Ron Barnette's Zeno's Coffeehouse Challenge #49 Result

Dear Zeno's patrons: Here are the results of our previous challenge. I have listed the original problem, with three winning results listed below. Thanks to all you Zeno's patrons who took up the challenge! Over 200 entries were received, with over 175 winning solutions. Thanks!

Ron Barnette

One late night at the Coffeehouse, a customer, Helen, posed the following challenge to Maggie and Charles. Helen removed four cards from her pocketbook and placed them on the table. She revealed (truthfully) that each card has a letter on one side and a number on the other. The cards below are what Maggie and Charles see. Helen then asked them to consider this statement:

"Of the four cards shown, those with vowels on one side have even numbers on the other"

Her challenge is this: what is the quickest way (i.e. the least number of card-flips) to deductively establish the truth of the above statement? And which card or cards would one flip over?

Must ALL be flipped? Fewer? Which ones? Obviously, Charles and Maggie must flip at least one card to see what's on the other side.

Results from three sample solutions:

Congratulations to James, Simon, and Gilles!

James Hess, Lafayette, LA, USA

comments: If all the cards shown are supposed to have a vowel on
one side have an even number on the other side, the card
with E facing up and the card with 5 facing up need to be
flipped to decide if it is so.

The opposite side of the 'E' card must have an even number
on the other side, then the statement can be true,
otherwise: if there is an odd number on the other side of
the 'E' card, the statement must be false.

The card with 'F' facing up does not need to be flipped,
because the only vowel sounds are the letters 'A', 'E',
'I', 'O', 'U', and since it is established one side of each
card is a letter and the other a number, the other side
cannot be a vowel.

The card with '2' facing up does not need to be flipped,
because it is an even number, and the other side must
be a letter: whether a vowel or a non-vowel is found
on the other side, this card does not violate the statement.

The card with '5' facing up needs to be flipped.
If something other than a vowel appears on the other side,
then the statement can be either true or false, but
if a vowel is found on the other side, the statement must
be false.

Having considered all possible ways for the statement
to be violated: if there is an even number on the
opposite side of the 'E' card and there is also something
other than a vowel on the opposite side of the '5' card,
then the statement is true, otherwise it is false.

Simon Lutterbie, St. Mary's City, MD, USA

comments: There need only be 2 flips.  First, you must flip the E, to make sure it has an even number on the other side.  You must also flip the 5, to make sure it doesn't have a vowel on the other side.

This problem can be simplified as an p-->q (If p, then q) situation.  To verify this, one must confirm the antecedent, which is done by showing that if a vowel is presented, an even number is on the flip side.  Second, one must deny the consequent, which means showing that, if an odd number is presented (~q), then a consonant will be on the other side. 

Gilles Gour, Montreal, Quebec, Canada

comments: You only have to flip the "E" card and the "5" card. Helen's states only that every card with a vowell on one side has an even number on the other. She does'nt say anything about cards with consonants on one side (the "F" card)and she does'nt say that cards with an even number on one side have a vowell on the other. If there waa an uneven number on the other side of the "E" card or a vowell on the other side of the "5" card, her statement would be untrue. But there could be an even or uneven number on the other side of the "F" card or a consonnant on the other side of the "2" card and that would not falsify her statement.
I rest my case.