Thread: logic puzzle for you

Look at a statement such as
P implies Q.
There are several different ways to say this
if P then Q.
Q if P.
Q is a necessary condition for P.
P is a sufficient condition for Q.

Now look at what conditions is P implies Q a true statement.

P is true and Q is true
P is false and Q is true
P is false and Q is false.

And it is a false statement when
P is true and Q is false. Thus using => to stand
for implies P=>Q is logically equivalent to ~P or Q for the truth tables of both statements is the same.

Now there are several ways to prove this kind
of statement. You can prove it directly:
assume P and then work with it until Q. Or indirectly by prove by contrapositive:
if ~Q=>~P or by prove by contradiction ~(P=>Q)=>c,
~(~P or Q)=>c, (P and ~Q)=>c . In words you want to show P and ~Q implies a contradiction. "~" stands for not, c for contradiction.

If we look at the rule
For all cards c, if c is an odd number then c is a vowel. There's no need to mention flipping it as a card cannot be an odd number and even. To prove a statement of this kind we must show that our implication: If c is an odd number then c is a vowel for all cards c. There is also a number of ways to prove a universal statement of this kind. The simpliest is just to go through each card and proving for each member c that If c is an odd number then c is a vowel. If we know that c is even then the implication is true and if we know that c is odd the contrapositive is true. Thus the only two cases where it is required to flip the card c is when c is odd and when c is even.

>> If c is an odd number then c is a vowel for all cards c.

This should be
if c is a vowel, then c is an odd number.

And it follows that we would have to flip a card if c is a vowel or c is even. This is because we either show
if c is a vowel, then c is an odd number or the indentical statement if c is an even number, then c is not a vowel.

OK. I've read around this and - hold on to your family jewels - it's from psychologists who have studied this.

There are two possible correct answers based on the original question:

A and 4 must be turned to prove the rule

OR

4 cards (can't remember which)

depending on the syntax of the original question.

For everyone like me who asked 'is the rule reversible?' apparently this is a common fallacy in solving logic problems of this nature. The reason that we do this is that we tend towards 'real-world' analogies in our problem solving methodologies and will therefore instinctively reject logical statements if they are bizarre! That's to say that logic can produce results that, although they are clearly bollocks, are still logically correct. Legal 'logic' may spring to mind in this context, but I assure that is something more related to what plops forth from the poop-tube of bull. In any event the 'is it reversible' question (from me anyway) was prompted by the analogy with: 4x3=3x4 (ie reversible rule) but 4/3!=3/4(ie rule NOT reversible) ; it's still wrong though. The other thing about 'real-world' problem solvers is that they invariably work in the real-world (now there's a surprise) and have a scientific background (read core sciences: physics, chemistry etc) where empirical observation is fairly mandatory. Pure (as opposed to Applied) Maths heads fare better presumably because they don't get out enough ( ) or that they can overcome the instinct to reject bizarre conclusions(?).

Betazep, I think has explored all possibilities except - all possibilities ie the quantum 'real-world' scenario where the opposite side of each card exists in a superposition of ALL posssibilities, however wierd, and therefore cannot be determined until each card is flipped and observation pins it down to one of the multiplicity of possibilities.

Nick's discussion/explanation is, apparently, 'classic' and generic in that it reduces the problem to it's boolean core and deals with it in that abstract way.

Other important things to note (from the shrinks) are:

1. Two-thirds of people cannot readily solve these kind of problems, irrespective of IQ.

2. Intelligence cannot be accurately measured from this kind of problem solving although they do figure highly in IQ tests. As an example, consider the creative imagination of Einstein who visualised how things might appear if he were riding on a ray of light and came up with relativity (ok, that's an oversimplification, but you get the picture). Or Kekule, with the ring structure for Benzene who had a 'vision' of a snake swalling its own tail.

3. The ability to solve problems of this kind do improve with age and experience. Clearly any programmer viewing Nick's algorithmic(?) approach should be better equipped to deal with future problems of this nature. I am not a real programmer and therefore exempt myself from having to re-read, study and understand his brain-cell fissioning discourse.

I'm going to hazard a guess that the correct answer is 2 cards - ie the A and the 4 because I believe that Ian did not state the question verbatim from the test paper.

That, of course, also means that Ian himself may well have entered the wrong answer in his test. I would be really interested to know when the results for that test come back whether two-thirds of the class got it wrong.

By the way, the text I ripped this info from is ~20 yrs old and everything I have reproduced here may have been superceded.

And yes, I did get it wrong. Both times.

...and I am 'unregistered' above with large, waffly monologue...but forget to...and can't edit....

This problem was worded badly.

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.

From this statement you know that
if one side is a vowel then the other side is a number and
if one side is not a number then the other side is not a vowel"

but I suppose you don't know
if one side is a number then the other side is a letter or
if one side is a letter then the other side is a number.

You logic doesn't take into account the distinct possiblity that there could a a vowel on the unseen side of a consanant. If an 'E' was on the other side of the 'K', does the rule pass?

No, this is not possible because if one side is a consanant and the
other side is a vowel then it violates "if one side is not a number then other side is not a vowel".

>>A and 4 must be turned to prove the rule

The only part i don't get is the 4.
Why does that need to be turned over?

If the answer is two then at some point on the test it must have been declared that each card has a number on one side and a letter on the other

you all care waaaayy to much about a silly logic question...