Twelve statements
From Rosetta Code
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.
Contents |
[edit] Ada
Here is the main program, using a generic package Logic. The expression function introduced by the new standard Ada 2012 are very handy for this task.
with Ada.Text_IO, Logic;
procedure Twelve_Statements is
package L is new Logic(Number_Of_Statements => 12); use L;
-- formally define the 12 statements as expression function predicates
function P01(T: Table) return Boolean is (T'Length = 12); -- list of 12 statements
function P02(T: Table) return Boolean is (Sum(T(7 .. 12)) = 3); -- three of last six
function P03(T: Table) return Boolean is (Sum(Half(T, Even)) = 2); -- two of the even
function P04(T: Table) return Boolean is (if T(5) then T(6) and T(7)); -- if 5 is true, then ...
function P05(T: Table) return Boolean is
( (not T(2)) and (not T(3)) and (not T(4)) ); -- none of preceding three
function P06(T: Table) return Boolean is (Sum(Half(T, Odd)) = 4); -- four of the odd
function P07(T: Table) return Boolean is (T(2) xor T(3)); -- either 2 or 3, not both
function P08(T: Table) return Boolean is (if T(7) then T(5) and T(6)); -- if 7 is true, then ...
function P09(T: Table) return Boolean is (Sum(T(1 .. 6)) = 3); -- three of first six
function P10(T: Table) return Boolean is (T(11) and T(12)); -- next two
function P11(T: Table) return Boolean is (Sum(T(7..9)) = 1); -- one of 7, 8, 9
function P12(T: Table) return Boolean is (Sum(T(1 .. 11)) = 4); -- four of the preding
-- define a global list of statements
Statement_List: constant Statements :=
(P01'Access, P02'Access, P03'Access, P04'Access, P05'Access, P06'Access,
P07'Access, P08'Access, P09'Access, P10'Access, P11'Access, P12'Access);
-- try out all 2^12 possible choices for the table
procedure Try(T: Table; Fail: Natural; Idx: Indices'Base := Indices'First) is
procedure Print_Table(T: Table) is
use Ada.Text_IO;
begin
Put(" ");
if Fail > 0 then
Put("(wrong at");
for J in T'Range loop
if Statement_List(J)(T) /= T(J) then
Put(Integer'Image(J) & (if J < 10 then ") " else ") "));
end if;
end loop;
end if;
if T = (1..12 => False) then
Put_Line("All false!");
else
Put("True are");
for J in T'Range loop
if T(J) then
Put(Integer'Image(J));
end if;
end loop;
New_Line;
end if;
end Print_Table;
Wrong_Entries: Natural := 0;
begin
if Idx <= T'Last then
Try(T(T'First .. Idx-1) & False & T(Idx+1 .. T'Last), Fail, Idx+1);
Try(T(T'First .. Idx-1) & True & T(Idx+1 .. T'Last), Fail, Idx+1);
else -- now Index > T'Last and we have one of the 2^12 choices to test
for J in T'Range loop
if Statement_List(J)(T) /= T(J) then
Wrong_Entries := Wrong_Entries + 1;
end if;
end loop;
if Wrong_Entries = Fail then
Print_Table(T);
end if;
end if;
end Try;
begin
Ada.Text_IO.Put_Line("Exact hits:");
Try(T => (1..12 => False), Fail => 0);
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line("Near Misses:");
Try(T => (1..12 => False), Fail => 1);
end Twelve_Statements;
- Output:
Exact hits:
True are 1 3 4 6 7 11
Near Misses:
(wrong at 1) True are 5 8 11
(wrong at 1) True are 5 8 10 11 12
(wrong at 1) True are 4 8 10 11 12
(wrong at 8) True are 1 5
(wrong at 11) True are 1 5 8
(wrong at 12) True are 1 5 8 11
(wrong at 12) True are 1 5 8 10 11 12
(wrong at 8) True are 1 5 6 9 11
(wrong at 8) True are 1 4
(wrong at 12) True are 1 4 8 10 11 12
(wrong at 6) True are 1 4 6 8 9
