Calculating pi using iterations in ruby - ruby

For a school's assignment I am trying to calculate pi using the Gauss Legendre algorithm to test cpu efficiency.
Therefore, I have written a program in Ruby.
This program should iterate 500000000 times and display the the time used for it. But everytime it executes within a second.
My question:
Is there a better way to iterate so it really does repeat 500 million times and display pi and the time?
include Math
a = 1
b = 1/sqrt(2)
t = 0.25
p = 1
i = 0
imax = 500000000
start = Time.now
until i = imax
an = (a/2) + (b/2)
bn = sqrt(a) * sqrt(b)
tn = t - p * ((a-an) * (a-an))
pn = 2 * p
a = an
b = bn
t = tn
p = pn
i +=1
PI = ((a+b)*(a+b))/(4*t)
end
finish = Time.now
time = finish - start
puts PI
puts time

Start by not making i equal imax right away:
until i = imax
Should be
until i == imax
Even better, just do
500000000.times do
Instead of that line.

In addition to the issues raised by #Nick and #sawa your algorithm is flawed: the square root of the product of a and b is not equal to the product of the square roots of a and b.
In ruby:
include Math
a, b, t, p = 1, 1/sqrt(2), 0.25, 1
imax = 5
imax.times do |i|
an = (a+b) / 2
bn = sqrt(a * b)
tn = t - p * ((a-an) * (a-an))
pn = 2 * p
a, b, t, p = an, bn, tn, pn
pi = ((a+b)*(a+b))/(4*t)
printf "%d : %10.60f\n", i, pi
end
Running this gives me:
0 : 3.140579250522168575088244324433617293834686279296875000000000
1 : 3.141592646213542838751209274050779640674591064453125000000000
2 : 3.141592653589794004176383168669417500495910644531250000000000
3 : 3.141592653589794004176383168669417500495910644531250000000000
4 : 3.141592653589794004176383168669417500495910644531250000000000
So clearly you need more accuracy, hence BigDecimal. As this is your homework assignment I'll leave that up to you :-). (If unsure which variables to change, try all except i and imax. Also check out http://www.ruby-doc.org/stdlib-1.9.3/libdoc/bigdecimal/rdoc/BigDecimal.html)

Another thing you are doing wrong is assigning a constant PI within a loop. Although it is possible to reassign a constant, it is not correct to do so. Either use a variable or move the assignment to outside of the loop so that it would be assigned only once.
Even if I remove the assignment and print out the result for each iteration like this:
include Math
a = 1
b = 1/sqrt(2)
t = 0.25
p = 1
i = 0
imax = 500000000
until i == imax
an = (a/2) + (b/2)
bn = sqrt(a) * sqrt(b)
tn = t - p * ((a-an) * (a-an))
pn = 2 * p
a = an
b = bn
t = tn
p = pn
i +=1
puts ((a+b)*(a+b))/(4*t)
end
I get the wrong result. It goes like this:
-2.1244311544725596
-1.1383928808463357
-1.1265990444799223
-1.1265961703346379
-1.126596170334544
-1.126596170334544
... # very long repetition of the same number
-1.126596170334544
-1.126596170334544
NaN
NaN
... # NaN forever
Something must be wrong with your algorithm.

