## KingGeorge Group Title [UNSOLVED, ANSWER GIVEN] KingGeorge's Challenge of the Month Suppose you have $$n$$ children sitting in a circle waiting for Santa to give each of them one present. When Santa finally arrives, he announces he'll pass out the presents in such a way that he'll go in a clockwise circle, and give a present to every other child who has not yet received a present starting with the second child. Find a closed formula to find which child got the $$n-1$$-th present. If, say, there were 6 children in a circle, the order in which they would get presents is 2, 4, 6, 3, 1, 5. So the 1st child got the $$6-1$$-th present. 2 years ago 2 years ago

1. KingGeorge Group Title

I should say that this is a rather difficult problem. I've also changed the wording so you can't google a solution. And even if you did manage to google the original problem, the one formula I found online for this, was incorrect.

2. KingGeorge Group Title

Also, if you want me to compute some test values for you, let me know what $$n$$ to use.

3. joemath314159 Group Title

im getting something like: let n be$n=3\cdot 2^k+r,0\le r\le 3\cdot 2^k$Then the answer is:$2r+1$Havent written out a proof yet though >.<

4. FoolForMath Group Title

This is a variation of Extended Josephus problem, where $$x=n-1$$ The recursive and the non-recursive formulas are discussed here: http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf

5. KingGeorge Group Title

@FoolForMath That was the formula I mentioned I was able to find. However, it gives the incorrect value.

6. FoolForMath Group Title

I think you are probably making mistake while implementing it. This looks correct to me.

7. KingGeorge Group Title

Their table is correct, but their formula is not. Try their formula for n=41, and x=40.

8. KingGeorge Group Title

If I've typed it into wolfram correctly, that formula gives you -10, while the solution should be 35. http://www.wolframalpha.com/input/?i=1%2B2%2841%29-%282%2841%29-2%2840%29%2B1%29+%282%5E%281%2Bceiling%28log_2%28ceiling%2841%2F%282%2841%29-2%2840%29%2B1%29%29%29%29%29+-sgn%28ceiling%28ceiling%2841%2F2%29%2F2%29%29%29

9. KingGeorge Group Title

@joemath314159 is close. I believe his formula gives you the number of the child that gets the last present.

10. FoolForMath Group Title

Well, as I said if you implement is right you will get 35.

11. FoolForMath Group Title

Do you understand C?

12. FoolForMath Group Title

The recursive solution given there is right. I can't find my implementation so I had to fix a broken program available in internet and got 35 for n = 41 and x=40. And please don't say a published paper is wrong just like that :)

13. KingGeorge Group Title

Maybe I'm missing something critical then. But where did I type it incorrectly into Wolfram? My personal opinion is that they just made a small typo in the formula because I was able to make a very small modification that gave me the correct answer.

14. FoolForMath Group Title

When we are dealing with decimals (log) it's really hard to hold the precision even for the wolf. This is the C (recursive) solution, I was talking about: http://ideone.com/iJkuK

15. KingGeorge Group Title

Even using their formula by hand, I get an incorrect value for n=6.

16. FoolForMath Group Title

Sorry, I haven't checked in the non-recursive version. But it's rather difficult to believe that the formula is incorrect :)

17. KingGeorge Group Title

Like I said, I believe it's a typo since it's easily changed so that I do always get the correct solution.

18. FoolForMath Group Title

The non-recursive version has little importance in computer science (programming) as it fails for sufficiently large $$n$$.

19. KingGeorge Group Title

For those just joining us, I know of two different closed form expressions that give me the correct solution. The first, is a modified form of the closed form expression in the link ffm provided above. http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf The second, is a formula more similar to what Joe had near the top. This is the formula I originally found on my own. Both formulas give the correct solution, although they look mildly different. I will accept either.

20. KingGeorge Group Title

Hint: Solve the problem for $$n=2^k$$. Now go back to the general case. Reduce that case to a $$2^k$$ case, and solve. In particular, what do you have to add in and then subtract?

21. Limitless Group Title

Could you please compute for values from $$1$$ to $$10$$?

22. Limitless Group Title

I believe $$1$$ is undefined, actually. Sorry.

23. KingGeorge Group Title

$n=5 \rightarrow 5$$n=6\rightarrow1$$n=7\rightarrow3$$n=8\rightarrow5$$n=9\rightarrow7$$n=10\rightarrow9$I'll let you do 2-4 since they're fairly simple

24. KingGeorge Group Title

$n=2^4=16\rightarrow9$$n=20\rightarrow17$

25. Ishaan94 Group Title

@KingGeorge Solution??

26. Ishaan94 Group Title

@Libniz

27. KingGeorge Group Title

First up, we have the modified form of the equation linked to above. This is $\Large 2n+1-(2n-2x+1)\left(2^{\left\lfloor \log_2 \left\lfloor \frac{n}{2n-2x+1}\right\rfloor\right\rfloor+1}-\text{sgn}\left(\left\lfloor \frac{\left\lceil\frac{n}{2}\right\rceil}{x}\right\rfloor\right)\right)$With $$x=n-1$$

28. KingGeorge Group Title

Sorry, at the end it's just $\Large -\text{sgn}\left(\left\lfloor \frac{\left\lceil\frac{n}{2}\right\rceil}{x}\right\rfloor\right)$

29. KingGeorge Group Title

The solution that I discovered on my own (both give the same values for all positive integers)$\Large 2\left(n-2^{\left\lfloor\log_2(n)\right\rfloor}\right)+2^{\left\lfloor\log_2(n)\right\rfloor+1}+1\pmod{2^{\left\lfloor\log_2(n)\right\rfloor-1}\cdot3}$

30. KingGeorge Group Title

Once again, the end is just $\Large \pmod{2^{\left\lfloor\log_2(n)\right\rfloor-1}\cdot3}$