(wrong at 7) True are 1 3 4 8 9
(wrong at 9) True are 1 3 4 6 7 9
(wrong at 12) True are 1 2 4 7 9 12
(wrong at 10) True are 1 2 4 7 9 10
(wrong at 8) True are 1 2 4 7 8 9
Here is the definition the package Logic:
generic
Number_Of_Statements: Positive;
package Logic is
--types
subtype Indices is Natural range 1 .. Number_Of_Statements;
type Table is array(Indices range <>) of Boolean;
type Predicate is access function(T: Table) return Boolean;
type Statements is array(Indices) of Predicate;
type Even_Odd is (Even, Odd);
-- convenience functions
function Sum(T: Table) return Natural;
function Half(T: Table; Which: Even_Odd) return Table;
end Logic;
And here is the implementation of the "convenience functions" in Logic:
package body Logic is
function Sum(T: Table) return Natural is
Result: Natural := 0;
begin
for I in T'Range loop
if T(I) then
Result := Result + 1;
end if;
end loop;
return Result;
end Sum;
function Half(T: Table; Which: Even_Odd) return Table is
Result: Table(T'Range);
Last: Natural := Result'First - 1;
begin
for I in T'Range loop
if I mod 2 = (if (Which=Odd) then 1 else 0) then
Last := Last+1;
Result(Last) := T(I);
end if;
end loop;
return Result(Result'First .. Last);
end Half;
end Logic;
[edit] BBC BASIC
nStatements% = 12
DIM Pass%(nStatements%), T%(nStatements%)
FOR try% = 0 TO 2^nStatements%-1
REM Postulate answer:
FOR stmt% = 1 TO 12
T%(stmt%) = (try% AND 2^(stmt%-1)) <> 0
NEXT
REM Test consistency:
Pass%(1) = T%(1) = (nStatements% = 12)
Pass%(2) = T%(2) = ((T%(7)+T%(8)+T%(9)+T%(10)+T%(11)+T%(12)) = -3)
Pass%(3) = T%(3) = ((T%(2)+T%(4)+T%(6)+T%(8)+T%(10)+T%(12)) = -2)
Pass%(4) = T%(4) = ((NOT T%(5) OR (T%(6) AND T%(7))))
Pass%(5) = T%(5) = (NOT T%(2) AND NOT T%(3) AND NOT T%(4))
Pass%(6) = T%(6) = ((T%(1)+T%(3)+T%(5)+T%(7)+T%(9)+T%(11)) = -4)
Pass%(7) = T%(7) = ((T%(2) EOR T%(3)))
Pass%(8) = T%(8) = ((NOT T%(7) OR (T%(5) AND T%(6))))
Pass%(9) = T%(9) = ((T%(1)+T%(2)+T%(3)+T%(4)+T%(5)+T%(6)) = -3)
Pass%(10) = T%(10) = (T%(11) AND T%(12))
Pass%(11) = T%(11) = ((T%(7)+T%(8)+T%(9)) = -1)
Pass%(12) = T%(12) = ((T%(1)+T%(2)+T%(3)+T%(4)+T%(5)+T%(6) + \
\ T%(7)+T%(8)+T%(9)+T%(10)+T%(11)) = -4)
CASE SUM(Pass%()) OF
WHEN -11:
PRINT "Near miss with statements ";
FOR stmt% = 1 TO 12
IF T%(stmt%) PRINT ; stmt% " ";
IF NOT Pass%(stmt%) miss% = stmt%
NEXT
PRINT "true (failed " ;miss% ")."
WHEN -12:
PRINT "Solution! with statements ";
FOR stmt% = 1 TO 12
IF T%(stmt%) PRINT ; stmt% " ";
NEXT
PRINT "true."
ENDCASE
NEXT try%
END
Output:
Near miss with statements 1 4 true (failed 8). Near miss with statements 1 5 true (failed 8). Near miss with statements 1 5 8 true (failed 11). Near miss with statements 1 3 4 6 7 9 true (failed 9). Near miss with statements 1 3 4 8 9 true (failed 7). Near miss with statements 1 4 6 8 9 true (failed 6). Near miss with statements 1 2 4 7 8 9 true (failed 8). Near miss with statements 1 2 4 7 9 10 true (failed 10). Solution! with statements 1 3 4 6 7 11 true. Near miss with statements 5 8 11 true (failed 1). Near miss with statements 1 5 8 11 true (failed 12). Near miss with statements 1 5 6 9 11 true (failed 8). Near miss with statements 1 2 4 7 9 12 true (failed 12). Near miss with statements 4 8 10 11 12 true (failed 1). Near miss with statements 1 4 8 10 11 12 true (failed 12). Near miss with statements 5 8 10 11 12 true (failed 1). Near miss with statements 1 5 8 10 11 12 true (failed 12).
