error loops, Gaussian elimination julia - system

the code below solves a linear system using Gauss method. when I run again, the error occurs:
LoadError: InexactError()
while loading In[176], in expression starting on line 4
in setindex! at array.jl:313
[inlined code] from In[176]:13
in anonymous at no file:12
Running more than 3 times the system is solved. what's up?
A = [4.1 -5 6.8;7.8 -8 -9;-17 4 1];
b = [1,2,3];
x = zeros(3);
m = 0;
al,ac = size(A)
for k= 1:(al-1)
#println("valor de k ",k)
for i = (k+1):al
#println("valor de i ",i)
m = A[i,k]/(A[k,k])
A[i,k] = 0
for j=(k+1):al
#println("valor de j",j)
A[i,j] = A[i,j] - m*A[k,j]
b[i]= b[i] - m*b[k]
end
end
end
x[al] = b[al]/(A[al,al])
for k = (al-1):-1:1
begin
s = 0;
for j = (k+1):al
s = s+A[k,j]*x[j]
end
x[k]=(b[k]-s)/A[k,k]
end
end
println(x)

The error is occurring mainly due to type mismatch between the variable and the value, consider the simple example,
julia> convert(Int64, 0.1)
ERROR: InexactError()
The InexactError() is thrown because it not possible to represent 0.1 as an integer.
julia>convert(Int64, 1.0)
1
As you can see the value 1.0 does not have problems being converted to Int64. So in your case since the type for array b is Array{Int64,1} and it sure cant hold floating point variables in due process and hence the error.
Considering #jeff 's inputs, it might be better to pakckthe code in a function,
julia>function f(A::Array{Float64,2}, b::Vector{Float64})
x = zeros(3);
m = 0.0;
al,ac = size(A)
for k= 1:(al-1)
#println("valor de k ",k)
for i = (k+1):al
#println("valor de i ",i)
m = A[i,k]/(A[k,k])
A[i,k] = 0
for j=(k+1):al
#println("valor de j",j)
A[i,j] = A[i,j] - m*A[k,j]
b[i]= b[i] - m*b[k]
end
end
end
x[al] = b[al]/(A[al,al])
for k = (al-1):-1:1
begin
s = 0;
for j = (k+1):al
s = s+A[k,j]*x[j]
end
x[k]=(b[k]-s)/A[k,k]
end
end
return x, A, b
end
f (generic function with 1 method)
julia> A = [4.1 -5 6.8;7.8 -8 -9;-17 4 1];
julia> b = [1,2,3.0];
This should result in the following transformation of matrix A into a upper triangular form,
julia> A
3×3 Array{Float64,2}:
4.1 -5.0 6.8
0.0 1.5122 -21.9366
0.0 0.0 -213.523

Related

Getting the charge of a single atom, per loop in MD Analysis

I have been trying to use the partial charge of one particular ion to go through a calculation within mdanalysis.
I have tried(This is just a snippet from the code that I know is throwing the error):
Cl = u.select_atoms('resname CLA and prop z <= 79.14')
Lz = 79.14 #Determined from system set-up
Q_sum = 0
COM = 38.42979431152344 #Determined from VMD
file_object1 = open(fors, 'a')
print(dcd, file = file_object1)
for ts in u.trajectory[200:]:
frame = u.trajectory.frame
time = u.trajectory.time
for coord in Cl.positions:
q= Cl.total_charge(Cl.position[coord][2])
coords = coord - (Lz/COM)
q_prof = q * (coords + (Lz / 2)) / Lz
Q_sum = Q_sum + q_prof
print(q)
But I keep getting an error associated with this.
How would I go about selecting this particular atom as it goes through the loop to get the charge of it in MD Analysis? Before I was setting q to equal a constant and the code ran fine so I know it is only this line that is throwing the error:
q = Cl.total_charge(Cl.position[coord][2])
Thanks for the help!
I figured it out with:
def Q_code(dcd, topo):
Lz = u.dimensions[2]
Q_sum = 0
count = 0
CLAs = u.select_atoms('segid IONS or segid PROA or segid PROB or segid MEMB')
ini_frames = -200
n_frames = len(u.trajectory[ini_frames:])
for ts in u.trajectory[ini_frames:]:
count += 1
membrane = u.select_atoms('segid PROA or segid PROB or segid MEMB')
COM = membrane.atoms.center_of_mass()[2]
q_prof = CLAs.atoms.charges * (CLAs.positions[:,2] + (Lz/2 - COM))/Lz
Q_instant = np.sum(q_prof)
Q_sum += Q_instant
Q_av = Q_sum / n_frames
with open('Q_av.txt', 'a') as f:
print('The Q_av for {} is {}'.format(s, Q_av), file = f)
return Q_av

