# logic puzzle for you

This is a discussion on logic puzzle for you within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Well, there are two possibilities. Either each card has a number one one side and a letter on the other, ...

1. Well, there are two possibilities. Either each card has a number one one side and a letter on the other, or each side has either a number or a letter.

The simplest way to remove the cards and focus on truths.

if A then B

Then divide the cards into those for which A is true, and B is possible but unknown, those for which B is true but A is possible ut unknown, and others for which either A is known to be false or B is known to be true. This is of course, dependant on the ambiguity I noted above and resolve below.

If each card had exactly one letter and one number:

Two cards, the A and the 4

A: Needs to be checked, because iff the number on the other side is odd, the rule is true

K: Does not need to be checked, because if the number on the back is even or odd, the rule is true.

3: Does not need to be checked, because the rule is not iff, so if the letter on the back is vowel or nonvowel, the rule is true.

M: Does not need to be checkd for the same reason as K

4: Needs to be checked because iff the letter on the back is a nonvowel the rule is true

If each side can have a leter or a number, and each card can thus have 0, 1, or two letters:

Four flips are needed, A, K, M, 4

A: Needs to be checked, because iff there is an odd number on the other side, the rule is true

K: Needs to be checked, because iff there is not a vowel on the other side, the rule is true

3: Does not need to be checked because if there is an odd number, an even number, a vowel, or a nonvowel on the other side, the rule is true

M: Needs to be checked for the same reason as K

4: Need to be checked for the same reason as K

Because data analysis traditionally resolves ambiguities of this type towards the chaotic (assuming lack of order unless order is defined), Four would most likely be the correct choice.

2. Exactly...

that is what I have been saying all along. And being that it doesn't say that there is a number on the opposite side (explicitly), I have to assume that the latter holds true... and it is indeed four cards that needs to be flipped.

Well stated. I wonder if anyone will come back and say...

I think you only need to flip the Ace.

3. I stand corrected.

4. Sorry... I was being a d*&(

You still may be right.

5. In order to prove

If the letter on side of a card is a vowel , the number on the other side is odd.

You must show that
1: For all vowels the number on the other side is odd.
Note that the contrapositive makes sense but would be
worded if the the other side is even, then the letter is not a vowel.

Now the quickest way would just to flip A as A is vowel.

There's a big difference if your rule was stated
The letter on side of a card is a vowel if and only if the number on the other side is odd. Sometimes in definitions the if and only if is implied but here it is not. Alot of times somebody will say A is a subset of S if for every x in A, x is in S. The if implies iff here.

6. I guess I over looked something... Most of what I said is correct but

You would want to flip over the A as it is vowel and
the 4 as it is even.

Then the rule is valid if A flipped is odd and 4 is flipped is not
a vowel.

7. the correct answer is two

the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant to the proof

8. >>>the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant to the proof<<<

Oh really... so what if there is a vowel behind the K? Would the rule stand?

It doesn't say that it has to be a number on the other side of a letter (and it wouldn't matter if it did. Because if it did explicitly say that and such wasn't the case... the rule would fail if there was a letter on the other side of a vowel.) So it could fail if condition one: an even is on the other side of a vowel... or condition two a letter is on the other side of a vowel.

I originally said two. Then changed it to four because either condition would make the rule fail and could be a possibility.

The only one that can absolutely 'not' fail is the 3. If you tell me that the cards absolutely 'must' have a letter on one side and a number on the other. Then I would tell you the answer is two because the 3, the M, and the K, absolutely could not cause the rule to fail, but that isn't the case. But you would have to tell me that outside of the rule.

What makes the rule fail as the cards lay? An even on the other side of the A. A vowel on the other side of the 4. A vowel on the other side of the K. A vowel on the other side of the M. Tell me that those do not make the rule fail, and I will look at you weird.

9. >>the correct answer is two

Damn i was certian it was either 1 or 4 and reading other posts i was more in favor of 4.
Maby i need to watch star trek to boost up my logic. . . .

10. Iain, I tend to agree with blue, as stated in my post and his, you cannot assume that the flip sides of K and M must be numbers. The situation did not define that.

11. yeah... I called my dad in AZ. He said the same thing. He added that the premise has to be in effect prior to the rule that each card has to have a number and a letter. So it can be either two or four cards... depending on the premise to the rule. (No premise=4 or premise of letter/number=2)

12. >>the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant
to the proof

I disagree. Only vowels have a rule not odd numbers, to state an odd has to have a vowel is not part of the rule.

The cards have to have two sides by definition.

The question says that only cards in the five (as you only have 5), showing a vowel on one side need to be checked for conformity ie odd on other side.
Cards with any other symbols may not fit the rule irregardless of the other side as the cards are not defined further.

The bit I disagree with you on.
As it does not state that the rule is reverseable, it is theoreticly possible to have a odd number with a number on the other side.

Therefore only one card need be turned.

The 'A', as the '3' could have a number on the other side and still fit the rule.

13. I don't think the question anymore is weather or not the rule is reversable (unless someone wants to argue it some more) but its more weather you must prove the rule by showing that there is no possibility of it being wrong (flipping 4) or weather you can just settle for proving the cards that will be affected by the rule (flipping one).
Something tells me (by that I'm mostly referring to induction) that its the showing that there is no possibility of it being wrong.

14. >>As it does not state that the rule is reverseable

What do you mean by reversable. It doesn't say that the number has to be on the other side of what you are looking at.

If one side of a card is a vowel, the number on the other side is a number.

It doesn't have to be the side that is facing you. Imagine this... I tell you that and ask you to verify that the rule is true with physical cards right in front of you.

You turn over the A and say, "Yes it is true."

I say, "Oh really." Then I turn over the four and there is a vowel on the other side. Is that one side of the card? Yes it is. The rule fails.

So the idea of one is definitely incorrect. It doesn't pass the rule. Two is supposed to be the answer, because the implications (as set by the rule itself) is that there is a number on the opposite side of a vowel.

Since it is set by the rule and not by a premise to the rule, I argue that it is actually four cards that need to be flipped....

15. >>What do you mean by reversable. It doesn't say that the number has to be on the other side of what you are looking at.

I think what is meant by reversable is if you can say that as the 3 is odd there should be a vowl on the other side.
Or thats what i think.

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