,"\n Non_transitive length-%d combinations found: %d\n",{N,l+rotations})
end if
for i=1 to l do
object ni = non_trans[i]
if sequence(ni) then
if i<5 then
printf(1,"\n")
for j=1 to length(ni) do
sequence d = dice[ni[j]]
printf(1," %s:%v\n",{Dnn(d),d})
end for
last = i
else
omissions += 1
end if
end if
end for
if omissions then
printf(1," (%d omitted)\n",omissions)
end if
if rotations then
printf(1," (%d rotations omitted)\n",rotations)
end if
if last then
printf(1,"\n")
if mx<=6 and mn=1 then
printf(1," More verbose comparison of last result:\n")
end if
printf(1," %s\n",{verbose_dice_cmp(dice,non_trans[last])})
printf(1,"\n ====\n")
end if
printf(1,"\n")
end procedure
sequence dice = possible_dice(4) for N=3 to 4 do
show_dice(dice,find_non_trans(dice,N),N)
end for
-- From the numberphile video (Efron's dice): mx = 6 mn = 0 dice = possible_dice(6) --show_dice(dice,find_non_trans(dice,6),6) -- ok, dunno about you but I'm not waiting for power(924,6) permutes... -- limit to the ones discussed, plus another 4 random ones -- (hopefully it'll just pick out the right ones...) printf(1,"\nEfron's dice\n") dice = {{0,0,4,4,4,4},
{1,1,1,1,1,1}, -- (rand)
{1,1,1,5,5,5},
{1,2,3,4,5,6}, -- (rand)
{2,2,2,2,6,6},
{5,5,5,6,6,6}, -- (rand)
{3,3,3,3,3,3},
{6,6,6,6,6,6}} -- (rand)
show_dice(dice,find_non_trans(dice,4),0)
-- and from wp: mx = 9 mn = 1 dice = possible_dice(6) printf(1,"\nFrom wp\n") dice = {{1,1,6,6,8,8},
{2,2,4,4,9,9},
{3,3,5,5,7,7}}
show_dice(dice,find_non_trans(dice,3),0)
-- Miwin's dice printf(1,"Miwin's dice\n") dice = {{1,2,5,6,7,9},
{1,3,4,5,8,9},
{2,3,4,6,7,8}}
show_dice(dice,find_non_trans(dice,3),0)</lang>
- Output:
There are 35 possible 1..4 4-sided dice
Non_transitive length-3 combinations found: 3
D16:{1,1,4,4}
D43:{1,3,3,3}
D88:{2,2,2,4}
(2 rotations omitted)
More verbose comparison of last result:
D16 > D43, D43 > D88, D16 < D88
====
Non_transitive length-4 combinations found: 4
D16:{1,1,4,4}
D43:{1,3,3,3}
D88:{2,2,2,4}
D91:{2,2,3,3}
(3 rotations omitted)
More verbose comparison of last result:
D16 > D43, D43 > D88, D88 < D91, D16 = D91
====
There are 924 possible 0..6 6-sided dice
Efron's dice
D1601:{0,0,4,4,4,4}
D19837:{1,1,1,5,5,5}
D39249:{2,2,2,2,6,6}
D58825:{3,3,3,3,3,3}
(3 rotations omitted)
D1601 < D19837, D19837 < D39249, D39249 < D58825, D1601 > D58825
====
There are 3003 possible 1..9 6-sided dice
From wp
D4121:{1,1,6,6,8,8}
D68121:{2,2,4,4,9,9}
D134521:{3,3,5,5,7,7}
(2 rotations omitted)
D4121 < D68121, D68121 < D134521, D4121 > D134521
====
Miwin's dice
D9945:{1,2,5,6,7,9}
D15705:{1,3,4,5,8,9}
D74825:{2,3,4,6,7,8}
(2 rotations omitted)
D9945 > D15705, D15705 > D74825, D9945 < D74825
====
Stretch: length-3 combinations of 1..6 6-sided dice
Actually, given recent optimisations and that this power(462,6) takes about 30s, the previously dismissed power(924,6) is probably doable, and maybe even power(3003,6)... <lang Phix>mx = 6 mn = 1 dice = possible_dice(6) show_dice(dice,find_non_trans(dice,3),3)</lang>
- Output:
There are 462 possible 1..6 6-sided dice
Non_transitive length-3 combinations found: 121998
D58:{1,1,1,2,4,4}
D87:{1,1,1,3,3,3}
D1556:{1,2,2,2,2,2}
D58:{1,1,1,2,4,4}
D88:{1,1,1,3,3,4}
D262:{1,1,2,2,2,4}
D58:{1,1,1,2,4,4}
D88:{1,1,1,3,3,4}
