Twelve statements

This puzzle is borrowed from here.

Twelve statements is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Given the following twelve statements, which of them are true?

1.  This is a numbered list of twelve statements.
2.  Exactly 3 of the last 6 statements are true.
3.  Exactly 2 of the even-numbered statements are true.
4.  If statement 5 is true, then statements 6 and 7 are both true.
5.  The 3 preceding statements are all false.
6.  Exactly 4 of the odd-numbered statements are true.
7.  Either statement 2 or 3 is true, but not both.
8.  If statement 7 is true, then 5 and 6 are both true.
9.  Exactly 3 of the first 6 statements are true.
10.  The next two statements are both true.
11.  Exactly 1 of statements 7, 8 and 9 are true.
12.  Exactly 4 of the preceding statements are true.

When you get tired of trying to figure it out in your head, write a program to solve it, and print the correct answer or answers.

Extra credit: also print out a table of near misses, that is, solutions that are contradicted by only a single statement.

Haskell

<lang haskell>tf [x] = [[True], [False]] tf (x:xs) = map (True:) s ++ map (False:) s where s = tf xs

sumbool = length . filter id wrongness b = sumbool . zipWith (/=) b . map (\f -> f b)

statements = [ (==12) . length, 3 ⊂ [length statements-6..], 2 ⊂ [1,3..], 4 → [4..6], 0 ⊂ [2..4], 4 ⊂ [0,2..], 1 ⊂ [1,2], 6 → [4..6], 3 ⊂ [0..5], 2 ⊂ [10,11], 1 ⊂ [6,7,8], 4 ⊂ [0..10] ] where (⊂) s x = \b -> s == (sumbool . map (b!!) . takeWhile (< length b)) x (→) a x = \b -> not (b!!a) || all (b!!) x

main = let t n s = filter ((==n).fst) [(wrongness b s, b) | b <- tf s] in do putStrLn "Answer" mapM_ print $ t 0 statements putStrLn "Near misses" mapM_ print $ t 1 statements</lang>

Output:

Answer
(0,[True,False,True,True,False,True,True,False,False,False,True,False])
Near misses
(1,[True,True,False,True,False,False,True,True,True,False,False,False])
(1,[True,True,False,True,False,False,True,False,True,True,False,False])
(1,[True,True,False,True,False,False,True,False,True,False,False,True])
(1,[True,False,True,True,False,True,True,False,True,False,False,False])
(1,[True,False,True,True,False,False,False,True,True,False,False,False])
(1,[True,False,False,True,False,True,False,True,True,False,False,False])
(1,[True,False,False,True,False,False,False,True,False,True,True,True])
(1,[True,False,False,True,False,False,False,False,False,False,False,False])
(1,[False,False,False,True,False,False,False,True,False,True,True,True])

Perl 6

<lang perl6>sub infix:<→> ($protasis,$apodosis) { !$protasis or $apodosis }

my @tests = { True }, # (there's no 0th statement)

   { all(.[1..12]) === any(True, False) },
   { 3 == [+] .[7..12] },
   { 2 == [+] .[2,4...12] },
   { .[5] → all .[6,7] },
   { none .[2,3,4] },
   { 4 == [+] .[1,3...11] },
   { one .[2,3] },
   { .[7] → all .[5,6] },
   { 3 == [+] .[1..6] },
   { all .[11,12] },
   { one .[7,8,9] },
   { 4 == [+] .[1..11] };

my @good; my @bad; my @ugly;

for reverse 0 ..^ 2**12 -> $i {

   my @b = $i.fmt("%012b").comb;
   my @assert = True, @b.map: { .so }
   my @result = @tests.map: { .(@assert).so }
   my @s = ( $_ if $_ and @assert[$_] for 1..12 );
   if @result eqv @assert {

push @good, "<{@s}> is consistent.";

   }
   else {

my @cons = gather for 1..12 { if @assert[$_] !eqv @result[$_] { take @result[$_] ?? $_ !! "¬$_"; } } my $mess = "<{@s}> implies {@cons}."; if @cons == 1 { push @bad, $mess } else { push @ugly, $mess }

   }

}

.say for @good; say "\nNear misses:"; .say for @bad;</lang>

Output:

<1 3 4 6 7 11> is consistent.

Near misses:
<1 2 4 7 8 9> implies ¬8.
<1 2 4 7 9 10> implies ¬10.
<1 2 4 7 9 12> implies ¬12.
<1 3 4 6 7 9> implies ¬9.
<1 3 4 8 9> implies 7.
<1 4 6 8 9> implies ¬6.
<1 4 8 10 11 12> implies ¬12.
<1 4> implies 8.
<1 5 6 9 11> implies 8.
<1 5 8 10 11 12> implies ¬12.
<1 5 8 11> implies 12.
<1 5 8> implies 11.
<1 5> implies 8.
<4 8 10 11 12> implies 1.
<5 8 10 11 12> implies 1.
<5 8 11> implies 1.