nextpart | R Documentation |
Given a partition, return the “next” one; or determine whether it is the last one.
nextpart(part, check=TRUE) islastpart(part) firstpart(n) nextdiffpart(part, check=TRUE) islastdiffpart(part) firstdiffpart(n) nextrestrictedpart(part, check=TRUE) islastrestrictedpart(part) firstrestrictedpart(n, m, include.zero=TRUE) nextblockpart(part, f, n=sum(part), include.fewer=FALSE, check=TRUE) islastblockpart(part, f, n=NULL , include.fewer=FALSE) firstblockpart( f, n=NULL , include.fewer=FALSE) nextcomposition(comp, restricted, include.zero=TRUE, check=TRUE) islastcomposition(comp, restricted, include.zero=TRUE) firstcomposition(n, m=NULL , include.zero=TRUE)
part,comp |
A partition or composition |
check |
Boolean, with default |
f, n, include.fewer, m, include.zero |
Other arguments as per the vectorized version |
restricted |
In function |
These functions are intended to enumerate partitions one at a time, eliminating the need to store a huge matrix. This is useful for optimization over large domains and makes it possible to investigate larger partitions than is possible with the vectorized codes.
The idea is to use a first...()
function to generate the first
partition, then iterate using a next...()
function, stopping when
the islast...()
function returns TRUE
.
An example is given below, in which the “scrabble” problem is
solved; note the small size of the sample space. More examples are
given in the tests/aab.R
file.
Functions nextpart()
and nextdiffpart()
require a vector
of the right length: they require and return a partition padded with
zeros. Functions nextrestrictedpart()
and
nextblockpart()
work with partitions of the specified length.
Function nextcomposition()
truncates any zeros at the end of
the composition. This behaviour is inherited from the C
code.
In functions nextcomposition()
and firstcomposition()
,
argument include.zero
is ignored if restricted
is
FALSE
.
I must say that the performance of these functions is terrible;
they are much much slower than their vectorized equivalents. The
magnitude of the difference is much larger than I expected. Heigh
ho. Frankly you would better off working directly in C
.
Robin K. S. Hankin
parts
# Do the optimization in scrabble vignette, one partition at a time: # (but with a smaller letter bag) scrabble <- c(a=9 , b=2 , c=2 , d=4 , e=12 , f=2 , g=3) f <- function(a){prod(choose(scrabble,a))/choose(sum(scrabble),7)} bestsofar <- 0 a <- firstblockpart(scrabble,7) while(!islastpart(a)){ jj <- f(a) if(jj>bestsofar){ bestsofar <- jj bestpart <- a } a <- nextblockpart(a,scrabble) }
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