[edit] D
import std.stdio, std.typecons, std.algorithm,std.range,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 sumi = curry!(reduce!q{a + b}, 0);
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 main() {
enum nStats = 12;
Tuple!(const(bool)[], const(bool)[])[] full, partial;
foreach (immutable n; 0 .. 2 ^^ nStats) {
const st = nStats.iota.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 (sols, isPartial; zip([full, partial], [false, true]))
foreach (const stm; sols) {
if (isPartial) {
immutable pos = stm[1].countUntil(false);
writefln(`Missed by statement %d: "%s"`,
pos + 1, texts[pos]);
} else
writeln("Solution:");
write(" ");
foreach (i, t; stm[0])
writef("%d:%s ", i + 1, t ? "T" : "F");
writeln;
}
}
- 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
[edit] 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
}
}
- 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
[edit] Groovy
Solution:
enum Rule {
r01( 1, { r()*.num == (1..12) }),
r02( 2, { r(7..12).count { it.truth } == 3 }),
r03( 3, { r(2..12, 2).count { it.truth } == 2 }),
r04( 4, { r(5).truth ? r(6).truth && r(7).truth : true }),
r05( 5, { r(2..4).count { it.truth } == 0 }),
r06( 6, { r(1..11, 2).count { it.truth } == 4 }),
r07( 7, { r(2).truth != r(3).truth }),
r08( 8, { r(7).truth ? r(5).truth && r(6).truth : true }),
r09( 9, { r(1..6).count { it.truth } == 3 }),
r10(10, { r(11).truth && r(12).truth }),
r11(11, { r(7..9).count { it.truth } == 1 }),
r12(12, { r(1..11).count { it.truth } == 4 });
final int num
final Closure statement
boolean truth
static final List<Rule> rules = [ null, r01, r02, r03, r04, r05, r06, r07, r08, r09, r10, r11, r12]
private Rule(num, statement) {
this.num = num
this.statement = statement
}
public static Rule r(int index) { rules[index] }
public static List<Rule> r() { rules[1..12] }
public static List<Rule> r(List<Integer> indices) { rules[indices] }
public static List<Rule> r(IntRange indices) { rules[indices] }
public static List<Rule> r(IntRange indices, int step) { r(indices.step(step)) }
public static void setAllTruth(int bits) {
(1..12).each { r(it).truth = !(bits & (1 << (12 - it))) }
}
public static void evaluate() {
def nearMisses = [:]
(0..<(2**12)).each { i ->
setAllTruth(i)
def truthCandidates = r().findAll { it.truth }
def truthMatchCount = r().count { it.statement() == it.truth }
if (truthMatchCount == 12) {
println ">Solution< ${truthCandidates*.num}"
} else if (truthMatchCount == 11) {
def miss = (1..12).find { r(it).statement() != r(it).truth }
nearMisses << [(truthCandidates): miss]
}
}
nearMisses.each { truths, miss ->
printf ("Near Miss: %-21s (failed %2d)\n", "${truths*.num}", miss)
}
}
}
Rule.evaluate()
Output:
>Solution< [1, 3, 4, 6, 7, 11] Near Miss: [1, 2, 4, 7, 8, 9] (failed 8) Near Miss: [1, 2, 4, 7, 9, 10] (failed 10) Near Miss: [1, 2, 4, 7, 9, 12] (failed 12) Near Miss: [1, 3, 4, 6, 7, 9] (failed 9) Near Miss: [1, 3, 4, 8, 9] (failed 7) Near Miss: [1, 4, 6, 8, 9] (failed 6) Near Miss: [1, 4, 8, 10, 11, 12] (failed 12) Near Miss: [1, 4] (failed 8) Near Miss: [1, 5, 6, 9, 11] (failed 8) Near Miss: [1, 5, 8, 10, 11, 12] (failed 12) Near Miss: [1, 5, 8, 11] (failed 12) Near Miss: [1, 5, 8] (failed 11) Near Miss: [1, 5] (failed 8) Near Miss: [4, 8, 10, 11, 12] (failed 1) Near Miss: [5, 8, 10, 11, 12] (failed 1) Near Miss: [5, 8, 11] (failed 1)
[edit] 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).
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
- 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])
[edit] J
In the following 'apply' is the foreign conjunction:
apply
128!:2
NB. example
'*:' apply 1 2 3
1 4 9
This enables us to apply strings (left argument) being verbs to the right argument, mostly a noun.