Related

Filling a matrix using parallel processing in Julia

I'm trying to speed up the solution time for a dynamic programming problem in Julia (v. 0.5.0), via parallel processing. The problem involves choosing the optimal values for every element of a 1073 x 19 matrix at every iteration, until successive matrix differences fall within a tolerance. I thought that, within each iteration, filling in the values for each element of the matrix could be parallelized. However, I'm seeing a huge performance degradation using SharedArray, and I'm wondering if there's a better way to approach parallel processing for this problem.
I construct the arguments for the function below:
est_params = [.788,.288,.0034,.1519,.1615,.0041,.0077,.2,0.005,.7196]
r = 0.015
tau = 0.35
rho =est_params[1]
sigma =est_params[2]
delta = 0.15
gamma =est_params[3]
a_capital =est_params[4]
lambda1 =est_params[5]
lambda2 =est_params[6]
s =est_params[7]
theta =est_params[8]
mu =est_params[9]
p_bar_k_ss =est_params[10]
beta = (1+r)^(-1)
sigma_range = 4
gz = 19
gp = 29
gk = 37
lnz=collect(linspace(-sigma_range*sigma,sigma_range*sigma,gz))
z=exp(lnz)
gk_m = fld(gk,2)
# Need to add mu somewhere to k_ss
k_ss = (theta*(1-tau)/(r+delta))^(1/(1-theta))
k=cat(1,map(i->k_ss*((1-delta)^i),collect(1:gk_m)),map(i->k_ss/((1-delta)^i),collect(1:gk_m)))
insert!(k,gk_m+1,k_ss)
sort!(k)
p_bar=p_bar_k_ss*k_ss
p = collect(linspace(-p_bar/2,p_bar,gp))
#Tauchen
N = length(z)
Z = zeros(N,1)
Zprob = zeros(Float32,N,N)
Z[N] = lnz[length(z)]
Z[1] = lnz[1]
zstep = (Z[N] - Z[1]) / (N - 1)
for i=2:(N-1)
Z[i] = Z[1] + zstep * (i - 1)
end
for a = 1 : N
for b = 1 : N
if b == 1
Zprob[a,b] = 0.5*erfc(-((Z[1] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2))
elseif b == N
Zprob[a,b] = 1 - 0.5*erfc(-((Z[N] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
else
Zprob[a,b] = 0.5*erfc(-((Z[b] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2)) -
0.5*erfc(-((Z[b] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
end
end
end
# Collecting tauchen results in a 2 element array of linspace and array; [2] gets array
# Zprob=collect(tauchen(gz, rho, sigma, mu, sigma_range))[2]
Zcumprob=zeros(Float32,gz,gz)
# 2 in cumsum! denotes the 2nd dimension, i.e. columns
cumsum!(Zcumprob, Zprob,2)
gm = gk * gp
control=zeros(gm,2)
for i=1:gk
control[(1+gp*(i-1)):(gp*i),1]=fill(k[i],(gp,1))
control[(1+gp*(i-1)):(gp*i),2]=p
end
endog=copy(control)
E=Array(Float32,gm,gm,gz)
for h=1:gm
for m=1:gm
for j=1:gz
# set the nonzero net debt indicator
if endog[h,2]<0
p_ind=1
else
p_ind=0
end
# set the investment indicator
if (control[m,1]-(1-delta)*endog[h,1])!=0
i_ind=1
else
i_ind=0
end
E[m,h,j] = (1-tau)*z[j]*(endog[h,1]^theta) + control[m,2]-endog[h,2]*(1+r*(1-tau)) +
delta*endog[h,1]*tau-(control[m,1]-(1-delta)*endog[h,1]) -
(i_ind*gamma*endog[h,1]+endog[h,1]*(a_capital/2)*(((control[m,1]-(1-delta)*endog[h,1])/endog[h,1])^2)) +
s*endog[h,2]*p_ind
elem = E[m,h,j]
if E[m,h,j]<0
E[m,h,j]=elem+lambda1*elem-.5*lambda2*elem^2
else
E[m,h,j]=elem
end
end
end
end
I then constructed the function with serial processing. The two for loops iterate through each element to find the largest value in a 1072-sized (=the gm scalar argument in the function) array:
function dynam_serial(E,gm,gz,beta,Zprob)
v = Array(Float32,gm,gz )
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
Tv = Array(Float32,gm,gz)
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1 # arbitrary initial value greater than convcrit
while diff>convcrit
exp_v=v*Zprob'
for h=1:gm
for j=1:gz
Tv[h,j]=findmax(E[:,h,j] + beta*exp_v[:,j])[1]
end
end
diff = maxabs(Tv - v)
v=copy(Tv)
end
end
Timing this, I get:
#time dynam_serial(E,gm,gz,beta,Zprob)
> 106.880008 seconds (91.70 M allocations: 203.233 GB, 15.22% gc time)
