Dominoes
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Take a box of dominoes give them a good shuffle and then arrange them in diverse orientations such that they form a rectangle with 7 rows of 8 columns. Make a tableau of the face values e.g.
05132231 05505246 43036620 06235126 11300245 21433466 64515414
Now torment your computer by making it identify where each domino is.
Do this for the above tableau and one of your own construction.
Extra credit:
How many ways are there to arrange dominoes in an 8x7 rectangle, first ignoring their values, then considering their values, and finally considering values but ignoring value symmetry, i.e. transposing 5 and 4.
F#
<lang fsharp> // Dominoes: Nigel Galloway. November 17th., 2021. let cP (n:seq<uint64 list * uint64>) g=seq{for y,n in n do for g in g do let l=n^^^g in if n+g=l then yield (g::y,l)} let rec fG n g=match g with h::t->fG(cP n h)t |_->fst(Seq.head n) let solve(N:int[])=let fG=let y=fG [([],0UL)]([for g in 0..47->((N.[g],N.[g+8]),(1UL<<<g)+(1UL<<<g+8))]@[for n in 0..6 do for g in n*8..n*8+6->((N.[g],N.[g+1]),(1UL<<<g)+(1UL<<<g+1))]
|>List.groupBy(fun((n,g),_)->(min n g,max n g))|>List.sort|>List.map(fun(_,n)->n|>List.map(fun(n,g)->g))) in (fun n g->if List.contains((1UL<<<n)+(1UL<<<g)) y then "+" else " ") N|>Array.chunkBySize 8|>Array.iteri(fun n g->let n=n*8 in [0..6]|>List.iter(fun y->printf $"%d{g.[y]}%s{fG(n+y)(n+y+1)}"); printfn $"%d{g.[7]}"; [0..7]|>List.iter(fun g->printf $"%s{fG(n+g)(n+g+8)} "); printfn "")
solve [|0;5;1;3;2;2;3;1;
0;5;5;0;5;2;4;6; 4;3;0;3;6;6;2;0; 0;6;2;3;5;1;2;6; 1;1;3;0;0;2;4;5; 2;1;4;3;3;4;6;6; 6;4;5;1;5;4;1;4|]
</lang>
- Output:
0+5 1+3 2 2+3 1 + + 0 5+5 0 5 2+4 6 + + 4 3 0 3 6+6 2 0 + + + + 0 6 2 3+5 1 2 6 + + 1 1 3 0+0 2 4 5 + + + + 2 1 4 3+3 4 6 6 + + 6 4+5 1+5 4 1+4
<lang fsharp> solve [|5;6;2;0;0;4;1;4;
3;6;1;3;0;4;2;2; 3;5;6;4;3;2;1;1; 3;5;1;1;3;0;0;5; 6;0;5;4;3;5;5;2; 4;4;1;3;6;6;0;2; 1;2;6;2;6;5;0;4|]
</lang>
- Output:
5+6 2+0 0 4 1+4 + + 3+6 1 3 0 4 2+2 + + 3 5 6 4 3+2 1 1 + + + + 3 5 1+1 3+0 0 5 6 0 5+4 3+5 5+2 + + 4 4 1+3 6 6+0 2 + + 1+2 6+2 6 5+0 4
Julia
<lang julia>const tableau = [
0 5 1 3 2 2 3 1; 0 5 5 0 5 2 4 6; 4 3 0 3 6 6 2 0; 0 6 2 3 5 1 2 6; 1 1 3 0 0 2 4 5; 2 1 4 3 3 4 6 6; 6 4 5 1 5 4 1 4
]
const dominoes = [(i, j) for i in 0:size(tableau)[1]-1, j in 0:size(tableau)[2]-1 if i <= j] sorted(dom) = first(dom) > last(dom) ? reverse(dom) : dom
