Twelve statements
You are encouraged to solve this task according to the task description, using any language you may know.
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.
D
<lang d>import std.stdio, std.typecons, std.algorithm, std.range,
std.array, std.functional;
immutable texts = [
"this is a numbered list of twelve statements", "exactly 3 of the last 6 statements are true", "exactly 2 of the even-numbered statements are true", "if statement 5 is true, then statements 6 and 7 are both true", "the 3 preceding statements are all false", "exactly 4 of the odd-numbered statements are true", "either statement 2 or 3 is true, but not both", "if statement 7 is true, then 5 and 6 are both true", "exactly 3 of the first 6 statements are true", "the next two statements are both true", "exactly 1 of statements 7, 8 and 9 are true", "exactly 4 of the preceding statements are true"];
alias curry!(reduce!q{a + b}, 0) sumi;
immutable bool function(in bool[])[] funcs = [
s => s.length == 12, s => sumi(s[$-6 .. $]) == 3, s => sumi(s[1 .. $].stride(2)) == 2, s => s[4] ? (s[5] && s[6]) : true, s => sumi(s[1 .. 4]) == 0, s => sumi(s[0 .. $].stride(2)) == 4, s => sumi(s[1 .. 3]) == 1, s => s[6] ? (s[4] && s[5]) : true, s => sumi(s[0 .. 6]) == 3, s => s[10] && s[11], s => sumi(s[6 .. 9]) == 1, s => sumi(s[0 .. 11]) == 4];
void show(in bool[] st, in bool[] matches, in bool isPartial) {
if (isPartial) {
immutable pos = matches.countUntil(false);
writefln(`Missed by statement %d: "%s"`, pos + 1, texts[pos]);
} else
writeln("Solution:");
write(" ");
foreach (i, t; st)
writef("%d:%s ", i + 1, t ? "T" : "F");
writeln();
}
void main() {
enum nStats = 12; assert(texts.length == nStats); assert(funcs.length == nStats); Tuple!(const bool[], const bool[])[] full, partial;
foreach (n; 0 .. 2 ^^ nStats) {
const st = iota(nStats).map!(i => !!(n & (2 ^^ i)))().array();
auto truths = funcs.map!(f => f(st))();
const matches = zip(st, truths)
.map!(s_t => s_t[0] == s_t[1])()
.array();
immutable mCount = matches.sumi();
if (mCount == nStats)
full ~= tuple(st, matches);
else if (mCount == nStats - 1)
partial ~= tuple(st, matches);
}
foreach (stm; full)
show(stm.tupleof, false);
foreach (stm; partial)
show(stm.tupleof, true);
}</lang>
- Output:
Solution: 1:T 2:F 3:T 4:T 5:F 6:T 7:T 8:F 9:F 10:F 11:T 12:F Missed by statement 8: "if statement 7 is true, then 5 && 6 are both true" 1:T 2:F 3:F 4:T 5:F 6:F 7:F 8:F 9:F 10:F 11:F 12:F Missed by statement 8: "if statement 7 is true, then 5 && 6 are both true" 1:T 2:F 3:F 4:F 5:T 6:F 7:F 8:F 9:F 10:F 11:F 12:F Missed by statement 11: "exactly 1 of statements 7, 8 && 9 are true" 1:T 2:F 3:F 4:F 5:T 6:F 7:F 8:T 9:F 10:F 11:F 12:F Missed by statement 9: "exactly 3 of the first 6 statements are true" 1:T 2:F 3:T 4:T 5:F 6:T 7:T 8:F 9:T 10:F 11:F 