Richard Wiseman's blog for this week has an interesting problem:

(paraphrasing): I think of one of the numbers: 1, 2, or 3. You ask me one question. I truthfully answer yes, no, or don't know. That's enough for you to know my number. The question is: what's your question?

My answer took me a few minutes to come up with. No idea if it is the same as the one Richard will post, but I am sure it works.

Now I am wondering about the next level up: Suppose I think of a number in the range 1 to 9. You get two questions; I answer each truthfully as before with yes, no, or don't know. Do you then know my number for sure?

I do not have an answer to that, though I have some thoughts about possible approaches. Can you do better than that?

Re: Not as easy as 1 2 3

Posted: Fri May 21, 2010 8:54 pm UTC

by campboy

Spoiler:

If you're allowed to add together that many odd primes, can you get every sufficiently large integer of the appropriate parity?

Re: Not as easy as 1 2 3

Posted: Fri May 21, 2010 10:05 pm UTC

by Vieto

Spoiler:

given two numbers, #2 and a random number that is in the range of 'three or higher', is one of those numbers your number?

Re: Not as easy as 1 2 3

Posted: Fri May 21, 2010 10:36 pm UTC

by lordatog

Spoiler:

"I'm thinking of a number. It is either 1.5 or 2.5. Is your number higher than my number?"

Re: Not as easy as 1 2 3

Posted: Fri May 21, 2010 10:48 pm UTC

by balr

I'll post my answer to the original, 1 2 3 question:

Spoiler:

I am thinking of a positive integer, N. If we raise the original secret number [1 2 or 3] to the power of N, does the result end in a 1?

Yes -- the original secret number must have been 1 [1**N always ends 1]

No -- it must have been 2 [2**N never ends in a 1]

Don't know -- it must have been 3 [3**N may or may not end in a 1]

Re: Not as easy as 1 2 3

Posted: Fri May 21, 2010 11:56 pm UTC

by Token

campboy wrote:

Spoiler:

If you're allowed to add together that many odd primes, can you get every sufficiently large integer of the appropriate parity?

Spoiler:

Aren't both 2 and 3 "don't know"s?

Re: Not as easy as 1 2 3

Posted: Sat May 22, 2010 12:14 am UTC

by Tiax

Now I am wondering about the next level up: Suppose I think of a number in the range 1 to 9. You get two questions; I answer each truthfully as before with yes, no, or don't know. Do you then know my number for sure?

Spoiler:

Surely if I can answer the first question, and guess 1, 2 or 3 in one question, then I can guess 1...3^n in n questions. I simply use my questions to guess the ternary expansion of the number, which is n-digits of 1,2,3. (Yes, 3^n has n+1 ternary digits, but since 0 isn't an option, you can work it out)

Re: Not as easy as 1 2 3

Posted: Sat May 22, 2010 6:10 am UTC

by Godskalken

Spoiler:

If I draw N (your number) points on a piece of paper, can you draw a straight line between them?

This and many of the answers here rely on you thinking or doing something randomly or unknown. I'm guessing the answer he wants is something along "is conjecture x true for your number", where x is some conjecture that has been proven true for one of the numbers, false for one, and is unsolved for the last one. Which is of course what campboy tried to do, though that answer doesn't work. Edit: reading how the question was originally stated, I'm guessing I was wrong about what kind of answer he wants.

Re: Not as easy as 1 2 3

Posted: Sat May 22, 2010 10:17 am UTC

by TheAlmightyEgg

Spoiler:

I'd say that I too am thinking of a number. My number is between 1.5 and 2.5. Is the number that you're thinking of larger than the number I'm thinking of?

Re: Not as easy as 1 2 3

Posted: Sat May 22, 2010 12:06 pm UTC

by Axidos

Godskalken wrote:

Spoiler:

If I draw N (your number) points on a piece of paper, can you draw a straight line between them?

This and many of the answers here rely on you thinking or doing something randomly or unknown. I'm guessing the answer he wants is something along "is conjecture x true for your number", where x is some conjecture that has been proven true for one of the numbers, false for one, and is unsolved for the last one. Which is of course what campboy tried to do, though that answer doesn't work. Edit: reading how the question was originally stated, I'm guessing I was wrong about what kind of answer he wants.

The answer to that question would be very ambiguous. You'd definitely get a "yes" for two, but you could get anything for one and three.

Re: Not as easy as 1 2 3

Posted: Sat May 22, 2010 3:51 pm UTC

by campboy

Token wrote:

campboy wrote:

Spoiler:

If you're allowed to add together that many odd primes, can you get every sufficiently large integer of the appropriate parity?

I am still pondering the 1,2,3,4,5,6,7,8,9 problem. Can we craft two questions answerable yes, no or don't know to uniquely identify one of those?

The second question is now easy....I think both mat and vieto's question allow us to distinguish any three distinct positive integers in one question.

