Richard O'Keefe pointed out that I was wrong; the distributions are in fact
different.
One (now!) "obvious" reason why is that for the order statistics, the
distribution of the max is clearly skewed. For my cumsum method, the max is
clearly uniform (on [0,55] for the specific example).
Context ma
... and here is a a simple 2-liner without a sort that I think is linear in
time and space (but please correct if this is wrong):
x <- cumsum(runif(10))
x/x[10] * runif(1, 0, 55) + seq.int(0, 45,5)
Question: Does this give the same distribution as Peter's method using the
order statistics? I bel
Richard:
"The "use an upper bound of 100 - (n+1)*5" and then "add back
cumsum(rep(5,n)) at the end" (or equivalent) trick ensures the gaps
are right but does nothing about the distribution.."
If I understand you correctly, I think the above is wrong. Here is a
one-line version of Peter's code for
The Bentley and Saxe paper answers the questions
(1) How can I draw a sample of n numbers in the range 0..1 such that
the numbers are returned in increasing order WITHOUT sorting and so
that all such sequences are equally likely in LINEAR time. That's the
code I showed in R.
(2) How can I do this
In Peter's solution, just sample without replacement from integers instead
of generating random uniforms. See ?sample.
-- Bert
On Wed, Jun 4, 2025 at 5:23 AM Brian Smith
wrote:
> Hi,
>
> I dont see from the solution pov they are different.
>
> One followup Q though, how can I extend this to dr
To answer my own question, yes they are different.
Peter's code can generate the solution 1 20 26 32 38 44 50 56 62 68. Mine
cannot.
So, again, context?
-- Bert
On Wed, Jun 4, 2025 at 4:43 AM Bert Gunter wrote:
> Is Peter's solution different then:
>
> diffs <- cumsum(runif(9, 5, 100/9))
> x
Hi!
It's possible I'm wrong. But for me it looks like:
n_sample <- 300
v <- round(runif(n_sample, 0, 100), 2)
m <- matrix(v, ncol = 10)
m_s <- t(apply(m, 1, sort))
m_d <- t(apply(m_s, 1, diff))
m_5 <- m_d>=5
f <- rowSums(m_5) == 9
sum(f)
res <- m_s[f, ]
Regards,
Sergiy
On Wed, Jun 4, 2025 at 1
Okay, I think I found the reason. This is due to accumulation of nine
5s in the cumsum. Thanks again for the elegant solution.
But I wonder, if the solution is simple then what is the significance
of the Research paper by Bentley and Saxe naming “Generating sorted
lists of random numbers” which Ri
Hi Peter,
Could you please help me to understand what is the basis of choosing
55 in runif(10,0,55))?
Thank you!
On Wed, 4 Jun 2025 at 02:45, peter dalgaard wrote:
>
> Can't you just generate 10 values in (0,55), sort them, generate the
> distances, add 5 and cumulate?
>
> > x <- sort(runif(10
Hi,
I dont see from the solution pov they are different.
One followup Q though, how can I extend this to draw only integer
mid-point between 0-100 while maintaining minimum difference as 5?
Also, are all generated points are equally likely?
Thanks for your time and suggestions.
Thanks and rega
Is Peter's solution different then:
diffs <- cumsum(runif(9, 5, 100/9))
x <-runif(1,0,100-diffs[9])
c(x, x+diffs)
I ask because:
1. If yes, this is why more context is needed;
2. If no, the above avoids a sort.
Cheers,
Bert
On Tue, Jun 3, 2025 at 2:15 PM peter dalgaard wrote:
> Can't you j
Can't you just generate 10 values in (0,55), sort them, generate the distances,
add 5 and cumulate?
> x <- sort(runif(10,0,55))
> d <- diff(x)+5
> cumsum(c(x[1],d))
[1] 12.27815 21.21060 26.37856 36.03812 41.97237 57.02945 67.86113
[8] 75.74085 81.28533 98.30792
> On 3 Jun 2025, at 09.21, Br
Hi Richard,
Thanks for your insight.
As I mentioned in one of my earlier emails to the group, I imposed a
constraint of accuracy up to two decimal places in order to obtain a
finite set of possible values. For instance, if I were to round values
to zero decimal places, the number of unique sequen
Hi JRG,
It is not a homwork, but a practical problem. I was unsure about the
theritical framework to generate such sequence.
I introduced the condition that accuracy up to second decimal number,
so that we would get a finite number of possible values. Otherwise I
thought there could be infinite p
On Saturday, May 31st, 2025 at 7:44 PM, Bert Gunter
wrote:
> JRG:
>
> I don't think your specification is correct -- perhaps just a thinko. I think
> a 10-tuple of "reals" (scare quotes because of computer precision) with your
> specifications is what is wanted.
>
> Bert
>
> "An educated perso
JRG:
I don't think your specification is correct -- perhaps just a thinko. I
think a 10-tuple of "reals" (scare quotes because of computer precision)
with your specifications is what is wanted.
Bert
"An educated person is one who can entertain new ideas, entertain others,
and entertain herself."
I'll second Bert's comments, also assuming this is not homework. In addition:
Your use of "mid-point" is not a standard one (in my world), nor perhaps is
that of "simulate".
Let me attempt to re-state your problem: You wish to choose 10-tuples of
integers 0 <= k <= 100 satisfying
1) 0 <= k_i
If this is a real problem and not homework, can you tell us the
context? It is not at all clear (to me) what you mean by "simulate",
i.e. what your target distribution is, which may depend on/be defined
by the context.
Bert
"An educated person is one who can entertain new ideas, entertain
others,
Your description is not clear at the point where you "project it to the
original space."
Anyway, this sounds like homework, as the solution to this particular example
is trivial when considered from the right point of view.
On May 31, 2025 12:06:49 PM PDT, Sergei Ko wrote:
>Hi!
>Just an idea.
Hi!
Just an idea. You randomly select the first point. Cut +/- 5 from the
original space [0;100]. Select second point. Project it to the original
space. Cut again. Repeat.
Simplified method without cutting: every time mark +/- 5 as a bad space and
select point again if it is in forbidden area.
Hope
Hi,
Let say I have a range [0, 100]
Now I need to simulate 1000 10 mid-points within the range with
accuracy upto second decimal number.
Let say, one simulated set is
X1, X2, ..., X10
Ofcourrse
X1 < X2 < ... https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide https:/
21 matches
Mail list logo
