# Its possible to use all cores or parallel cores to do some project/calculations?

sorry, it is ok now

At first glance, I see you do a lot of redundant calculations. For example:

``````function Φll(r,sigma)
((-(3*sigma.^3)/(r^3*(1+r)^2))-(6*sigma.^3)/(r^2*(1+r)^3)
-(9*sigma.^3)/(r*(1+r)^4)+(12*sigma.^3)/(1+r)^5
+(3*sigma.^9)/(10*r^3*(1+r)^8)
+(12*sigma.^9)/(5*r^2*(1+r)^9)
+(54*sigma.^9)/(5*r*(1+r)^10)
-(12*sigma.^9)/(1+r)^11+(9*sigma.^3)/((r-1)^4*r)
-(54*sigma.^9)/(5*(r-1)^10*r)
+(6*sigma.^3)/((r-1)^3*r^2)
-(12*sigma.^9)/(5*(r-1)^9*r^2)
+(3*sigma.^3)/((r-1)^2*r^3)
-(3*sigma.^9)/(10*(r-1)^8*r^3)
-(12*sigma.^3)/(r-1)^5+(12*sigma.^9)/(r-1)^11)
end
``````

Along the lines of a better algorithm, if you actually are using quadgk, perhaps see http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.7.3656&rep=rep1&type=pdf. The authors claim it is faster and amenable to parallelization.

it is a complicated problem …
i have big and big equations … this in a expanded equation to simplify calculations …
my problem in to calculate the cross section Qb … it takes too long … i am running for 2 days the program … at least will take 10 days … can you parrallelize the Qb function ? or make it faster ?

you can easily simplify the problem if you use

`function interpQb()`

`````` ag = 0.01:0.05:3                       # use ag = 0.1:0.1:0.2 to go faster to test
aσ = 0.03 : 0.025 : 0.63           #  use aσ = 0.03 : 0.01 : 0.04  to go faster to test
aα = 0:0.1:0.4                           # use aα = 0:0.1:0.2  to go faster to test
aT= 0.1:0.05:0.4                        # use  aT= 0.1:0.1:0.2  to go faster to test``````

Looking at `interpQb`, the bulk of the runtime is spent looping over `Qd`, so let’s start by optimizing that at a single representative point in the provided range:

``````julia> @btime Qd(.1, .1, .1, .1)
853.414 ms (139498 allocations: 8.26 MiB)
5.978500817052537
``````

You’re solving some nested integrals to low precision, and there’s root-finding within the integrand. The quickest speedup comes from switching from your `brent.jl` to Roots.jl’s built-in `Brent()` method. Beyond that, `@fastmath` helps, and there’s some minor benefit to pre-computing `σ3 = σ^3` and `σ9 = σ^9` in `Φl` and `Φll`. With those changes, I get

``````julia> @btime Qd(.1, .1, .1, .1)
70.482 ms (139303 allocations: 8.26 MiB)
5.978500817052536
``````

Using `Threads.@threads` to distribute work,

``````function calcQb()
ag = 0.01:0.05:3    # teste
aσ = 0.03 : 0.025 : 0.63
aα = 0:0.1:0.4
aT= 0.1:0.05:0.4
a = collect(Iterators.product(ag, aσ, aα, aT))
Qb = zeros(size(a))
Qb[i] = Qd(a[i]...)
end
return Qb
end
``````

Iterating over the full range of 52500 elements took 1056 seconds, or 20 ms/element, which is already a ~42x speedup over the original code.

There’s an additional opportunity for speedup here:

``````function Qd(g,σ,α,T)
rc0, bc0, gc0 = rc0bc0gc0(σ)
if g > gc0
int = quadgk(b->dQ(b,g,σ), 0, Inf, atol=1e-3)
else ...
``````

You can calculate `rc0bc0gc0(σ)` over your range of `σ` from the top-level function, and determine which items satisfy `g > gc0`, in which case the integral doesn’t depend on `α` or `T`. That allows you to skip a bunch of redundant calculations, and skipping is much faster than even the fastest algorithm.

7 Likes

MWE (note that it’s possible to trim down the code to about a third the size of what you shared): https://gist.github.com/stillyslalom/6d44489240a85e8148c4cc5c65b305db

1 Like  Great solution!

I will replace the code … but, another question

what about my `interpolate` funcion ? the `.dat` that i have do save in my computer to use further ?
i will not use the