Now, I try using Shared Arrays to benefit from parallel processing. Note that I reconfigured the iteration so that I only have one for loop, rather than two. I also use v=deepcopy(Tv); otherwise, v is copied as an Array object, rather than a SharedArray:
function dynam_parallel(E,gm,gz,beta,Zprob)
v = SharedArray(Float32,(gm,gz),init = S -> S[Base.localindexes(S)] = myid() )
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1 # arbitrary initial value greater than convcrit
while diff>convcrit
exp_v=v*Zprob'
Tv = SharedArray(Float32,gm,gz,init = S -> S[Base.localindexes(S)] = myid() )
#sync #parallel for hj=1:(gm*gz)
j=cld(hj,gm)
h=mod(hj,gm)
if h==0;h=gm;end;
#async Tv[h,j]=findmax(E[:,h,j] + beta*exp_v[:,j])[1]
end
diff = maxabs(Tv - v)
v=deepcopy(Tv)
end
end
Timing the parallel version; and using a 4-core 2.5 GHz I7 processor with 16GB of memory, I get:
addprocs(3)
#time dynam_parallel(E,gm,gz,beta,Zprob)
> 164.237208 seconds (2.64 M allocations: 201.812 MB, 0.04% gc time)
Am I doing something incorrect here? Or is there a better way to approach parallel processing in Julia for this particular problem? I've considered using Distributed Arrays, but it's difficult for me to see how to apply them to the present problem.
UPDATE:
Per #DanGetz and his helpful comments, I turned instead to trying to speed up the serial processing version. I was able to get performance down to 53.469780 seconds (67.36 M allocations: 103.419 GiB, 19.12% gc time) through:
1) Upgrading to 0.6.0 (saved about 25 seconds), which includes the helpful #views macro.
2) Preallocating the main array I'm trying to fill in (Tv), per the section on Preallocating Outputs in the Julia Performance Tips: https://docs.julialang.org/en/latest/manual/performance-tips/. (saved another 25 or so seconds)
The biggest remaining slow-down seems to be coming from the add_vecs function, which sums together subarrays of two larger matrices. I've tried devectorizing and using BLAS functions, but haven't been able to produce better performance.
In any event, the improved code for dynam_serial is below:
function add_vecs(r::Array{Float32},h::Int,j::Int,E::Array{Float32},exp_v::Array{Float32},beta::Float32)
#views r=E[:,h,j] + beta*exp_v[:,j]
return r
end
function dynam_serial(E::Array{Float32},gm::Int,gz::Int,beta::Float32,Zprob::Array{Float32})
v = Array{Float32}(gm,gz)
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
Tv = Array{Float32}(gm,gz)
r = Array{Float32}(gm)
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1 # arbitrary initial value greater than convcrit
while diff>convcrit
exp_v=v*Zprob'
for h=1:gm
for j=1:gz
#views Tv[h,j]=findmax(add_vecs(r,h,j,E,exp_v,beta))[1]
end
end
diff = maximum(abs,Tv - v)
v=copy(Tv)
end
return Tv
end
If add_vecs seems to be the critical function, writing an explicit for loop could offer more optimization. How does the following benchmark:
function add_vecs!(r::Array{Float32},h::Int,j::Int,E::Array{Float32},
exp_v::Array{Float32},beta::Float32)
#inbounds for i=1:size(E,1)
r[i]=E[i,h,j] + beta*exp_v[i,j]
end
return r
end
UPDATE
To continue optimizing dynam_serial I have tried to remove more allocations. The result is:
function add_vecs_and_max!(gm::Int,r::Array{Float64},h::Int,j::Int,E::Array{Float64},
exp_v::Array{Float64},beta::Float64)
#inbounds for i=1:gm
r[i] = E[i,h,j]+beta*exp_v[i,j]
end
return findmax(r)[1]
end
function dynam_serial(E::Array{Float64},gm::Int,gz::Int,
beta::Float64,Zprob::Array{Float64})
v = Array{Float64}(gm,gz)
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
r = Array{Float64}(gm)
exp_v = Array{Float64}(gm,gz)
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1.0 # arbitrary initial value greater than convcrit
while diff>convcrit
A_mul_Bt!(exp_v,v,Zprob)
diff = -Inf
for h=1:gm
for j=1:gz
oldv = v[h,j]
newv = add_vecs_and_max!(gm,r,h,j,E,exp_v,beta)
v[h,j]= newv
