Hi Ted.

you've found a bug in the documentation for blockparts().
It should read "0 <= ai <= yi".  I'll fix it before the next
major release (which will include sampling without replacement from
a multiset, Insha'Allah).

Best wishes

rksh


(Ted Harding) wrote:
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 ------------------------------


--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877

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