``````if first_time_using
mQ = [Qi(g,sigma) for g in ag, sigma in aσ]
writedlm("Qc_termo2020.dat",mQ)
else
``````
``````iQ = interpolate(mQ, BSpline(Cubic(Line())), OnGrid())    # Interpola na grade
sQ = scale(iQ, ag, aσ)
(x,y)->sQ[x,y]
``````

how can i substitute that you right in my own program to have save the file ?

The interpolation should be orders of magnitude faster than the initial calculation, so I didn’t spend any time optimizing it. `calcQb()` should do the same thing as `[Qd(g,sigma,α,T) for g in ag, sigma in aσ, α in aα, T in aT]`, so you can just replace that array comprehension with a function call.

So i used interp instead of calcQb() ?

like this ``````using  NLsolve, Plots, QuadGK, Interpolations, QuadGK, Roots, ProgressMeter

include("brent.jl")

aRe=  [0.5;1;1.5;2;2.5;3;3.5;4;4.5;5;5.5;6;6.5;7;7.5;8;8.5;9;9.5;10]
aTe=[0.1;0.2;0.3;0.4]

lenR = length(aRe)
aσ = 0.388./aRe

Qb = []     # Variável onde será atribuída a função Qb(g,α,sigma)

First_time_using = true

brent(f::F, a, b) where {F} = find_zero(f, (a, b), Roots.Brent())

@fastmath function Φ(r,σ)
return (2/15*σ.^9*(1/(r-1)^9-1/(r+1)^9-9/8/r*(1/(r-1)^8-1/(r+1)^8))-
σ.^3*(1/(r-1)^3-1/(r+1)^3-3/2/r*(1/(r-1)^2-1/(r+1)^2)))
end

@fastmath function Φl(r,σ)
σ3 = σ^3
σ9 = σ^9
((3*σ3)/(2*r^2*(1+r)^2)+(3*σ3)/(r*(1+r)^3)-(3*σ3)/(1+r)^4
-(3*σ9)/(20*r^2*(1+r)^8)
-(6*σ9)/(5*r*(1+r)^9)
+(6*σ9)/(5*(1+r)^10)
-(3*σ3)/((r-1)^3*r)
+(6*σ9)/(5*(r-1)^9*r)
-(3*σ3)/(2*(r-1)^2*r^2)
+(3*σ9)/(20*(r-1)^8*r^2)
+(3*σ3)/(r-1)^4-(6*σ9)/(5*(r-1)^10))
end

@fastmath function Φll(r,σ)
σ3 = σ^3
σ9 = σ^9
((-(3*σ3)/(r^3*(1+r)^2))-(6*σ3)/(r^2*(1+r)^3)
-(9*σ3)/(r*(1+r)^4)+(12*σ3)/(1+r)^5
+(3*σ9)/(10*r^3*(1+r)^8)
+(12*σ9)/(5*r^2*(1+r)^9)
+(54*σ9)/(5*r*(1+r)^10)
-(12*σ9)/(1+r)^11+(9*σ3)/((r-1)^4*r)
-(54*σ9)/(5*(r-1)^10*r)
+(6*σ3)/((r-1)^3*r^2)
-(12*σ9)/(5*(r-1)^9*r^2)
+(3*σ3)/((r-1)^2*r^3)
-(3*σ9)/(10*(r-1)^8*r^3)
-(12*σ3)/(r-1)^5+(12*σ9)/(r-1)^11)
end

@fastmath Veff(r,b,g,σ) = Φ(r,σ)+b^2*g^2/r^2
@fastmath Veffl(r,b,g,σ) = Φl(r,σ)-2b^2*g^2/r^3
@fastmath Veffll(r,b,g,σ) = Φll(r,σ)+6b^2*g^2/r^4

@fastmath function rc0bc0gc0(σ)
rc0 = brent(r->3Φl(r,σ)+r*Φll(r,σ),1.02,5)
gc0= sqrt(Φ(rc0,σ)+rc0*Φl(rc0,σ)/2)
bc0 = rc0*sqrt(gc0^2-Φ(rc0,σ))/gc0
rc0, bc0, gc0
end

@fastmath function frbc(g, σ)
rc0,bc0,gc0 = rc0bc0gc0(σ)
if g == gc0 return rc0, bc0 end
if g > gc0 || g == 0 error("g = \$g, there is no critical b") end
rc = rc0
f(r,σ) = Φ(r,σ)+r*Φl(r,σ)/2-g^2
while f(rc0,σ)*f(rc,σ)>0 rc = rc+1 end
rc = brent(r->f(r,σ),rc0,rc)
bc = sqrt(rc^2*(g^2-Φ(rc,σ))/g^2)
rc, bc
end

@fastmath function Rmin(b,g,σ)
if b > 100    # Este caso é quando Φ -> 0 e g^2r^2-b^2g^2+Φ=0
rmin = b
return rmin
end
rc0, bc0, gc0 = rc0bc0gc0(σ)
if g == gc0 return rc0 end    # É o caso de b e g críticos
