2009-05-05 00:29:20 +00:00
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-- The Computer Language Benchmarks Game
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2008-07-14 17:58:37 +00:00
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-- http://shootout.alioth.debian.org/
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-- contributed by Mike Pall
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local function fannkuch(n)
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local p, q, s, odd, check, maxflips = {}, {}, {}, true, 0, 0
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2009-05-05 00:29:20 +00:00
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for i=1,n do p[i] = i; q[i] = i; s[i] = i end
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2008-07-14 17:58:37 +00:00
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repeat
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-- Print max. 30 permutations.
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if check < 30 then
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if not p[n] then return maxflips end -- Catch n = 0, 1, 2.
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2012-09-07 14:05:41 +00:00
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io.write(table.unpack(p)); io.write("\n")
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2008-07-14 17:58:37 +00:00
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check = check + 1
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end
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-- Copy and flip.
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local q1 = p[1] -- Cache 1st element.
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if p[n] ~= n and q1 ~= 1 then -- Avoid useless work.
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for i=2,n do q[i] = p[i] end -- Work on a copy.
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for flips=1,1000000 do -- Flip ...
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local qq = q[q1]
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if qq == 1 then -- ... until 1st element is 1.
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if flips > maxflips then maxflips = flips end -- New maximum?
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break
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end
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q[q1] = q1
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2009-05-05 00:29:20 +00:00
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if q1 >= 4 then
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local i, j = 2, q1 - 1
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repeat q[i], q[j] = q[j], q[i]; i = i + 1; j = j - 1; until i >= j
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end
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2008-07-14 17:58:37 +00:00
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q1 = qq
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end
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end
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-- Permute.
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if odd then
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p[2], p[1] = p[1], p[2]; odd = false -- Rotate 1<-2.
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else
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p[2], p[3] = p[3], p[2]; odd = true -- Rotate 1<-2 and 1<-2<-3.
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for i=3,n do
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local sx = s[i]
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2009-05-05 00:29:20 +00:00
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if sx ~= 1 then s[i] = sx-1; break end
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2008-07-14 17:58:37 +00:00
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if i == n then return maxflips end -- Out of permutations.
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s[i] = i
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-- Rotate 1<-...<-i+1.
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local t = p[1]; for j=1,i do p[j] = p[j+1] end; p[i+1] = t
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end
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end
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until false
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end
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2009-05-05 00:29:20 +00:00
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local n = tonumber(arg and arg[1]) or 1
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io.write("Pfannkuchen(", n, ") = ", fannkuch(n), "\n")
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