But what is the first question? tiax hints he has one, but no one has posted one yet --- and yes, I do not have one either.

Re: Not as easy as 1 2 3

Posted: Mon May 24, 2010 3:39 pm UTC

by oscarlevin

Spoiler:

First ask: Assuming that I am thinking of either 3 or 6, is your number greater than mine? If 'yes' - they are thinking of 7, 8, or 9. If 'no' - then they are thinking of 1, 2, or 3. If 'I don't know' - they are thinking of 4, 5, or 6. Then ask: Assuming that I am thinking of either 7 or 8 (or 1 or 2, or 4 or 4, depending on the results of question 1), is your number greater than mine? Using the same reasoning as above, you have your answer.

Alternatively, you can first ask: Assuming I'm thinking of either 1 or 2, is my number greater than the remainder of your number when divided by 3? Then ask the first question above to determine if the secrete number is in {1,2,3}, {4,5,6}, or {7,8,9}.

Re: Not as easy as 1 2 3

Posted: Mon May 24, 2010 4:29 pm UTC

by redrogue

oscarlevin wrote:

Spoiler:

First ask: Assuming that I am thinking of either 3 or 6, is your number greater than mine? If 'yes' - they are thinking of 7, 8, or 9. If 'no' - then they are thinking of 1, 2, or 3. If 'I don't know' - they are thinking of 4, 5, or 6. Then ask: Assuming that I am thinking of either 7 or 8 (or 1 or 2, or 4 or 4, depending on the results of question 1), is your number greater than mine? Using the same reasoning as above, you have your answer.

Alternatively, you can first ask: Assuming I'm thinking of either 1 or 2, is my number greater than the remainder of your number when divided by 3? Then ask the first question above to determine if the secrete number is in {1,2,3}, {4,5,6}, or {7,8,9}.

Spoiler:

This is surprisingly similar to my method for 1-9 in two questions:

First Question: I'm thinking of either 3.5 or 6.5. is your number greater than mine?

Second Question: If their answer to the first is yes, it falls within 7, 8, 9, so ask again with 7.5 and 8.5 If their answer to the first is dunno, it falls within 4, 5, 6, so ask again with 4.5 and 5.5 If their answer to the first is no, it falls within 1, 2, 3, so ask again with 1.5 and 2.5

Re: Not as easy as 1 2 3

Posted: Mon May 24, 2010 8:56 pm UTC

by thc

Spoiler:

It is easy to generalize by hashing each answer, which can easily be encoded in the question that you ask:

Then you only need to ask about the hash, and you don't even need to know anything about the real answers.

Re: Not as easy as 1 2 3

Posted: Tue May 25, 2010 12:55 am UTC

by ConMan

Spoiler:

In an encoding where 1 is represented by "yes", 2 by "no", and 3 by "don't know", what is the encoding of your number?

Almost certainly not what he was looking for, but it works.

Re: Not as easy as 1 2 3

Posted: Tue May 25, 2010 3:41 pm UTC

by levantis

Spoiler:

It`s primitive. Suppose it`s number x. you take (x-1)/3 (using integer division) and (x%3)+1. Then apply any of the previous questions to each, reducing this problem to an already solved one

Re: Not as easy as 1 2 3

Posted: Tue May 25, 2010 9:54 pm UTC

by BurningLed

Now what about questions in which only integers may be used?

Re: Not as easy as 1 2 3

Posted: Wed May 26, 2010 5:02 am UTC

by jestingrabbit

BurningLed wrote:Now what about questions in which only integers may be used?

I think oscarlevin had a perfectly good question which only used integers.

Re: Not as easy as 1 2 3

Posted: Wed May 26, 2010 5:17 am UTC

by phlip

Spoiler:

Right, the only reason for saying "greater than 3.5" instead of "greater than 3" is to forcibly remove all chance of ambiguity about strictly-greater-than and greater-than-or-equal.

Still, a generic way of doing this, basically the same as thc's answer... to distinguish at most 3^{n} options by n questions, enumerate the answers by 0 to 3^{n}, written in ternary (that is, every n-digit ternary number). Then, your questions are:

1) If the correct option is written as an n-digit ternary number according to this enumeration, is the first digit in the set S, where S contains 2 and may or may not contain 1, but does not contain 0? 2) If the correct option is written as an n-digit ternary number according to this enumeration, is the second digit in the set S, where S contains 2 and may or may not contain 1, but does not contain 0? 3) If the correct option is written as an n-digit ternary number according to this enumeration, is the third digit in the set S, where S contains 2 and may or may not contain 1, but does not contain 0? etc

Re: Not as easy as 1 2 3

Posted: Thu May 27, 2010 3:49 am UTC

by julian_del_cuatro

Here are a few decidedly unmathematical solutions.

