If I think of a number between 1 and 1,000, guessing it in 10 yesno questions is easy... so that's not the puzzle!
Guessing it in 10 yesno questions, that must be all asked in advance, is also relatively easy... so that's not the puzzle either!
How many questions would you need to guess my number, if you had to ask all questions beforehand, and I also could lie once?
(In reply to
No Subject by Leming)
I think it could be possible to elude question 15, if you accept that you can give instructions before the questions, as I say in may precedent mail.
Anyway if this is not accepted, you can try to express instructions in the form or complex questions to know the column of the digit in which he has lie, or if he is instead lying in the questions 1114. You can ask like this:
The amount (15log (base 2) of the number of the column in which you have lie), or in the case this is not a natural number, the number of the column in which you have lie, or number 0 in the case you haven't lie, has a 0 as first digit expressed in 4binary?
If he has lied in column 1, (or 2, or 4 or 8), he has been asked about if the number 15 (or 14, or 13 or 12) has a 0 as first digit; if he has lied in one of the other columns, he has been asked about if the first digit of the number of the column where he lied is a 0; if he hasn't lied he has been asked about if the first binary 4digit of 0 is a 0.
Repeating this "the amount (15log... has as second digit expressed in binary a 0?" (same for third and four digit) we obtain:
If he lied on the firstten digits we obtain as answer the binary number of the column where he lies (or the binary of 12, 13, 14, 15 in the case of colums 1, 2, 4, and 8); instead if he is lying in this four last questions we will obtain as answer 1, 2, 4, 8 expressed in binary. Finally, if we obbtain a 0 (0,0,0,0) it means he choose not to lie.
Edited on April 15, 2005, 9:06 am
Edited on April 15, 2005, 9:09 am

Posted by armando
on 20050415 09:04:03 