Division by Zero error in calculating series

I am trying to compute a series, and I am running into an issue that I don't know why is occurring.
"RuntimeWarning: divide by zero encountered in double_scalars"
When I checked the code, it didn't seem to have any singularities, so I am confused. Here is the code currently(log stands for natural logarithm)(edit: extending code if that helps):
from numpy import pi, log
#Create functions to calculate the sums
def phi(z: int):
k = 0
phi = 0
#Loop through 1000 times to try to approximate the series value as if it went to infinity
while k <= 100:
phi += ((1/(k+1)) - (1/(k+(2*z))))
k += 1
return phi
def psi(z: int):
psi = 0
k = 1
while k <= 101:
psi += ((log(k))/( k**(2*z)))
k += 1
return psi
def sig(z: int):
sig = 0
k = 1
while k <= 101:
sig += ((log(k))**2)/(k^(2*z))
k += 1
return sig
def beta(z: int):
beta = 0
k = 1
while k <= 101:
beta += (1/(((2*z)+k)^2))
k += 1
return beta
#Create the formula to approximate the value. For higher accuracy, either calculate more derivatives of Bernoulli numbers or increase the boundry of k.
def Bern(z :int):
#Define Euler–Mascheroni constant
c = 0.577215664901532860606512
#Begin computations (only approximation)
B = (pi/6) * (phi(1) - c - 2 * log(2 * pi) - 1) - z * ((pi/6) * ((phi(1)- c - (2 * log(2 * pi)) - 1) * (phi(1) - c) + beta(1) - 2 * psi(1)) - 2 * (psi(1) * (phi(1) - c) + sig(1) + 2 * psi(1) * log(2 * pi)))
#output
return B
A = int(input("Choose any value: "))
print("The answer is", Bern(A + 1))
Any help would be much appreciated.
are you sure you need a ^ bitwise exclusive or operator instead of **? I've tried to run your code with input parameter z = 1. And on a second iteration the result of k^(2*z) was equal to 0, so where is from zero division error come from (2^2*1 = 0).

Understanding issue in Scilab for my code

I am doing a projectile motion where i need to plot curves between position x and y for various angles but the scilab shows only one plot. I am confused.
My code below
function[H,R,T]=projectile(m,r,h,c,rho,theta,v0,x0,y0,t0)
g=9.8
A=%pi*r^2
k=c*rho*A/2;
i=1
t(i)=t0
x(i)=x0
y(i)=y0
for j=0:5
thetha=theta+j*15;
vx(i)=v0*cos(thetha*%pi/180);
vy(i)=v0*sin(thetha*%pi/180);
while (y(i)>=0)
v=sqrt(vx(i)^2+vy(i)^2);
t(i+1)=t(i)+h;
vx(i+1)=vx(i)-h*(k*v*vx(i)/m);
vy(i+1)=vy(i)-h*(g+k*v*vy(i)/m);
x(i+1)=x(i)+h*vx(i)
y(i+1)=y(i)+h*vy(i)
i=i+1;
end
plot(x(i),y(i),'.');
end
n=i-1
R=x(n)-x(1);
T=t(n);
H=max(y)
endfunction
You should use vectors to improve compacity and readability of your code. Here is my proposition of improved (and working) code:
function [H,R,T] = projectile(m,r,h,c,rho,theta0,v0,x0,y0,t0)
g = 9.81
A = %pi*r^2
k = c*rho*A/2;
for theta = theta0 + (0:15:75)
v = v0*[cos(theta*%pi/180); sin(theta*%pi/180)];
t = t0
xy = [x0;y0]
i = 1
while xy(2,i) >= 0
t(i+1) = t(i)+h;
v = v + h*([0;-g] - k*norm(v)*v/m);
xy(:,i+1) = xy(:,i) + h*v;
i = i+1;
end
plot(xy(1,:), xy(2,:));
end
R = xy(1,$) - xy(1,1);
T = t($);
H = max(xy(2,:))
endfunction
clf
[H,R,T] = projectile(1,0.1,0.001,2,1000,5,1,0,0,0)

