On Wed, Feb 25, 2009 at 9:25 AM, Fox, Gordon g...@cas.usf.edu wrote:
The tricky part isn't finding the common factors -- we knew how to do
that, though not in so concise a fashion as some of these suggestions.
It was finding all their products without what I (as a recovered Fortran
programmer)
Sorry - I didn't express myself very clearly. Yes, we're looking for
common factors of A and B. One way to get there is by my initial
approach: find the prime factors of each, select those that are in
common, and then take pairwise products. My initial question was about
the last step only.
Fox, Gordon gfox at cas.usf.edu writes:
Example: 40 and 80 have these factors: c(1,2,2,2,5) and c(1,2,2,2,2,5).
We can use match() to get the common factors c(1,2,2,2,5). What we want
to be able to get from this is the list of all the possible products,
which should be the concatenation of
Dear Gordon,
Try also,
unique(apply(expand.grid(commfac,commfac),1,prod))
[1] 1 2 5 4 10 25
HTH,
Jorge
On Mon, Feb 23, 2009 at 9:52 PM, Fox, Gordon g...@cas.usf.edu wrote:
This is a seemingly simple problem - hopefully someone can help.
Problem: we have two integers. We want (1) all
On Mon, Feb 23, 2009 at 9:52 PM, Fox, Gordon g...@cas.usf.edu wrote:
This is a seemingly simple problem - hopefully someone can help.
Problem: we have two integers. We want (1) all the common factors, and
(2) all the possible products of these factors. We know how to get (1),
but can't figure
L'esprit de l'escalier strikes again
An even simpler statement of your original problem:
Find the factors that A and B have in common.
If A and B are fairly small ( 1e7, say), a very direct approach is:
which( ! (A %% 1:min(A,B)) !(B %% 1:min(A,B)) )
Is this brute force?
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