# 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; after finishing my joyful maths assignment (all 8 hours and nine pages of it) i though i would post one ...

1. ## logic puzzle for you

after finishing my joyful maths assignment (all 8 hours and nine pages of it) i though i would post one question for you> It is alogic and reasoning question (and by about 600% was the simplest question on the paper)Note: hand in date was two days ago - so i am not requesting assignment completion

You have 5 cards
You must prove the rul by turning over as few cards as possible.
the rule is : If the letter on side of a card is a vowel , the number on the other side is odd.

A K 3 M 4

2. hmm... ok, my logic sucks...

3. it really is simple when you think about it

4. I'd start by noting that there is only one vowel card, and only one odd number card.

So you only have to worry about when the vowel card is turned over (at turn n)
If the card turned over at turn n+2 is not the odd card, the rule is broken.

On the other hand, if you see the odd card at turn n, and you haven't seen the vowel card, the rule is broken at turn n.

5. yeah but you have to state the mininum to either fully prove or disprove! I shall post the answer tomorrow

6. I know the answer! tra la la la la! It will take just one person to ask for me to tell everyone! And i promise..... when you see the answer, you be as duley uninpressed with it, as you were with the question!

7. I know this:

You ask the first guy what the other guy would say and then you do the opposite.

No. It's that 4-move check-mate thing in chess that catches me out everytime because it actually takes three moves.

If the letter on side of a card is a vowel , the number on the other side is odd.
You only need turn over two cards to prove the rule as stated. was my initial answer. (A and 3).

Is this statement commutative/associative/distributive(? help me i'm drowning) ie true both ways? Not asked - just asked to prove 'if vowel then odd.'

So 1 card: turn over letter 'A' if the reverse is odd the rule is proved. That, as they say, is my final answer.

So, am I 'Spock' and therefore illogical or am I 'Data' and therefore a machine?

8. I say you only have to flip the ace.
The only specification of the rule is for it ta vowel, the odd part is only a consequence of this (what i'm trying to say is that it doesn't matter whats on the back of the 3, as the rule only says what must be on cards with vowels).
The rest don't matter at all.
I'm not sure if thats what Ken Fitlike was trying to say but if it was then I agree with him.

9. i think the minimum's five. if any of the cards yield unexpected results, the rule is broken, but you don't know that until all 5 cards are tested.

If the letter on side of a card is a vowel , the number on the other side is odd.
A K 3 M 4

i can't assume that the cards are playing cards, nor can i assume that if a card has a letter on one side, it must have a number on the other side. for all i know each side has one of 36 combinations(assuming single digit numbers, english language letters).

i think the rule needs to be broken to be disproven. the only way to prove the rule is to test all cases.

10. I would say two. You only have to deal with the 'A' which is a vowel... so the other side should be odd, and the 4 to see if the other side is a vowel.

The rule only states that if one side is a vowel the other side is odd. It doesn't matter if one side is not a vowel and the other side is odd. So the 3 is of no consequence whether or not is is a vowel and the M and the K are not vowels... so they do not matter either.

11. Actually... I take that back. The rule doesn't state that there has to be a number on the other side. So there may be a vowel under the M or the K and that would break the rule. The three still doesn't matter because it is the only one that cannot break the rule no matter what the other side is...

so I am now changing my answer to 4.

12. shtarker is correct. No where does it state that the converse of the statement has to be true. Therefore, with those cards, the only one that has to be flipped is the ace.

13. ummm no. First off, they aren't a deck of playing cards... notice the M. So flipping the 'ace' as you call it isn't logical.

There is a problem with the statement anyway.

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

The problem is right between the on and the side. A word is missing. That word can have some serious implications. If the word was "this", you would be right. You look at this side... see a vowel... and know that you only need to flip the "A" over.

But more likely the word missing is an "a" or a "the". If either is the case, it can't only be the "A". Taken to the extreme it can only not be the 3. If it said this...

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

You could rule out the M and the K because the rule is implicitly stating that there is a number on the other side. But we do not have that... we have...

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

So it can be argued that there may not be a number on the side of the M or the K, but very well may be a vowel. This would disprove the rule... and if you didn't check for vowels under those cards, you would be wrong.

So some clarification of the actual rule 'word for word' needs to be set. Otherwise four is the only answer. Give me a set of index cards and I will trick you every time if you just pick the "A".

14. let me change my answer to 1. only one card needs to be flipped to disprove the rule: the A
to iain: tell us the answer! i cant wait

15. Hmm... I say two, the A and the 4. The A to make sure that the other side is odd, and the 4 to make sure the other side isn't a vowel.

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