if g > gc0        # Neste caso não tem extremos
rinicial = nextfloat(1.0)
rfinal = rinicial
while (g^2-Veff(rinicial,b,g,σ))*(g^2-Veff(rfinal,b,g,σ)) > 0 rfinal = rfinal+1 end
rmin = brent(r->g^2-Veff(r,b,g,σ),rinicial,rfinal)
return rmin
end
# Neste caso Veff tem máximo e mínimo e temos b crítico
rc, bc = frbc(g, σ)
if b == bc return rc end
if b < bc
rinicial = nextfloat(1.0)
rfinal = rinicial
while (g^2-Veff(rinicial,b,g,σ))*(g^2-Veff(rfinal,b,g,σ)) > 0 rfinal = rfinal+1 end
rmin = brent(r->g^2-Veff(r,b,g,σ),rinicial,rfinal)
else
rfinal = rc
while (g^2-Veff(rc,b,g,σ))*(g^2-Veff(rfinal,b,g,σ)) > 0 rfinal = rfinal+1 end
rmin = brent(r->g^2-Veff(r,b,g,σ),rc,rfinal)
end
rmin
end

@fastmath function dX(r,b,g,σ)
if g == 0 return 0 end
rmin = Rmin(b,g,σ)
den1 = g^2 - Veff(r,b,g,σ)
den2 = -Veffl(rmin,b,g,σ)*(r-rmin)
# if den1<0 || den2 < 0
#       warn("g^2-Veff = \$den1 -Veffl*(r-rmin) = \$den2, r = \$r, b = \$b, g = \$g")
#  end
den1 = abs(den1)
den2 = abs(den2)
# The lines below are
# 2b*g*/r^2(1/sqrt(den1)-1/sqrt(den2))
t0 = 2*b*g/r^2
t1 = 1/sqrt(den1)
t2 = 1/sqrt(den2)
if abs(t2/t1) < 0.5
res = t0*t1*exp(log1p(-t2/t1))
else
res = t0*(t1-t2)
end
res
end

@fastmath function X(b,g,σ)
if g > 1e5 return 0 end
if g == 0 || b == 0 return π end
rc0, bc0, gc0 = rc0bc0gc0(σ)
if b == bc0 && g == gc0 return -Inf end
if 0 < g <= gc0
rc, bc = frbc(g, σ)
if b == bc return -Inf end
end
rmin = Rmin(b,g,σ)
if b > 100 return 0 end
if g < 0.2 && b > bc+0.2
return -4π*σ^3/g^2/b^6
end
int = 2b*g*pi/(2rmin^(3/2)*sqrt(abs(Veffl(rmin,b,g,σ))))
try
int = int + quadgk(r->dX(r,b,g,σ), rmin, Inf, atol=1e-3)
catch
println("b=\$b, g=\$g")
int = int + quadgk(r->dX(r,b,g,σ), BigFloat(rmin), Inf, atol=1e-3)
end
pi-int
end

@fastmath function dQ(b,g,σ)
2*(1-cos(X(b,g,σ)))*b
end

@fastmath function dQd(b,g,σ,α,T)
2*(1+1/sqrt(1+α)*sqrt(π*T)/2g*sin(X(b,g,σ)/2))*b
end

@fastmath function Qd(g,σ,α,T)
if g > 1e5 return 0 end
if g == 0 return Inf end
rc0, bc0, gc0 = rc0bc0gc0(σ)
if g > gc0
int = quadgk(b->dQ(b,g,σ), 0, Inf, atol=1e-3)
else
rc, bc = frbc(g, σ)
int = quadgk(b->dQd(b,g,σ,α,T), 0, bc-1e-3, atol=1e-3)
int = int + quadgk(b->dQ(b,g,σ), bc+1e-3, Inf, atol=1e-3)
end
int
end

function interpQb()

ag = 0.01:0.05:3
aσ = 0.03 : 0.025 : 0.63
aα = 0:0.1:0.4
aT= 0.1:0.05:0.4

if First_time_using
mQ = [Qd(g,sigma,α,T) for g in ag, sigma in aσ, α in aα, T in aT]
writedlm("Qb_termo2020.dat",mQ)
else
mQ = reshape(mQ,(length(ag),length(aσ),length(aα),length(aT)))
end

# Matriz com os valores de Q na grade
iQ = interpolate(mQ, BSpline(Cubic(Line())), OnGrid())    # Interpola na grade
sQ = scale(iQ, ag, aσ, aα,aT)
(x,y,z,w) -> sQ[x,y,z,w]
end

time0 = time()
println("Generating  Qb(g,α,sigma,T) ")
Qb = interpQb()
tempod = (time() - time0)/60
println("Tempo decorrido: \$tempod min")
``````