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&^)@#
The output follows the Haskell convention: true/false bitstring followed by the index of a contradiction
All true
(#~0=#@{::~&_1"1) testall S
+------------------------+
¦1 0 1 1 0 1 1 0 0 0 1 0¦¦
+------------------------+
Near misses
(#~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 ¦
+--------------------------+
Iterative for all true
In fact a repeat while true construction: x f^:(p)^:_ y
(-N)&{. #: S <:@]^:((]-.@-:(apply&><)"1) (-N)&{.@#:@])^:(_) 2^N=.#S
1 0 1 1 0 1 1 0 0 0 1 0
[edit] Java
The following Java code uses brute force. It tries to translate the logical statements as naturally as possible. The run time is almost zero.
public class LogicPuzzle
{
boolean S[] = new boolean[13];
int Count = 0;
public boolean check2 ()
{
int count = 0;
for (int k = 7; k <= 12; k++)
if (S[k]) count++;
return S[2] == (count == 3);
}
public boolean check3 ()
{
int count = 0;
for (int k = 2; k <= 12; k += 2)
if (S[k]) count++;
return S[3] == (count == 2);
}
public boolean check4 ()
{
return S[4] == ( !S[5] || S[6] && S[7]);
}
public boolean check5 ()
{
return S[5] == ( !S[2] && !S[3] && !S[4]);
}
public boolean check6 ()
{
int count = 0;
for (int k = 1; k <= 11; k += 2)
if (S[k]) count++;
return S[6] == (count == 4);
}
public boolean check7 ()
{
return S[7] == ((S[2] || S[3]) && !(S[2] && S[3]));
}
public boolean check8 ()
{
return S[8] == ( !S[7] || S[5] && S[6]);
}
public boolean check9 ()
{
int count = 0;
for (int k = 1; k <= 6; k++)
if (S[k]) count++;
return S[9] == (count == 3);
}
public boolean check10 ()
{
return S[10] == (S[11] && S[12]);
}
public boolean check11 ()
{
int count = 0;
for (int k = 7; k <= 9; k++)
if (S[k]) count++;
return S[11] == (count == 1);
}
public boolean check12 ()
{
int count = 0;
for (int k = 1; k <= 11; k++)
if (S[k]) count++;
return S[12] == (count == 4);
}
public void check ()
{
if (check2() && check3() && check4() && check5() && check6()
&& check7() && check8() && check9() && check10() && check11()
&& check12())
{
for (int k = 1; k <= 12; k++)
if (S[k]) System.out.print(k + " ");
System.out.println();
Count++;
}
}
public void recurseAll (int k)
{
if (k == 13)
check();
else
{
S[k] = false;
recurseAll(k + 1);
S[k] = true;
recurseAll(k + 1);
}
}
public static void main (String args[])
{
LogicPuzzle P = new LogicPuzzle();
P.S[1] = true;
P.recurseAll(2);
System.out.println();
System.out.println(P.Count + " Solutions found.");
}
}
- Output:
1 3 4 6 7 11 1 Solutions found.
[edit] Mathematica
Print["Answer:\n", Column@Cases[#, {s_, 0} :> s], "\nNear misses:\n",
Column@Cases[#, {s_, 1} :> s]] &[{#,
Count[Boole /@ {Length@# == 12, Total@#[[7 ;;]] == 3,
Total@#[[2 ;; 12 ;; 2]] == 2, #[[5]] (#[[6]] + #[[7]] - 2) ==
0, Total@#[[2 ;; 4]] == 0,
Total@#[[1 ;; 11 ;; 2]] == 4, #[[2]] + #[[3]] ==
1, #[[7]] (#[[5]] + #[[6]] - 2) == 0,
Total@#[[;; 6]] == 3, #[[11]] + #[[12]] == 2,
Total@#[[7 ;; 9]] == 1, Total@#[[;; 11]] == 4} - #,
Except[0]]} & /@ Tuples[{1, 0}, 12]]
- 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}
{1,1,0,1,0,0,1,0,1,1,0,0}
{1,1,0,1,0,0,1,0,1,0,0,1}
{1,0,1,1,0,1,1,0,1,0,0,0}
{1,0,1,1,0,0,0,1,1,0,0,0}
{1,0,0,1,0,1,0,1,1,0,0,0}
{1,0,0,1,0,0,0,1,0,1,1,1}
{1,0,0,1,0,0,0,0,0,0,0,0}
{1,0,0,0,1,1,0,0,1,0,1,0}
{1,0,0,0,1,0,0,1,0,1,1,1}
{1,0,0,0,1,0,0,1,0,0,1,0}
{1,0,0,0,1,0,0,1,0,0,0,0}
{1,0,0,0,1,0,0,0,0,0,0,0}
{0,0,0,1,0,0,0,1,0,1,1,1}
{0,0,0,0,1,0,0,1,0,1,1,1}
{0,0,0,0,1,0,0,1,0,0,1,0}
[edit] Perl 6
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;
- 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.
[edit] Prolog
Works with SWI-Prolog and library(clpfd).
puzzle :-
% 1. This is a numbered list of twelve statements.
L = [A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12],
L ins 0..1,
element(1, L, 1),
% 2. Exactly 3 of the last 6 statements are true.