""" `patterns` contains solution(s), each containing a partially completed grid, the dominos used, and steps taken to get to that point in the grid. Proceed via iterating through possible tile placements from upper left to lower right, adding horizontal and vertical tile placements, dropping those that require more than one of the same domino. Consolidate in `patterns`` the newly lengthened layouts each step as moves are added. """ function findlayouts(tab = tableau, doms = dominoes)
nrows, ncols = size(tab) patterns = [(zero(tab) .- 1, Tuple{Int, Int}[], Int[])] while true newpat = empty(patterns) for (ut, ud, up) in patterns pos = findfirst(x -> x == -1, ut) pos == nothing && continue row, col = Tuple(pos) if row < nrows && ut[row + 1, col] == -1 && !(sorted((tab[row, col], tab[row + 1, col])) in ud) newut = copy(ut) newut[row:row+1, col] .= tab[row:row+1, col] push!(newpat, (newut, [ud; sorted((tab[row, col], tab[row + 1, col]))], [up; [row, col, row + 1, col]])) end if col < ncols && ut[row, col + 1] == -1 && !(sorted((tab[row, col], tab[row, col + 1])) in ud) newut = copy(ut) newut[row, col:col+1] .= tab[row, col:col+1] push!(newpat, (newut, [ud; sorted((tab[row, col], tab[row, col + 1]))], [up; [row, col, row, col + 1]])) end end isempty(newpat) && break patterns = newpat length(last(first(patterns))) == length(doms) && break end return patterns
end
function printlayout(pattern)
tab, dom, pos = pattern bytes = [[UInt8(' ') for _ in 1:(size(tab)[2] * 2 - 1)] for _ in 1:size(tab)[1]*2] for idx in 1:4:length(pos)-1 x1, y1, x2, y2 = pos[idx:idx+3] n1, n2 = tab[x1, y1], tab[x2, y2] bytes[x1 * 2 - 1][y1 * 2 - 1] = Char(n1 + '0') bytes[x2 * 2 - 1][y2 * 2 - 1] = Char(n2 + '0') if x1 == x2 # horizontal bytes[x1 * 2 - 1][y1 * 2] = Char('+') elseif y1 == y2 # vertical bytes[x1 * 2][y1 * 2 - 1] = Char('+') end end println(join(String.(bytes), "\n"))
end
for pat in findlayouts()
printlayout(pat)
end @time findlayouts()
const t2 = [
6 4 2 2 0 6 5 0; 1 6 2 3 4 1 4 3; 2 1 0 2 3 5 5 1; 1 3 5 0 5 6 1 0; 4 2 6 0 4 0 1 1; 4 4 2 0 5 3 6 3; 6 6 5 2 5 3 3 4
] @time lays = findlayouts(t2, dominoes) printlayout(first(lays)) println(length(lays), " layouts found.")
</lang>
- Output:
0+5 1+3 2 2+3 1 + + 0 5+5 0 5 2+4 6 + + 4 3 0 3 6+6 2 0 + + + + 0 6 2 3+5 1 2 6 + + 1 1 3 0+0 2 4 5 + + + + 2 1 4 3+3 4 6 6 + + 6 4+5 1+5 4 1+4 0.000507 seconds (6.06 k allocations: 1.715 MiB) 0.023503 seconds (92.66 k allocations: 35.817 MiB) 6 4 2 2 0 6+5 0 + + + + + + 1 6 2 3 4 1+4 3 2 1 0 2 3+5 5 1 + + + + + + 1 3 5 0 5 6 1 0 + + 4 2 6 0 4 0 1+1 + + + + 4 4 2 0 5 3 6 3 + + + + 6+6 5+2 5 3 3 4 2025 layouts found.
