On 2025-04-24 3:25 p.m., Bert Gunter wrote:
Folks:
Unless my wee old brain errs (always a danger), uniform sampling from an
n-vector <xi> for which 0 <= ai <= xi <= bi and SUM(xi) = k, a constant,
where to ensure that the constraints are consistent (and nontrivial),
SUM(ai)< k and SUM(bi) > k, is a simple linear transformation (details left
to the reader) of uniform sampling from a standard n-1 dim simplex, for
which a web search yielded:
https://cs.stackexchange.com/questions/3227/uniform-sampling-from-a-simplex
I think for n > 2 it is sometimes a simplex, and sometimes a different
shape. For example, with n = 3 and all a_i = 0 and all b_i = 1, it's a
simplex if k < 1 or k > 2, but not for 1 < k < 2, where it's a hexagon.
You can visualize this with the following code:
x <- matrix(runif(30000), ncol = 3) # the full cube
x0 <- x[1.45 < sums & sums < 1.55,] # points near k = 1.5
rgl::plot3d(x0)
The proposed "solutions" in this thread which rejection sample sequentially
from uniforms do *not* produce a uniform distribution on the simplex, as
the link and references therein explain.
My solution didn't do rejection sampling, but it won't give a uniform
density. I didn't realize that was requested.
Apologies if I have misunderstood the problem and proposed solutions,
although there does seem to be some confusion in the thread about exactly
what was sought.
Yes, that's true.
Duncan Murdoch
Cheers,
Bert
"An educated person is one who can entertain new ideas, entertain others,
and entertain herself."
On Thu, Apr 24, 2025 at 10:33 AM Brian Smith <briansmith199...@gmail.com>
wrote:
Hi Rui,
This code is able to generate absolutely correct random sample vector
based on the applicable constraints.
I have one question though.
If I understood the R code correctly then, the first element is
drawing random number from Uniform distribution unconditionally,
however drawing of sample point for the second element is conditional
to the first one. Therefore if we have large vector size instead of
current 2, I guess the feasible region for the last few elements will
be very small.
Will that be any problem? does there any algorithm exist where all
(n-1) elements would be drawn unconditionally assuming our vector has
n elements?
Thanks and regards,
On Wed, 23 Apr 2025 at 10:57, Rui Barradas <ruipbarra...@sapo.pt> wrote:
Hello,
Here are your tests and the random numbers' histograms.
one_vec <- function(a, b, s) {
repeat{
repeat{
u <- runif(1, a[1], b[1])
if(s - u > 0) break
}
v <- s - u
if(a[2] < v && v < b[2]) break
}
c(u, v)
}
gen_mat <- function(m, a, b, s) {
replicate(m, one_vec(a, b, s))
}
a <- c(0.015, 0.005)
b <- c(0.070, 0.045)
s <- 0.05528650577311
m <- 10000L
set.seed(2025)
res <- gen_mat(m, a, b, s)
apply(res, 1, min) > a
#> [1] TRUE TRUE
apply(res, 1, max) < b
#> [1] TRUE TRUE
# plot histograms of one million 2d vectors
system.time(
res1mil <- gen_mat(1e6, a, b, s)
)
#> user system elapsed
#> 3.01 0.06 3.86
old_par <- par(mfrow = c(1, 2))
hist(res1mil[1L,])
hist(res1mil[2L,])
par(old_par)
Hope this helps,
Rui Barradas
Às 23:31 de 22/04/2025, Rui Barradas escreveu:
Hello,
Inline.
Às 17:55 de 22/04/2025, Brian Smith escreveu:
i.e. we should have
all elements of Reduce("+", res) should be equal to s =
0.05528650577311
My assertion is that it is not happing here.
You are right, that's not what is happening. The output is n vectors of
2 elements each. It's each of these vectors that add up to s.
Appparently I misunderstood the problem.
Maybe this is what you want?
(There is no n argument, the matrix is always 2*m)
one_vec <- function(a, b, s) {
repeat{
repeat{
u <- runif(1, a[1], b[1])
if(s - u > 0) break
}
v <- s - u
if(a[2] < v && v < b[2]) break
}
c(u, v)
}
gen_mat <- function(m, a, b, s) {
replicate(m, one_vec(a, b, s))
}
set.seed(2025)
res <- gen_mat(10000, a, b, s)
colSums(res)
Hope this helps,
Rui Barradas
On Tue, 22 Apr 2025 at 22:20, Brian Smith <briansmith199...@gmail.com
wrote:
Hi Rui,
Thanks for the explanation.
But in this case, are we looking at the correct solution at all?
My goal is to generate random vector where:
1) the first element is bounded by (a[1], b[1]) and second element is
bounded by (a[2], b[2])
2) sum of the element is s
According to the outcome,
The first matrix values are bounded by c(a[1], b[1]) & second matrix
values are bounded by c(a[2], b[2])
But,
regarding the sum, I think we should have sum (element-wise) sum
should be equal to s = 0.05528650577311.
How could we achieve that then?
On Tue, 22 Apr 2025 at 22:03, Rui Barradas <ruipbarra...@sapo.pt>
wrote:
Às 12:39 de 22/04/2025, Brian Smith escreveu:
Hi Rui,
Many thanks for your time and insight.
However, I am not sure if I could understand the code. Below is
what I
tried based on your code
library(Surrogate)
a <- c(0.015, 0.005)
b <- c(0.070, 0.045)
set.seed(2025)
res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m),
MoreArgs = list(s = 0.05528650577311, n = 2, m =
10000), a, b)
res1 = res[[1]]
res2 = res[[2]]
apply(res1, 1, min) > a ## [1] TRUE TRUE
apply(res2, 1, min) > a ## [1] FALSE TRUE
I could not understand what basically 2 blocks of res signify?