Spoiler:

Call your number N. You and I are playing a chess game. We have set it up and taken our respective sides. Is it possible for you to win in N moves?

Spoiler:

I am a male. There are three kids in my family, one of which is a girl. Does your number exceed the number of boys born in my family?

Spoiler:

Call your number N. A basketball team is down by N points and makes a 2 point basket as time expires. Do they win the game?

Re: Not as easy as 1 2 3

Posted: Thu May 27, 2010 10:14 pm UTC

by Token

julian_del_cuatro wrote:

Spoiler:

Call your number N. You and I are playing a chess game. We have set it up and taken our respective sides. Is it possible for you to win in N moves?

Nice one.

Re: Not as easy as 1 2 3

Posted: Fri May 28, 2010 4:36 pm UTC

by redrogue

Call your number N. Do I have at least N lungs?

Re: Not as easy as 1 2 3

Posted: Fri May 28, 2010 4:55 pm UTC

by WarDaft

Another way of selecting amongst 1/2/3

Spoiler:

Where N is your number, is (N-1)^(N-1) odd?

Re: Not as easy as 1 2 3

Posted: Fri May 28, 2010 7:00 pm UTC

by rigwarl

Wardaft, wouldn't that give you yes/yes/no?

Re: Not as easy as 1 2 3

Posted: Sat May 29, 2010 3:49 am UTC

by phlip

Yes, but this would work:

Spoiler:

if f and g are functions such that the lim_{x -> 0} f(x) = lim_{x -> 0} g(x) = N - 1, where N is your number, is lim_{x -> 0} f(x)^{g(x)} odd?

Re: Not as easy as 1 2 3

Posted: Sat May 29, 2010 7:11 am UTC

by BlackSails

For {1,2,3}, choose a number N.

Is the busy beaver function of 2N even or odd?

(BB(2)=6, BB(4)=107, BB(6) is unknown)

Re: Not as easy as 1 2 3

Posted: Sat May 29, 2010 1:14 pm UTC

by tastelikecoke

Spoiler:

Is it 3 or is it the 100000000th digit of pi but not 2?

Re: Not as easy as 1 2 3

Posted: Sat May 29, 2010 9:54 pm UTC

by whaatt

oscarlevin wrote:

Spoiler:

First ask: Assuming that I am thinking of either 3 or 6, is your number greater than mine? If 'yes' - they are thinking of 7, 8, or 9. If 'no' - then they are thinking of 1, 2, or 3. If 'I don't know' - they are thinking of 4, 5, or 6. Then ask: Assuming that I am thinking of either 7 or 8 (or 1 or 2, or 4 or 4, depending on the results of question 1), is your number greater than mine? Using the same reasoning as above, you have your answer.

Alternatively, you can first ask: Assuming I'm thinking of either 1 or 2, is my number greater than the remainder of your number when divided by 3? Then ask the first question above to determine if the secrete number is in {1,2,3}, {4,5,6}, or {7,8,9}.

My first question is similar to your first question #2:

Spoiler:

Question 1: Let us assume that if your number is {1,2,3} it has value "1', if it is {4,5,6} it is a "2", and if it is {7,8,9} it is a "3". The number you are thinking of is in a group with a certain value. Courtesy Mat: If I were to pick one of the other two values at random, would it be greater than your value? 1-yes, 2-dont know, 3-no (with the particular number corresponding to one of the groups of three.)

Re: Not as easy as 1 2 3

Posted: Sat May 29, 2010 11:46 pm UTC

by phlip

I'm not a fan of the answers where the "I don't know" is due to it simply being some mathematical fact that hasn't been proven yet. Especially if one of the other two options is rather obscure. In the unlikely event that the person you're talking to has a proof of, say, the value of BB(6), then it won't work. On the other hand, if the person you're talking to doesn't know, say, the value of BB(4) then they're going to answer "I don't know" even if Mathematics in general knows.

An ideal question ensures that the answer will be right by making sure the "yes" and "no" options are very obvious (limiting the chance that the askee won't know) and the "I don't know" option is unknowable (because the question doesn't have enough information).

Re: Not as easy as 1 2 3

Posted: Sun May 30, 2010 4:35 pm UTC

by freddyfish

doesnt this work?

Spoiler:

I am thinking of a number that is either 3.5 or 6.5. Is your number greater than mine? Yes: his number is 7,8,9 No: his number is 1,2,3 maybe: his number is 4,5,6

Then ask the same question using 1.5 and 2.5, 4.5 and 5.5, or 7.5 and 8.5 depending on which range you are in to get the number.

Re: Not as easy as 1 2 3

Posted: Thu Jun 03, 2010 2:33 am UTC

by Random_person

Given that the solution to the original question is simple, can't you work from that?

