%function [flag] =verify(area)
[FileName,PathName,FilterIndex]= uigetfile('*.tif','Select the signature file');
display(PathName)
m=[PathName,FileName];
area=nor_area(m);
%display(area)
%area=0.8707;
class(area)
flag=0;
extract=xlsread('D:\Project\Image_processing\important\best.xlsx', 'CW4:CW17');
c=numel(extract);
display(c)
l=extract(1);
class(l)
display(l)
for k = 1:c
%x=extract(k);
if (l==area && flag==0)
% display(extract(k));
flag=1;
display(flag)
end
end
display(flag)
The above is my code for verification, i am not able to compare "l==area", even if the values are same am not able to enter inside the loop. If i try passing the value assume l=0.9999 and the area that i obtain to be the same , if i sent l value explicitly it works..!! but if i try using some function and pass the same value it wont work. I have tried checking the type by using class, both returns double.
Can anyone please help me out with this and if this approach is not good, suggest any alternative that may be used.
It is not generally a good idea to compare floats like you are doing (with the == operator) since floats, unlike integer values are subject to round off. See here and here for a discussion on comparing floats in MATLAB.
Essentially you have to check that two floats are 'close enough' rather than exactly equal, which is what == checks for. MATLAB has a built in function eps for determining the floating point precision on your machine, so use that function when comparing floats. See its documentation for more information.
In most cases it is not wise to compare floating point numbers by a == b. Use abs(a-b)<epsilon where epsilonis some small tolerance like 1e-10 instead.
Related
The value is absolute integer, not a floating point to be doubted, also, it is not about an overflow since a double value can hold until 2^1024.
fprintf('%f',realmax)
179769313486231570000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
The problem I am facing in nchoosek function that it doesn't produce exact values
fprintf('%f\n',nchoosek(55,24));
2488589544741302.000000
While it is a percentage error of 2 regarding that binomian(n,m)=binomial(n-1,m)+binomial(n-1,m-1) as follows
fprintf('%f',nchoosek(55-1,24)+nchoosek(55-1,24-1))
2488589544741301.000000
Ps: The exact value is 2488589544741300
this demo shows
What is wrong with MATLAB?
Your understanding of the realmax function is wrong. It's the maximum value which can be stored, but with such large numbers you have a floating point precision error far above 1. The first integer which can not be stored in a double value is 2^53+1, try 2^53==2^53+1 for a simple example.
If the symbolic toolbox is available, the easiest to implement solution is using it:
>> nchoosek(sym(55),sym(24))
ans =
2488589544741300
There is a difference between something that looks like an integer (55) and something that's actually an integer (in terms of variable type).
The way you're calculating it, your values are stored as floating point (which is what realmax is pointing you to - the largest positive floating point number - check intmax('int64') for the largest possible integer value), so you can get floating point errors. An absolute difference of 2 in a large value is not that unexpected - the actual percentage error is tiny.
Plus, you're using %f in your format string - e.g. asking it to display as floating point.
For nchoosek specifically, from the docs, the output is returned as a nonnegative scalar value, of the same type as inputs n and k, or, if they are different types, of the non-double type (you can only have different input types if one is a double).
In Matlab, when you type a number directly into a function input, it generally defaults to a float. You have to force it to be an integer.
Try instead:
fprintf('%d\n',nchoosek(int64(55),int64(24)));
Note: %d not %f, converting both inputs to specifically integer. The output of nchoosek here should be of type int64.
I don't have access to MATLAB, but since you're obviously okay working with Octave I'll post my observations based on that.