diff = max(diff, oldv-newv, newv-oldv)
end
end
end
return v
end
Switching the functions to use Float64 should increase speed (as CPUs are inherently optimized for 64-bit word lengths). Also, using the mutating A_mul_Bt! directly saves another allocation. Avoiding the copy(...) by switching the arrays v and Tv.
How do these optimizations improve your running time?
2nd UPDATE
Updated the code in the UPDATE section to use findmax. Also, changed dynam_serial to use v without Tv, as there was no need to save the old version except for the diff calculation, which is now done inside the loop.
Here's the code I copied-and-pasted, provided by Dan Getz above. I include the array and scalar definitions exactly as I ran them. Performance was: 39.507005 seconds (11 allocations: 486.891 KiB) when running #time dynam_serial(E,gm,gz,beta,Zprob).
using SpecialFunctions
est_params = [.788,.288,.0034,.1519,.1615,.0041,.0077,.2,0.005,.7196]
r = 0.015
tau = 0.35
rho =est_params[1]
sigma =est_params[2]
delta = 0.15
gamma =est_params[3]
a_capital =est_params[4]
lambda1 =est_params[5]
lambda2 =est_params[6]
s =est_params[7]
theta =est_params[8]
mu =est_params[9]
p_bar_k_ss =est_params[10]
beta = (1+r)^(-1)
sigma_range = 4
gz = 19 #15 #19
gp = 29 #19 #29
gk = 37 #25 #37
lnz=collect(linspace(-sigma_range*sigma,sigma_range*sigma,gz))
z=exp.(lnz)
gk_m = fld(gk,2)
# Need to add mu somewhere to k_ss
k_ss = (theta*(1-tau)/(r+delta))^(1/(1-theta))
k=cat(1,map(i->k_ss*((1-delta)^i),collect(1:gk_m)),map(i->k_ss/((1-delta)^i),collect(1:gk_m)))
insert!(k,gk_m+1,k_ss)
sort!(k)
p_bar=p_bar_k_ss*k_ss
p = collect(linspace(-p_bar/2,p_bar,gp))
#Tauchen
N = length(z)
Z = zeros(N,1)
Zprob = zeros(Float64,N,N)
Z[N] = lnz[length(z)]
Z[1] = lnz[1]
zstep = (Z[N] - Z[1]) / (N - 1)
for i=2:(N-1)
Z[i] = Z[1] + zstep * (i - 1)
end
for a = 1 : N
for b = 1 : N
if b == 1
Zprob[a,b] = 0.5*erfc(-((Z[1] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2))
elseif b == N
Zprob[a,b] = 1 - 0.5*erfc(-((Z[N] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
else
Zprob[a,b] = 0.5*erfc(-((Z[b] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2)) -
0.5*erfc(-((Z[b] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
end
end
end
# Collecting tauchen results in a 2 element array of linspace and array; [2] gets array
# Zprob=collect(tauchen(gz, rho, sigma, mu, sigma_range))[2]
Zcumprob=zeros(Float64,gz,gz)
# 2 in cumsum! denotes the 2nd dimension, i.e. columns
cumsum!(Zcumprob, Zprob,2)
gm = gk * gp
control=zeros(gm,2)
for i=1:gk
control[(1+gp*(i-1)):(gp*i),1]=fill(k[i],(gp,1))
control[(1+gp*(i-1)):(gp*i),2]=p
end
endog=copy(control)
E=Array(Float64,gm,gm,gz)
for h=1:gm
for m=1:gm
for j=1:gz
# set the nonzero net debt indicator
if endog[h,2]<0
p_ind=1
else
p_ind=0
end
# set the investment indicator
if (control[m,1]-(1-delta)*endog[h,1])!=0
i_ind=1
else
i_ind=0
end
E[m,h,j] = (1-tau)*z[j]*(endog[h,1]^theta) + control[m,2]-endog[h,2]*(1+r*(1-tau)) +
delta*endog[h,1]*tau-(control[m,1]-(1-delta)*endog[h,1]) -
(i_ind*gamma*endog[h,1]+endog[h,1]*(a_capital/2)*(((control[m,1]-(1-delta)*endog[h,1])/endog[h,1])^2)) +
s*endog[h,2]*p_ind
elem = E[m,h,j]
if E[m,h,j]<0
E[m,h,j]=elem+lambda1*elem-.5*lambda2*elem^2
else
E[m,h,j]=elem
end
end
end
end
function add_vecs_and_max!(gm::Int,r::Array{Float64},h::Int,j::Int,E::Array{Float64},
exp_v::Array{Float64},beta::Float64)
maxr = -Inf
#inbounds for i=1:gm r[i] = E[i,h,j]+beta*exp_v[i,j]
maxr = max(r[i],maxr)
end
return maxr
end
function dynam_serial(E::Array{Float64},gm::Int,gz::Int,
beta::Float64,Zprob::Array{Float64})
v = Array{Float64}(gm,gz)
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
Tv = Array{Float64}(gm,gz)
r = Array{Float64}(gm)
exp_v = Array{Float64}(gm,gz)
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1.0 # arbitrary initial value greater than convcrit
