This also has a closed form solution: > choose(16+8-1,7) - choose(7+8-1, 7) - 7*choose(6+8-1,7) [1] 229713
-- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -----Original Message----- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- > project.org] On Behalf Of Brian Diggs > Sent: Thursday, December 31, 2009 3:08 PM > To: baptiste auguie; David Winsemius > Cc: r-help > Subject: Re: [R] expand.grid game > > baptiste auguie wrote: > > 2009/12/19 David Winsemius <dwinsem...@comcast.net>: > >> On Dec 19, 2009, at 9:06 AM, baptiste auguie wrote: > >> > >>> Dear list, > >>> > >>> In a little numbers game, I've hit a performance snag and I'm not > sure > >>> how to code this in C. > >>> > >>> The game is the following: how many 8-digit numbers have the sum of > >>> their digits equal to 17? > >> And are you considering the number "00000089" to be in the > acceptable set? > >> Or is the range of possible numbers in 10000079:98000000 ? > >> > > > > The latter, the first digit should not be 0. But if you have an > > interesting solution for the other case, let me know anyway. > > > > I should also stress that this is only for entertainment and > curiosity's sake. > > > > baptiste > > > > I realize I'm late coming to this, but I was reading it in my post- > vacation catch-up and it sounded interesting so I thought I'd give it a > shot. > > After coding a couple of solutions that were exponential in time (for > the number of digits), I rearranged things and came up with something > that is linear in time (for the number of digits) and gives the count > of numbers for all sums at once: > > library(plyr) > nsum3 <- function(digits) { > digits <- as.integer(digits)[[1L]] > if (digits==1) { > rep(1,9) > } else { > dm1 <- nsum3(digits-1) > Reduce("+", llply(0:9, function(x) {c(rep(0,x),dm1,rep(0,9-x))})) > } > } > > nsums <- llply(1:8, nsum3) > nsums[[5]][17] > # [1] 3675 > nsums[[8]][17] > # [1] 229713 > > The whole thing runs in well under a second on my machine (a several > years old dual core Windows machine). In the results of nsum3, the i- > th element is the number of "numbers" whose digits sum to i. The basic > idea is recursion on the number of digits; if n_{t,d} is the number of > d-digit "numbers" that sum to t, then n_{t,d} = \sum_{i\in(0,9)} n_{t- > i,d-1}. (Adding the digit i to each of those "numbers" makes their sum > t and increases the digits to d). When digits==1, then 0 isn't a valid > choice and that also implies the sum of digits can't be 0, which fits > well with the 1 indexing of arrays. > > -- > Brian Diggs, Ph.D. > Senior Research Associate, Department of Surgery, Oregon Health & > Science University > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.