A2 #<==> A7 + A8 + A9 + A10 + A11 + A12 #= 3,
% 3. Exactly 2 of the even-numbered statements are true.
A3 #<==> A2 + A4 + A6 + A8 + A10 + A12 #= 2,
% 4. If statement 5 is true, then statements 6 and 7 are both true.
A4 #<==> (A5 #==> (A6 #/\ A7)),
% 5. The 3 preceding statements are all false.
A5 #<==> A2 + A3 + A4 #= 0,
% 6. Exactly 4 of the odd-numbered statements are true.
A6 #==> A1 + A3 + A5 + A7 + A9 + A11 #= 4,
% 7. Either statement 2 or 3 is true, but not both.
A7 #<==> A2 + A3 #= 1,
% 8. If statement 7 is true, then 5 and 6 are both true.
A8 #<==> (A7 #==> A5 #/\ A6),
% 9. Exactly 3 of the first 6 statements are true.
A9 #<==> A1 + A2 + A3 + A4 + A5 + A6 #= 3,
% 10. The next two statements are both true.
A10 #<==> A11 #/\ A12,
% 11. Exactly 1 of statements 7, 8 and 9 are true.
A11 #<==> A7 + A8 + A9 #= 1,
% 12. Exactly 4 of the preceding statements are true.
A12 #<==> A1 + A2 + A3 + A4 + A5 + A6 + A7 +A8 + A9 + A10 + A11 #= 4,
label(L),
numlist(1, 12, NL),
write('Statements '),
maplist(my_write, NL, L),
writeln('are true').
my_write(N, 1) :-
format('~w ', [N]).
my_write(_N, 0).
Output :
?- puzzle. Statements 1 3 4 6 7 11 are true true .
[edit] 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. Python's boolean type boolis a subclass of int, so boolean values True, False can be used as integers (1, 0, respectively) in numerical contexts. This fact is used in the lambda expressions that use function sum.
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('Full match:')
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)
- 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
[edit] Racket
This question really begs to be done with amb
#lang racket
;; A quick `amb' implementation
(define failures null)
(define (fail)
(if (pair? failures) ((first failures)) (error "no more choices!")))
(define (amb/thunks choices)
(let/cc k (set! failures (cons k failures)))
(if (pair? choices)
(let ([choice (first choices)]) (set! choices (rest choices)) (choice))
(begin (set! failures (rest failures)) (fail))))
(define-syntax-rule (amb E ...) (amb/thunks (list (lambda () E) ...)))
(define (assert condition) (unless condition (fail)))
;; just to make things more fun
(define (⇔ x y) (assert (eq? x y)))
(require (only-in racket [and ∧] [or ∨] [implies ⇒] [xor ⊻] [not ¬]))
(define (count xs)
(let loop ([n 0] [xs xs])
(if (null? xs) n (loop (if (car xs) (add1 n) n) (cdr xs)))))
;; even more fun, make []s infix
(require (only-in racket [#%app r:app]))
(define-syntax (#%app stx)
(if (not (eq? #\[ (syntax-property stx 'paren-shape)))
(syntax-case stx () [(_ x ...) #'(r:app x ...)])
(syntax-case stx ()
;; extreme hack on next two cases, so it works for macros too.
[(_ x op y) (syntax-property #'(op x y) 'paren-shape #f)]
[(_ x op y op1 z) (free-identifier=? #'op #'op1)
(syntax-property #'(op x y z) 'paren-shape #f)])))
;; might as well do more
(define-syntax-rule (define-booleans all x ...)
(begin (define x (amb #t #f)) ...
(define all (list x ...))))
(define (puzzle)
(define-booleans all q1 q2 q3 q4 q5 q6 q7 q8 q9 q10 q11 q12)
;; 1. This is a numbered list of twelve statements.
[q1 ⇔ [12 = (length all)]]
;; 2. Exactly 3 of the last 6 statements are true.
[q2 ⇔ [3 = (count (take-right all 6))]]
;; 3. Exactly 2 of the even-numbered statements are true.
[q3 ⇔ [2 = (count (list q2 q4 q6 q8 q10 q12))]]
;; 4. If statement 5 is true, then statements 6 and 7 are both true.
[q4 ⇔ [q5 ⇒ [q6 ∧ q7]]]
;; 5. The 3 preceding statements are all false.
[q5 ⇔ (¬ [q2 ∨ q3 ∨ q4])]
;; 6. Exactly 4 of the odd-numbered statements are true.
[q6 ⇔ [4 = (count (list q1 q3 q5 q7 q9 q11))]]
;; 7. Either statement 2 or 3 is true, but not both.
[q7 ⇔ [q2 ⊻ q3]]
;; 8. If statement 7 is true, then 5 and 6 are both true.