D1556:{1,2,2,2,2,2}
D58:{1,1,1,2,4,4}
D88:{1,1,1,3,3,4}
D1557:{1,2,2,2,2,3}
(40662 omitted)
(81332 rotations omitted)
More verbose comparison of last result:
D58 > D88, D88 > D1557, D58 < D1557
====
Python
<lang python>from collections import namedtuple from itertools import permutations, product from functools import lru_cache
Die = namedtuple('Die', 'name, faces')
@lru_cache(maxsize=None) def cmpd(die1, die2):
'compares two die returning 1, -1 or 0 for >, < =='
# Numbers of times one die wins against the other for all combinations
# cmp(x, y) is `(x > y) - (y > x)` to return 1, 0, or -1 for numbers
tot = [0, 0, 0]
for d1, d2 in product(die1.faces, die2.faces):
tot[1 + (d1 > d2) - (d2 > d1)] += 1
win2, _, win1 = tot
return (win1 > win2) - (win2 > win1)
def is_non_trans(dice):
"Check if ordering of die in dice is non-transitive returning dice or None"
check = (all(cmpd(c1, c2) == -1
for c1, c2 in zip(dice, dice[1:])) # Dn < Dn+1
and cmpd(dice[0], dice[-1]) == 1) # But D[0] > D[-1]
return dice if check else False
def find_non_trans(alldice, n=3):
return [perm for perm in permutations(alldice, n)
if is_non_trans(perm)]
def possible_dice(sides, mx):
print(f"\nAll possible 1..{mx} {sides}-sided dice")
dice = [Die(f"D{n+1}", faces)
for n, faces in enumerate(product(range(1, mx+1), repeat=sides))]
print(f' Created {len(dice)} dice')
print(' Remove duplicate with same bag of numbers on different faces')
found = set()
filtered = []
for d in dice:
count = tuple(sorted(d.faces))
if count not in found:
found.add(count)
filtered.append(d)
l = len(filtered)
print(f' Return {l} filtered dice')
return filtered
- %% more verbose extra checks
def verbose_cmp(die1, die2):
'compares two die returning their relationship of their names as a string'
# Numbers of times one die wins against the other for all combinations
win1 = sum(d1 > d2 for d1, d2 in product(die1.faces, die2.faces))
win2 = sum(d2 > d1 for d1, d2 in product(die1.faces, die2.faces))
n1, n2 = die1.name, die2.name
return f'{n1} > {n2}' if win1 > win2 else (f'{n1} < {n2}' if win1 < win2 else f'{n1} = {n2}')
def verbose_dice_cmp(dice):
c = [verbose_cmp(x, y) for x, y in zip(dice, dice[1:])] c += [verbose_cmp(dice[0], dice[-1])] return ', '.join(c)
- %% Use
if __name__ == '__main__':
dice = possible_dice(sides=4, mx=4)
for N in (3, 4): # length of non-transitive group of dice searched for
non_trans = find_non_trans(dice, N)
print(f'\n Non_transitive length-{N} combinations found: {len(non_trans)}')
for lst in non_trans:
print()
for i, die in enumerate(lst):
print(f" {' ' if i else '['}{die}{',' if i < N-1 else ']'}")
if non_trans:
print('\n More verbose comparison of last non_transitive result:')
print(' ', verbose_dice_cmp(non_trans[-1]))
print('\n ====')</lang>
- Output:
All possible 1..4 4-sided dice
Created 256 dice
Remove duplicate with same bag of numbers on different faces
Return 35 filtered dice
Non_transitive length-3 combinations found: 3
[Die(name='D16', faces=(1, 1, 4, 4)),
Die(name='D88', faces=(2, 2, 2, 4)),
Die(name='D43', faces=(1, 3, 3, 3))]
[Die(name='D43', faces=(1, 3, 3, 3)),
Die(name='D16', faces=(1, 1, 4, 4)),
Die(name='D88', faces=(2, 2, 2, 4))]
[Die(name='D88', faces=(2, 2, 2, 4)),
Die(name='D43', faces=(1, 3, 3, 3)),
