if only i had more time to figure out how to effectively overcome that possibility of you lying...
the "intelligent" and computing-intensive "way" is just too meh because one would have to think of 11 matrices containing up to 1024 numbers each...
the "dumb" way requires the question to be asked subsequently after you answering each question... trial and error basically, and the easiest way is to divide the value in half with each iteration...
gah binaries and regula falsi...
1. is it in this list (all the numbers between 1-2000 which has 1 on its right tail in binary)?
2. is it in this list (all the numbers between 1-2000 which has 1 on its 2nd value from the right in binary)?
3. is it in this list (all the numbers between 1-2000 which has 1 on its 3rd value from the right binary)?
4. is it in this list (all the numbers between 1-2000 which has 1 its 4th value from the right binary)?
and so on until Question 11:
11. is it in this list (all the numbers between 1-2000 which has 1 its 11th value from the right binary)?
then questions 12-15 would have to spot the lie... again in binaries...
12. did you lie on the 1st, 3rd, 5th, 7th, 9th, 11th question?
13. did you lie on the 2nd, 3rd, 6th, 7th, 10th, 11th question?
14. did you lie on the 4th, 5th, 6th, 7th question?
15. did you lie on the 8th, 9th, 10th, 11th question?
the last 4 questions will pinpoint on which of the first 11 questions you lied (if any) and because it's in binary code (yes-no) the corroborated values would pinpoint the secret number. However, this is still not foolproof as you can lie on one of the last four questions and really throw the calculations off...
-w-