Trying to do a gaussian bell in Scilab

I'm trying to do a Gaussian bell using the data I am obtaining from a matrix but everytime I try to run the program I obtain this message:
"Error: syntax error, unexpected identifier, expecting end"
The data used to obtain the gaussina bell is a matrix which includes the last point of every n displacements, which are the last position of a particle. I want to know if there is an easier way to obtain the gaussian bell in scilab because I have to also do a fit with an histogram using the same data.
function bla7()
t=4000
n=1000
l=0.067
p=%pi*2
w1=zeros(t,1);
w2=zeros(t,1);
for I=1:t
a=(grand(n,1,"unf",0,p));
x=l*cos(a)
y=l*sin(a)
z1=zeros(n,1);
z2=zeros(n,1);
for i=2:n
z1(i)=z1(i-1)+x(i);
z2(i)=z2(i-1)+y(i);
end
w1(I)=z1($)
w2(I)=z2($)
end
n=10000
w10=zeros(t,1);
w20=zeros(t,1);
for I=1:t
a=(grand(n,1,"unf",0,p));
x=l*cos(a)
y=l*sin(a)
z1=zeros(n,1);
z2=zeros(n,1);
for i=2:n
z1(i)=z1(i-1)+x(i);
z2(i)=z2(i-1)+y(i);
end
w10(I)=z1($)
w20(I)=z2($)
end
n=100
w100=zeros(t,1);
w200=zeros(t,1);
for I=1:t
a=(grand(n,1,"unf",0,p));
x=l*cos(a)
y=l*sin(a)
z1=zeros(n,1);
z2=zeros(n,1);
for i=2:n
z1(i)=z1(i-1)+x(i);
z2(i)=z2(i-1)+y(i);
end
w100(I)=z1($)
w200(I)=z2($)
end
k=70
v=12/k
c1=zeros(k,1)
for r=1:t
c=w1(r)
m=-6+v
n=-6
for g=1:k
if (c<m & c>=n) then
c1(g)=c1(g)+1
m=m+v
n=n+v
else
m=m+v
n=n+v
end
end
end
c2=zeros(k,1)
c2(1)=-6+(6/k)
for b=2:k
c2(b)=c2(b-1)+v
end
y = stdev(w1)
normal1=zeros(k,1)
normal2=zeros(k,1)
bb=-6
bc=-6+v
for wa=1:k
bd=(bb+bc)/2
gauss1=(1/(y*sqrt(2*%pi)))exp(-0.5(bb/y)^2)
gauss2=(1/(y*sqrt(2*%pi)))exp(-0.5(bc/y)^2)
gauss3=(1/(y*sqrt(2*%pi)))exp(-0.5(bd/y)^2)
gauss4=((bc-bb)/6)*(gauss1+gauss2+4*gauss3)
bb=bb+v
bc=bc+v
normal2(wa,1)=gauss4
end
normal3=normal2*4000
k=100
v=24/k
c10=zeros(k,1)
for r=1:t
c=w10(r)
m=-12+v
n=-12
for g=1:k
if (c<m & c>=n) then
c10(g)=c10(g)+1
m=m+v
n=n+v
else
m=m+v
n=n+v
end
end
end
c20=zeros(k,1)
c20(1)=-12+(12/k)
for b=2:k
c20(b)=c20(b-1)+v
end
y = stdev(w10)
normal10=zeros(k,1)
normal20=zeros(k,1)
bb=-12
bc=-12+v
for wa=1:k
bd=(bb+bc)/2
gauss10=(1/(y*sqrt(2*%pi)))exp(-0.5(bb/y)^2)
gauss20=(1/(y*sqrt(2*%pi)))exp(-0.5(bc/y)^2)
gauss30=(1/(y*sqrt(2*%pi)))exp(-0.5(bd/y)^2)
gauss40=((bc-bb)/6)*(gauss10+gauss20+4*gauss30)
bb=bb+v
bc=bc+v
normal20(wa,1)=gauss40
end
normal30=normal20*4000
k=70
v=12/k
c100=zeros(k,1)
for r=1:t
c=w100(r)
m=-6+v
n=-6
for g=1:k
if (c<m & c>=n) then
c100(g)=c100(g)+1
m=m+v
n=n+v
else
m=m+v
n=n+v
end
end
end
c200=zeros(k,1)
c200(1)=-6+(6/k)
for b=2:k
c200(b)=c200(b-1)+v
end
y = stdev(w100)