calc(Qb) is not faster than interpolationQb, ok ?
but calc(Qb) does not generate a .dat file to save my calculations

I’m confused - what’s your question?

I will keep with

``````function interpQb()

ag = 0.01:0.05:3
aσ = 0.03 : 0.025 : 0.63
aα = 0:0.1:0.4
aT= 0.1:0.05:0.4

if First_time_using
mQ = [Qd(g,sigma,α,T) for g in ag, sigma in aσ, α in aα, T in aT]
writedlm("Qb_termo2020.dat",mQ)
else
mQ = reshape(mQ,(length(ag),length(aσ),length(aα),length(aT)))
end

# Matriz com os valores de Q na grade
iQ = interpolate(mQ, BSpline(Cubic(Line())), OnGrid())    # Interpola na grade
sQ = scale(iQ, ag, aσ, aα,aT)
(x,y,z,w) -> sQ[x,y,z,w]
end

time0 = time()
println("Generating  Qb(g,α,sigma,T) ")
Qb = interpQb()
tempod = (time() - time0)/60
println("Tempo decorrido: \$tempod min")
``````

and do not use this :

``````function calcQb()
ag = 0.01:0.05:3    # teste
aσ = 0.03 : 0.025 : 0.63
aα = 0:0.1:0.4
aT= 0.1:0.05:0.4
a = collect(Iterators.product(ag, aσ, aα, aT))
out = zeros(size(a))
pm = Progress(length(out))

out[i] = Qd(a[i]...)
next!(pm)
end

return out
end
``````

because in the `function calcQb()` you use more threads and do not save any file `.dat` for me.

Well, you wanted to speed things up, so I’d recommend sticking to `Threads.@threads`. You need to recognize that `calcQb()` does the same thing as `[Qd(g,sigma,α,T) for g in ag, sigma in aσ, α in aα, T in aT]`, so you can just replace the latter with a function call to the former, and keep the rest of `interpQb()` as is.

In the end of the program, how can i combine my text and yours, to use `calcQb()` to save a file `.dat` for me with all my calculations, to use it further?

I explained how to combine them in my previous answers. It’s analogous to going from

``````Q(a, b, c) = atan(a, b)/c

function f()
a = 0:.1:1
b = 2:.2:4
c = 4:.1:5

datpath = "abc.dat"
if ispath(datpath) # check if file already exists
abc = reshape(abc0, size(Iterators.product(a, b, c)))
else
abc = [Q(ai, bi, ci) for ai in a, bi in b, ci in c]
writedlm(datpath, abc)
end

iabc = interpolate(abc, BSpline(Cubic(Line(OnGrid()))))
sabc = scale(iabc, a, b, c)
(x, y, z) -> sabc[x, y, z]
end
``````

to

``````function Qthreaded(a, b, c)
abcrange = collect(Iterators.product(a, b, c))
out = zeros(size(abcrange))
out[i] = Q(abcrange[i]...)
end

return out
end

function g(a, b, c, datpath)
if ispath(datpath) # check if file already exists
abc = reshape(abc0, size(Iterators.product(a, b, c)))
else
writedlm(datpath, abc)
end

iabc = interpolate(abc, BSpline(Cubic(Line(OnGrid()))))
sabc = scale(iabc, a, b, c)
(x, y, z) -> sabc[x, y, z]
end
``````

I can show you how exactly to do it, but I can’t make you understand it - that’s your job. If there are parts you don’t understand, ask about them.

1 Like

Hello man … I cannot change my problem to adequate your solution.
Because i do not understand abc. Is abc = g sigma alfa ?
My problem is to evaluate a table of `Qd(g,sigma,α,T)` to integrate further.
But i try for hours to read your program changes and adequate to mine, using variables g, sigma, alfa and T for the values

``````     ag = 0.01:0.05:3    g in ag
aσ = 0.03 : 0.025 : 0.63  σ in aσ
aα = 0:0.1:0.4    α in aα
aT= 0.1:0.05:0.4  T in aT
``````

This command `````` if First_time_using
mQ = [Qd(g,sigma,α,T) for g in ag, sigma in aσ, α in aα, T in aT]
writedlm("Qb_termo2020.dat",mQ)
else
mQ = reshape(mQ,(length(ag),length(aσ),length(aα),length(aT)))
end
``````

and this :