12:F Missed by statement 7: "either statement 2 or 3 is true, but not both" 1:T 2:F 3:T 4:T 5:F 6:F 7:F 8:T 9:T 10:F 11:F 12:F Missed by statement 6: "exactly 4 of the odd-numbered statements are true" 1:T 2:F 3:F 4:T 5:F 6:T 7:F 8:T 9:T 10:F 11:F 12:F Missed by statement 8: "if statement 7 is true, then 5 && 6 are both true" 1:T 2:T 3:F 4:T 5:F 6:F 7:T 8:T 9:T 10:F 11:F 12:F Missed by statement 10: "the next two statements are both true" 1:T 2:T 3:F 4:T 5:F 6:F 7:T 8:F 9:T 10:T 11:F 12:F Missed by statement 1: "this is a numbered list of twelve statements" 1:F 2:F 3:F 4:F 5:T 6:F 7:F 8:T 9:F 10:F 11:T 12:F Missed by statement 12: "exactly 4 of the preceding statements are true" 1:T 2:F 3:F 4:F 5:T 6:F 7:F 8:T 9:F 10:F 11:T 12:F Missed by statement 8: "if statement 7 is true, then 5 && 6 are both true" 1:T 2:F 3:F 4:F 5:T 6:T 7:F 8:F 9:T 10:F 11:T 12:F Missed by statement 12: "exactly 4 of the preceding statements are true" 1:T 2:T 3:F 4:T 5:F 6:F 7:T 8:F 9:T 10:F 11:F 12:T Missed by statement 1: "this is a numbered list of twelve statements" 1:F 2:F 3:F 4:T 5:F 6:F 7:F 8:T 9:F 10:T 11:T 12:T Missed by statement 12: "exactly 4 of the preceding statements are true" 1:T 2:F 3:F 4:T 5:F 6:F 7:F 8:T 9:F 10:T 11:T 12:T Missed by statement 1: "this is a numbered list of twelve statements" 1:F 2:F 3:F 4:F 5:T 6:F 7:F 8:T 9:F 10:T 11:T 12:T Missed by statement 12: "exactly 4 of the preceding statements are true" 1:T 2:F 3:F 4:F 5:T 6:F 7:F 8:T 9:F 10:T 11:T 12:T
Go
<lang go>package main
import "fmt"
// its' not too much more work to check all the permutations concurrently var solution = make(chan int) var nearMiss = make(chan int) var done = make(chan bool)
func main() {
// iterate and use the bits as the permutation
for i := 0; i < 4096; i++ {
go checkPerm(i)
}
// collect the misses and list them after the complete solution(s)
var ms []int
for i := 0; i < 4096; {
select {
case <-done:
i++
case s := <-solution:
print12("solution", s)
case m := <-nearMiss:
ms = append(ms, m)
}
}
for _, m := range ms {
print12("near miss", m)
}
}
func print12(label string, bits int) {
fmt.Print(label, ":")
for i := 1; i <= 12; i++ {
if bits&1 == 1 {
fmt.Print(" ", i)
}
bits >>= 1
}
fmt.Println()
}
func checkPerm(tz int) {
// closure returns true if tz bit corresponding to
// 1-based statement number is 1.
ts := func(n uint) bool {
return tz>>(n-1)&1 == 1
}
// variadic closure returns number of statements listed as arguments
// which have corresponding tz bit == 1.
ntrue := func(xs ...uint) int {
nt := 0
for _, x := range xs {
if ts(x) {
nt++
}
}
return nt
}
// a flag used on repeated calls to test.
// set to true when first contradiction is found.
// if another is found, this function (checkPerm) can "short circuit"
// and return immediately without checking additional statements.
