How to generate a random array of 1*N matrix in which sum of all elements is 1 and numbers generate should be upto 1 decimlal place only.

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Abhinav
Abhinav 2014-11-13
评论： Roger Stafford 2014-11-13
For eg. [0.4 0.3 0.3] It should be generated randomly.

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Roger Stafford
Roger Stafford 2014-11-13
编辑：Roger Stafford 2014-11-13
diff([0,sort(randi([0,10],1,N-1)),10])/10; % <-- Corrected
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Abhinav
Abhinav 2014-11-13
编辑：Abhinav 2014-11-13
Thanks a lot. I have one small issue with it. Sometimes it is assigning 0 to an element. I don't want 0 in my array. Any non zero number is acceptable. And if you can explain the code, it will be very helpful.
Roger Stafford
Roger Stafford 2014-11-13
To avoid zeros you can do this, Abhinav:
x = (diff([0,sort(randi([0,10-N],1,N-1)),10-N])+ones(1,N))/10;
Note that with this restriction, N cannot be greater than 10. Otherwise there will be error messages.
As for an explanation, first, if your N numbers are each multiplied by ten, then they are integers and their sum must always be 10, which explains the division by 10 at the last step. Next, if 1 is subtracted from each integer, then their sum is 10-N, and they range from 0 to 10-N, which explains the addition of "ones(1,N)".
So now the equivalent problem is to find random integers ranging from 0 to 10-N whose sum is 10-N. The call "randi([0,10-N],1,N-1)" gives N-1 integers in this range and 'sort' arranges them in ascending order. The row vector
[0,sort(randi([0,10-N],1,N-1)),10-N]
consists of N+1 ascending integers which start with 0 and end with 10-N. If we perform a 'diff' on these, the resulting integers will all necessarily have a sum of 10-N because 0 and 10-N are the two end values of that vector. Also all the resulting integer differences must lie between 0 and 10-N. That is what was required in the above equivalent version. Therefore problem solved.
To get a better feeling for this solution you can separate out the parts of the code:
t1 = randi([0,10-N],1,N-1);
t2 = sort(t1);
t3 = [0,t2,10-N];
t4 = diff(t3);
t5 = t4 + ones(1,N);
t6 = t5/10;
and experiment with each step of the computation to see how it proceeds to a solution.

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