Find the missing permutation: Difference between revisions

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=={{header|Phix}}==

<lang Phix>constant perms = {"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",

with javascript_semantics

"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",

constant perms = {"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",

"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"}

"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",

"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"}

-- 1: sum of letters

sequence r = repeat(0,4)

-- 1: sum of letters

for i=1 to length(perms) do

sequence r = repeat(0,4)

r = sq_add(r,perms[i])

for i=1 to length(perms) do

end for

r = sq_add(r,perms[i])

r = sq_sub(max(r)+'A',r)

end for

puts(1,r&'\n')

r = sq_sub(max(r)+'A',r)

⚫

-- based on the notion that missing = sum(full)-sum(partial) would be true,

printf(1,"%s\n",{r})

⚫

-- and that sum(full) would be like {M,M,M,M} rather than a mix of numbers.

⚫

-- based on the notion that missing = sum(full)-sum(partial) would be true,

-- the final step is equivalent to eg {1528,1530,1531,1529}

⚫

-- and that sum(full) would be like {M,M,M,M} rather than a mix of numbers.

⚫

-- max-r[i] -> { 3, 1, 0, 2}

-- to ~~chars ->~~ { D, B, A, C}

-- the final step is equivalent to eg {1528,1530,1531,1529}

⚫

-- max-r[i] -> { 3, 1, 0, 2}

⚫

-- (but obviously both done in one line)

-- to chars -> { D, B, A, C}

⚫

-- (but obviously both done in one line)

⚫

-- 2: the xor trick

r = repeat(0,4)

⚫

-- 2: the xor trick

for i=1 to length(perms) do

r = repeat(0,4)

r = sq_xor_bits(r,perms[i])

for i=1 to length(perms) do

end for

r = sq_xor_bits(r,perms[i])

puts(1,r&'\n')

end for

⚫

-- (relies on the missing chars being present an odd number of times, non-missing chars an even number of times)

printf(1,"%s\n",{r})

⚫

-- (relies on the missing chars being present an odd number of times, non-missing chars an even number of times)

⚫

-- 3: find least frequent letters

r = " "

⚫

-- 3: find least frequent letters

for i=1 to length(r) do

r = " "

sequence count = repeat(0,4)

for i=1 to length(r) do

for j=1 to length(perms) do

sequence count = repeat(0,4)

count[perms[j][i]-'A'+1] += 1

for j=1 to length(perms) do

end for

integer cdx = perms[j][i]-'A'+1

r[i] = smallest(count,1)+'A'-1

count[cdx] += 1

end for

end for

puts(1,r&'\n')

r[i] = smallest(count,1)+'A'-1

⚫

-- (relies on the assumption that a full set would have each letter occurring the same number of times in each position)

end for

⚫

-- (smallest(count,1) returns the index position of the smallest, rather than it's value)

printf(1,"%s\n",{r})

⚫

-- (relies on the assumption that a full set would have each letter occurring the same number of times in each position)

⚫

-- 4: test all permutations

⚫

-- (smallest(count,1) returns the index position of the smallest, rather than it's value)

for i=1 to factorial(4) do

r = permute(i,"ABCD")

⚫

-- 4: test all permutations

if not find(r,perms) then exit end if

for i=1 to factorial(4) do

end for

r = permute(i,"ABCD")

puts(1,r&'\n')

if not find(r,perms) then exit end if

⚫

-- (relies on brute force(!) - but this is the only method that could be made to cope with >1 omission)</~~lang~~>

end for

printf(1,"%s\n",{r})

⚫

-- (relies on brute force(!) - but this is the only method that could be made to cope with >1 omission)

<pre>