ok, this is a lame attempt - can someone explain if it's correct, or
why not :
int limited_rand() {
return rand() % 8 + 1;
}
value = limited_rand() % 3 + limited_rand();
On Jan 31, 7:29 am, Ming \(Amos\) Zhang [EMAIL PROTECTED] wrote:
It's not uniformly distributed, suppose the given
I think the method given by Paul is practical.
the precise expression of F(n) is
F(n) = K * (x1^n - x2^n ) where K = 1/sqrt(5), x1 = (1+sqrt(5))/2 and
x2 = (1-sqrt(5))/2
because the x2 is small (about -0.61803...), the bigger n is ,the
smaller the x2^n is.
so, we consider K*(x1^2) is similar to
Hi there,
How to write a c program to list all combinations of C(n,k), where 1
= k =n and n is a variable.
For example, n=3, C(3,k).
k=1, {1},{2},{3}
k=2, {1,2}, {1,3}, {2,3}
k=3, {1,2,3}
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recur( start , filled){
if start k then print
else {
for i=start to n
}
}
On Jan 31, 9:34 pm, sofin [EMAIL PROTECTED] wrote:
Hi there,
How to write a c program to list all combinations of C(n,k), where 1
= k =n and n is a variable.
For example, n=3, C(3,k).
k=1, {1},{2},{3}
Here's my lame attempt. Output is uniformly distributed if input is so.
1 - 20.065%
2 - 19.986%
3 - 19.989%
4 - 19.964%
5 - 19.996%
1 - 14.352%
2 - 14.302%
3 - 14.281%
4 - 14.285%
5 - 14.295%
6 - 14.252%
7 - 14.233%
/* */
#includestdio.h
#includestdlib.h
#includestring.h
int
recur( start , fill){
if start k then print
else {
for i=start to n {
a[fill] = i
recur ( start+1, fill+1)
}
}
}
array a, n k are global variables.
and call the function as recur(1,1)
On Jan 31, 9:57 pm, Sandesh [EMAIL PROTECTED] wrote:
recur( start , filled){
On Jan 30, 8:57 pm, Jialin [EMAIL PROTECTED] wrote:
Question:
Given a program which can generate one of {1, 2, 3, 4, 5} randomly.
How can we get another generator which can generate one of
{1,2,3,4,5,6,7} randomly?
Thank you!
If you generate random 1 to 5 twice, there are 25 equally
Why can't we simply take the 1..5 random number, multiply by 7 and
divide by 5.
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Then you will get only the following numbers
(1*7)/5 = 1
(2*7)/5 = 2
(3*7)/5 = 4
(4*7)/5 = 5
(5*7)/5 = 7
What you will do to get 3 and 6?
On 2/1/07, pramod [EMAIL PROTECTED] wrote:
Why can't we simply take the 1..5 random number, multiply by 7 and
divide by 5.
