Just an unsolicited tip here.
Try not to "over type" your functions. Your function here would be
cumbersome to use for someone on a 32-bit system for example. Unless you
have a good reason, prefer Int instead of explicit Int32 / Int64.
On Thursday, January 7, 2016 at 7:18:31 PM UTC+1, Asbjørn Nilsen Riseth
wrote:
>
> For future reference, I used the following:
>
> function traverse(numpoints::Int64, β::Array{Int64}, level::Int64 = 1)
> N = length(β)
> @assert 0 < level < N
>
> for i = 0:numpoints-sum(β[1:level-1])
> β[level] = i
> if level < N-2
> traverse(numpoints, β, level+1)
> else
> for j = 0:numpoints-sum(β[1:N-2])
> β[N-1] = j
> β[N] = numpoints - sum(β[1:N-1])
> @show β/numpoints
> end
> end
> end
> end
> traverse(5, Array{Int64}(4))
>
> Asbjørn
>
>
> On Thursday, 7 January 2016 17:14:03 UTC, Asbjørn Nilsen Riseth wrote:
>>
>> Thank you both. I'll have a look at a recursion approach.
>>
>>
>> On Thursday, 7 January 2016 13:47:02 UTC, Tim Holy wrote:
>>>
>>> This is a problem for recursion. See, for example, the LightGraphs or
>>> Graphs
>>> packages.
>>>
>>> Best,
>>> --Tim
>>>
>>> On Thursday, January 07, 2016 02:10:18 AM Asbjørn Nilsen Riseth wrote:
>>> > I would like to loop over all the \beta_i values of the following
>>> tree.
>>> >
>>> > <
>>> https://lh3.googleusercontent.com/-EBjanFs-BC4/Vo409IFAiTI/AAAAAAAAMEw/U7iq
>>> > Ev9nPmU/s1600/tree.png>
>>> >
>>> > What is the best Julia way of doing this? I'm also interested in
>>> > generalising this to N levels.
>>> >
>>> >
>>> > For the given tree, the following works. *Is there a way to use
>>> @nloops to
>>> > shorten this?*
>>> > n = Int(1/0.2)
>>> > for i_3 = 0:n
>>> > for i_2 = 0:n-i_3
>>> > for i_1 = 0:n-i_2-i_3
>>> > i_0 = n-sum(@ntuple 3 i)
>>> > β = [i_3, i_2, i_1, i_0]/n
>>> > @show β
>>> > end
>>> > end
>>> > end
>>>
>>>
