I'd made a couple of mistakes (I'd specifically tuned my code for the previous set of scores and hadn't re-tuned it for these, so it missed some combos), also somehow I missed out the second 24. Also I stopped at 14 darts instead of running to 17.

So yeah, NOW I also get [SPOILER]82339371[/SPOILER].

Unfortunately I have no idea how FFTs work, so I don't know what specific methods you're refering to. However, I will read around and have a think as to how I could speed up my code (this last run 93 mins :shock:).

Somewhat optimized results: :whistle:

[code]
kossy@leela:~/darts2> time ./darts2
Value for N=100 is [spoiler]82339371.[/spoiler]
Peak at [spoiler]N=667: 17496397018818748.
Peak at N=668: 17496397018818748.
[/spoiler]
real 0m0.002s
user 0m0.001s
sys 0m0.001s
kossy@leela:~/darts2>
[/code]Code might run faster with mprime stopped, but ... nah. :geek:

Where did the 1072 come from, again? :unsure:

[QUOTE=ckdo;125358]Value for N=100 is [spoiler]82339371.[/spoiler]
Peak at [spoiler]N=667: 17496397018818748.
Peak at N=668: 17496397018818748.
[/spoiler][/QUOTE]Yes, I concur with that.[QUOTE=ckdo;125358]Where did the 1072 come from, again? :unsure:[/QUOTE]1072 was for the original problem with 43 unique targets. Your answer is for 62 unique targets with maximal sum of 1335, hence the peaks at floor[1335/2] and ceil[1335/2].

[quote=retina;125243]You need a better algorithm. It is a trivial Q actually. I ran all sums up to 1072 in <1 second. Think about using [spoiler]divide and conquer methods similar to FFT[/spoiler]

[edit]For all area scores (62 areas will bullseye) I get [spoiler]82339371[/spoiler] for N=100[/edit][/quote]

Ah, yes. It escaped me that the 1072 was in the unedited part of the message and I didn't conclude you were talking about the original problem. :blush:

[QUOTE=retina;125243][edit]For all area scores (62 areas will bullseye) I get [spoiler]82339371[/spoiler] for N=100[/edit][/QUOTE]Oops, that should be ... (62 areas [b]with[/b] bullseye) ... :blush:

[quote=R.D. Silverman;125172]Compute the coefficient of x^100 in the generating function
1/[(1-x)(2-x)(3-x)(4-x)............(60-x)][/quote]

Is this the right generating function?
Shouldn't it be the coefficient of x^100 in
(1+x)(1+x^2)(1+x^3)(1+x^4).....(1+x^57)(1+x^60) [43 terms]?

And then if we want to allow for one dart per area,
just raise the appropriate term to the power = #areas with that value,
e.g. (1+x^6)^3 because 6 is the value of three areas.
(Which yields 62 terms.)

I think Davar is right.

If you wanted up to three terms, I believe the term would be

(1 + x^6 + x^12 + x^18) rather than (1+x^6)^3

An interesting link, slightly related

[url]http://www.maa.org/mathland/mathland_5_19.html[/url]

I'm still intrigued by Davar's choice of 43 numbers.

[quote=davieddy;125738]I'm still intrigued by Davar's choice of 43 numbers.[/quote]

Intrigued by something other than the fact they are a list of all possible point values one can score by throwing one dart?

Doh:blush:
[quote=bsquared;125739]Intrigued by something other than the fact they are a list of all possible point values one can score by throwing one dart?[/quote]