View Full Version : 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

test your logic....

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

it really is simple when you think about it

Salem

02-01-2002, 01:54 PM

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.

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

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!

Ken Fitlike

02-01-2002, 02:55 PM

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?

shtarker

02-01-2002, 06:12 PM

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.

ygfperson

02-01-2002, 09:39 PM

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.

Betazep

02-01-2002, 09:48 PM

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.

Betazep

02-01-2002, 09:53 PM

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. :D

golfinguy4

02-01-2002, 10:52 PM

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.

Betazep

02-02-2002, 03:54 AM

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".

ygfperson

02-02-2002, 08:39 AM

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

QuestionC

02-02-2002, 04:17 PM

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.

Imperito

02-02-2002, 04:59 PM

Well, there are two possibilities. Either each card has a number one one side and a letter on the other, or each side has either a number or a letter.

The simplest way to remove the cards and focus on truths.

if A then B

Then divide the cards into those for which A is true, and B is possible but unknown, those for which B is true but A is possible ut unknown, and others for which either A is known to be false or B is known to be true. This is of course, dependant on the ambiguity I noted above and resolve below.

If each card had exactly one letter and one number:

Two cards, the A and the 4

A: Needs to be checked, because iff the number on the other side is odd, the rule is true

K: Does not need to be checked, because if the number on the back is even or odd, the rule is true.

3: Does not need to be checked, because the rule is not iff, so if the letter on the back is vowel or nonvowel, the rule is true.

M: Does not need to be checkd for the same reason as K

4: Needs to be checked because iff the letter on the back is a nonvowel the rule is true

If each side can have a leter or a number, and each card can thus have 0, 1, or two letters:

Four flips are needed, A, K, M, 4

A: Needs to be checked, because iff there is an odd number on the other side, the rule is true

K: Needs to be checked, because iff there is not a vowel on the other side, the rule is true

3: Does not need to be checked because if there is an odd number, an even number, a vowel, or a nonvowel on the other side, the rule is true

M: Needs to be checked for the same reason as K

4: Need to be checked for the same reason as K

Because data analysis traditionally resolves ambiguities of this type towards the chaotic (assuming lack of order unless order is defined), Four would most likely be the correct choice.

Betazep

02-02-2002, 06:14 PM

Exactly...

that is what I have been saying all along. And being that it doesn't say that there is a number on the opposite side (explicitly), I have to assume that the latter holds true... and it is indeed four cards that needs to be flipped.

Well stated. I wonder if anyone will come back and say...

I think you only need to flip the Ace. ;)

golfinguy4

02-02-2002, 10:03 PM

I stand corrected.

Betazep

02-02-2002, 10:41 PM

Sorry... I was being a d*&(

You still may be right. ;)

In order to prove

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

You must show that

1: For all vowels the number on the other side is odd.

Note that the contrapositive makes sense but would be

worded if the the other side is even, then the letter is not a vowel.

Now the quickest way would just to flip A as A is vowel.

There's a big difference if your rule was stated

The letter on side of a card is a vowel if and only if the number on the other side is odd. Sometimes in definitions the if and only if is implied but here it is not. Alot of times somebody will say A is a subset of S if for every x in A, x is in S. The if implies iff here.

I guess I over looked something... Most of what I said is correct but

You would want to flip over the A as it is vowel and

the 4 as it is even.

Then the rule is valid if A flipped is odd and 4 is flipped is not

a vowel.

the correct answer is two

the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant to the proof

Betazep

02-03-2002, 03:26 PM

>>>the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant to the proof<<<

Oh really... so what if there is a vowel behind the K? Would the rule stand?

It doesn't say that it has to be a number on the other side of a letter (and it wouldn't matter if it did. Because if it did explicitly say that and such wasn't the case... the rule would fail if there was a letter on the other side of a vowel.) So it could fail if condition one: an even is on the other side of a vowel... or condition two a letter is on the other side of a vowel.

I originally said two. Then changed it to four because either condition would make the rule fail and could be a possibility.

The only one that can absolutely 'not' fail is the 3. If you tell me that the cards absolutely 'must' have a letter on one side and a number on the other. Then I would tell you the answer is two because the 3, the M, and the K, absolutely could not cause the rule to fail, but that isn't the case. But you would have to tell me that outside of the rule.

What makes the rule fail as the cards lay? An even on the other side of the A. A vowel on the other side of the 4. A vowel on the other side of the K. A vowel on the other side of the M. Tell me that those do not make the rule fail, and I will look at you weird.

shtarker

02-03-2002, 03:36 PM

>>the correct answer is two

Damn i was certian it was either 1 or 4 and reading other posts i was more in favor of 4.

Maby i need to watch star trek to boost up my logic. . . .

Imperito

02-03-2002, 07:48 PM

Iain, I tend to agree with blue, as stated in my post and his, you cannot assume that the flip sides of K and M must be numbers. The situation did not define that.

Betazep

02-03-2002, 07:50 PM

yeah... I called my dad in AZ. He said the same thing. He added that the premise has to be in effect prior to the rule that each card has to have a number and a letter. So it can be either two or four cards... depending on the premise to the rule. (No premise=4 or premise of letter/number=2)

novacain

02-04-2002, 03:58 AM

>>the vowel and the odd number, as corectly pointed out it is a one way rule and what is on the reverse of the other cards is irrelevant

to the proof

I disagree. Only vowels have a rule not odd numbers, to state an odd has to have a vowel is not part of the rule.

The cards have to have two sides by definition.

The question says that only cards in the five (as you only have 5), showing a vowel on one side need to be checked for conformity ie odd on other side.

