# Thread: logic puzzle for you

1. 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.
See the statement implies that
"if the *letter* on side of a card is a vowel, the *number*
on the other side is odd.

So the contrapositive of the statement is

If the *number* on side of a card is not odd, then the *letter* on side of a card is not a vowel.

So your right a "card" with one side a vowel and one side a letter
would not follow the rule. But it also would not meet the definition of a card because the rule assumes to that a card has a letter on one side and a letter on the other side.

2. <<who has a phd in maths and in computer science>>?

What, don't phd's make errors? Ever?

Still, give the guy/girl his/her due: to paraphrase Betazep, why don't you send him/her over to this forum where a diverse cross-section of the planetary population with a multiplicity of experiences and knowledge can discuss this and any other 'logic tests' further.

And note that 'diverse cross-section of the planetary population with a multiplicity of experiences and knowledge' is as meaningless as 'who has a phd in maths and in computer science'.

Errare humanum est.

(and yes, I was still wrong )

3. If it was led by "assumptions" then one card could be a viable solution as well.

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

You could assume or better yet presume that "the other side" is the side you do not see. Therefore, only one has a vowel with "the other side" having to be an odd.... only one needs to be flipped.

I stated this earlier "on____side of a"... put a 'this' in the blank... it can be one card. Put a 'the' or an 'a' in the blank, 2 cards with a premise (presumption, assumption) that letters are opposite of numbers, 4 cards with no provided presumption/premise.

+++++

The physical known:

a card has two sides.

The front sides have letters and numbers.

The side not facing you is the 'other side' of the side facing you.

The side facing you is the 'other side' of the side not facing you.

Any falsity in the rule, causes the rule to be invalid.

No presumtions can be accessed from the rule as the rule is the testing point. All rules that govern the rule must be supplied outside of the rule, else they are subject to testing... assumed, implied or otherwise.

Without any premise or assumption, the unknown:

What is on the other side of the cards from the side that faces you.

++++++

From that information... four is a very viable answer because we were not told to assume, presume, or otherwise anything about the context of the cards, except for in the rule, which is being tested for validity.

If the validity of a number on the other side of a vowel does not pass... the rule fails.

If the validity of an odd number on the other side of a vowel does not pass... the rule fails.

That is the best I can do to explain that... I give up.

edit>>>

if (one=1)
return 0;
else
{
while (1)
one = 1;
}

You could assume that one was set to a 1 upon instantiation because it says one... but that would be a bad idea if you were wrong.

The premise outside of that rule has to be: one must be set to a value of an integer 1 to avoid looping.

<<<