while diff>convcrit
A_mul_Bt!(exp_v,v,Zprob)
diff = -Inf
for h=1:gm
for j=1:gz
Tv[h,j]=add_vecs_and_max!(gm,r,h,j,E,exp_v,beta)
diff = max(abs(Tv[h,j]-v[h,j]),diff)
end
end
(v,Tv)=(Tv,v)
end
return v
end
Now, here's another version of the algorithm and inputs. The functions are similar to what Dan Getz suggested, except that I use findmax rather than an iterated max function to find the array maximum. In the input construction, I am using both Float32 and mixing different bit-types together. However, I've consistently achieved better performance this way: 24.905569 seconds (1.81 k allocations: 46.829 MiB, 0.01% gc time). But it's not clear at all why.
using SpecialFunctions
est_params = [.788,.288,.0034,.1519,.1615,.0041,.0077,.2,0.005,.7196]
r = 0.015
tau = 0.35
rho =est_params[1]
sigma =est_params[2]
delta = 0.15
gamma =est_params[3]
a_capital =est_params[4]
lambda1 =est_params[5]
lambda2 =est_params[6]
s =est_params[7]
theta =est_params[8]
mu =est_params[9]
p_bar_k_ss =est_params[10]
beta = Float32((1+r)^(-1))
sigma_range = 4
gz = 19
gp = 29
gk = 37
lnz=collect(linspace(-sigma_range*sigma,sigma_range*sigma,gz))
z=exp(lnz)
gk_m = fld(gk,2)
# Need to add mu somewhere to k_ss
k_ss = (theta*(1-tau)/(r+delta))^(1/(1-theta))
k=cat(1,map(i->k_ss*((1-delta)^i),collect(1:gk_m)),map(i->k_ss/((1-delta)^i),collect(1:gk_m)))
insert!(k,gk_m+1,k_ss)
sort!(k)
p_bar=p_bar_k_ss*k_ss
p = collect(linspace(-p_bar/2,p_bar,gp))
#Tauchen
N = length(z)
Z = zeros(N,1)
Zprob = zeros(Float32,N,N)
Z[N] = lnz[length(z)]
Z[1] = lnz[1]
zstep = (Z[N] - Z[1]) / (N - 1)
for i=2:(N-1)
Z[i] = Z[1] + zstep * (i - 1)
end
for a = 1 : N
for b = 1 : N
if b == 1
Zprob[a,b] = 0.5*erfc(-((Z[1] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2))
elseif b == N
Zprob[a,b] = 1 - 0.5*erfc(-((Z[N] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
else
Zprob[a,b] = 0.5*erfc(-((Z[b] - mu - rho * Z[a] + zstep / 2) / sigma)/sqrt(2)) -
0.5*erfc(-((Z[b] - mu - rho * Z[a] - zstep / 2) / sigma)/sqrt(2))
end
end
end
# Collecting tauchen results in a 2 element array of linspace and array; [2] gets array
# Zprob=collect(tauchen(gz, rho, sigma, mu, sigma_range))[2]
Zcumprob=zeros(Float32,gz,gz)
# 2 in cumsum! denotes the 2nd dimension, i.e. columns
cumsum!(Zcumprob, Zprob,2)
gm = gk * gp
control=zeros(gm,2)
for i=1:gk
control[(1+gp*(i-1)):(gp*i),1]=fill(k[i],(gp,1))
control[(1+gp*(i-1)):(gp*i),2]=p
end
endog=copy(control)
E=Array(Float32,gm,gm,gz)
for h=1:gm
for m=1:gm
for j=1:gz
# set the nonzero net debt indicator
if endog[h,2]<0
p_ind=1
else
p_ind=0
end
# set the investment indicator
if (control[m,1]-(1-delta)*endog[h,1])!=0
i_ind=1
else
i_ind=0
end
E[m,h,j] = (1-tau)*z[j]*(endog[h,1]^theta) + control[m,2]-endog[h,2]*(1+r*(1-tau)) +
delta*endog[h,1]*tau-(control[m,1]-(1-delta)*endog[h,1]) -
(i_ind*gamma*endog[h,1]+endog[h,1]*(a_capital/2)*(((control[m,1]-(1-delta)*endog[h,1])/endog[h,1])^2)) +
s*endog[h,2]*p_ind
elem = E[m,h,j]
if E[m,h,j]<0
E[m,h,j]=elem+lambda1*elem-.5*lambda2*elem^2
else
E[m,h,j]=elem
end
end
end
end
function add_vecs!(gm::Int,r::Array{Float32},h::Int,j::Int,E::Array{Float32},
exp_v::Array{Float32},beta::Float32)
#inbounds #views for i=1:gm
r[i]=E[i,h,j] + beta*exp_v[i,j]
end
return r
end
function dynam_serial(E::Array{Float32},gm::Int,gz::Int,beta::Float32,Zprob::Array{Float32})
v = Array{Float32}(gm,gz)
fill!(v,E[cld(gm,2),cld(gm,2),cld(gz,2)])
Tv = Array{Float32}(gm,gz)
# Set parameters for the loop
convcrit = 0.0001 # chosen convergence criterion
diff = 1.00000 # arbitrary initial value greater than convcrit
iter=0
exp_v=Array{Float32}(gm,gz)
r=Array{Float32}(gm)
while diff>convcrit
A_mul_Bt!(exp_v,v,Zprob)
for h=1:gm
for j=1:gz
Tv[h,j]=findmax(add_vecs!(gm,r,h,j,E,exp_v,beta))[1]
end
end
diff = maximum(abs,Tv - v)
(v,Tv)=(Tv,v)
end
return v
end