Extra credit task
<lang julia>""" From https://en.wikipedia.org/wiki/Domino_tiling#Counting_tilings_of_regions The number of ways to cover an m X n rectangle with m * n / 2 dominoes, calculated independently by Temperley & Fisher (1961) and Kasteleyn (1961), is given by """ function dominotilingcount(m, n)
return BigInt( floor( prod([ prod([ big"4.0" * (cospi(j / (m + 1)))^2 + 4 * (cospi(k / (n + 1)))^2 for k in 1:(n+1)÷2 ]) for j in 1:(m+1)÷2 ]), ), )
end
arrang = dominotilingcount(7, 8) perms = factorial(big"28") flips = 2^28
println("Arrangements ignoring values: $arrang") println("Permutations of 28 dominos: ", perms) println("Permuted arrangements ignoring flipping dominos: ", arrang * perms) println("Possible flip configurations: $flips") println("Possible permuted arrangements with flips: ", flips * arrang * perms)
</lang>
- Output:
Arrangements ignoring values: 1292697 Permutations of 28 dominos: 304888344611713860501504000000 Permuted arrangements ignoring flipping dominos: 394128248414528672328712716288000000 Possible flip configurations: 268435456 Possible permuted arrangements with flips: 105797996085635281181632579889767907328000000
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Dominoes use warnings;
my $gap = qr/(.{15}) (.{15})/s; my $grid = <<END; 0 5 1 3 2 2 3 1
0 5 5 0 5 2 4 6
4 3 0 3 6 6 2 0
0 6 2 3 5 1 2 6
1 1 3 0 0 2 4 5
2 1 4 3 3 4 6 6
6 4 5 1 5 4 1 4 END eval { find( 0, 0, $grid ) };
$grid = <<END; 0 0 0 1 1 1 0 2
1 2 2 2 0 3 1 3
2 3 3 3 0 4 1 4
2 4 3 4 4 4 0 5
1 5 2 5 3 5 4 5
5 5 0 6 1 6 2 6
3 6 4 6 5 6 6 6 END eval { find( 0, 0, $grid ) };
sub find
{ my ($x, $y, $try) = @_; if( $x > $y ) { $x = 0; if( ++$y > 6 ) # solved { print "\nfound:\n\n", $grid | $try; die; } } while( $try =~ /(?=(?|$x$gap$y|$y$gap$x))/g ) # vertical { my $new = $try; substr $new, $-[0], 33, " $1+$2 "; find( $x + 1, $y, $new ); } while( $try =~ /(?=$x $y|$y $x)/g ) # horizontal { my $new = $try; substr $new, $-[0], 3, ' + '; find( $x + 1, $y, $new ); } }</lang>
- Output:
found: 0+5 1+3 2 2+3 1 + + 0 5+5 0 5 2+4 6 + + 4 3 0 3 6+6 2 0 + + + + 0 6 2 3+5 1 2 6 + + 1 1 3 0+0 2 4 5 + + + + 2 1 4 3+3 4 6 6 + + 6 4+5 1+5 4 1+4 found: 0+0 0+1 1+1 0+2 1 2+2 2 0+3 1+3 + + 2 3+3 3 0 4 1+4 + + 2+4 3+4 4 4 0+5 1 5 2+5 3+5 4+5 + + 5 5 0+6 1+6 2 6 + + 3+6 4+6 5+6 6 6
Phix
with javascript_semantics function domino_set() sequence set = {} for i=0 to 6 do for j=i to 6 do set = append(set,{i,j}) end for end for return set end function sequence set = domino_set(), used = repeat(repeat(false,7),7), tags = shuffle(tagset(length(set))), grid function unpack(sequence s) s = split(s,' ') s = apply(true,join,{s,'?'}) s = join(s,"\n? ? ? ? ? ? ? ?\n") s = split(s,'\n') return s end function function clear(integer r, c, sequence s) if grid[r][c]!='?' then ?9/0 end if grid[r][c]='+' sequence res = {{r,c}} for i=1 to length(s) do {r,c} = s[i] if r>=1 and r<=13 and c>=1 and c<=15 then integer prev = grid[r][c] if prev='?' then grid[r][c] = ' ' res = append(res,{r,c}) elsif prev!