Hi Stephane,
Many thanks for the code pointers - graphics look fab!
Cheers
Lester
On Thu, 14 Apr 2022 at 15:29, Stéphane Mottelet
wrote:
> You can also enjoy the 3d version:
>
> function X=anglePath3(r, th, ph)
> cumth = cumsum(th);
> cumph = cumsum(ph);
> X = cumsum([0 r.*sin(cump
You can also enjoy the 3d version:
function X=anglePath3(r, th, ph)
cumth = cumsum(th);
cumph = cumsum(ph);
X = cumsum([0 r.*sin(cumph)
0 r.*sin(cumth).*cos(cumph)
0 r.*cos(cumth).*cos(cumph)], 2);
end
function N=collatz(n)
N = n;
Hi,
Le 14/04/2022 à 13:12, Claus Futtrup a écrit :
Dear Scilabers
I hope you can help me out. My combinatorics is a bit rusty.
So, the spouse has purchased a lock and I wondered how many
combinations are available?
The lock has 10 push buttons, they are numbered 1-2-3-4-5-6-7-8-9-0.
From
Dear Scilabers
I hope you can help me out. My combinatorics is a bit rusty.
So, the spouse has purchased a lock and I wondered how many combinations
are available?
The lock has 10 push buttons, they are numbered 1-2-3-4-5-6-7-8-9-0.
From a programming point of view, any of the numbers can be
Hi again,
Here is a small code doing the graphic job:
function X=anglePath(r, th)
cumth = cumsum(th);
X = cumsum([0 r.*cos(cumth);0 r.*sin(cumth)], 2);
end
function N=collatz(n)
N = n;
while n<>1
if modulo(n,2) == 0
n = n/2;
else