Vectorized code slower than loops? MATLAB

In the problem Im working on there is such a part of code, as shown below. The definition part is just to show you the sizes of arrays. Below I pasted vectorized version - and it is >2x slower. Why it happens so? I know that i happens if vectorization requiers large temporary variables, but (it seems) it is not true here.
And generally, what (other than parfor, with I already use) can I do to speed up this code?
maxN = 100;
levels = maxN+1;
xElements = 101;
umn = complex(zeros(levels, levels));
umn2 = umn;
bessels = ones(xElements, xElements, levels); % 1.09 GB
posMcontainer = ones(xElements, xElements, maxN);
tic
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
nn = n + 1;
mm = 1;
for m = 1 : 2 : n
umn(nn, mm) = bessels(i, j, nn) * posMcontainer(i, j, m);
mm = mm + 1;
end
end
end
end
toc % 0.520594 seconds
tic
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
umn2(nn, 1:numOfEl) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
end
end
toc % 1.275926 seconds
sum(sum(umn-umn2)) % veryfying, if all done right
Best regards,
Alex
From the profiler:
Edit:
In reply to #Jason answer, this alternative takes the same time:
for n = 1:2:maxN
nn(n) = n + 1;
numOfEl(n) = ceil(n/2);
end
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
umn2(nn(n), 1:numOfEl(n)) = bessels(i, j, nn(n)) * posMcontainer(i, j, 1:2:n);
end
end
end
Edit2:
In reply to #EBH :
The point is to do the following:
parfor i = 1 : xElements
for j = 1 : xElements
umn = complex(zeros(levels, levels)); % cleaning
for n = 0:maxN
mm = 1;
for m = -n:2:n
nn = n + 1; % for indexing
if m < 0
umn(nn, mm) = bessels(i, j, nn) * negMcontainer(i, j, abs(m));
end
if m > 0
umn(nn, mm) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
if m == 0
umn(nn, mm) = bessels(i, j, nn);
end
mm = mm + 1; % for indexing
end % m
end % n
beta1 = sum(sum(Aj1.*umn));
betaSumSq1(i, j) = abs(beta1).^2;
beta2 = sum(sum(Aj2.*umn));
betaSumSq2(i, j) = abs(beta2).^2;
end % j
end % i
I speeded it up as much, as I was able to. What you have written is taking only the last bessels and posMcontainer values, so it does not produce the same result. In the real code, those two containers are filled not with 1, but with some precalculated values.
After your edit, I can see that umn is just a temporary variable for another calculation. It still can be mostly vectorizable:
betaSumSq1 = zeros(xElements); % preallocating
betaSumSq2 = zeros(xElements); % preallocating
% an index matrix to fetch the right values from negMcontainer and
% posMcontainer:
indmat = tril(repmat([0 1;1 0],ceil((maxN+1)/2),floor(levels/2)));
indmat(end,:) = [];
% an index matrix to fetch the values in correct order for umn:
b_ind = repmat([1;0],ceil((maxN+1)/2),1);
b_ind(end) = [];
tempind = logical([fliplr(indmat) b_ind indmat+triu(ones(size(indmat)))]);
% permute the arrays to prevent squeeze:
PM = permute(posMcontainer,[3 1 2]);
NM = permute(negMcontainer,[3 1 2]);
B = permute(bessels,[3 1 2]);
for k = 1 : maxN+1 % third dim
for jj = 1 : xElements % columns
b = B(:,jj,k); % get one vector of B
% perform b*NM for every row of NM*indmat, than flip the result:
neg = fliplr(bsxfun(#times,bsxfun(#times,indmat,NM(:,jj,k).'),b));
% perform b*PM for every row of PM*indmat:
pos = bsxfun(#times,bsxfun(#times,indmat,PM(:,jj,k).'),b);
temp = [neg mod(1:levels,2).'.*b pos].'; % concat neg and pos
% assign them to the right place in umn:
umn = reshape(temp(tempind.'),[levels levels]).';
beta1 = Aj1.*umn;
betaSumSq1(jj,k) = abs(sum(beta1(:))).^2;
beta2 = Aj2.*umn;
betaSumSq2(jj,k) = abs(sum(beta2(:))).^2;
end
end
This reduce running time from ~95 seconds to less 3 seconds (both without parfor), so it improves in almost 97%.
I would suspect it is memory allocation. You are re-allocating the m array in a 3 deep loop.
try rearranging the code:
tic
for n = 1 : 2 : maxN
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
for j = 1 : xElements
for i = 1 : xElements
umn2(nn, 1:numOfEl) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
end
end
toc % 1.275926 seconds
I was trying this in Igor pro, which a similar language, but with different optimizations. So the direct translations don't time the same way as Matlab (vectorized was slightly faster in Igor). But reordering the loops did speed up the vectorized form.
In your second part of the code, that is setting umn2, inside the loops, you have:
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
Those 3 lines don't require any input from the i and j loops, they only use the n loop. So reordering the loops such that i and j are inside the n loop will mean that those 3 lines are done xElements^2 (100^2) times less often. I suspect it is that m = 1:2:n line that takes time, since that is allocating an array.

