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[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 \(n1\)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 \(61\)th present.
 one year ago
 one year ago
[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 \(n1\)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 \(61\)th present.
 one year ago
 one year ago

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KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Also, if you want me to compute some test values for you, let me know what \(n\) to use.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.0
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 >.<
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
This is a variation of Extended Josephus problem, where \( x=n1 \) The recursive and the nonrecursive formulas are discussed here: http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
@FoolForMath That was the formula I mentioned I was able to find. However, it gives the incorrect value.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
I think you are probably making mistake while implementing it. This looks correct to me.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Their table is correct, but their formula is not. Try their formula for n=41, and x=40.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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%292%2840%29%2B1%29+%282%5E%281%2Bceiling%28log_2%28ceiling%2841%2F%282%2841%292%2840%29%2B1%29%29%29%29%29+sgn%28ceiling%28ceiling%2841%2F2%29%2F2%29%29%29
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
@joemath314159 is close. I believe his formula gives you the number of the child that gets the last present.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Well, as I said if you implement is right you will get 35.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Do you understand C?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
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
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Even using their formula by hand, I get an incorrect value for n=6.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Sorry, I haven't checked in the nonrecursive version. But it's rather difficult to believe that the formula is incorrect :)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Like I said, I believe it's a typo since it's easily changed so that I do always get the correct solution.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
The nonrecursive version has little importance in computer science (programming) as it fails for sufficiently large \(n \).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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?
 one year ago

LimitlessBest ResponseYou've already chosen the best response.0
Could you please compute for values from \(1\) to \(10\)?
 one year ago

LimitlessBest ResponseYou've already chosen the best response.0
I believe \(1\) is undefined, actually. Sorry.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
\[n=5 \rightarrow 5\]\[n=6\rightarrow1\]\[n=7\rightarrow3\]\[n=8\rightarrow5\]\[n=9\rightarrow7\]\[n=10\rightarrow9\]I'll let you do 24 since they're fairly simple
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
\[n=2^4=16\rightarrow9\]\[n=20\rightarrow17\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
@KingGeorge Solution??
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
First up, we have the modified form of the equation linked to above. This is \[\Large 2n+1(2n2x+1)\left(2^{\left\lfloor \log_2 \left\lfloor \frac{n}{2n2x+1}\right\rfloor\right\rfloor+1}\text{sgn}\left(\left\lfloor \frac{\left\lceil\frac{n}{2}\right\rceil}{x}\right\rfloor\right)\right)\]With \(x=n1\)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Sorry, at the end it's just \[\Large \text{sgn}\left(\left\lfloor \frac{\left\lceil\frac{n}{2}\right\rceil}{x}\right\rfloor\right)\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
The solution that I discovered on my own (both give the same values for all positive integers)\[\Large 2\left(n2^{\left\lfloor\log_2(n)\right\rfloor}\right)+2^{\left\lfloor\log_2(n)\right\rfloor+1}+1\pmod{2^{\left\lfloor\log_2(n)\right\rfloor1}\cdot3}\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Once again, the end is just \[\Large \pmod{2^{\left\lfloor\log_2(n)\right\rfloor1}\cdot3}\]
 one year ago
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