If you can solve this puzzle, then who was phone?

If you can solve this puzzle, then who was phone?

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It's impossible!

Suppose you have a way to determinate. Let's name the first answer (Da or Ja) you get by R, and the other by L. so there are 4 possibilities to the answers you get: RRR,RRL,RLR,RLL. You won't get any other information from the lolcats.

There are also 6 possibilities for the cats: TRUE##FALSE##RANDOM may be ABC,ACB,BAC,BCA,CAB,CBA.

So, you should identify 1 of 6 possibilities by datum that has only 4 options, which is, of course, impossible.

By the way, First!!!!!!!!!!!!!!11111

HAHAHA NO.

Don't you know, that true heroes index their arrays starting by zero ?

Anyway, I'm first :P

You wish.

The indexes start from zero, but the elements are still first,second,etc

(So arr[n] is the (n+1)th element)

*Indices

Thanks.

(I'm not a native speaker of English, and I'm always happy to learn)

Both 'indexes' and 'indices' are perfectly fine plurals of 'index,' though if you use the first, some will correct you, and if you use the second, some will think you are an elitist snob. So try to use whichever feels appropriate for your current audience, or whichever you prefer.

A former colleague of mine had the opinion that you should start counting from 0 and that "first" meant "foremost", "second" meant "following" and "3rd", "4th", etc. meant "indexed by the number 3", "indexed by the number 4", etc.

So it goes first, second, twoth, third, fourth, ...

Or 0st, 1nd, 2th, 3rd, 4th, ...

Or, you simply say 'element 0, element 1, element 2' which is really well-defined.

Not necessarily. In solving the puzzle, you don't have to determine what Da and Ja mean, merely who A, B, and C are. So there are 8 possible outcomes of the yes/no questions and 6 possible permutations of A, B, and C.

Pigeon hole principle. Nice.

It's possible if they answer Yes/No, but I don't think its possibly with them answering Da/Ja since that doubles the number of possibilities.

ah noes, looked it up and it is possible! http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

Yes, you are right.

My mistake was the statement "You won't get any other information from the lolcats." - I didn't think about questions like "what does 'JA' mean?"

The problem is a variation of the well-known Half-Liar Paradox. The solution is known.

Hint: my approach to the solution was to chart out the possibilities. Brute force but it worked.

You can figure out their identities in three questions. Unfortunately, that's not enough to figure out what Ja and Da actually mean :-(

Ask each cat, "Would Random answer Ja if Random liked pie?" As Random's answer will be true or false by random, it would be impossible for True to answer truly or for False to answer falsely. It is only possible for Random to answer this question so the cat that replies to this question is Random.

Then for you use your 50/50 life line to determine the identity of the last two cats :P

The question should be "Would Random answer Ja if I asked Random if he or she liked pie?"

Should read the comment before attempting an answer. There is a solution

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

of course you can just ask the first two cats & work out the 3rd

And Joshu answered 'mew'.

I think I saw an equivalent puzzle (or one very like it) linked from the xkcd forums recently. I didn't realise it also had a wikipedia entry. :)

"da" is yes in russian, and "ja" is yes in swedish. Coincidence?

nyet.

Nein

Well, what will happen if I ask "Is your answer to this question 'is your answer 'Da' to the question that I'm asking you now' the same with the one to this question 'is your answer 'Ja' to the question that I'm asking you now'?" Ask this question 3 times and the gods may disappear, nothing to worry about now!

Ok guys, the gardener has it... :D

It is easy to solve this. True = 1, but lolcats refer to it as nawt sekkond. Which, using just the low order bit, means you are limited to zero and not zero. Since no = 0, True Cat will either use a word once or twice, as will Random Cat, but False Cat will always use both.

If we have 'r' random lolcats, 'f' liars and 't' lolcats that allways answer the truth, and if 'r' is small enough, (log_3(t+f+r) or less, for exemple), it's possible to solve the problem in at most (r + log_2(number of possibilities)).