normal100=zeros(k,1)
normal200=zeros(k,1)
bb=-6
bc=-6+v
for wa=1:k
bd=(bb+bc)/2
gauss100=(1/(y*sqrt(2*%pi)))exp(-0.5(bb/y)^2)
gauss200=(1/(y*sqrt(2*%pi)))exp(-0.5(bc/y)^2)
gauss300=(1/(y*sqrt(2*%pi)))exp(-0.5(bd/y)^2)
gauss400=((bc-bb)/6)*(gauss100+gauss200+4*gauss300)
bb=bb+v
bc=bc+v
normal200(wa,1)=gauss400
end
normal300=normal200*4000
bar(c20,c10,1.0,'white')
plot(c20, normal30, 'b-')
bar(c2,c1,1.0,'white')
plot(c2, normal3, 'r-')
bar(c200,c100,1.0,'white')
plot(c200, normal300, 'm-')
poly1.thickness=3;
xlabel(["x / um"]);
ylabel("molecules");
gcf().axes_size=[500,500]
a=gca();
a.zoom_box=[-12,12;0,600];
a.font_size=4;
a.labels_font_size=5;
a.x_label.font_size = 5;
a.y_label.font_size = 5;
ticks = a.x_ticks
ticks.labels =["-12";"-10";"-8";"-6";"-4";"-2";"0";"2";"4";"6";"8";"10";"12"]
ticks.locations = [-12;-10;-8;-6;-4;-2;0;2;4;6;8;10;12]
a.x_ticks = ticks
endfunction
Each and every one of your gauss variables are missing the multiplication operator in two places. Check every line at it will run. For example, this:
gauss1=(1/(y*sqrt(2*%pi)))exp(-0.5(bb/y)^2)
should be this:
gauss1=(1/(y*sqrt(2*%pi))) * exp(-0.5 * (bb/y)^2)
As for the Gaussian bell, there is no standard function in Scilab. However, you could define a new function to make things more clear in your case:
function x = myGauss(s,b_)
x = (1/(s*sqrt(2*%pi)))*exp(-0.5*(b_/s)^2)
endfunction
Actually, while we're at it, your whole code is really difficult to read. You should define functions instead of repeating code: it helps clarify what you mean, and if there is a mistake, you need to fix only one place. Also, I personally do not recommend that you enclose everything in a function like bla7() because it makes things harder to debug. Your example could be rewritten like this:
The myGauss function;
A function w_ to calculate w1, w2, w10, w20, w100 and w200;
A function c_ to calculate c1, c2, c10, c20, c100 and c200;
A function normal_ to calculate normal1, normal2, normal10, normal20, normal100 and normal200;
Call all four functions as many times as needed with different inputs for different results.
If you do that, your could will look like this:
function x = myGauss(s,b_)
x = (1 / (s * sqrt(2 * %pi))) * exp(-0.5 * (b_/s)^2);
endfunction
function [w1_,w2_] = w_(t_,l_,n_,p_)
w1_ = zeros(t_,1);
w2_ = zeros(t_,1);
for I = 1 : t_
a = (grand(n_,1,"unf",0,p_));
x = l_ * cos(a);
y = l_ * sin(a);
z1 = zeros(n_,1);
z2 = zeros(n_,1);
for i = 2 : n_
z1(i) = z1(i-1) + x(i);
z2(i) = z2(i-1) + y(i);
end
w1_(I) = z1($);
w2_(I) = z2($);
end
