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I'm newby in Matlab. I have took the work code with complex if-statement condition and need to rewrite it. This code should prepare some initial data to solve an optimization task. This if-statement condition looks like:
x=[784.8 959.2 468 572 279 341 139.5 170.5 76.5 93.5 45 55];
a=nchoosek(x,6); % all possible combinations from 6 elements of x
n=length(a);
q=[];
for i=1:n
if( ((a(i,1)==x(1)) & (a(i,2)==x(2))) |
((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) |
((a(i,1)==x(7)) & (a(i,2)==x(8))) |
((a(i,2)==x(3)) & (a(i,3)==x(4))) |
((a(i,2)==x(5)) & (a(i,3)==x(6))) |
((a(i,2)==x(7)) & (a(i,3)==x(8))) |
((a(i,3)==x(3)) & (a(i,4)==x(4))) |
((a(i,3)==x(5)) & (a(i,4)==x(6))) |
((a(i,3)==x(7)) & (a(i,4)==x(8))) |
((a(i,3)==x(9)) & (a(i,4)==x(10)))|
((a(i,4)==x(5)) & (a(i,5)==x(6))) |
((a(i,4)==x(7)) & (a(i,5)==x(8))) |
((a(i,4)==x(9)) & (a(i,5)==x(10)))|
((a(i,5)==x(5)) & (a(i,6)==x(6))) |
((a(i,5)==x(7)) & (a(i,6)==x(8))) |
((a(i,5)==x(9)) & (a(i,6)==x(10)) |
((a(i,5)==x(11)) & (a(i,6)==x(12)))))
q(i,:)=a(i,:);
end;
end;
q;
R1=a-q;
R1(~any(R1,2),:) = [];
R1(:, ~any(R1)) = [];
Question: Could anyone give an idea how to rewrite if-statement to improve readability of code?
If I understood you correctly, what the convoluted if statement basically says
If "x(1) x(2)" or "x(3) x(4)" or ... "x(11) x(12)" appears anywhere consecutively in row i
Think about it:
((a(i,1)==x(1)) & (a(i,2)==x(2))) | ((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) | ((a(i,1)==x(7)) & (a(i,2)==x(8)))
is no different from:
((a(i,1)==x(1)) & (a(i,2)==x(2))) | ((a(i,1)==x(3)) & (a(i,2)==x(4))) |
((a(i,1)==x(5)) & (a(i,2)==x(6))) | ((a(i,1)==x(7)) & (a(i,2)==x(8))) |
((a(i,1)==x(9)) & (a(i,2)==x(10))) | ((a(i,1)==x(11)) & (a(i,2)==x(12)))
since [x(9) x(10)] and [x(11) x(12)] will never appear at a(i, 1:2), so the line I added is always false and does not change the result of the chain of OR's. But if makes the logic much easier to understand. Same logic applies to a(i,2:3), a(i,3:4)..., complete those cases too and then you will get the first statement I made in this answer.
Then, instead of generating a directly from x, you should generate a from the INDEX of x, i.e. [1:12], as such:
a = nchoosek(1:length(x), 6);
Why? You said x consists of real numbers, and using == on real numbers does not guarantee success, and is a very bad practice in general.
Then, your target becomes:
find if sequence `[1 2]` or `[3 4]` or `[5 6]` ... exists in each line of `a`
which is equivalent to:
find if there is any odd number n followed by n+1
This logic can be represented as:
success = any (mod(a(:,1:end-1), 2) & diff(a,1,2)==1, 2)
Now success(i) will be true/false for the every a(i) that your statement evaluates to the same value. This method is better than your statement because it is very concise, automatically adapts to different sizes of x and does not need to run in a loop.
And if you want to get the actual combination of x values, just do
x(a(i)); % Get the ith permutation of x
x(a); % Get all permutation of x
x(a(success,:)); % Get all permutation of x that satisfy the requirement.
EDIT:
q = a; % q is basically a copy of a
q(~success,:) = 0; % except the `non-success` rows are zero
x(q) - x(a) % suppose q and a store index, this will give you the substraction.
I have two versions of code - I wrote the second (more explicit) loop when the first didn't do what I wanted.
Where did I go wrong? I suspect a slicing problem (as in, I'm not correctly slicing the data out)
The first version, which doesn't do what I want, is commented out above the loop:
rBool = false(h.numDirs, h.numTimes, h.numR);
for d = 1:h.numDirs
U_first = h.data(d,1,:);
U_first = U_first{1};
for t = 2:h.numTimes
U = h.data(d,t,:);
U = U{1};
dU = abs(U-U_first);
%rBool(d,t,:) = (dU > (smallVal*U_first) | rBool(d,t-1));
for r=1:h.numR
rBool(d,t,r) = (dU(r) > (smallVal*U_first(r))| rBool(d,t-1,r));
end
end
end
You are missing the third index of the second rBool in your commented line:
rBool(d,t,:) = (dU > (smallVal*U_first) | rBool(d,t-1,:));
Although I'd parenthesize it like this:
rBool(d,t,:) = (dU > (smallVal*U_first)) | rBool(d,t-1,:);
The version you originally had implicitly assumed r==1, I think.