If we have a lot of lolcats that answer random (for exemple all but one), it's not possible to solve the problem with probality 1 with any finite number of questions.

A good question would be: on which conditions on r, t and f is the problem solvable, and what is the minimal number of question needed?

My suggestion:

1. Ask every cat: “Ye alwayz tellz liez, cat?”

2. The one that exlodes is the false cat (due to the effect of the Liar paradox).

3. Further steps are left as an exercise to the reader.

P.S. Da stands for “Yes” in Russian, “Ja” stands for “Yes” in German...

Actually, it will just say no in its language

ha you crazy

The new Zune browser is surprisingly good, but not as good as the iPod's. It works well, but isn't as fast as Safari, and has a clunkier interface. If you occasionally plan on using the web browser that's not an issue, but if you're planning to browse the web alot from your PMP then the iPod's larger screen and better browser may be important.

It's True, False and Random. There is no phone.

Updated epsilon>0 times per month?

This site is dead. It's a shame ...

I seldom write remarks, however i did some searching and wound up here The Hardest Lolgic Puzzle Evar!

- Spiked Math. And I actually do have a few questions for you if you

do not mind. Could it be just me or does it give the impression like some of these comments appear like left by brain dead folks?

:-P And, if you are posting at other social sites, I'd like to follow anything fresh you have to post. Would you list of all of your communal pages like your Facebook page, twitter feed, or linkedin profile?

I have a weird theory about what would happen. As I'm not a mathematician xor English speaker, I will use simple language.

First: I will put an order for the cats places. Just for counting

1º 2º 3º

Then I will ask 1º, 2º and 3º (in that order): The same question "If I asked BOTH your fellow lolcats the question "Is 1º the cat known as True?" what would the answer be?"

This leads to one of the following cases:

1º is True: True would say 0 (No, Ja xor Da) and False would give the same answer (as False would have said the negative True would have answered to the first question). And Random would say 0 xor 1

1º is False:

True would have answered 1 (Yes, Ja xor Da), False would have answered the same. Random would answer 0 xor 1

1º is Random:

As to "Is 1º the cat known as True?" True would have said 0, False 1, and Random 1 xor 0.

So "the other two" for True would be 1 & (1 xor 0)->1

For False Not(0&(1 xor 0))-> 1

But as this are simultaneous events, we conclude that Random would be silent to that question in order to avoid a paradox (as True would have spoken falsely xor False would have spoken truth). As he can only speak one of both, by Reduction to Absurd, we should conclude he would not answer.

Maybe we already know Random for giving a different answer. If not, we just ask him the paradoxical question "Is 1º the cat known as True?" And iff he doesn't answer, he's Random.

So I already know Random. With just 3 to 6 question, 1-2 each lolcat.

I then place the cats in a circle facing each other positioned as R-X1-X2-R-X1...when I move in the circle.

So I ask to them: "Is the cat in your right the cat known as True?"

I ignore Random. True would always say no. False would say no only if true is on his right. So (R-0-0 is R-F-T; and R-0-1 is R-T-F).

So I know wich lolcat is True, False and Random.

With the remaining questions I may have I ask them "Would you like tuna for dinner?" As they will always want tuna for dinner I will also know which Ja or Da is Yes or No.

Also, after so much logic tables it's the last time I look after Mr. Schrodinger's cats.

I challenge you to do it askimg all 3 questions to the same cat -it can be done, as I did it by accident having misread th question. Solution ro be rebealed if no-one gets it by 6 Oct my bday

Simple. Call the one on your left A, the middle B, the right C. Ask A if B lies more often than C. If he says yes, ask B the next question, otherwise ask it to C. Suppose you ask B. Ask which of the other two lies more often. The one B does not answer is Random. Then ask if he exists. If B says yes he is True, otherwise he is False.

Edit: forgot about DA and JA. Ask if the cat answers Da to the above questions.