var con bool
// closure called to test each statement
test := func(statement uint, b bool) {
switch {
case ts(statement) == b:
case con:
panic("bail")
default:
con = true
}
}
// short circuit mechanism
defer func() {
if x := recover(); x != nil {
if msg, ok := x.(string); !ok && msg != "bail" {
panic(x)
}
}
done <- true
}()
// 1. This is a numbered list of twelve statements. test(1, true)
// 2. Exactly 3 of the last 6 statements are true. test(2, ntrue(7, 8, 9, 10, 11, 12) == 3)
// 3. Exactly 2 of the even-numbered statements are true. test(3, ntrue(2, 4, 6, 8, 10, 12) == 2)
// 4. If statement 5 is true, then statements 6 and 7 are both true. test(4, !ts(5) || ts(6) && ts(7))
// 5. The 3 preceding statements are all false. test(5, !ts(4) && !ts(3) && !ts(2))
// 6. Exactly 4 of the odd-numbered statements are true. test(6, ntrue(1, 3, 5, 7, 9, 11) == 4)
// 7. Either statement 2 or 3 is true, but not both. test(7, ts(2) != ts(3))
// 8. If statement 7 is true, then 5 and 6 are both true. test(8, !ts(7) || ts(5) && ts(6))
// 9. Exactly 3 of the first 6 statements are true. test(9, ntrue(1, 2, 3, 4, 5, 6) == 3) // 10. The next two statements are both true. test(10, ts(11) && ts(12)) // 11. Exactly 1 of statements 7, 8 and 9 are true. test(11, ntrue(7, 8, 9) == 1) // 12. Exactly 4 of the preceding statements are true. test(12, ntrue(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) == 4)
// no short circuit? send permutation as either near miss or solution
if con {
nearMiss <- tz
} else {
solution <- tz
}
}</lang>
- Output:
solution: 1 3 4 6 7 11 near miss: 1 4 near miss: 1 5 near miss: 1 5 8 near miss: 1 3 4 6 7 9 near miss: 1 3 4 8 9 near miss: 1 4 6 8 9 near miss: 1 2 4 7 8 9 near miss: 1 2 4 7 9 10 near miss: 5 8 11 near miss: 1 5 8 11 near miss: 1 5 6 9 11 near miss: 1 2 4 7 9 12 near miss: 1 4 8 10 11 12 near miss: 4 8 10 11 12 near miss: 5 8 10 11 12 near miss: 1 5 8 10 11 12
Haskell
Shows answers with 1 for true, followed by list of indices of contradicting elements in 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
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.
Python
Note: we choose to adapt the statement numbering to zero-based indexing in the constraintinfo lambda expressions but convert back to one-based on output.
The program uses brute force to generate all possible boolean values of the twelve statements then checks if the actual value of the statements matches the proposed or matches apart from exactly one deviation. Pythons' boolean values True, False are converted to the integer values one, zero in numerical contexts. This fact is used in the lambda expressions that use function sum. <lang python> from itertools import product
- from pprint import pprint as pp
constraintinfo = (
(lambda st: len(st) == 12 ,(1, 'This is a numbered list of twelve statements')), (lambda st: sum(st[-6:]) == 3 ,(2, 'Exactly 3 of the last 6 statements are true')), (lambda st: sum(st[1::2]) == 2 ,(3, 'Exactly 2 of the even-numbered statements are true')), (lambda st: (st[5]&st[6]) if st[4] else 1 ,(4, 'If statement 5 is true, then statements 6 and 7 are both true')), (lambda st: sum(st[1:4]) == 0 ,(5, 'The 3 preceding statements are all false')), (lambda st: sum(st[0::2]) == 4 ,(6, 'Exactly 4 of the odd-numbered statements are true')), (lambda st: sum(st[1:3]) == 1 ,(7, 'Either statement 2 or 3 is true, but not both')), (lambda st: (st[4]&st[5]) if st[6] else 1 ,(8, 'If statement 7 is true, then 5 and 6 are both true')), (lambda st: sum(st[:6]) == 3 ,(9, 'Exactly 3 of the first 6 statements are true')), (lambda st: (st[10]&st[11]) ,(10, 'The next two statements are both true')), (lambda st: sum(st[6:9]) == 1 ,(11, 'Exactly 1 of statements 7, 8 and 9 are true')), (lambda st: sum(st[0:11]) == 4 ,(12, 'Exactly 4 of the preceding statements are true')),
