While it's quiet

You have been sentenced to death along with, say, another eleven prisoners.The method of execution is as follows:

Each of you are allocated a number and ordered to stand in a circle - in sequence.

Starting at no.1 and moving clockwise, the executioner shoots every other man in turn until only one is left - he is reprieved.

If you could choose your number, what would it be?
What if there were thirty-seven prisoners?
Or ninety-three?

What if there were X no.

A tenner to the Charity designated by the person who posts the briefest correct answer to the last.

edit: didn't read properly lol

x-1 ?

I think lep got it.

no, he shoots every EVEN person and keeps going around the circle.

Ok, so you want to be x-1'th if there's an odd number of people, and x'th if there are an even number of people.

x-1 was my first answer so I was half right I guess. :smile:

http://www.lepra.org.uk/ Is my charity of choice, so send a fiver! :biggrin:

No. Sorry. The sequence would run (in the case of 12 people)

1 x 3 x 5 x 7 x 9 x 11 x 2 x 6 x 10 x 4 x 12 - leaving 8
Shoot one-miss one-shoot one-miss one...

flashman said:
What if there were thirty-seven prisoners?
Or ninety-three?

This is the classic Josephus problem, isn't it? There's a formula I used to use for this one... I think it is:

Survivors position = 1 + 2X - 2^floor(1+lg(X))

lg(X) is to the base 2.

For 37, it's 11. For 93, it's 123. (Edit: Should be 58. I'm stupid)

? posted by fraggard

Survivors position = 1 + 2X - 2^floor(1+lg(X))

lg(X) is to the base 2.

For 37, it's 11. For 93, it's 123.

The answer is always an even number - All odd numbers are taken out in the first circuit.

Ok, so 2n - 2^floor(1+log<SUB>2</SUB>(n)) for n>2 then :razz:

Leperous said:
Ok, so 2n - 2^floor(1+log<SUB>2</SUB>(n)) for n>2 then :razz:

What he said.

And the number for 93 is wrong... It should be 58. (59 by the forumla, 58 for your answer)

Awesome riddle.
1....1

2....2

3....2
4....4

5....2
6....4
7....6
8....8

9....2
10...4
11...6
12...8
13...10
14...12
15...14
16...16

17...2
18...4
[...]
31...30
32...32

33...2
[...]
64...64

65...2
[...]
128..128

I was looking around, and found another solution...

Apparently, if you convert X to binary, and then rotate all the bits left by 1 place, you get the final position to be in.

If X = X1[size=13]X2[size=13]X3[size=13]X4[size=13]X5...[size=13]Xn-1Xn [/size][/size][/size][/size][/size]In binary, then Rotating it to become [size=10][size=10][size=13]X2[size=13]X3[size=13]X4...[size=13]XnX1 [/size][/size][/size][/size][/size][/size]gives you the final position. (For this particular case, you have to subtract 1 from that value, I think.)