Winners, losers and liars
This is probably my last on this theme.
A, B, C and D run in a race. When interviewed A, B and C made two statments each:
A: C won; B was second
B: C was second; D was third
C: D was last; A was second
If each told one true statement and one porky, who won the race?
A, B, C and D run in a race. When interviewed A, B and C made two statments each:
A: C won; B was second
B: C was second; D was third
C: D was last; A was second
If each told one true statement and one porky, who won the race?
Labels: logic
posted by Chris at
4:44 PM
11 Comments:
You may find the following suffices useful ₁ ₂ ₃ ₄,thanks Zaux.
First post!
Race results:
A - 1st
B - 2nd
D - 3rd
C - 4th
nah ... that's wrong
That's right Zaux (you were wrong - snigger).
1st: C
2nd: A
3rd: D
4th: B
Here's my final answer:
C - 1st place
A - 2nd place
D - 3rd place
B - 4th place
Have to run ... maybe someone will provide explanation
Graham and I agree
Fortunately you both agree with me as well.
Long-winded solution coming up soon ;)
Let A1, A2, A3, A4 represent A first, second third or fourth etc.
The sets of statements considered as truth statements are:
C1.!B2 + !C1.B2 = 1
C2.!D3 + !C2.D3 = 1
D4.!A2 + !D4.A2 = 1
in above am simply saying either first statement true
and second false OR vice-versa.
Note A1 + A2 + A3 + A4 = 1 etc.
A1 + B1 + C1 + D1 = 1 etc.
A1.A2 = 0, A1.B1 = 0, !A1.A1 = 0 etc.
Combine the first three statements:
[C1.!B2 + !C1.B2].[C2.!D3 + !C2.D3].[D4.!A2 + !D4.A2] = 1
Multiplying that lot out (initially without simplifying) =>
C1.!B2.C2.!D3.D4.!A2 + C1.!B2.C2.!D3.!D4.A2
+ C1.!B2.!C2.D3.D4.!A2 + C1.!B2.!C2.D3.!D4.A2
+ !C1.B2.C2.!D3.D4.!A2 + !C1.B2.C2.!D3.!D4.A2
+ !C1.B2.!C2.D3.D4.!A2 + !C1.B2.!C2.D3.!D4.A2 = 1
Now simplify using the obvious contradictions, leaves only
C1.!B2.!C2.D3.!D4.A2 = 1
Each factor must be 1 (true), so:
C1, D3, A2 are all true. So realise B4 is true by exhaustion.
i.e. they came in, in the order C, A, D and B.
If anyone's interested, I've added a bunch more info (and ravings) about this logical approach on the "Who's who" problem posted yesterday.
But unless I see an especially good problem, I'm moving on from it (mainly because it's quite tedious to do).
Later that year...
When doing a more complex problem, I found it useful to extend the
rules. For instance C1∙!C2 = C1∙(C1+C3+C4) = C1
Another rule A1 = A1∙(B1+B2+B3+B4) = A1∙(B2+B3+B4) = A1∙!B1
In fact, using just the latter rule (3 times) =>
C1∙!B2∙!C2∙D3∙!D4∙A2 = 1 => C1∙D3∙A2 = 1
Just to be a devil:
C1∙D3∙A2 = (B1+B2+B3+B4)∙C1∙D3∙A2 = B4∙C1∙D3∙A2 = 1
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