[q8 ⇔ [q7 ⇒ (and q5 q6)]]
;; 9. Exactly 3 of the first 6 statements are true.
[q9 ⇔ [3 = (count (take all 3))]]
;; 10. The next two statements are both true.
[q10 ⇔ [q11 ∧ q12]]
;; 11. Exactly 1 of statements 7, 8 and 9 are true.
[q11 ⇔ [1 = (count (list q7 q8 q9))]]
;; 12. Exactly 4 of the preceding statements are true.
[q12 ⇔ [4 = (count (drop-right all 1))]]
;; done
(for/list ([i (in-naturals 1)] [q all] #:when q) i))
(puzzle)
;; -> '(1 3 4 6 7 11)
[edit] REXX
[edit] generalized logic
/*REXX program to solve the "Twelve Statement Puzzle". */
q=12; @stmt=right('statement',20) /*number of statements in puzzle.*/
m=0
do pass=1 for 2 /*find the maximum number trues. */
/*statement 1 is TRUE by fiat. */
do e=0 for 2**(q-1); n='1'right(x2b(d2x(e)), q-1, 0)
do b=1 for q /*define the various bits. */
@.b=substr(n,b,1) /*define a particular @ bit. */
end /*b*/
if @.1 then if yeses(1,1) \==1 then iterate
if @.2 then if yeses(7,12) \==3 then iterate
if @.3 then if yeses(2,12,2) \==2 then iterate
if @.4 then if yeses(5,5) then if yeses(6,7) \==2 then iterate
if @.5 then if yeses(2,4) \==0 then iterate
if @.6 then if yeses(1,12,2) \==4 then iterate
if @.7 then if yeses(2,3) \==1 then iterate
if @.8 then if yeses(7,7) then if yeses(5,6) \==2 then iterate
if @.9 then if yeses(1,6) \==3 then iterate
if @.10 then if yeses(11,12) \==2 then iterate
if @.11 then if yeses(7,9) \==1 then iterate
if @.12 then if yeses(1,11) \==4 then iterate
_=yeses(1,12)
if pass==1 then do; m=max(m,_); iterate; end
else if _\==m then iterate
do j=1 for q; _=substr(n,j,1)
if _ then say @stmt right(j,2) " is " word('false true',1+_)
end /*tell*/
end /*e*/
end /*pass*/
exit /*stick a fork in it, we're done.*/
/*----------------------------------YESES subroutine--------------------*/
yeses: parse arg L,H,B; #=0
do i=L to H by word(B 1,1); #=#+@.i; end /*i*/
return #
output
statement 1 is true
statement 3 is true
statement 4 is true
statement 6 is true
statement 7 is true
statement 11 is true
[edit] discrete logic
/*REXX program to solve the "Twelve Statement Puzzle". */
q=12; @stmt=right('statement',20) /*number of statements in puzzle.*/
m=0
do pass=1 for 2 /*find the maximum number trues. */
/*statement 1 is TRUE by fiat. */
do e=0 for 2**(q-1); n='1'right(x2b(d2x(e)), q-1, 0)
do b=1 for q /*define the various bits. */
@.b=substr(n,b,1) /*define a particular @ bit. */
end /*b*/
if @.1 then if \ @.1 then iterate
if @.2 then if @.7+@.8+@.9+@.10+@.11+@.12 \==3 then iterate
if @.3 then if @.2+@.4+@.6+@.8+@.10+@.12 \==2 then iterate
if @.4 then if @.5 then if \(@.6 & @.7) then iterate
if @.5 then if @.2 | @.3 | @.4 then iterate
if @.6 then if @.1+@.3+@.5+@.7+@.9+@.11 \==4 then iterate
if @.7 then if \ (@.2 && @.3 ) then iterate
if @.8 then if @.7 then if \(@.5 & @.6) then iterate
if @.9 then if @.1+@.2+@.3+@.4+@.5+@.6 \==3 then iterate
if @.10 then if \ (@.11 & @.12) then iterate
if @.11 then if @.7+@.8+@.9 \==1 then iterate
_=@.1+@.2+@.3+@.4+@.5+@.6+@.7+@.8+@.9+@.10+@.11
if @.12 then if _ \==4 then iterate
_=_+@.12
if pass==1 then do; m=max(m,_); iterate; end
else if _\==m then iterate
do j=1 for q
if @.j then say @stmt right(j,2) " is " word('false true',1+@.j)
end /*j*/
end /*e*/
end /*pass*/
/*stick a fork in it, we're done.*/
output is the same as the 1st version.