Cards with any other symbols may not fit the rule irregardless of the other side as the cards are not defined further.

The bit I disagree with you on.

As it does not state that the rule is reverseable, it is theoreticly possible to have a odd number with a number on the other side.

Therefore only one card need be turned.

The 'A', as the '3' could have a number on the other side and still fit the rule.

shtarker

02-04-2002, 08:04 AM

I don't think the question anymore is weather or not the rule is reversable (unless someone wants to argue it some more) but its more weather you must prove the rule by showing that there is no possibility of it being wrong (flipping 4) or weather you can just settle for proving the cards that will be affected by the rule (flipping one).

Something tells me (by that I'm mostly referring to induction) that its the showing that there is no possibility of it being wrong.

Betazep

02-04-2002, 10:17 AM

>>As it does not state that the rule is reverseable

What do you mean by reversable. It doesn't say that the number has to be on the other side of what you are looking at.

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

It doesn't have to be the side that is facing you. Imagine this... I tell you that and ask you to verify that the rule is true with physical cards right in front of you.

You turn over the A and say, "Yes it is true."

I say, "Oh really." Then I turn over the four and there is a vowel on the other side. Is that one side of the card? Yes it is. The rule fails.

So the idea of one is definitely incorrect. It doesn't pass the rule. Two is supposed to be the answer, because the implications (as set by the rule itself) is that there is a number on the opposite side of a vowel.

Since it is set by the rule and not by a premise to the rule, I argue that it is actually four cards that need to be flipped....

shtarker

02-04-2002, 04:14 PM

>>What do you mean by reversable. It doesn't say that the number has to be on the other side of what you are looking at.

I think what is meant by reversable is if you can say that as the 3 is odd there should be a vowl on the other side.

Or thats what i think.

novacain

02-04-2002, 08:48 PM

>>If one side of a card is a vowel, the number on the other side is a number.

In most cases numbers are numbers.

By reversable I ment the rule. eg that an odd number has a vowel on the other side.

You can not assume that each card has a letter on one side and a number on the other.

Just as you can not assume that the cards do not have a face and a back (ie different sides like playing cards). If so then you would not need to turn the three as it has the 'wrong' side showing.

Yes, it all comes down to 'do I have to remove all chance of the rule being wrong'.

In which case you must turn two.

[to prove the 3 still fits the rule (ie has a number or vowel on the other side)]

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.

novacain

02-04-2002, 10:02 PM

Huh??

>>Thus the only two cases where it is required to flip the card c is when c is odd and when c is even.

Huh??? What about letters? The card is odd and 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.

Betazep

02-05-2002, 01:25 PM

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?

The only answer to that question is "no it doesn't."

But... you may argue that the rule states that there is a number on the other side. Exactly! If there isn't a number on the other side of a consonant or a vowel, and that letter is a vowel... the rule fails.

Same rule and here are my cards...

3 K M 4 A

How many do you need to flip to prove the rule? Two you say...

Here is the back of those cards.

A E I O U

Hmmmm... seems like four cards made it fail. Too bad you only flipped two... ;)

Unregistered

02-05-2002, 05:36 PM

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.

Ken Fitlike

02-05-2002, 05:40 PM

...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".

shtarker

02-06-2002, 04:19 AM

>>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?

the answer is posted.

the answer has been confirmed with the author of the test, however if you would like to argue the point withe someone who has a phd in maths and in computer science

Unregistered

02-06-2002, 01:38 PM

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

Betazep

02-06-2002, 02:00 PM

>>>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".<<<

EXACTLY!!! And if it violates the rule, the rule is not proved!!!!

On top of that... you are adding a premise. You cannot add a premise to the rule. You are assuming that there cannot be a letter on the other side of a letter because that would be a logical premise that you are gathering from the rule (therin lies the problem... you cannot trust the rule). No such premise was made.

In the rule itself, it states that there is a number (an odd one at that) on the other side of a vowel. So if there isn't a number on the other side of a vowel, then the rule fails. If there isn't an odd number on the other side of a vowel, then the rule fails.

I understand where you are coming from... I originally stated two because I, like you, assumed that the premise of the entire puzzle is that a number is on the other side of a letter (be it consonant or a vowel).

But we cannot justify that without such a premise being available outside of the rule... and it is not.

That rule as written can fail, as I showed above, if there is either an even number on one side of a vowel, or if there is a letter on the other side of a vowel. (hell it could fail if there is a symbol on the other side of a vowel...)

The A fails if anything other than an odd number is on the other side (even, letter, symbol, or nothing at all).

The 3 passes for any number, letter, or symbol, or nothing at all.

The 4 fails if there is a vowel.

The M fails if there is a vowel.

The K fails if there is a vowel.

There is really nothing you can argue about that, though many seem to try... all the arguements come down to, "but he meant... but it implies a premise of... blah blah blah".

The rule as stated can fail as I showed above. If it can fail as such... you must flip the cards that have the possibility of making the rule fail.... four cards.

>>>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 is very very likely... and I bet that the original question did only leave room for two.

Betazep

02-06-2002, 02:14 PM

>>the answer has been confirmed with the author of the test, however if you would like to argue the point withe someone who has a phd in maths and in computer science<<

Yeah... have him read these posts, and email me. betazep@hawaii.rr.com

>>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<<<

I am thinking so...

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

Betazep

02-06-2002, 02:49 PM

LOL... yeah, I am about done trying to explain.

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.

Ken Fitlike

02-06-2002, 04:09 PM

<<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 ;) )

Betazep

02-06-2002, 04:09 PM

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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