I wonder whether this answers Baptiste's question as asked. 1: An 8-digit number can have some digits equal to 0; see Baptiste's comment "maxi <- 9 # digits from 0 to 9"
2: According to the man-page fror blockparts in partitions, "all sets of a=(a1,...,an) satisfying Sum[ai] = n subject to 0 < ai <= yi are given in lexicographical order." So it would seem that blockparts would not count 8-digit numbers which have some zero digits. One could presumably "fake" it by looping over the number of non-zero digits, from 2 to 8 -- something like: all <- 0 for(i in (2:8)){ jj <- blockparts(rep(9,8),17) all <- all + dim(jj) } Or am I missing something?! Ted. On 21-Dec-09 07:57:32, Robin Hankin wrote: > Hi > library(partitions) > jj <- blockparts(rep(9,8),17) > dim(jj) > > gives 318648 > > HTH > rksh > > > > 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? >> The brute-force answer could be: >> >> maxi <- 9 # digits from 0 to 9 >> N <- 5 # 8 is too large >> test <- 17 # for example's sake >> >> sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi), >> N-1)))) == test) >> ## 3675 >> >> Now, if I make N = 8, R stalls for some time and finally gives up >> with: >> Error: cannot allocate vector of size 343.3 Mb >> >> I thought I could get around this using Reduce() to recursively apply >> rowSum to intermediate results, but it doesn't seem to help, >> >> >> a=list(1:maxi) >> b=rep(list(0:maxi), N-1) >> >> foo <- function(a, b, fun="sum", ...){ >> switch(fun, >> 'sum' = rowSums(do.call(expand.grid, c(a, b))), >> 'mean' = rowMeans(do.call(expand.grid, c(a, b))), >> apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic >> case >> } >> >> sum(Reduce(foo, list(b), init=a) == test) >> ## 3675 # OK >> >> Same problem with N=8. >> >> Now, if N was fixed I could write a little C code to do this >> calculation term-by-term, something along those lines, >> >> test = 0; >> >> for (i1=1, i1=9, i1++) { >> for (i2=0, i2=9, i2++) { >> >> [... other nested loops ] >> >> test = test + (i1 + i2 + [...] == 17); >> >> } [...] >> } >> >> but here the number of for loops, N, should remain a variable. >> >> In despair I coded this in R as a wicked eval(parse()) construct, and >> it does produce the expected result after an awfully long time. >> >> makeNestedLoops <- function(N=3){ >> >> startLoop <- function(ii, start=1, end=9){ >> paste("for (i", ii, " in seq(",start,", ",end,")) {\n", sep="") >> } >> >> first <- startLoop(1) >> inner.start <- lapply(seq(2, N), startLoop, start=0) >> calculation <- paste("test <- test + (", paste("i", seq(1, N), >> sep="", collapse="+"), "==17 )\n") >> end <- replicate(N, "}\n") >> code.to.run <- do.call(paste, c(list(first), inner.start, >> calculation, end)) >> cat(code.to.run) >> invisible(code.to.run) >> } >> >> test <- 0 >> eval(parse(text = makeNestedLoops(8)) ) >> ## 229713 >> >> I hope I have missed a better way to do this in R. Otherwise, I >> believe what I'm after is some kind of C or C++ macro expansion, >> because the number of loops should not be hard coded. >> >> Thanks for any tips you may have! >> >> Best regards, >> baptiste -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk> Fax-to-email: +44 (0)870 094 0861 Date: 21-Dec-09 Time: 08:45:09 ------------------------------ XFMail ------------------------------ ______________________________________________ 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.