[edit] optimized
/*REXX program to solve the "Twelve Statement Puzzle". */
q=12; @stmt=right('statement',20) /*number of statements in puzzle.*/
m=0
do pass=1 for 2 /*find the maximum number trues. */
/*statement 1 is TRUE by fiat. */
do e=0 for 2**(q-1); n='1'right(x2b(d2x(e)), q-1, 0)
parse var n @1 2 @2 3 @3 4 @4 5 @5 6 @6 7 @7 8 @8 9 @9 10 @10 11 @11 12 @12
/*¦¦¦ if @1 then if \ @1 then iterate ¦¦¦*/
if @2 then if @7+@8+@9+@10+@11+@12 \==3 then iterate
if @3 then if @2+@4+@6+@8+@10+@12 \==2 then iterate
if @4 then if @5 then if \(@6 & @7) then iterate
if @5 then if @2 | @3 | @4 then iterate
if @6 then if @1+@3+@5+@7+@9+@11 \==4 then iterate
if @7 then if \ (@2 && @3 ) then iterate
if @8 then if @7 then if \(@5 & @6) then iterate
if @9 then if @1+@2+@3+@4+@5+@6 \==3 then iterate
if @10 then if \ (@11 & @12) then iterate
if @11 then if @7+@8+@9 \==1 then iterate
_=@1+@2+@3+@4+@5+@6+@7+@8+@9+@10+@11 /*shortcut*/
if @12 then if _ \==4 then iterate
_=_+@12
if pass==1 then do; m=max(m,_); iterate; end
else if _\==m then iterate
do j=1 for q; _=substr(n,j,1)
if _ then say @stmt right(j,2) " is " word('false true',1+_)
end /*j*/
end /*e*/
end /*pass*/
/*stick a fork in it, we're done.*/
output is the same as the 1st version.
[edit] Ruby
constraints = [
->(st) { st.size == 12 },
->(st) { st[-6,6].count(true) == 3 },
->(st) { st.each_slice(2).map(&:last).count(true) == 2 },
->(st) { st[4] ? (st[5] & st[6]) : true },
->(st) { st[1..3].map(&:!).all? },
->(st) { st.each_slice(2).map(&:first).count(true) == 4 },
->(st) { st[1] ^ st[2] },
->(st) { st[6] ? (st[4] & st[5]) : true },
->(st) { st[0,6].count(true) == 3 },
->(st) { st[10] & st[11] },
->(st) { st[6..8].count(true) == 1 },
->(st) { st[0,11].count(true) == 4 },
]
Result = Struct.new(:truths, :consistency)
results = [true, false].repeated_permutation(12).map do |truths|
Result.new(truths, constraints.zip(truths).map {|cn,truth| cn[truths] == truth })
end
puts "solution:",
results.find {|r| r.consistency.all? }.truths.inspect
puts "near misses: "
near_misses = results.select {|r| r.consistency.count(false) == 1 }
near_misses.each do |r|
puts "missed by statement #{r.consistency.index(false) + 1}"
puts r.truths.inspect
end
- Output:
solution: [true, false, true, true, false, true, true, false, false, false, true, false] near misses: missed by statement 8 [true, true, false, true, false, false, true, true, true, false, false, false] missed by statement 10 [true, true, false, true, false, false, true, false, true, true, false, false] missed by statement 12 [true, true, false, true, false, false, true, false, true, false, false, true] missed by statement 9 [true, false, true, true, false, true, true, false, true, false, false, false] missed by statement 7 [true, false, true, true, false, false, false, true, true, false, false, false] missed by statement 6 [true, false, false, true, false, true, false, true, true, false, false, false] missed by statement 12 [true, false, false, true, false, false, false, true, false, true, true, true] missed by statement 8 [true, false, false, true, false, false, false, false, false, false, false, false] missed by statement 8 [true, false, false, false, true, true, false, false, true, false, true, false] missed by statement 12 [true, false, false, false, true, false, false, true, false, true, true, true] missed by statement 12 [true, false, false, false, true, false, false, true, false, false, true, false] missed by statement 11 [true, false, false, false, true, false, false, true, false, false, false, false] missed by statement 8 [true, false, false, false, true, false, false, false, false, false, false, false] missed by statement 1 [false, false, false, true, false, false, false, true, false, true, true, true] missed by statement 1 [false, false, false, false, true, false, false, true, false, true, true, true] missed by statement 1 [false, false, false, false, true, false, false, true, false, false, true, false]
[edit] Tcl
package require Tcl 8.6
# Function to evaluate the truth of a statement
proc tcl::mathfunc::S {idx} {
upvar 1 state s
apply [lindex $s [expr {$idx - 1}]] $s
}
# Procedure to count the number of statements which are true
proc S+ args {
upvar 1 state state
tcl::mathop::+ {*}[lmap i $args {expr {S($i)}}]