If you look at the Octave source code using edit nchoosek or here you'll see that the equation for calculating the binomial coefficient is quite simple:
A = round (prod ((v-k+1:v)./(1:k)));
As you can see, there are k divisions, each with the possibility of introducing some small error. The next line attempts to be helpful and warn you of the possibility of loss of precision:
if (A*2*k*eps >= 0.5)
warning ("nchoosek", "nchoosek: possible loss of precision");
So, if I may slightly modify your final question, what is wrong with Octave? I would say nothing is wrong. The authors obviously knew of the possibility of imprecision and included a check to warn users when that possibility arises. So the function is working as intended. If you require greater precision for your application than the built-in function provides, it looks as though you'll need to code (or find) something that calculates the intermediate results with greater precision.
I am trying to solve a non-linear system of equations using the Newton-Raphson iterative method, and in order to explore the parameter space of my variables, it is useful to store the previous solutions and use them as my first initial guess so that I stay in the basin of attraction.
I currently save my solutions in a structure array that I store in a .mat file, in about this way:
load('solutions.mat','sol');
str = struct('a',Param1,'b',Param2,'solution',SolutionVector);
sol=[sol;str];
save('solutions.mat','sol');
Now, I do another run, in which I need the above solution for different parameters NewParam1 and NewParam2. If Param1 = NewParam1-deltaParam1, and Param2 = NewParam2 - deltaParam2, then
load('solutions.mat','sol');
index = [sol.a]== NewParam1 - deltaParam1 & [sol.b]== NewParam2 - deltaParam2;
% logical index to find solution from first block
SolutionVector = sol(index).solution;
I sometimes get an error message saying that no such solution exists. The problem lies in the double precisions of my parameters, since 2-1 ~= 1 can happen in Matlab, but I can't seem to find an alternative way to achieve the same result. I have tried changing the numerical parameters to strings in the saving process, but then I ran into problems with logical indexing with strings.
Ideally, I would like to avoid multiplying my parameters by a power of 10 to make them integers as this will make the code quite messy to understand due to the number of parameters. Other than that, any help will be greatly appreciated. Thanks!
You should never use == when comparing double precision numbers in MATLAB. The reason is, as you state in the the question, that some numbers can't be represented precisely using binary numbers the same way 1/3 can't be written precisely using decimal numbers.
What you should do is something like this:
index = abs([sol.a] - (NewParam1 - deltaParam1)) < 1e-10 & ...
abs([sol.b] - (NewParam2 - deltaParam2)) < 1e-10;
I actually recommend not using eps, as it's so small that it might actually fail in some situations. You can however use a smaller number than 1e-10 if you need a very high level of accuracy (but how often do we work with numbers less than 1e-10)?
I am using a while loop with an index t starting from 1 and increasing with each loop.
I'm having problems with this index in the following bit of code within the loop:
dt = 100000^(-1);
t = 1;
equi = false;
while equi==false
***some code that populates the arrays S(t) and I(t)***
t=t+1;
if (t>2/dt)
n = [S(t) I(t)];
np = [S(t-1/dt) I(t-1/dt)];
if sum((n-np).^2)<1e-5
equi=true;
end
end
First, the code in the "if" statement is accessed at t==200000 instead of at t==200001.
Second, the expression S(t-1/dt) results in the error message "Subscript indices must either be real positive integers or logicals", even though (t-1/dt) is whole and equals 1.0000e+005 .
I guess I can solve this using "round", but this worked before and suddenly doesn't work and I'd like to figure out why.
Thanks!
the expression S(t-1/dt) results in the error message "Subscript indices must either be real positive integers or logicals", even though (t-1/dt) is whole and equals 1.0000e+005
Is it really? ;)
mod(200000 - 1/dt, 1)
%ans = 1.455191522836685e-11
Your index is not an integer. This is one of the things to be aware of when working with floating point arithmetic. I suggest reading this excellent resource: "What every computer scientist should know about floating-point Arithmetic".
You can either use round as you did, or store 1/dt as a separate variable (many options exist).
