Rounded random numbers with fixed sum

I'm trying to generate a 1xK array with 'random' numbers that sum to 1, but that are also rounded to either 1 or 2 decimal places.

For example, if K=3, and I want the numbers to 2 decimal places, this is what I would expect:

array = [0.27, 0.45, 0.27].

So far I am unable to achieve this. Here is my code.

K = 3;
r = rand(1,K);
r = round(r/sum(r),2);

Without rounding the numbers sum to 1, but once rounded the numbers only to sum to 1 approximately 50% of the time.

Any help would be greatly appreciated, thanks!

[madhan ravi]

sum(array) is not exactly one?

[Thomas Coppin]

No, once you round the values (after dividing by the sum) they no longer sum to exactly one.

Answer 1

[Walter Roberson]

https://www.mathworks.com/matlabcentral/answers/327656-conditional-random-number-generation#answer_257296

Multiply your desired total by 10 to the number of digits you need. Generate using Roger's code. Divide everything by 10 to the number of digits you need.

The sum of these will not in general be exactly the required value due to round off when adding floating-point numbers. In the test I just did of 1000 integers from 1 to 999, sum(v) /10 was different than sum(v/10) by 2*eps() of the value.

Answer 2

[Bruno Luong]

k = 3;
r = (diff([0 sort(randperm(100+k-1,k-1)) 100+k])-1)/100

[Thomas Coppin]

Thank you so much, Bruno. This is perfect!

[Thomas Coppin]

I kind of understand your algorithm, however is there a way to modify it to constrain the random numbers between an upper and lower bound? For example, 0 and 0.5?

[Bruno Luong]

No but you can use Roger's RANDFIXEDSUM

k = 3;

x = round([0; cumsum(randfixedsum(k, 1, 100, 0, 0.5*100))]); 
x(end) = 100; % prevent floating point error
r = diff(x')/100

The distribution might be not perfectly uniform but possibly close enough and suitable for what ever you want to do with it.

[Thomas Coppin]

Thanks Bruno, that does exactly what I require.