endfunction
function [c1_,c2_] = c_(t_,k_,v_,w1_,x_)
c1_ = zeros(k_,1)
for r = 1 : t_
c = w1_(r);
m = -x_ + v_;
n = -x_;
for g = 1 : k_
if (c < m & c >= n) then
c1_(g) = c1_(g) + 1;
m = m + v_;
n = n + v_;
else
m = m + v_;
n = n + v_;
end
end
end
c2_ = zeros(k_,1);
c2_(1) = -x_ + (x_/k_);
for b = 2 : k_
c2_(b) = c2_(b-1) + v_;
end
endfunction
function [normal1_,normal2_,normal3_] = normal_(k_,bb_,bc_,v_,w1_)
y = stdev(w1_);
normal1_ = zeros(k_,1);
normal2_ = zeros(k_,1);
for wa = 1 : k_
bd_ = (bb_ + bc_) / 2;
gauss1 = myGauss(y,bb_);
gauss2 = myGauss(y,bc_);
gauss3 = myGauss(y,bd_);
gauss4 = ((bc_ - bb_) / 6) * (gauss1 + gauss2 + 4 * gauss3);
bb_ = bb_ + v_;
bc_ = bc_ + v_;
normal2_(wa,1) = gauss4;
end
normal3_ = normal2_ * 4000;
endfunction
t = 4000;
l = 0.067;
p = 2 * %pi;
n = 1000;
k = 70;
v = 12 / k;
x = 6;
bb = -x;
bc = -x + v;
[w1,w2] = w_(t,l,n,p);
[c1,c2] = c_(t,k,v,w1,x);
[normal1,normal2,normal3] = normal_(k,bb,bc,v,w1);
bar(c2,c1,1.0,'white');
plot(c2, normal3, 'r-');
n = 10000;
k = 100;
v = 24 / k;
x = 12;
bb = -x;
bc = -x + v;
[w10,w20] = w_(t,l,n,p);
[c10,c20] = c_(t,k,v,w10,x);
[normal10,normal20,normal30] = normal_(k,bb,bc,v,w10);
bar(c20,c10,1.0,'white');
plot(c20, normal30, 'b-');
n = 100;
k = 70;
v = 12 / k;
x = 6;
bb = -x;
bc = -x + v;
[w100,w200] = w_(t,l,n,p);
[c100,c200] = c_(t,k,v,w100,x);
[normal100,normal200,normal300] = normal_(k,bb,bc,v,w100);
bar(c200,c100,1.0,'white');
plot(c200, normal300, 'm-');
poly1.thickness=3;
xlabel(["x / um"]);
ylabel("molecules");
gcf().axes_size=[500,500]
a=gca();
a.zoom_box=[-12,12;0,600];
a.font_size=4;
a.labels_font_size=5;
a.x_label.font_size = 5;
a.y_label.font_size = 5;
ticks = a.x_ticks
ticks.labels =["-12";"-10";"-8";"-6";"-4";"-2";"0";"2";"4";"6";"8";"10";"12"]
ticks.locations = [-12;-10;-8;-6;-4;-2;0;2;4;6;8;10;12]
a.x_ticks = ticks

Index was outside the bounds of the array VB

This is my code and when I run it appears index was outside the bounds of the array in line b = Asc(y(j + m)). I have tried Try and Catch and it didn't work out.
Public Function SMITH(x, y, SX, SY)
Dim a, b, j As Integer
result = 0
m = x.Length
n = y.Length
preBmBc(x)
preQsBc(x)
j = 0
While (j <= (n - m))
If (SX = SY.ToString.Substring(j, m)) Then
result = 1
End If
a = Asc(y(j + (m - 1)))
b = Asc(y(j + m))
j = j + Math.Max(bmBc(a), qsBc(b))
End While
Return result
End Function
Have you tried making m = "m-1" for b also? You really need to step through the code using breakpoints and the debugger to figure out when the program throws the OutOfRangeException.