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White/Black Triangle/Circle

I really liked this one! (and the accompanying odds/vowel problem.) Raymond Smullyan's books are full of those.

This is a great site for when you're waiting for a pointless meeting to start...

Daniel Earwicker
Monday, March 25, 2002

Spoilers (possible) ahead...

Couldn't find an answer to this solution, so figured I'd post what I think.

Perhaps they are all "frobby" - you go from WT -> WC, WT -> BT, BT (or WC) -> BC ?

Or have I misunderstood the question?

On part II - I say we turn 2 cards - 2 (to prove it has a vowel because of 5 nearby) and E (to prove that it doesn't have an odd number, because of an F nearby).

There is no need to turn around 5 or F then.

But I'm sure I'm wrong on these - corrections are appreciated.

Tuesday, March 26, 2002

If a white triangle is frobby, and therefore only one of of "white" and "triangle" is correct, then the opposite, black circle, would also be frobby, assuming that the questioner can only think of white/black/circle/triangle. The others wouldn't, because they're either both correct, or neither correct, thus not frobby.

I think you have to check out 5 and F, since with 2 you know that the number is not odd, and with E you know it's a vowel. In other words, only 5 and F could disprove the rule.


Norman Spears
Tuesday, March 26, 2002


If the white triangle is frobby, it must be because of either the colour or the shape (it can't be both.)

Our friend is either thinking of white circle or black triangle.

And then you think, well, I don't know which of those it is, and its obviously a symmetrical situation, so I'm never going to know...

Doesn't matter. If you only have two choices, then it's not too much effort to use "brute force" - think through the consequences of white circle, then do the same for black triangle. If you get a different answer each way, then you know you can't get anywhere.

If you get the same answer both ways, then clearly it doesn't matter which one our friend is thinking of!

The quick answer is to think about XOR...

The card question is about realising the rule only has to apply to those four cards, not the whole alphabet or all the numbers. You can eliminate two cards from your enquires because they clearly aren't covered by the rule. It's more of a trick question, if you ask me.

Daniel Earwicker
Tuesday, March 26, 2002

Oops, I said exactly the same thing as Norman. Memo to self: read other messages more carefully.

Daniel Earwicker
Tuesday, March 26, 2002

Agree with Daniel and Norman on Part I.  Part II, answer is turn over one card, the 5.  The rule is that a card with an odd number on one side must have a vowel on the other.  It does not say anything about what is on the other side of even numbers or letters(vowel or consonants).

Terry Roe
Tuesday, March 26, 2002


"there are four cards which have a letter on one side and a number on the other side.  i lay them out and the cards appear as 2 5 F E.  the rule is that a card with an odd number on one side must have a vowel on the other.  what is the minimum number of cards you should turn over to prove the rule is true (and which cards would they be)?"

But if the F had an odd number on the other side it would break the rule....

Norman Spears
Tuesday, March 26, 2002

That's right, Norman.  Both 5 and F have to be checked.

In classic logic p->q is equivalent to (!p or q).  That disjunction ("disjunction" is the name given to an or-expression) must hold for all cards.  For 2 and E, you know one of !p and q is satisfied; 2 satisfies "not odd", and E satisfies "vowel".  5 and F don't satisfy either "not odd" or "vowel", and so you have to check them.

Paul Brinkley
Tuesday, March 26, 2002

Exclusive disjunction: XOR
Inclusive disjunction: OR

Norman Spears
Tuesday, March 26, 2002

where do you get these questions from?

Thursday, April 18, 2002

Try the book list, now on the front page.  ;-)

Paul Brinkley
Friday, April 19, 2002

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