speeding up some for loops in matlab

Basically I am trying to solve a 2nd order differential equation with the forward euler method. I have some for loops inside my code, which take considerable time to solve and I would like to speed things up a bit. Does anyone have any suggestions how could I do this?
And also when looking at the time it takes, I notice that my end at line 14 takes 45 % of my total time. What is end actually doing and why is it taking so much time?
Here is my simplified code:
t = 0:0.01:100;
dt = t(2)-t(1);
B = 3.5 * t;
F0 = 2 * t;
BB=zeros(1,length(t)); % Preallocation
x = 2; % Initial value
u = 0; % Initial value
for ii = 1:length(t)
for kk = 1:ii
BB(ii) = BB(ii) + B(kk) * u(ii-kk+1)*dt; % This line takes the most time
end % This end takes 45% of the other time
x(ii+1) = x(ii) + dt*u(ii);
u(ii+1) = u(ii) + dt * (F0(ii) - BB(ii));
end
Running the code it takes me 8.552 sec.
You can remove the inner loop, I think:
for ii = 1:length(t)
for kk = 1:ii
BB(ii) = BB(ii) + B(kk) * u(ii-kk+1)*dt; % This line takes the most time
end % This end takes 45% of the other time
x(ii+1) = x(ii) + dt*u(ii);
u(ii+1) = u(ii) + dt * (F0(ii) - BB(ii));
end
So BB(ii) = BB(ii) (zero at initalisation) + sum for 1 to ii of BB(kk)* u(ii-kk+1).dt
but kk = 1:ii, so for a given ii, ii-kk+1 → ii-(1:ii) + 1 → ii:-1:1
So I think this is equivalent to:
for ii = 1:length(t)
BB(ii) = sum(B(1:ii).*u(ii:-1:1)*dt);
x(ii+1) = x(ii) + dt*u(ii);
u(ii+1) = u(ii) + dt * (F0(ii) - BB(ii));
end
It doesn't take as long as 8 seconds for me using either method, but the version with only one loop is about 2x as fast (the output of BB appears to be the same).
Is the sum loop of B(kk) * u(ii-kk+1) just conv(B(1:ii),u(1:ii),'same')
The best way to speed up loops in matlab is to try to avoid them. Try if you are able to perform a matrix operation instead of the inner loop. For example try to break the calculation you do there in small parts, then decide, if there are parts you can perform in advance without knowing the results of the next iteration of the loop.
to your secound part of the question, my guess:: The end contains the check if the loop runs for another round and this check by it self is not that long but called 50.015.001 times!