=' ' then ?9/0 end if end if end for return res end function procedure restore(sequence s) for i=1 to length(s) do integer {r,c} = s[i] grid[r][c] = '?' end for end procedure function rand_grid(integer rem=28) if rem=0 then for r=1 to 13 by 2 do for c=2 to 14 by 2 do grid[r][c] = '?' end for end for for r=2 to 12 by 2 do for c=1 to 15 by 2 do grid[r][c] = '?' end for end for return grid end if for r=1 to 13 by 2 do for c=1 to 15 by 2 do bool flat = (c<15 and grid[r,c+1]='?'), vert = (r<13 and grid[r+1,c]='?') sequence res = {}, opt = {} if flat then opt = append(opt,{{r,c+2,r,c+1},{{r,c-1},{r,c+3},{r-1,c},{r-1,c+2},{r+1,c},{r+1,c+2}}}) end if if vert then opt = append(opt,{{r+2,c,r+1,c},{{r-1,c},{r,c-1},{r+2,c-1},{r,c+1},{r+2,c+1},{r+3,c}}}) end if opt = shuffle(opt) for i=1 to length(opt) do integer {r2,c2,r3,c3} = opt[i][1] sequence tile = shuffle(set[tags[rem]]) grid[r][c] = tile[1]+'0' grid[r2][c2] = tile[2]+'0' sequence reset = clear(r3,c3,opt[i][2]) res = rand_grid(rem-1) if length(res) then return res end if restore(reset) end for if flat or vert then return {} end if end for end for return {} end function string soln1 = "" string solnn = "" function solve(integer rem=28) if rem=0 then solnn = join(grid,'\n')&"\n\n\n" if soln1 = "" then soln1 = solnn end if return 1 end if for r=1 to 13 by 2 do for c=1 to 15 by 2 do bool flat = (c<15 and grid[r,c+1]='?'), vert = (r<13 and grid[r+1,c]='?') integer count = 0 if flat then integer {R,C} = sort({grid[r][c]-'0'+1,grid[r][c+2]-'0'+1}) if not used[R][C] then used[R][C] = true sequence reset = clear(r,c+1,{{r,c-1},{r,c+3},{r-1,c},{r-1,c+2},{r+1,c},{r+1,c+2}}) count += solve(rem-1) restore(reset) used[R][C] = false end if end if if vert then integer {R,C} = sort({grid[r][c]-'0'+1,grid[r+2][c]-'0'+1}) if not used[R][C] then used[R][C] = true sequence reset = clear(r+1,c,{{r-1,c},{r,c-1},{r+2,c-1},{r,c+1},{r+2,c+1},{r+3,c}}) count += solve(rem-1) restore(reset) used[R][C] = false end if end if if flat or vert then return count -- (may still be 0) end if end for end for return 0 end function procedure test(sequence g) grid = g g = {} soln1 = "" solnn = "" atom t0 = time() integer n = solve() puts(1,soln1) if n>1 then if n>2 then puts(1,"...\n\n\n") end if puts(1,solnn) end if printf(1,"%d solution%s found (%s)\n\n\n",{n,iff(n=1?"":"s"),elapsed(time()-t0)}) end procedure test(unpack("05132231 05505246 43036620 06235126 11300245 21433466 64515414")) test(rand_grid())
- Output:
0+5 1+3 2 2+3 1 + + 0 5+5 0 5 2+4 6 + + 4 3 0 3 6+6 2 0 + + + + 0 6 2 3+5 1 2 6 + + 1 1 3 0+0 2 4 5 + + + + 2 1 4 3+3 4 6 6 + + 6 4+5 1+5 4 1+4 1 solution found (0.2s) 6+4 2+2 0+6 5 0 + + 1+6 2+3 4+1 4 3 2+1 0+2 3+5 5+1 1+3 5+0 5+6 1+0 4+2 6 0 4+0 1+1 + + 4+4 2 0 5 3+6 3 + + 6+6 5+2 5 3+3 4 ... 6 4 2 2 0 6+5 0 + + + + + + 1 6 2 3 4 1+4 3 2 1 0 2 3+5 5 1 + + + + + + 1 3 5 0 5 6 1 0 + + 4 2 6 0 4 0 1+1 + + + + 4 4 2 0 5 3 6 3 + + + + 6+6 5+2 5 3 3 4 2025 solutions found (0.1s)
Note that 2025 is not the maximum number of solutions or anything like that, just a higher than average result.