)
def printer(st, matches):
if False in matches:
print('Missed by one statement: %i, %s' % docs[matches.index(False)])
else:
print('Statements:')
print(' ' + ', '.join('%i:%s' % (i, 'T' if t else 'F') for i, t in enumerate(st, 1)))
funcs, docs = zip(*constraintinfo)
full, partial = [], []
for st in product( *([(False, True)] * 12) ):
truths = [bool(func(st)) for func in funcs]
matches = [s == t for s,t in zip(st, truths)]
mcount = sum(matches)
if mcount == 12:
full.append((st, matches))
elif mcount == 11:
partial.append((st, matches))
for stm in full + partial:
printer(*stm)</lang>
- Output:
Full match: 1:T, 2:F, 3:T, 4:T, 5:F, 6:T, 7:T, 8:F, 9:F, 10:F, 11:T, 12:F Missed by one statement: 1, This is a numbered list of twelve statements: 1:F, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:T, 9:F, 10:F, 11:T, 12:F Missed by one statement: 1, This is a numbered list of twelve statements: 1:F, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:T, 9:F, 10:T, 11:T, 12:T Missed by one statement: 1, This is a numbered list of twelve statements: 1:F, 2:F, 3:F, 4:T, 5:F, 6:F, 7:F, 8:T, 9:F, 10:T, 11:T, 12:T Missed by one statement: 8, If statement 7 is true, then 5 and 6 are both true: 1:T, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:F, 9:F, 10:F, 11:F, 12:F Missed by one statement: 11, Exactly 1 of statements 7, 8 and 9 are true: 1:T, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:T, 9:F, 10:F, 11:F, 12:F Missed by one statement: 12, Exactly 4 of the preceding statements are true: 1:T, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:T, 9:F, 10:F, 11:T, 12:F Missed by one statement: 12, Exactly 4 of the preceding statements are true: 1:T, 2:F, 3:F, 4:F, 5:T, 6:F, 7:F, 8:T, 9:F, 10:T, 11:T, 12:T Missed by one statement: 8, If statement 7 is true, then 5 and 6 are both true: 1:T, 2:F, 3:F, 4:F, 5:T, 6:T, 7:F, 8:F, 9:T, 10:F, 11:T, 12:F Missed by one statement: 8, If statement 7 is true, then 5 and 6 are both true: 1:T, 2:F, 3:F, 4:T, 5:F, 6:F, 7:F, 8:F, 9:F, 10:F, 11:F, 12:F Missed by one statement: 12, Exactly 4 of the preceding statements are true: 1:T, 2:F, 3:F, 4:T, 5:F, 6:F, 7:F, 8:T, 9:F, 10:T, 11:T, 12:T Missed by one statement: 6, Exactly 4 of the odd-numbered statements are true: 1:T, 2:F, 3:F, 4:T, 5:F, 6:T, 7:F, 8:T, 9:T, 10:F, 11:F, 12:F Missed by one statement: 7, Either statement 2 or 3 is true, but not both: 1:T, 2:F, 3:T, 4:T, 5:F, 6:F, 7:F, 8:T, 9:T, 10:F, 11:F, 12:F Missed by one statement: 9, Exactly 3 of the first 6 statements are true: 1:T, 2:F, 3:T, 4:T, 5:F, 6:T, 7:T, 8:F, 9:T, 10:F, 11:F, 12:F Missed by one statement: 12, Exactly 4 of the preceding statements are true: 1:T, 2:T, 3:F, 4:T, 5:F, 6:F, 7:T, 8:F, 9:T, 10:F, 11:F, 12:T Missed by one statement: 10, The next two statements are both true: 1:T, 2:T, 3:F, 4:T, 5:F, 6:F, 7:T, 8:F, 9:T, 10:T, 11:F, 12:F Missed by one statement: 8, If statement 7 is true, then 5 and 6 are both true: 1:T, 2:T, 3:F, 4:T, 5:F, 6:F, 7:T, 8:T, 9:T, 10:F, 11:F, 12:F