Which
one I should take as final simulation of the vector? If I take the
first block then the lower bound condition is fulfilled, but not
with
the second block. However with the both blocks, the total equals s
is
satisfying.
I appreciate your further insight.
Thanks and regards,
On Mon, 21 Apr 2025 at 20:43, Rui Barradas <ruipbarra...@sapo.pt>
wrote:
Hello,
Inline.
Às 16:08 de 21/04/2025, Rui Barradas escreveu:
Às 15:27 de 21/04/2025, Brian Smith escreveu:
Hi,
There is a function called RandVec in the package Surrogate
which can
generate andom vectors (continuous number) with a fixed sum
The help page of this function states that:
a
The function RandVec generates an n by m matrix x. Each of the m
columns contain n random values lying in the interval [a,b]. The
argument a specifies the lower limit of the interval. Default 0.
b
The argument b specifies the upper limit of the interval.
Default 1.
However in my case, the lower and upper limits are not same. For
example, if I need to draw a pair of number x, y, such that x +
y = 1,
then the lower and upper limits are different.
I tried with below code
library(Surrogate)
RandVec(a=c(0.1, 0.2), b=c(0.2, 0.8), s=1, n=2,
m=5)$RandVecOutput
This generates error with message
Error in if (b - a == 0) { : the condition has length > 1
Is there any way to generate the numbers with different lower
and
upper limits?
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code.
Hello,
Use ?mapply to cycle through all values of a and b.
Note that the output matrices are transposed, the random vectors
are the
rows.
Sorry, this is not true. The columns are the random vectors, as
documented. An example setting the RNG seed, for reproducibility.
library(Surrogate)
a <- c(0.1, 0.2)
b <- c(0.2, 0.8)
set.seed(2025)
res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m),
MoreArgs = list(s = 1, n = 2, m = 5), a, b)
res
#> $RandVecOutput
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.146079 0.1649319 0.1413759 0.257086 0.1715478
#> [2,] 0.253921 0.2350681 0.2586241 0.142914 0.2284522
#>
#> $RandVecOutput
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.5930918 0.2154583 0.6915523 0.7167089 0.3617918
#> [2,] 0.4069082 0.7845417 0.3084477 0.2832911 0.6382082
lapply(res, colSums)
#> $RandVecOutput
#> [1] 0.4 0.4 0.4 0.4 0.4
#>
#> $RandVecOutput
#> [1] 1 1 1 1 1
Hope this helps,
Rui Barradas
library(Surrogate)
a <- c(0.1, 0.2)
b <- c(0.2, 0.8)
mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m),
MoreArgs = list(s = 1, n = 2, m = 5), a, b)
#> $RandVecOutput
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.2004363 0.1552328 0.2391742 0.1744857 0.1949236
#> [2,] 0.1995637 0.2447672 0.1608258 0.2255143 0.2050764
#>
#> $RandVecOutput
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0.2157416 0.4691191 0.5067447 0.7749258 0.7728955
#> [2,] 0.7842584 0.5308809 0.4932553 0.2250742 0.2271045
Hope this helps,
Rui Barradas
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Hello,
The two blocks of res are the two random matrices, one for each
combination of (a,b). mapply passes each of the values in its
arguments
list (the ellipses in the help page) and computes the anonymous
function
with the pairs (a[1], b[1]), (a[2], b[2]).
Since a and b are two elements vectors the output res is a two
members
named list. Your error is to compare the result of apply(res2, 1,
min)
to a, when you should compare to a[2]. See the code below.
library(Surrogate)
a <- c(0.015, 0.005)
b <- c(0.070, 0.045)
set.seed(2025)
res <- mapply(\(a, b, s, n, m) RandVec(a, b, s, n, m),
MoreArgs = list(s = 0.05528650577311, n = 2, m =
10000),
a, b)
res1 = res[[1]]
res2 = res[[2]]
# first check that the sums are correct
# these sums should be s = 0.05528650577311, up to floating-point
accuracy
lapply(res, \(x) colSums(x[, 1:5]) |> print(digits = 14L))
#> [1] 0.05528650577311 0.05528650577311 0.05528650577311
0.05528650577311
#> [5] 0.05528650577311
#> [1] 0.05528650577311 0.05528650577311 0.05528650577311
0.05528650577311
#> [5] 0.05528650577311
#> $RandVecOutput
#> [1] 0.05528651 0.05528651 0.05528651 0.05528651 0.05528651
#>
#> $RandVecOutput
#> [1] 0.05528651 0.05528651 0.05528651 0.05528651 0.05528651
# now check the min and max
apply(res1, 1, min) > a[1L] ## [1] TRUE TRUE
#> [1] TRUE TRUE
apply(res2, 1, min) > a[2L] ## [1] TRUE TRUE
#> [1] TRUE TRUE
apply(res1, 1, max) < b[1L] ## [1] TRUE TRUE
#> [1] TRUE TRUE
apply(res2, 1, max) < b[2L] ## [1] TRUE TRUE
#> [1] TRUE TRUE
Which one should you take as final simulation of the vector? Both.
The first matrix values are bounded by c(a[1], b[1]) with column
sums
equal to s.
The second matrix values are bounded by c(a[2], b[2]) with column
sums
also equal to s.
Hoep this helps,
Rui Barradas
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and provide commented, minimal, self-contained, reproducible code.
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