Extra credit
Pretty dumb brute force approach, dreadfully slow.
without js -- too slow enum IGNORE, CONSIDER, NOSYM function count(integer what, rem=28, doubles=6) if rem=0 then return 1 end if atom total = 0 for r=1 to 13 by 2 do for c=1 to 15 by 2 do bool flat = (c<15 and grid[r,c+1]='?'), vert = (r<13 and grid[r+1,c]='?') sequence res = {}, opt = {} if flat then opt = append(opt,{{r,c+2,r,c+1},{{r,c-1},{r,c+3},{r-1,c},{r-1,c+2},{r+1,c},{r+1,c+2}}}) end if if vert then opt = append(opt,{{r+2,c,r+1,c},{{r-1,c},{r,c-1},{r+2,c-1},{r,c+1},{r+2,c+1},{r+3,c}}}) end if for i=1 to length(opt) do integer {r2,c2,r3,c3} = opt[i][1] sequence reset = clear(r3,c3,opt[i][2]) if what=IGNORE then total += count(what,rem-1) elsif what=CONSIDER then if doubles then total += doubles*count(what,rem-1,doubles-1) end if if rem>doubles then total += 2*(rem-doubles)*count(what,rem-1,doubles) end if else -- NOSYM total += 2*rem*count(what,rem-1) end if restore(reset) end for if flat or vert then return total end if end for end for return total end function atom t0 = time() printf(1,"Arrangements ignoring values: %,d\n",count(IGNORE)) --printf(1,"Arrangements considering values: %d\n",count(CONSIDER)) -- too slow printf(1,"Arrangements ignoring symmetry: %g\n",count(NOSYM)) ?elapsed(time()-t0)
- Output:
Arrangements ignoring values: 1,292,697 Arrangements ignoring symmetry: 1.05798e+44 "2 minutes and 37s"
Much faster
with javascript_semantics include mpfr.e function domino_tiling_count(integer m=7, n=8) atom prod = 1 for j=1 to ceil(m/2) do for k=1 to ceil(n/2) do atom cm = cos(PI * (j / (m + 1))), cn = cos(PI * (k / (n + 1))) prod *= (cm*cm + cn*cn) * 4 end for end for return floor(prod) end function atom start = time() integer arrang = domino_tiling_count(), flips = power(2,28) mpz perms = mpz_init() mpz_fac_ui(perms,28) printf(1,"Arrangements ignoring values: %,d\n", arrang) printf(1,"Permutations of 28 dominos: %s\n", mpz_get_str(perms,10,true)) mpz_mul_si(perms,perms,arrang) printf(1,"Permuted arrangements ignoring flipping dominos: %s\n", mpz_get_str(perms,10,true)) printf(1,"Possible flip configurations: %,d\n", flips) mpz_mul_si(perms,perms,flips) printf(1,"Possible permuted arrangements with flips: %s\n", mpz_get_str(perms,10,true)) printf(1,"Took %s\n",elapsed(time()-start))
- Output:
Arrangements ignoring values: 1,292,697 Permutations of 28 dominos: 304,888,344,611,713,860,501,504,000,000 Permuted arrangements ignoring flipping dominos: 394,128,248,414,528,672,328,712,716,288,000,000 Possible flip configurations: 268,435,456 Possible permuted arrangements with flips: 105,797,996,085,635,281,181,632,579,889,767,907,328,000,000 Took 0.1s
Wren
Basic task
<lang ecmascript>var tableau = [
[0, 5, 1, 3, 2, 2, 3, 1], [0, 5, 5, 0, 5, 2, 4, 6], [4, 3, 0, 3, 6, 6, 2, 0], [0, 6, 2, 3, 5, 1, 2, 6], [1, 1, 3, 0, 0, 2, 4, 5], [2, 1, 4, 3, 3, 4, 6, 6], [6, 4, 5, 1, 5, 4, 1, 4]
]
var tableau2 = [
[6, 4, 2, 2, 0, 6, 5, 0], [1, 6, 2, 3, 4, 1, 4, 3], [2, 1, 0, 2, 3, 5, 5, 1], [1, 3, 5, 0, 5, 6, 1, 0], [4, 2, 6, 0, 4, 0, 1, 1], [4, 4, 2, 0, 5, 3, 6, 3], [6, 6, 5, 2, 5, 3, 3, 4]
]
var dominoes = [] for (j in 0...tableau[0].count) {
