pouët.net

As Doom suggested, it must be a number which after two steps leads to a number in which 0 appears.

anes: I'm afraid I really don't follow. IF doesn't enter into it: IF = ->, IFF = <-> (short for "if and only if, which is used in a logical BIconditional").

And "Is this the right door if and only if you are the guy who always tells the truth?" IS a yes/no question. The truth value of the biconditional is negated if that of the second part is negated, but if it's negated then you are talking to the guy who negates all his answers, so voila. Effectively you get the truthful yes/no answer to "is this the right door?". There's nothing wrong with the logic.

Or the question could be rephrased if you don't like the use of iff: "Is the statement 'this is the right door' as true as the statement 'you always tell the truth'?"

Maybe Texel wants a less convoluted solution?

A classic one:

You choose two arbitrary numbers out of {1,2,3,4,5,6,7,8,9} (may be the same ones). Now you tell Paul the product of these two numbers, and Susan the sum of them.

Now the following dialogue unfolds:
Paul: "I don't know the numbers"
Susan: "I don't know the numbers"

Paul: "I don't know the numbers"
Susan: "I don't know the numbers"

Paul: "I don't know the numbers"
Susan: "I don't know the numbers"

Paul: "I don't know the numbers"
Susan: "I don't know the numbers"

Paul: "Now i know them".

What are the numbers?

I already solved kb_'s puzzle some years ago...

thought so... that's why i wrote "classic". It just came to my mind when i read this thread.

Doom, supposedly the problem is to get rid of the arbitrary guy in the first question.

You ask to 1: Is 2 more lier than 3? (or "use to lie more often")

if you asked to the truth guy, then a "yes" answer will make 2 the lier and 3 the arbitrary. A "no" the opossite.
If you asked the lier, a "yes" answer will make 2 the truth guy and 3 the arbitrary.
If you asked the arbitrary, whatever the answer, 2 and 3 won't be the arbitrary.

So, in all the cases, in a "yes", you ask the second question to 2, and in a "no" you ask to 3. This way you will be sure your second question won't be to the arbitrary guy.

Then you proceed in a similar fashion to the original problem with only 2 guys. You can ask: "Would you say this is the door I should pass if you were your friend of absolutely opposite way of answering of you?" or something like that...

...no os dnA

.llew sa tuo delur si )3,2( os ,)3,2( srebmun eht wonk dluow eh neht esuaceb 6 eb tonnac tcudorp eht taht snaem osla tI .oot ,tuo delur si )2,2( taht snaem sihT .)2,2( srebmun eht wonk dluow eh neht esuaceb 4 eb tonnac tcudorp eht snaem siht os ,srebmun eht wonk t'nseod llits eh syas luaP

.tuo delur si )4,1( ,esac suoiverp eht ot suogolana os ,})3,2( ,)4,1({ snoitulos elbissop eht era ereht ,5 roF .tuo delur osla si )3,1( ,rewsna eht wonk t'nseod luaP taht swonk nasuS ecniS .})2,2( ,)3,1({ rof esac eht si hcihw ,4 si mus eht enigami tuB .tuo delur era })2,1( ,)1,1({ os ,rewsna eht wonk t'nseod nasuS

.tuo delur era seitilibissop eseht os ,rewsna eht wonk dluow luaP neht ,})3,2( ,)7,1( ,)5,1( ,)3,1( ,)2,1( ,)1,1({ gniwollof eht fo eno erew riap rebmun eht fI

:srennigeb rof pleH

Adok, that has been too freaky for me...

2*8. Great problem kb, thanks so much for the fun :D

Now the problem is... how the fuck read what adok wrote without getting crazy?

texel: You shouldn't try to crack the secret mensa code. It messes with yer brain.

doom: i'm fine with iff. i mean, "is a if b" is not a "yes-no" question, it's a "yes-no-hmm how the hell can i answer that?" question.

texel: you didn't tell the guys know how the other two are replying.

adok: using a computer doesn't mean trying all blind possibilities that can fit into 31bits.

anes: shut the hell up.

here's kb's problem solved (beware: spoiler;)

it's more fun to do this problem in excel than on math paper, because you can copy-paste instead of trying to sqeeze 10 letters into one cell ;)

texel: i recall that dos navigator could switch letter order from right-to-left to left-to-right. but it's also like 5 mins of code in c if it's that annoying to you to read it directly ;)

adok: i'm pretty sure that paul knows that 3*2=6*1

anes: Well stuff like "what would you say to X if you replied in an opposite manner of how this other guy is most likely to reply, etc." is also kind of a "yes-no-hmm" question. It's a very artificial situation to begin with anyway, so the rules should state if there's a limit to the complexity or unnaturalness of the questions you ask.

But anyway, I think a solution that allows for the three guys to be unaware of each other is simpler and nicer.

iq:

on your first problem (orthogonal vectors). It's a mathematical theorem (usually called the "hairy ball theorem") that there is no continuous function which assign to unit vector in 3D to another unit vector in 3D such that the two are orthogonal. Of course the same holds if you replace the word "unit" to "nonzero". This basically means that the answer to you question is that no, you can't do that without conditionals.

on the second question, i think again that you cannot avoid doing approximations: it seems to me that the answer cannot be expressed in terms of "usual" functions like sin, cos, sinh, exp, whatever. In any case, your computer use approximation to calculate these too, so you won't lose much by doing some kind of approximation.

ahh, these days even wikipedia knows about combing hedgehods: http://en.wikipedia.org/wiki/Hairy_ball_theorem

yes, anes pointed out already to that theorem.

222..222 (51 2's) has a mp of 3. i demand a challenge :)

Anes, it is not the biggest number... I want the biggest one

And just to simplify things a bit... it has more than 1 hundred decimal digits...

i found another one that is five more to it. (222...227)

i'm not going one by one of course. but the damn code crashes after more digits :/

It is not difficult to find the number without demostrating it is the biggest, but as I said the code need to work with big numbers and you have to think a bit to search for it in a fast way. Searching for it is a good start to understand the nature of this problem. Demostrating it is a bit harder, but the previous work in the code is good for giving ideas.

is there a chance we're talking about more than some tens of thousands of digits (that's where i've come if don't do smt bad) :)

bah this goes on like forever, i won't come back with a demonstration again.

you are doing something bad. Now you have enough information: more than 100 digits and less than 10.000 digits. I won't say anything more about it.