Non-transitive dice: Difference between revisions

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=={{header|F_Sharp|F#}}==

===The task (4 sided die)===

// Non-transitive dice. Nigel Galloway: August 9th., 2020

let die=[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 n4=seq{for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do for d4 in N.[d3] do if List.contains d1 N.[d4] then yield (d1,d2,d3,d4)}

n3|>Seq.iter(fun(d1,d2,d3)->printfn "%A<%A; %A<%A; %A>%A\n" d1 d2 d2 d3 d1 d3)

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<pre>

===Extra credit (6 sided die)===

;Sides numbered 1..6

// Non-transitive diceNigel Galloway: August 9th., 2020

let die=[for n0 in [1..6] do for n1 in [n0..6] do for n2 in [n1..6] do for n3 in [n2..6] do for n4 in [n3..6] do for n5 in [n4..6]->[n0;n1;n2;n3;n4;n5]]

let n3=[for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do if List.contains d1 N.[d3] then yield (d1,d2,d3)]

let d1,d2,d3=List.last n3 in printfn "Solutions found = %d\nLast solution found is %A<%A; %A<%A; %A>%A" (n3.Length) d1 d2 d2 d3 d1 d3

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<pre>

</pre>

;Sides numbered 1..7

// Non-transitive diceNigel Galloway: August 9th., 2020

let die=[for n0 in [1..7] do for n1 in [n0..7] do for n2 in [n1..7] do for n3 in [n2..7] do for n4 in [n3..7] do for n5 in [n4..7]->[n0;n1;n2;n3;n4;n5]]

let n3=[for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do if List.contains d1 N.[d3] then yield (d1,d2,d3)]

let d1,d2,d3=List.last n3 in printfn "Solutions found = %d\nLast solution found is %A<%A; %A<%A; %A>%A" (n3.Length) d1 d2 d2 d3 d1 d3

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<pre>

 

=={{header|Factor}}==

prettyprint sequences ;

 

 

"All ordered lists of 4 non-transitive dice with 4 sides:" print

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<pre>

=={{header|Go}}==

{{trans|Wren}}

 

import (

fmt.Println(a)

}

 

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[[2 2 3 3] [1 3 3 3] [1 1 4 4] [2 2 2 4]]

</pre>

 

=={{header|Julia}}==

{{trans|Python}}

using Memoize, Combinatorics

@time testnontransitivedice(5)

@time testnontransitivedice(6)

<pre>

All possible 1..3 3-sided dice

123.182432 seconds (553.51 M allocations: 392.013 GiB, 17.02% gc time

</pre>

 

=={{header|MiniZinc}}==

===The model===

%Non transitive dice. Nigel Galloway, September 14th., 2020

int: die; int: faces; set of int: values;

constraint pN(g[die,1..faces],g[1,1..faces]);

output[join(" ",[show(g[n,1..faces])++"<"++show(g[n+1,1..faces]) | n in 1..die-1])," "++show(g[die,1..faces])++">"++show(g[1,1..faces])];

===The task(4 sided die)===

;3 die values=1..4

{{trans|Python}}

{{libheader|itertools}}

import itertools

 

 

func cmp(die1, die2: Die): int =

var tot: array[3, int]

for d in product(die1.faces, die2.faces):

 

func verboseCmp(die1, die2: Die): string =

var win1, win2 = 0

for (d1, d2) in product(die1.faces, die2.faces):

echo &"\n More verbose comparison of last non-transitive result:"

echo " ", verboseDiceCmp(nonTrans[^1])

 

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====</pre>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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<pre>

With the recent optimisations I can just about stretch to power(924,3), but I think I was right to skip power(924,4) above, it would probably take over two and a half ''days''.<br>

I have also estimated that power(3003,3) (ie 1..9 on the sides) would probably take around 4 hours (assuming such attempts didn't run out of memory).

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<pre>

 

=={{header|Python}}==

from itertools import permutations, product

from functools import lru_cache

print('\n More verbose comparison of last non_transitive result:')

print(' ', verbose_dice_cmp(non_trans[-1]))

 

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Out[45]: CacheInfo(hits=2148761, misses=1190, maxsize=None, currsize=1190)

</pre>

 

=={{header|Raku}}==

Thanks to Thundergnat for the nice "less-is-more" tweaks now the 4 dice portion takes around 10 (down from 17 ) minutes to run ..

{{trans|Go}}

 

my @dicepool = ^4 xx 4 ;

say +@output, " ordered lists of $_ non-transitive dice found, namely:";

.say for @output;

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<pre>Number of eligible 4-faced dice : 35

=={{header|Wren}}==

{{libheader|Wren-sort}}

 

var fourFaceCombs = Fn.new {

it = findIntransitive4.call(combs)

System.print("\n%(it.count) ordered lists of 4 non-transitive dice found, namely:")

 

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