hello guys!
which way would be the easiest way to repeat a sequence 1000 times ?
this is my code
-------------------------------------------------------
x0=[zeros(1,276) ones(1,290)];
x=x0(randperm(566));
b=(0:566);
d=[x 0]-[0 x];
l=(b(d==(-1))-b(d==1))
L=max(l)
-------------------------------
i want 1000 results of L, i've tried searching for something that might work, but nothing that works.
any ideas?

 Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 16 Apr. 2014

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ones(1,1000)*L

4 Kommentare

Rasmus
Rasmus am 16 Apr. 2014
I accepted to early, the answer is good though.
But what if i want L to be different every time? I want the squence to generate a new x all the time. By doing it this way, L is always the same.
Azzi Abdelmalek
Azzi Abdelmalek am 16 Apr. 2014
This is not clear
Rasmus
Rasmus am 16 Apr. 2014
Bearbeitet: Azzi Abdelmalek am 16 Apr. 2014
x=x0(randperm(566)); generates a new vector(1,566) of ones and zeroes in a different order.
l=(b(d==(-1))-b(d==1)); gives me the chains of ones of the vector above
L=max(l); gives me the value of longest chain of ones in l
(you of cause know this i am sure)
what i want is to have 1000 different generated results of L, so i can put it into a histogram - This is for probability calculation.
is it possible to make code that gives me this?
Rasmus
Rasmus am 16 Apr. 2014
I want to make a loop of this sequence so it repeats it self 1000 times.

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