On 14/01/2020 10:07 a.m., peter dalgaard wrote:
Yep, that looks wrong (probably want to continue discussion over on R-devel)
I think the culprit is here (in src/nmath/choose.c)
if (k < k_small_max) {
int j;
if(n-k < k && n >= 0 && R_IS_INT(n)) k = n-k; /* <- Symmetry */
if (k < 0) return 0.;
if (k == 0) return 1.;
/* else: k >= 1 */
if n is a near-integer, then k can become non-integer and negative. In your
case,
n == 4 - 1e-7
k == 4
n - k == -1e-7 < 4
n >= 0
R_IS_INT(n) = TRUE (relative diff < 1e-7 is allowed)
so k gets set to
n - k == -1e-7
which is less than 0, so we return 0. However, as you point out, 1 would be more
reasonable and in accordance with the limit as n -> 4, e.g.
factorial(4 - 1e-10)/factorial(1e-10)/factorial(4) -1
[1] -9.289025e-11
I guess that the fix could be as simple as replacing n by R_forceint(n) in the
k = n - k step.
I think that would break symmetry: you want choose(n, k) to equal
choose(n, n-k) when n is very close to an integer. So I'd suggest the
replacement whenever R_IS_INT(n) is true.
Duncan Murdoch
-pd
On 14 Jan 2020, at 00:33 , Wright, Erik Scott <eswri...@pitt.edu> wrote:
This struck me as incorrect:
choose(3.999999, 4)
[1] 0.9999979
choose(3.9999999, 4)
[1] 0
choose(4, 4)
[1] 1
choose(4.0000001, 4)
[1] 4
choose(4.000001, 4)
[1] 1.000002
Should base::choose(n, k) check whether n is within machine precision of k and
return 1?
Thanks,
Erik
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