}
# Turn a list of expressions into a list of lambda terms
proc lambdas items {lmap x $items {list state [list expr $x]}}
# Find the truth assignment that produces consistency. And those that are
# near misses too.
proc findTruthMatch {statements} {
set n [llength $statements]
for {set i 0} {$i < 2**$n} {incr i} {
set state [split [format %0.*b $n $i] ""]
set truths [lmap f $statements {apply $f [lambdas $state]}]
set counteq [tcl::mathop::+ {*}[lmap s $state t $truths {expr {
$s == $t
}}]]
if {$counteq == $n} {
lappend exact $state
} elseif {$counteq == $n-1} {
set j 0
foreach s $state t $truths {
incr j
if {$s != $t} {
lappend differ $state $j
break
}
}
}
}
return [list $exact $differ]
}
# Rendering code
proc renderstate state {
return ([join [lmap s $state {
incr i
expr {$s ? "S($i)" : "\u00acS($i)"}
}] "\u22c0"])
}
# The statements, encoded as expressions
set statements {
{[llength $state] == 12}
{[S+ 7 8 9 10 11 12] == 3}
{[S+ 2 4 6 8 10 12] == 2}
{S(5) ? S(6) && S(7) : 1}
{[S+ 2 3 4] == 0}
{[S+ 1 3 5 7 9 11] == 4}
{S(2) != S(3)}
{S(7) ? S(5) && S(6) : 1}
{[S+ 1 2 3 4 5 6] == 3}
{S(11) && S(12)}
{[S+ 7 8 9] == 1}
{[S+ 1 2 3 4 5 6 7 8 9 10 11] == 4}
}
# Find the truth assignment(s) that give consistency
lassign [findTruthMatch [lambdas $statements]] exact differ
# Print the results
foreach state $exact {
puts "exact match\t[renderstate $state ]"
}
foreach {state j} $differ {
puts "almost found\t[renderstate $state] \u21d2 [expr {[lindex $state $j-1]?"\u00ac":{}}]S($j)"
}
- Output:
exact match (S(1)?¬S(2)?S(3)?S(4)?¬S(5)?S(6)?S(7)?¬S(8)?¬S(9)?¬S(10)?S(11)?¬S(12)) almost found (¬S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?¬S(10)?S(11)?¬S(12)) ? S(1) almost found (¬S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?S(10)?S(11)?S(12)) ? S(1) almost found (¬S(1)?¬S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?S(10)?S(11)?S(12)) ? S(1) almost found (S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?¬S(8)?¬S(9)?¬S(10)?¬S(11)?¬S(12)) ? S(8) almost found (S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?¬S(10)?¬S(11)?¬S(12)) ? S(11) almost found (S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?¬S(10)?S(11)?¬S(12)) ? S(12) almost found (S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?S(10)?S(11)?S(12)) ? ¬S(12) almost found (S(1)?¬S(2)?¬S(3)?¬S(4)?S(5)?S(6)?¬S(7)?¬S(8)?S(9)?¬S(10)?S(11)?¬S(12)) ? S(8) almost found (S(1)?¬S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?¬S(7)?¬S(8)?¬S(9)?¬S(10)?¬S(11)?¬S(12)) ? S(8) almost found (S(1)?¬S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?¬S(7)?S(8)?¬S(9)?S(10)?S(11)?S(12)) ? ¬S(12) almost found (S(1)?¬S(2)?¬S(3)?S(4)?¬S(5)?S(6)?¬S(7)?S(8)?S(9)?¬S(10)?¬S(11)?¬S(12)) ? ¬S(6) almost found (S(1)?¬S(2)?S(3)?S(4)?¬S(5)?¬S(6)?¬S(7)?S(8)?S(9)?¬S(10)?¬S(11)?¬S(12)) ? S(7) almost found (S(1)?¬S(2)?S(3)?S(4)?¬S(5)?S(6)?S(7)?¬S(8)?S(9)?¬S(10)?¬S(11)?¬S(12)) ? ¬S(9) almost found (S(1)?S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?S(7)?¬S(8)?S(9)?¬S(10)?¬S(11)?S(12)) ? ¬S(12) almost found (S(1)?S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?S(7)?¬S(8)?S(9)?S(10)?¬S(11)?¬S(12)) ? ¬S(10) almost found (S(1)?S(2)?¬S(3)?S(4)?¬S(5)?¬S(6)?S(7)?S(8)?S(9)?¬S(10)?¬S(11)?¬S(12)) ? ¬S(8)