for (i in 0...tableau.count) if (i <= j) dominoes.add([i, j])
}
var containsDom = Fn.new { |l, m, n| // assumes m <= n
for (i in 0...l.count) { var d = l[i] if (d[0] == m && d[1] == n) return true } return false
}
var copyTab = Fn.new { |t|
var c = List.filled(t.count, null) for (r in 0...t.count) c[r] = t[r].toList return c
}
var sorted = Fn.new { |dom| (dom[0] > dom[1]) ? [dom[1], dom[0]] : dom }
var findLayouts = Fn.new { |tab, doms|
var nrows = tab.count var ncols = tab[0].count var m = List.filled(nrows, null) for (i in 0...nrows) m[i] = List.filled(ncols, -1) var patterns = [ [m, [], []] ] var count = 0 while (true) { var newpat = [] for (pat in patterns) { var ut = pat[0] var ud = pat[1] var up = pat[2] var pos = null for (j in 0...ncols) { var breakOuter = false for (i in 0...nrows) { if (ut[i][j] == -1) { pos = [i, j] breakOuter = true break } } if (breakOuter) break } if (!pos) continue var row = pos[0] var col = pos[1] if (row < nrows - 1 && ut[row+1][col] == -1) { var dom = sorted.call([tab[row][col], tab[row+1][col]]) if (!containsDom.call(ud, dom[0], dom[1])) { var newut = copyTab.call(ut) newut[row][col] = tab[row][col] newut[row+1][col] = tab[row+1][col] newpat.add([newut, ud + [sorted.call( [tab[row][col], tab[row+1][col]])], up + [row, col, row+1, col]]) } } if (col < ncols - 1 && ut[row][col+1] == -1) { var dom = sorted.call([tab[row][col], tab[row][col+1]]) if (!containsDom.call(ud, dom[0], dom[1])) { var newut = copyTab.call(ut) newut[row][col] = tab[row][col] newut[row][col+1] = tab[row][col+1] newpat.add([newut, ud + [sorted.call([tab[row][col], tab[row][col+1]])], up + [row, col, row, col+1]]) } } } if (newpat.count == 0) break patterns = newpat if (patterns[0][-1].count == doms.count) break } return patterns
}
var printLayout = Fn.new { |pattern|
var tab = pattern[0] var dom = pattern[1] var pos = pattern[2] var bytes = List.filled(tab.count*2, null) for (i in 0...bytes.count) bytes[i] = List.filled(tab[0].count*2 - 1, " ") var idx = 0 while (idx < pos.count-1) { var p = pos[idx..idx+3] var x1 = p[0] var y1 = p[1] var x2 = p[2] var y2 = p[3] var n1 = tab[x1][y1] var n2 = tab[x2][y2] bytes[x1*2][y1*2] = String.fromByte(48+n1) bytes[x2*2][y2*2] = String.fromByte(48+n2) if (x1 == x2) { // horizontal bytes[x1*2][y1*2 + 1] = "+" } else if (y1 == y2) { // vertical bytes[x1*2 + 1][y1*2] = "+" } idx = idx + 4 }
for (i in 0...bytes.count) { System.print(bytes[i].join()) }
}
for (t in [tableau, tableau2]) {
var start = System.clock var lays = findLayouts.call(t, dominoes) printLayout.call(lays[0]) var lc = lays.count var pl = (lc > 1) ? "s" : "" var fo = (lc > 1) ? " (first one shown)" : "" System.print("%(lays.count) layout%(pl) found%(fo).") System.print("Took %(System.clock - start) seconds.\n")
}</lang>
- Output:
0+5 1+3 2 2+3 1 + + 0 5+5 0 5 2+4 6 + + 4 3 0 3 6+6 2 0 + + + + 0 6 2 3+5 1 2 6 + + 1 1 3 0+0 2 4 5 + + + + 2 1 4 3+3 4 6 6 + + 6 4+5 1+5 4 1+4 1 layout found. Took 0.014597 seconds. 6 4 2 2 0 6+5 0 + + + + + + 1 6 2 3 4 1+4 3 2 1 0 2 3+5 5 1 + + + + + + 1 3 5 0 5 6 1 0 + + 4 2 6 0 4 0 1+1 + + + + 4 4 2 0 5 3 6 3 + + + + 6+6 5+2 5 3 3 4 2025 layouts found (first one shown). Took 0.217176 seconds.
