Memory allocation in a fixed point algorithm - algorithm

Question

I need to find the fixed point of a function f. The algorithm is very simple:
Given X, compute f(X)
If ||X-f(X)|| is lower than a certain tolerance, exit and return X,
otherwise set X equal to f(X) and go back to 1.
I'd like to be sure I'm not allocating memory for a new object at every iteration
For now, the algorithm looks like this:
iter1 = function(x::Vector{Float64})
for iter in 1:max_it
oldx = copy(x)
g1(x)
delta = vnormdiff(x, oldx, 2)
if delta < tolerance
break
end
end
end
Here g1(x) is a function that sets x to f(x)
But it seems this loop allocates a new vector at every loop (see below).
Another way to write the algorithm is the following:
iter2 = function(x::Vector{Float64})
oldx = similar(x)
for iter in 1:max_it
(oldx, x) = (x, oldx)
g2(x, oldx)
delta = vnormdiff(oldx, x, 2)
if delta < tolerance
break
end
end
end
where g2(x1, x2) is a function that sets x1 to f(x2).
Is thi the most efficient and natural way to write this kind of iteration problem?
Edit1: timing shows that the second code is faster:
using NumericExtensions
max_it = 1000
tolerance = 1e-8
max_it = 100
g1 = function(x::Vector{Float64})
for i in 1:length(x)
x[i] = x[i]/2
end
end
g2 = function(newx::Vector{Float64}, x::Vector{Float64})
for i in 1:length(x)
newx[i] = x[i]/2
end
end
x = fill(1e7, int(1e7))
#time iter1(x)
# elapsed time: 4.688103075 seconds (4960117840 bytes allocated, 29.72% gc time)
x = fill(1e7, int(1e7))
#time iter2(x)
# elapsed time: 2.187916177 seconds (80199676 bytes allocated, 0.74% gc time)
Edit2: using copy!
iter3 = function(x::Vector{Float64})
oldx = similar(x)
for iter in 1:max_it
copy!(oldx, x)
g1(x)
delta = vnormdiff(x, oldx, 2)
if delta < tolerance
break
end
end
end
x = fill(1e7, int(1e7))
#time iter3(x)
# elapsed time: 2.745350176 seconds (80008088 bytes allocated, 1.11% gc time)

Answer 1

I think replacing the following lines in the first code
for iter = 1:max_it
oldx = copy( x )
...
by
oldx = zeros( N )
for iter = 1:max_it
oldx[:] = x # or copy!( oldx, x )
...
will be more efficient because no array is allocated. Also, the code can be made more efficient by writing for-loops explicitly. This can be seen, for example, from the following comparison
function test()
N = 1000000
a = zeros( N )
b = zeros( N )
#time c = copy( a )
#time b[:] = a
#time copy!( b, a )
#time \
for i = 1:length(a)
b[i] = a[i]
end
#time \
for i in eachindex(a)
b[i] = a[i]
end
end
test()
The result obtained with Julia0.4.0 on Linux(x86_64) is
elapsed time: 0.003955609 seconds (7 MB allocated)
elapsed time: 0.001279142 seconds (0 bytes allocated)
elapsed time: 0.000836167 seconds (0 bytes allocated)
elapsed time: 1.19e-7 seconds (0 bytes allocated)
elapsed time: 1.28e-7 seconds (0 bytes allocated)
It seems that copy!() is faster than using [:] in the left-hand side,
though the difference becomes marginal in repeated calculations (there seems to be
some overhead for the first [:] calculation). Btw, the last example using eachindex() is very convenient for looping over multi-dimensional arrays.
Similar comparison can be made for vnormdiff(), where use of norm( x - oldx ) etc is slower than an explicit loop for vector norm, because the former allocates one temporary array for x - oldx.