And you can simplify your code by setting
U = h.data{d,t,1};
instead of cutting out a cell vector and choosing the first element.
What does the statement for if=ilow:ihigh mean in this program?
function [d]=for_taup(m,dt,h,q,N,flow,fhigh);
nt= max(size(m));
nh = max(size(h));
M = fft(m,[],1);
D = zeros(nt,nh);
i = sqrt(-1);
ilow = floor(flow*dt*nt)+1; if ilow<1; ilow=1;end;
ihigh = floor(fhigh*dt*nt)+1;
if ihigh>floor(nt/2)+1; ihigh=floor(nt/2)+1;end
for if=ilow:ihigh
f = 2.*pi*(if-1)/nt/dt;
L = exp(i*f*(h.^N)’*q);
x = M(if,:)’;
y = L * x;
D(if,:) = y’;
D(nt+2-if,:) = conj(y)’;
end
D(nt/2+1,:) = zeros(1,nh);
d = real(ifft(D,[],1));
return;
if is used as a variable name. I am surprised that this does not raise a syntax error: most languages would forbid the use of "reserved" keywords. Maybe it would be a good idea to replace if with a different name in order to clarify your code and avoid confusion.
As far as MATLAB is concerned, this code doesn't really mean anything, because it's just a syntax error. if is reserved keyword, and you can't create a variable called if. As such, it just instantly errors and won't run.
You should probably replace all occurrences of the variable if (although not the keyword if in lines 8 and 10) with some other variable name. Avoid i, since you're using that as the imaginary unit.
I stumbled across this problem while working with Jacket.
I use a compiled function (compiled with gcompile) within a gfor loop. This is meant to be supported as far as I know: http://wiki.accelereyes.com/wiki/index.php/GCOMPILE
But I observed that while the uncompiled function delivers the correct results, the compiled function gives the same output for all the gfor-iterations:
%================
% function[C] = test(A,B)
% C = A+B;
% end
%================
testing = gcompile('test.m');
A = gdouble(1:1:10);
B = gdouble(2:2:20);
C1 = gzeros(10,1);
C2 = gzeros(10,1);
gfor l=1:10
C1(l) = test(A(l),B(l));
C2(l) = testing(A(l),B(l));
gend
The output is:
C1 = [ 3,6,9,12,15,18,21,24,27,30]
(correct result)
C2 = [ 3,3,3,3,3,3,3,3,3,3]
Can you verify/rebut my results?
What am I doing wrong?
Cheers,
Angela
I was able to reproduce this behavior by running Jacket on MATLAB. It appears that gcompile does not work over GFOR as it should, and the documentation was in error. Sorry about that.
I need to simplify this Boolean expression with De Morgan's laws.
¬c xor (¬b ∨ c)
Could someone help me?
(accidentally made two accounts, so just responding with this one)
Ive found the best way to visualize a logic formula you do not understand is to make a table for it.
In the case of XOR, it represents One variable or another, but not both. So, lets make a table for A XOR B
A | B | Result
T | T | F *1
T | F | T *2
F | T | T *3
F | F | F *4
To generate the smallest possible result from the above table we can first take the most complex result that takes into account each option. Converting each line into a logical statement is fairly easy.
First, throw out anything that results in a False, Then take those that result in true, and convert them into a logical statement separated by 'OR's. In this case, 1 and 4 are false, and 2 and 3 are true. This means we only need to create logical statements for 2 and 3. I think how to do so would be best explained by example
Lets say X, Y, and Z are our variables, and the table gave us the following rows as true:
T | T | F - X & Y & ¬Z
F | T | F - ¬X & Y & ¬Z
F | F | F - ¬X & ¬Y & ¬Z
then to complete, we simply 'OR' them together
(X & Y & ¬Z) V (¬X & Y & ¬Z) V (¬X & ¬Y & ¬Z)
as you can see, where the variable is true, you put the variable directly in, and where it is false, you put a '¬' before the variable. The statement above basically says...
(True when X=T,Y=T,Z=F: False otherwise) OR (True when X=F,Y=T,Z=F: False otherwise) OR (True when X=F,Y=F,Z=F: False otherwise)
So finally bringing it back to our XOR the table rows are...
*2 A & ¬B
*3 ¬A & B
and are combined to be...
(A & ¬B) V (¬A & B)
So, now that you have an explanation of what to do with xor, you can apply this example to your problem, and come up with a logical statement you can use De Morgan's laws on to simplify.
first you have to split up xor into its basic form.
XOR represents A or B where A != B. If you can do that you should have more luck using demorgans on the whole equation