No frills ... just Logic ...
Correctly associate each letter and number:
If A is 1, then B is not 3.
If B is not 1, then D is 4.
If B is 1, then C is 4.
If C is 3, then D is not 2.
If C is not 2, then D is 2.
If D is 3, then A is not 4.
If A is 1, then B is not 3.
If B is not 1, then D is 4.
If B is 1, then C is 4.
If C is 3, then D is not 2.
If C is not 2, then D is 2.
If D is 3, then A is not 4.
posted by Zaux at
6:48 PM
9 Comments:
A=3
B=1
C=4
D=2
Cam
No frills....
Assume only numbers 1 to 4 allowed and no repeats.
#1) If A is 1, then B is not 3.
#2) If B is not 1, then D is 4.
#3) If B is 1, then C is 4.
#4) If C is 3, then D is not 2.
#5) If C is not 2, then D is 2.
#6) If D is 3, then A is not 4.
look at B
If B=1
by #3 C=4
by #5) D=2
by elimination A=3
If B=2
by #2 D=4
By eliminate A=1,C=3 OR C=1,A=3
Try A=1,C=3
#1 is ok, #2 is ok, #3 is ok, #4 is ok, #5 CONFLICT, #6 is ok
A=1 C=3 is invalid
Try C=1, A=3
#1 is ok, #2 is ok, #3 is ok, #4 is ok, #5 CONFLICT, #6 is ok
C=1, A=3 is invalid
B=2 is invalid
If B=3
by #2 D=4
By eliminate A=1,C=2 OR A=2,C=1
Try A=1,C=2
#1 CONFLICT
A=1 C=2 is invalid
Try A=2,C=1
#1 is ok, #2 is ok, #3 is ok, #4 is ok, #5 CONFLICT, #6 is ok
A=3, C=1 is invalid
B=3 is invalid
If B=4
by #2 D=4 !CONFLICT
B=4 is invalid
Thus B can only=1
Answer:
A=3
B=1
C=4
D=2
Cam
Cam ... right again ...
it's a pleasure to follow your analyses ...
Also works for
A=1
B=2
C=3
D=4
regards, Curtis
No that's also wrong Curtis
there's a conflics on hint #5 w/ your solution
This post has been removed by the author.
This post has been removed by the author.
Just for the heck of it, I'll do this without trial and error.
If A is 1, then B is not 3.
If B is not 1, then D is 4.
If B is 1, then C is 4.
If C is 3, then D is not 2.
If C is not 2, then D is 2.
If D is 3, then A is not 4.
NB "." = logical AND, "+" = logical OR and "!" is logical NOT
A statement such as: if a then b => !a + b = 1
I'm using 0 and 1 to represent false and true.
Let A1 denote A has value 1 , etc. The 6 statements are all true
simultaneously, so the logical AND of them is true also.
Put that together and we get:
(!A1 +!B3).(B1 +D4).(!B1 +C4).(!C3 +!D2).(C2 +D2).(!D3 +!A4) = 1
We'll need to use the following rules (some are e.g.s only)
I've combined the fundamental rules into R1 to save space.
R1': a+0 = a, a+1 = 1, a+a = a, a+!a = 1,
R1": a.0 = 0, a.1 = a, a.a = a, a.!a = 0
R2: a+a.b = a.(1+a) = a.1 = a
The above are just standard statements of logic
R3: !A1 = A2+A3+A4
R4: A1+A2+A3+A4 = 1
R5: A1.A2 = 0
R6: A1.B1 = 0
R7: A1.!A2 = A1.(A1+A3+A4) = A1
R8: A1 = A1.(B1 +!B1) = A1.!B1
R9: !A1.!A2 = A3+A4 (and !A1.!A2.!A3 = A4)
-- OK that's the logic basics dealt with
Need to fully expand the 6 factor equation. But as that is
tedious, we'll simplfy it a bit first:
(B1 +D4).(!B1 +C4) = B1.!B1 +B1.C4 +D4.!B1 +D4.C4
R1, R6 => B1.C4 +!B1.D4
Also (!C3 +!D2).(C2 +D2) = !C3.C2 +!C3.D2 +!D2.C2 +!D2.D2
R1 => !D2.D2 = 0, R8 => !C3.C2 = C2, leaves C2 +!C3.D2 +!D2.C2
R1 => C2 +!C3.D2
That leaves us with:
(!A1 +!B3).(B1.C4 +!B1.D4).(C2 +!C3.D2).(!D3 +!A4) = 1
Expanding the first two and the second two factors =>
(!A1.B1.C4 +!A1.!B1.D4 +!B3.B1.C4 +!B3.!B1.D4)
.(C2.!D3 +C2.!A4 +!C3.D2.!D3 +!C3.D2.!A4) = 1
R7, R8, R9 => (B1.C4 +!A1.!B1.D4 +B1.C4 +!B3.!B1.D4)
.(C2.!D3 +C2.!A4 +!C3.D2 +!C3.D2.!A4) = 1
R1 => (!A1.!B1.D4 + B1.C4 + !B3.!B1.D4)
.(C2.!D3 + C2.!A4 + !C3.D2) = 1
R9, R6 => (!B3+!B1).D4 = (B2+B4).D4 = B2.D4
=> (!A1.!B1.D4 + B1.C4 + B2.D4)
.(C2.!D3 + C2.!A4 + !C3.D2) = 1
Expanding =>
!A1.!B1.D4.C2.!D3 +!A1.!B1.D4.C2.!A4 +!A1.!B1.D4.!C3.D2
+B1.C4.C2.!D3 + B1.C4.C2.!A4 + B1.C4.!C3.D2
+B2.D4.C2.D3 +B2.D4.C2.!A4 +B2.D4.C2.!C3.D2 = 1
Simplifying => !A1.!B1.D4.C2 +B1.C4.D2 = 1
R3 =>(A2+A3+A4).!B1.D4.C2 +B1.C4.D2 = 1
R1, R6 => A3.!B1.D4.C2 +B1.C4.D2 = 1
R3 => A3.(B2+B3+B4).D4.C2 +B1.C4.D2 = 1
R6 => B1.C4.D2 = 1
So B = 1, C = 4, D = 2 and A = 3 (by elimination ;)
Of course, assuming the problem had a unique answer, I'd have have started with "If B is 1 then C is 4" (as it is the only fully positive statement) and got there on the first go, just as Cam did.
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