Die(name='D16', faces=(1, 1, 4, 4))]
More verbose comparison of last non_transitive result:
D88 < D43, D43 < D16, D88 > D16
====
Non_transitive length-4 combinations found: 4
[Die(name='D16', faces=(1, 1, 4, 4)),
Die(name='D88', faces=(2, 2, 2, 4)),
Die(name='D91', faces=(2, 2, 3, 3)),
Die(name='D43', faces=(1, 3, 3, 3))]
[Die(name='D43', faces=(1, 3, 3, 3)),
Die(name='D16', faces=(1, 1, 4, 4)),
Die(name='D88', faces=(2, 2, 2, 4)),
Die(name='D91', faces=(2, 2, 3, 3))]
[Die(name='D88', faces=(2, 2, 2, 4)),
Die(name='D91', faces=(2, 2, 3, 3)),
Die(name='D43', faces=(1, 3, 3, 3)),
Die(name='D16', faces=(1, 1, 4, 4))]
[Die(name='D91', faces=(2, 2, 3, 3)),
Die(name='D43', faces=(1, 3, 3, 3)),
Die(name='D16', faces=(1, 1, 4, 4)),
Die(name='D88', faces=(2, 2, 2, 4))]
More verbose comparison of last non_transitive result:
D91 < D43, D43 < D16, D16 < D88, D91 > D88
====
- Caching information
In [45]: cmpd.cache_info() Out[45]: CacheInfo(hits=2148761, misses=1190, maxsize=None, currsize=1190)
Wren
<lang ecmascript>import "/sort" for Sort
var fourFaceCombs = Fn.new {
var res = []
var found = List.filled(256, false)
for (i in 1..4) {
for (j in 1..4) {
for (k in 1..4) {
for (l in 1..4) {
var c = [i, j, k, l]
Sort.insertion(c)
var key = 64*(c[0]-1) + 16*(c[1]-1) + 4*(c[2]-1) + (c[3]-1)
if (!found[key]) {
found[key] = true
res.add(c)
}
}
}
}
}
return res
}
var cmp = Fn.new { |x, y|
var xw = 0
var yw = 0
for (i in 0..3) {
for (j in 0..3) {
if (x[i] > y[j]) {
xw = xw + 1
} else if (y[j] > x[i]) {
yw = yw + 1
}
}
}
return (xw - yw).sign
}
var findIntransitive3 = Fn.new { |cs|
var c = cs.count
var res = []
for (i in 0...c) {
for (j in 0...c) {
for (k in 0...c) {
var first = cmp.call(cs[i], cs[j])
if (first == -1) {
var second = cmp.call(cs[j], cs[k])
if (second == -1) {
var third = cmp.call(cs[i], cs[k])
if (third == 1) res.add([cs[i], cs[j], cs[k]])
}
}
}
}
}
return res
}
var findIntransitive4 = Fn.new { |cs|
var c = cs.count
var res = []
for (i in 0...c) {
for (j in 0...c) {
for (k in 0...c) {
for (l in 0...c) {
var first = cmp.call(cs[i], cs[j])
if (first == -1) {
var second = cmp.call(cs[j], cs[k])
if (second == -1) {
var third = cmp.call(cs[k], cs[l])
if (third == -1) {
var fourth = cmp.call(cs[i], cs[l])
if (fourth == 1) res.add([cs[i], cs[j], cs[k], cs[l]])
}
}
}
}
}
}
}
return res
}
var combs = fourFaceCombs.call() System.print("Number of eligible 4-faced dice: %(combs.count)") var it = findIntransitive3.call(combs) System.print("\n%(it.count) ordered lists of 3 non-transitive dice found, namely:") System.print(it.join("\n")) it = findIntransitive4.call(combs) System.print("\n%(it.count) ordered lists of 4 non-transitive dice found, namely:") System.print(it.join("\n"))</lang>
- Output:
Number of eligible 4-faced dice: 35 3 ordered lists of 3 non-transitive dice found, namely: [[1, 1, 4, 4], [2, 2, 2, 4], [1, 3, 3, 3]] [[1, 3, 3, 3], [1, 1, 4, 4], [2, 2, 2, 4]] [[2, 2, 2, 4], [1, 3, 3, 3], [1, 1, 4, 4]] 4 ordered lists of 4 non-transitive dice found, namely: [[1, 1, 4, 4], [2, 2, 2, 4], [2, 2, 3, 3], [1, 3, 3, 3]] [[1, 3, 3, 3], [1, 1, 4, 4], [2, 2, 2, 4], [2, 2, 3, 3]] [[2, 2, 2, 4], [2, 2, 3, 3], [1, 3, 3, 3], [1, 1, 4, 4]] [[2, 2, 3, 3], [1, 3, 3, 3], [1, 1, 4, 4], [2, 2, 2, 4]]