Thread: logic puzzle for you

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.

Known about the rule

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.

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