Counter logic

X is BLUE
Y is RED
Z is WHITE

The colors are as in the figure: X is red, Y is blue, Z is White.

Hoping my logic holds up:

1) If X is Red true then Y is not red also true: Contradicts initial premise that only one of given statements is true. Hence X is not Red. X is Blue or White.

2) If Y is not Red then it is Blue or White. If Z is not Blue then it is Red or White.

3) If Y not Red is true then Z not Blue is false as only one of these statements can be true. In which case Z is blue and 2) above means Y is white. However this would mean X is red which contradicts 1) above.

4) With Z not blue true, Y not Red is false. Y is red, 2) gives Z white and 1) gives X blue.

Lines up with statements given as follows:

X is Red - False. X is Blue.
Y is not Red - False. Y is Red
Z is not Blue - True. Z is White

X is blue
Y is red
Z is white

X is blue
Y is red
Z is white

I just couldn't figure out how to solve this with truth tables and values. My reasoning went:

If "X is red" is true, then "Y is not red" must be false because only one statement can be true, so Y is red, so "X is red" is false after all. Since X is not red, X is either blue or white. If X is blue, then "Z is not blue" is true, so the only remaining colour Z can be is white.

Conclusion: X is blue, Y is red, Z is white. The first two statements in the original question are false, the third is true. Right?

If the above approach can be described in contrast to Boole's as "trial and improvement", it isn't too laboriously so I hope, since at least I didn't go through every possible combination of colours for the counters and truth values for the statements. I only tried out two statements "X is red" and then "X is blue" in the course of homing in. (Though admittedly it took a lot of cogitation before deciding on that course in the first place). I really look forward to an explanation of how the answer can be calculated using AND and OR Boolean style without this process of elimination.

Interesting that Boole mathematised logic, but later thinkers in the 19th and early 20th century tried to logicise maths.

Chris G

Z is White, X is Blue and Y is Red

X is BLUE
Y is RED
Z is WHITE

Z is not blue is TRUE.

X is Blue
Y is Red
Z is White

X is blue, Y is red and Z is white. Enjoyed the puzzle, thanks.

Tag these statements A,B,C. Here is a direct solution.

A implies B so it cannot be the only true statement. Therefore, X cannot be red.
B cannot be true because if "Y is not Red" then Z must be red, rendering both B and C true. So B is False and thus Y=Red.
C remains the only option for the true statement.
Z cannot be Blue or Red so it Z=White. This leaves X=Blue.

Check it out: X=Blue, Y=Red, Z=White renders A, B False while C is True.

I must say I'm impatient for an answer to this puzzle. Well, not for the answer itself, which I've already sent in though it hasn't been posted yet (just to repeat: X blue, Y red, Z white), but to be enlightened as to how to get it Boolean style by adding and multiplying some 1's and 0's. Two weeks now ought to be long enough eh?

Chris G

X is red
Y is Blue
Z is white

Good one ..
Conditions
1) each one has different colours
2) Exactly one of the statement can be correct

My arguments is like this
If X is Red is True then Y is not Red will also be true which violates second condition .
So X is RED not possible
Hence X cannot be true ( so X is White or Blue)

Now Y is not Red is True imply Y is White or Blue
Suppose Y is Blue then third statement will be True Not possible
Y is White then X will become Blue again third statement also become true.
Hence Y is not red is False So Y is RED...
Z can be White or Blue But since only one statement is True Z is Blue Hence X is White
X=White
Y= Red
Z= Blue

Answer :

y is red
x is blue
z is white

x is red ? false, x is blue
y is not red ? false, y is red
z is not blue ? true, z is white

X is red
Y is blue
Z is white

X is Blue
Y is red
Z is white

Straightforward: 1st or 2nd statement can't be true as this would lead to a contradiction -> last statement is true meaning:
x is not red
y is red
z is not blue -> x is blue, y is red, z is white

Great puzzle. My approach to it was this:
1. The 3 statements are MUTUALLY EXCLUSIVE (one true implies others all false).
2. They are also EXHAUSTIVE (one of them must be true, or in other words they can't all be false).
3. There are only three of them.
4. Therefore assume in turn that each is true and the others false, and seek in each case a contradiction to eliminate the possibility of the adopted premise (reductio ad absurdum).

Most commentators seem to have followed this approach. But I also liked the post which viewed the sculpture from a different angle.

Mutual exclusivity and exhaustiveness don't just apply to the statements to be logically analysed. They also apply to the value of, say, Z (it must be either red, blue or white; and it can't be more than one of these) and they apply to the colours (red, say, must be one of the counters, X, Y or Z; and it can't be the colour of more than one of them).

Similar duality in the lines of attack apply in puzzles like Sudoku. A move can be either position-led (what value goes into this cell?) or value-led (where can 7 be placed in this column or row or box)? In my experience, once you get up to speed, most Sudoku solutions are about 80% value-led and 20% position-led, though it varies a bit from puzzle to puzzle. I suspect it also varies from solver to solver, as people develop their own preferred attack styles and special personal 'hooks'.

I often finish off Sudoku puzzles abandoned by others left on tube or pub newspapers, and try to work out their reasoning errors. A lot of less adept Sudoku solvers adopt poor notation practices (or none at all) and confuse statements about values with statements about positions. They are actually quite different, and a statement about one tells us sweet jack about the other.

Let:
S1 = X is red
S2 = Y is not red
S3 = Z is not blue

if S1 is TRUE, then S2 will also be TRUE - this cannot be, so S1 is FALSE
Now if S2 is TRUE, then Z is Red => S3 is TRUE - this also cannot be, S2 is FALSE
This leaves S3 is TRUE
=> Z is Red or White
Since S2 is FALSE Y is Red => Z is White => X is Blue

Therefore:
X = Blue
Y = Red
Z = White

X= Red
Y= Blue
Z= White

This one can be done by assuming each one to be true, as you can then test for a logical contradiction. If there is a logical contradiction when something is assumed to be true, it must be false. With this logic, I found that Z is true, so X is Blue, Y is Red and Z is White.

X is blue.
Y is red.
Z is white.
Since, the option 3 suggests that
Z is not blue than it could be red or white.

As option 2 is wrong as per the question's condition; therefore Y is red.
Resulting Z as White as the only option.
And hence, X is blue.

There is only one row where only one of S1, S2 and S3 is true. (S1 = X is red, S2 = Y is not red, S3 = Z is not blue).

X | Y | Z || S1 || S2 || S3
------------------------
R | B | W || T || T || T
R | W | B || T || T || F
B | W | R || F || T || T
B | R | W || F || F || T
W | R | B || F || F || F
W | B | R || F || T || T