Matlab is lying to you. You're running into floating point inaccuracies and Matlab does not have an honest printing policy. Try printing the numbers with full precision:
dt = 100000^(-1);
t = 200000;
fprintf('2/dt == %.12f\n',2/dt) % 199999.999999999971
fprintf('t - 1/dt == %.12f\n',t - 1/dt) % 100000.000000000015
While powers of 10 are very nice for us to type and read, 1e-5 (your dt) cannot be represented exactly as a floating point number. That's why your resulting calculations aren't coming out as even integers.
The statement
S(t-1/dt)
can be replaced by
S(uint32(t-1/dt))
And similarly for I.
Also you might want to save 1/dt hardcoded as 100000 as suggested above.
I reckon this will improve the comparison.
I have used MatLab to write the following code for Horner's Algorithm
function [answer ] = Simple( a,x )
%Simple takes two arguments that are in order and returns the value of the
%polynomial p(x). Simple is called by typing Simple(a,x)
% a is a row vector
%x is the associated scalar value
n=length(a);
result=a(n);
for j=n-1:-1:1 %for loop working backwards through the vector a
result=x*result+a(j);
end
answer=result;
end
I now need to add an error check to ensure the caller uses integer values in the row vector a.
For previous integer checks I have used
if(n~=floor(n))
error(...
But this was for a single value, I am unsure how to do this check for each of the elements in a.
You've got (at least) two options.
1) Use any:
if (any(n ~= floor(n)))
error('Bummer. At least one wasn''t an integer.')
end
Or even more succinctly...
assert(all(n == floor(n)), 'Bummer. At least one wasn''t an integer.')
2) Use the much more capable validateattributes:
validateattributes(n, {'double'}, {'integer'})
This function can check for more than a dozen other things, too.
Same math will work, but is now checking each element. Try this:
if any(n~=floor(n))
error(...
This question may initially appear similar to this other question but my situation is a little bit different.
I have a function 'deriv' that takes a symbolic expression as its input, then takes the first derivative of that symbolic expression. That derivative is then converted into an anonymous function using matlabFunction(), and is then evaluated over an array of points. I also use the anonymous function later in some other code.
The problem I'm having is that sometimes the input symbolic expression happens to be linear, and thus the derivative is constant; therefore the anonymous function is also a constant. When I evaluate the anonymous function over the array of points, I only get one output instead of an array of outputs.
Here's some code showing what I'm doing. For the sake of simplicity here, let's assume that the symbolic input expressions will involve only one symbolic variable called q.
function[derivFun,derivVals] = deriv(input)
derivSym = diff(input,q);
derivFun = matlabFunction(derivSym,'vars',q);
evalPoints = [1;2;3;4;5]; %in my true application, a much larger array
derivVals = derivFun(evalPoints);
end
So if the input is q^2, then the output derivVals will be [2;4;6;8;10]. But if the input happens to be, say, 3*q, then derivVals will be 3 (just a single scalar). What I'd like is for derivVals to be [3;3;3;3;3].
That is, I'd like derivVals to be the same size as evalPoints even if the input function happens to be linear (or constant). And I don't know ahead of time what the input expression will be.
Can anyone give suggestions for a scheme that would do that? I understand that a constant anonymous function will just return a single constant scalar, regardless of the size of its input. What I'm hoping for is perhaps some way to recognize when the anonymous function is constant and then still cause derivVals to be the same size as evalPoints.
I know that I could use a for loop to evaluate derivFun for every row of evalPoints, but I'd like to avoid using such a loop if possible.
Thank you for your time and consideration.
I think that this is a slightly simpler solution. The issue is that you're using matlabFunction, which simplifies down the equations and doesn't allow much customization. However, you can create an anonymous function of an anonymous function. Just add the this line right after your matlabFunction line:
derivFun = #(evalPoints)derivFun(evalPoints)+zeros(size(evalPoints));
This only evaluates the original derivFun once. However, I do like you symvar solution (just remember that adding zeros is always better than multiplying ones).
Not 100% sure I got the problem correctly.
Would this solve your issue?:
if isscalar(derivVals)
derivVals = repmat(derivVals, size(evalPoints));
end