Spoiler:

I am thinking of a number which is either two or three. Is your number, modulo three, less than mine? Your next question depends on their answer. If they say no, then your second question is: I am thinking of a number which is either six or nine. Is your number less than mine? If they say yes to the first question, you ask: I am thinking of a number which is either four or seven. Is your number less than mine? If they don't know the first answer, then: I am thinking of a number which is either five or eight. Is your number less than mine?

Re: Not as easy as 1 2 3

Posted: Thu Jun 03, 2010 4:41 am UTC

by BlackSails

phlip wrote:I'm not a fan of the answers where the "I don't know" is due to it simply being some mathematical fact that hasn't been proven yet. \

It is unlikely it will ever be proven though.

And you can demand a large enough number that storing it in a reasonable space would cause black hole formation.

Re: Not as easy as 1 2 3

Posted: Thu Jun 03, 2010 6:41 pm UTC

by Ddanndt

Spoiler:

Your number is N Q1 : Minus 1 from your number and take the quotient by 3. Now multiply the result with a random number that I have chosen between 1 and 3. Is the result greater than 1? No(1,2,3) Yes(7,8,9) I don't know (4,5,6)

Q2: I take the remainder of N / 3 and I multiply it with a random number between 1 and 3. Is the number greater than 1? Yes(2,5,8) No(3,6,9) I dont know(1,4,7)

Then we have Q1 Q2 1 N IDK 2 N Y 3 N N 4 IDK IDK 5 IDK Y 6 IDK N 7 Y IDK 8 Y Y 9 Y N

Re: Not as easy as 1 2 3

Posted: Fri Jun 04, 2010 1:32 pm UTC

by di gama

BlackSails wrote:

phlip wrote:I'm not a fan of the answers where the "I don't know" is due to it simply being some mathematical fact that hasn't been proven yet. \

It is unlikely it will ever be proven though.

And you can demand a large enough number that storing it in a reasonable space would cause black hole formation.

But you're just asking if it is even or odd. Although 3^5^7^11^13 may, in fact, cause black holes if stored, I can tell you now that it is odd, and a similar fact may arise in research of BB(6).

EDIT: If you look at the original blog, the setting is a prisoner asking a guard a question about 1, 2, or 3. I find it funny that anyone would trust a guard to know about Goldbach's Conjecture and Vinogradov's theorem.

Re: Not as easy as 1 2 3

Posted: Sat Jun 05, 2010 10:26 am UTC

by t1mm01994

I still like freddy fish's solution. Apologies for the lack of use of math, I don't quite know how to use that. I'll look it up soon. To generalize: I think of one of the numbers: 1, 2, 3, ... 3^n You ask me n questions. I truthfully answer yes, no, or don't know. That's enough for you to know my number. The question is: what questions are you going to ask?

Spoiler:

Your first question: I have a number in mind, that is either 3^n-1+0,1, or 2*3^n-1+0,1. Is your number greater than mine? This leaves an interval of 3^n-1 numbers. First, you tell him to make those numbers 1, 2, 3, .... 3^n-1 (substracting 3^n-1 once or twice). After that, repeat first question, just with n one lower. In one question, it lowers the power of 3 by one. You need to have 3^0, which is 3^n-n, so can be done in n questions.

So, I think that yields us a solution for every power of 3. Second question: Can we do something similar for every number n, instead of just 3^n?

Re: Not as easy as 1 2 3

Posted: Sat Jun 05, 2010 11:24 am UTC

by oagersnap

As shown by the many different possible solutions, this is very very easy. One of the simplest ways to ask it IMO is, for 1-3 and one question:

Spoiler:

On a piece of paper, I have written two numbers. One of the numbers I have written is 1; I have definitely not written the number 3. Have I written your number?

And for 1-9 and two questions:

Spoiler:

On a piece of paper, I have written four numbers. I have written the numbers 1, 2 and 3; I have definitely not written either of the numbers 7, 8 and 9. Have I written your number? The second question is like the question for numbers 1-3, but obviously using the numbers you get from the first question.

Re: Not as easy as 1 2 3

Posted: Mon Jul 11, 2011 9:18 pm UTC

by Ankit1010

I didn't look at all the answers, but this looks significantly simpler than most:

Spoiler:

We need a bijection between 1,2,3 and yes,no,maybe.

Bijection: "Does the number divide any given odd number?" 1 - Yes 2 - No 3 - Maybe

Re: Not as easy as 1 2 3

Posted: Mon Jul 11, 2011 9:59 pm UTC

by zemerick

tastelikecoke wrote:

Spoiler:

Is it 3 or is it the 100000000th digit of pi but not 2?

Spoiler:

Well, the 100 millionth digit of Pi is 9, so: 1) No 2) No 3) Yes

The 100 millionth digit of pi is indeed known, so there is no "I don't know". Sure, a given person is pretty unlikely to know it..but it's not a certainty.