``````time0 = time()
println("Generating  Qb(g,α,sigma,T) ")
Qb = interpQb()
tempod = (time() - time0)/60
println("Tempo decorrido: \$tempod min")
``````

Gives me the .dat file that i want …

but i cannot understand you g function, f function and Qthread function …

I cannot use this and place this news functions to solve my problem … Can you help me with that ?

Yes, `a, b, c` in my example are analogous to `ag, aσ, aα, aT` in your code. You’re using an array comprehension (like `[func(x, y) for x in X, y in Y]`) to fill in the elements of `Qb`, and comprehensions can’t currently be multithreaded. To allow threading, you have to think about what the loop comprehension is actually doing, and figure out how to do it with threads-compatible constructs like `for` loops. I’ll walk you through my thought process.

First, I recognized that an array comprehension can always be rewritten as a for loop. Let’s check:

``````julia> X, Y = 1:3, 10:30 # variables are just an example, nothing to do w/ your code
(1:3, 10:30)

julia> [x + y for x in X, y in Y] # using a comprehension
3×3 Array{Int64,2}:
11  21  31
12  22  32
13  23  33

julia> out = zeros(Int, length(X), length(Y)); # preallocate output array

julia> for i in eachindex(X) # using for loops
for j in eachindex(Y)
out[i, j] = x[i] + y[j]
end
end

julia> out
3×3 Array{Int64,2}:
11  21  31
12  22  32
13  23  33
``````

Then the question is how to make it multithreaded. I could just put a `Threads.@threads` on any one of the nested loops, but that could be inefficient if the number of threads is close (but not equal) to the number of elements, which might lead to some threads sitting idle while the others work. Instead, I went with `Iterators.product`, which allows you to get elements one-by-one, and also has the benefit of having the same output shape as an array comprehension.

``````julia> prodXY = Iterators.product(X, Y)
Base.Iterators.ProductIterator{Tuple{UnitRange{Int64},StepRange{Int64,Int64}}}((1:3, 10:10:30))

julia> size(prodXY)
(3, 3)

julia> for p in prodXY
@show p
end
p = (1, 10)
p = (2, 10)
p = (3, 10)
p = (1, 20)
p = (2, 20)
p = (3, 20)
p = (1, 30)
p = (2, 30)
p = (3, 30)
``````

Unfortunately, we can’t yet get individual elements from an iterator like we can with an array, so I had to use `collect` to turn `prodXY` into an array.

``````julia> prodXY[1, 1]
ERROR: MethodError: no method matching getindex(::Base.Iterators.ProductIterator{Tuple{UnitRange{Int64},StepRange{Int64,Int64}}}, ::Int64, ::Int64)

julia> cXY = collect(prodXY)
3×3 Array{Tuple{Int64,Int64},2}:
(1, 10)  (1, 20)  (1, 30)
(2, 10)  (2, 20)  (2, 30)
(3, 10)  (3, 20)  (3, 30)
``````
``````julia> Threads.@threads for i in eachindex(cXY)
x, y = cXY[i]
out[i] += x + y
end

julia> out
3×3 Array{Int64,2}:
22  42  62
24  44  64
26  46  66
``````

Great, it’s multithreaded! The same principle applies to your problem - first, create the `product` iterator and preallocate an output array.

``````    ag = 0.01:0.05:3    # teste
aσ = 0.03 : 0.025 : 0.63
aα = 0:0.1:0.4
aT= 0.1:0.05:0.4
a = collect(Iterators.product(ag, aσ, aα, aT))
mQ = zeros(size(a))
``````

Now use `Threads.@threads` to iterate over each element of the inputs, calling `Qd(g, σ, α, T)` and putting its output into your preallocated output array.

``````   Threads.@threads for i in eachindex(a)
g, σ, α, T = a[i]
mQ[i] = Qd(g, σ, α, T)
end
``````

Now that the output array is filled, you can save it to disk with `writedlm("Qb_termo2020.dat, mQ)`.

1 Like

Great. And in the final, how can I interpolate the terms ? to have a continuous values of Qb …
like this:

```````writedlm("Qb_termo2020.dat, mQ)` .
iQ = interpolate(mQ, BSpline(Cubic(Line())), OnGrid())
sQ = scale(iQ, ag, aσ, aα,aT)
(x,y,z,w) -> sQ[x,y,z,w]

end

time0 = time()
println("Generating Qb(g,α,sigma,T) ")
Qb = interpQb()
tempod = (time() - time0)/60
println("total time: \$tempod min")
``````

That part should be no different from your original function.