Extra credit (Cli)
<lang ecmascript>import "./big" for BigInt import "./fmt" for Fmt
var dominoTilingCount = Fn.new { |m, n|
var prod = 1 for (j in 1..(m/2).ceil) { for (k in 1..(n/2).ceil) { var cm = (Num.pi * (j / (m + 1))).cos var cn = (Num.pi * (k / (n + 1))).cos prod = prod * ((cm*cm + cn*cn) * 4) } } return prod.floor
}
var start = System.clock var arrang = dominoTilingCount.call(7, 8) var perms = BigInt.factorial(28) var flips = 2.pow(28)
Fmt.print("Arrangements ignoring values: $,i", arrang) Fmt.print("Permutations of 28 dominos: $,i", perms) Fmt.print("Permuted arrangements ignoring flipping dominos: $,i", perms * arrang) Fmt.print("Possible flip configurations: $,i", flips) Fmt.print("Possible permuted arrangements with flips: $,i", perms * flips * arrang) System.print("\nTook %(System.clock - start) seconds.")</lang>
- Output:
Arrangements ignoring values: 1,292,697 Permutations of 28 dominos: 304,888,344,611,713,860,501,504,000,000 Permuted arrangements ignoring flipping dominos: 394,128,248,414,528,672,328,712,716,288,000,000 Possible flip configurations: 268,435,456 Possible permuted arrangements with flips: 105,797,996,085,635,281,181,632,579,889,767,907,328,000,000 Took 0.00046 seconds.
Extra credit (Embedded)
This is just to give what will probably be a rare outing to the Mpf class though (despite their usage in the Julia example) we don't need 'big floats' here, just 'big ints'. Slightly slower than the Wren-cli example as a result. <lang ecmascript>import "./gmp" for Mpz, Mpf import "./fmt" for Fmt
var dominoTilingCount = Fn.new { |m, n|
var prec = 128 var prod = Mpf.from(1, prec) for (j in 1..(m/2).ceil) { for (k in 1..(n/2).ceil) { var cm = Mpf.pi(prec).mul(Mpf.from(j / (m + 1))).cos.square var cn = Mpf.pi(prec).mul(Mpf.from(k / (n + 1))).cos.square prod.mul(cm.add(cn).mul(4)) } } return Mpz.from(prod.floor)
}
var start = System.clock var arrang = dominoTilingCount.call(7, 8) var perms = Mpz.new().factorial(28) var flips = 2.pow(28)
Fmt.print("Arrangements ignoring values: $,i", arrang) Fmt.print("Permutations of 28 dominos: $,i", perms) Fmt.print("Permuted arrangements ignoring flipping dominos: $,i", perms * arrang) Fmt.print("Possible flip configurations: $,i", flips) Fmt.print("Possible permuted arrangements with flips: $,i", perms * flips * arrang) System.print("\nTook %(System.clock - start) seconds.")</lang>
- Output:
Arrangements ignoring values: 1,292,697 Permutations of 28 dominos: 304,888,344,611,713,860,501,504,000,000 Permuted arrangements ignoring flipping dominos: 394,128,248,414,528,672,328,712,716,288,000,000 Possible flip configurations: 268,435,456 Possible permuted arrangements with flips: 105,797,996,085,635,281,181,632,579,889,767,907,328,000,000 Took 0.00058 seconds.