Twelve statements
This puzzle is borrowed from here.
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
Shows answers with 1 for true, followed by list of indices of incorrect elements each set of 1/0s (index is 0-based).
<lang haskell>import Data.List (findIndices)
tf = mapM (\_ -> [1,0])
wrongness b = findIndices id . zipWith (/=) b . map (fromEnum . ($ b))
statements = [ (==12) . length, 3 ⊂ [length statements-6..], 2 ⊂ [1,3..], 4 → [4..6], 0 ⊂ [1..3], 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 == (sum . map (b!!) . takeWhile (< length b)) x (a → x) b = (b!!a == 0) || all ((==1).(b!!)) x
testall s n = [(b, w) | b <- tf s, w <- [wrongness b s], length w == n]
main = let t = testall statements in do putStrLn "Answer" mapM_ print $ t 0 putStrLn "Near misses" mapM_ print $ t 1</lang>
- Output:
Answer ([1,0,1,1,0,1,1,0,0,0,1,0],[]) Near misses ([1,1,0,1,0,0,1,1,1,0,0,0],[7]) ([1,1,0,1,0,0,1,0,1,1,0,0],[9]) ([1,1,0,1,0,0,1,0,1,0,0,1],[11]) ([1,0,1,1,0,1,1,0,1,0,0,0],[8]) ([1,0,1,1,0,0,0,1,1,0,0,0],[6]) ([1,0,0,1,0,1,0,1,1,0,0,0],[5]) ([1,0,0,1,0,0,0,1,0,1,1,1],[11]) ([1,0,0,1,0,0,0,0,0,0,0,0],[7]) ([1,0,0,0,1,1,0,0,1,0,1,0],[7]) ([1,0,0,0,1,0,0,1,0,1,1,1],[11]) ([1,0,0,0,1,0,0,1,0,0,1,0],[11]) ([1,0,0,0,1,0,0,1,0,0,0,0],[10]) ([1,0,0,0,1,0,0,0,0,0,0,0],[7]) ([0,0,0,1,0,0,0,1,0,1,1,1],[0]) ([0,0,0,0,1,0,0,1,0,1,1,1],[0]) ([0,0,0,0,1,0,0,1,0,0,1,0],[0])
J
Follows the Haskell solution.
In the following 'apply' is a foreign conjunction: <lang j> apply 128!:2
NB. example
'*:' apply 1 2 3
1 4 9</lang> This enables us to apply strings (left argument) being verbs to the right argument, mostly a noun. <lang j>S=: <;._2 (0 :0) 12&=@# 3=+/@:{.~&_6 2= +/@:{~&1 3 5 7 9 11 4&{=*./@:{~&4 5 6 0=+/@:{~&1 2 3 4=+/@:{~&0 2 4 6 8 10 1=+/@:{~&1 2 6&{=*./@:{~&4 5 6 3=+/@:{.~&6 2=+/@:{~&10 11 1=+/@:{~&6 7 8 4=+/@:{.~&11 )
testall=: (];"1 0<@I.@:(]~:(apply&><))"1) #:@i.@(2&^)@#</lang>
All true <lang j> (#~0=#@{::~&_1"1) testall S ┌───────────────────────┬┐ │1 0 1 1 0 1 1 0 0 0 1 0││ └───────────────────────┴┘</lang> Near misses <lang j> (#~1=#@{::~&_1"1) testall S ┌───────────────────────┬──┐ │0 0 0 0 1 0 0 1 0 0 1 0│0 │ ├───────────────────────┼──┤ │0 0 0 0 1 0 0 1 0 1 1 1│0 │ ├───────────────────────┼──┤ │0 0 0 1 0 0 0 1 0 1 1 1│0 │ ├───────────────────────┼──┤ │1 0 0 0 1 0 0 0 0 0 0 0│7 │ ├───────────────────────┼──┤ │1 0 0 0 1 0 0 1 0 0 0 0│10│ ├───────────────────────┼──┤ │1 0 0 0 1 0 0 1 0 0 1 0│11│ ├───────────────────────┼──┤ │1 0 0 0 1 0 0 1 0 1 1 1│11│ ├───────────────────────┼──┤ │1 0 0 0 1 1 0 0 1 0 1 0│7 │ ├───────────────────────┼──┤ │1 0 0 1 0 0 0 0 0 0 0 0│7 │ ├───────────────────────┼──┤ │1 0 0 1 0 0 0 1 0 1 1 1│11│ ├───────────────────────┼──┤ │1 0 0 1 0 1 0 1 1 0 0 0│5 │ ├───────────────────────┼──┤ │1 0 1 1 0 0 0 1 1 0 0 0│6 │ ├───────────────────────┼──┤ │1 0 1 1 0 1 1 0 1 0 0 0│8 │ ├───────────────────────┼──┤ │1 1 0 1 0 0 1 0 1 0 0 1│11│ ├───────────────────────┼──┤ │1 1 0 1 0 0 1 0 1 1 0 0│9 │ ├───────────────────────┼──┤ │1 1 0 1 0 0 1 1 1 0 0 0│7 │ └───────────────────────┴──┘</lang>
Iterative for all true <lang j> (-N)&{. #: S <:@]^:((]-.@-:(apply&><)"1) (-N)&{.@#:@])^:(_) 2^N=.#S 1 0 1 1 0 1 1 0 0 0 1 0</lang>
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.