Using matrix structure to speed up matlab

Suppose that I have an N-by-K matrix A, N-by-P matrix B. I want to do the following calculations to get my final N-by-P matrix X.
X(n,p) = B(n,p) - dot(gamma(p,:),A(n,:))
where
gamma(p,k) = dot(A(:,k),B(:,p))/sum( A(:,k).^2 )
In MATLAB, I have my code like
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
which are highly inefficient since it uses three for loops! Is there a good way to speed up this code?
Use bsxfun for the division and matrix multiplication for the loops:
gamma = bsxfun(#rdivide, B.'*A, sum(A.^2));
x = B - A*gamma.';
And here is a test script
N = 3;
K = 4;
P = 5;
A = rand(N, K);
B = rand(N, P);
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
gamma2 = bsxfun(#rdivide, B.'*A, sum(A.^2));
X2 = B - A*gamma2.';
isequal(x, X2)
isequal(gamma, gamma2)
which returns
ans =
1
ans =
1
It looks to me like you can hoist the gamma calculations out of the loop; at least, I don't see any dependencies on N in the gamma calculations.
So something like this:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
end
for p = 1:P
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
I'm not familiar enough with your code (or matlab) to really know if you can merge the two loops, but if you can:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
bxfun is slow...
How about something like the following (I might have a transpose wrong)
modA = A * (1./sum(A.^2,2)) * ones(1,k);
gamma = B' * modA;
x = B - A * gamma';

Matlab - Speeding up Nested For-Loops

I'm working on a function with three nested for loops that is way too slow for its intended use. The bottleneck is clearly the looping part - almost 100 % of the execution time is spent in the innermost loop.
The function takes a 2d matrix called rM as input and returns a 3d matrix called ec:
rows = size(rM, 1);
cols = size(rM, 2);
%preallocate.
ec = zeros(rows+1, cols, numRiskLevels);
ec(1, :, :) = 100;
for risk = minRisk:stepRisk:maxRisk;
for c = 1:cols,
for r = 2:rows+1,
ec(r, c, risk) = ec(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
Any help on speeding up the for loops would be appreciated...
The problem is, that the inner loop is slowest, while it is also near-impossible to vectorize. As every iteration directly depends on the previous one.
The outer two are possible:
clc;
rM = rand(50);
rows = size(rM, 1);
cols = size(rM, 2);
minRisk = 1;
stepRisk = 1;
maxRisk = 100;
numRiskLevels = maxRisk/stepRisk;
%preallocate.
ec = zeros(rows+1, cols, numRiskLevels);
ec(1, :, :) = 100;
riskArray = (minRisk:stepRisk:maxRisk)';
tic
for r = 2:rows+1
tmp = riskArray * rM(r-1, :);
tmp = permute(tmp, [3 2 1]);
ec(r, :, :) = ec(r-1, :, :) .* (1 + tmp);
end
toc
%preallocate.
ec2 = zeros(rows+1, cols, numRiskLevels);
ec2(1, :, :) = 100;
tic
for risk = minRisk:stepRisk:maxRisk;
for c = 1:cols
for r = 2:rows+1
ec2(r, c, risk) = ec2(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
toc
all(all(all(ec == ec2)))
But to my surprise, the vectorized code is indeed slower. (But maybe someone can improve the code, so I figured I leave it her for you.)
I have just tried to vectorize the outer loop, and actually noticed a significant speed increase. Of course it is hard to judge the speed of a script without knowing (the size of) the inputs but I would say this is a good starting point:
% Here you can change the input parameters
riskVec = 1:3:120;
rM = rand(50);
%preallocate and calculate non vectorized solution
ec2 = zeros(size(rM,2)+1, size(rM,1), max(riskVec));
ec2(1, :, :) = 100;
tic
for risk = riskVec
for c = 1:size(rM,2)
for r = 2:size(rM,1)+1
ec2(r, c, risk) = ec2(r-1, c, risk) * (1 + risk * rM(r-1, c));
end
end
end
t1=toc;
%preallocate and calculate vectorized solution
ec = zeros(size(rM,2)+1, size(rM,1), max(riskVec));
ec(1, :, :) = 100;
tic
for c = 1:size(rM,2)
for r = 2:size(rM,1)+1
ec(r, c, riskVec) = ec(r-1, c, riskVec) .* reshape(1 + riskVec * rM(r-1, c),[1 1 length(riskVec)]);
end
end
t2=toc;
% Check whether the vectorization is done correctly and show the timing results
if ec(:) == ec2(:)
t1
t2
end
The given output is:
t1 =
0.1288
t2 =
0.0408
So for this riskVec and rM it is about 3 times as fast as the non-vectorized solution.

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