Well, this sounds ridiculous but its now more than ten times I am trying to see what I am doing wrong in this example. I have seen the tutorial on the web, and the example is shown below:
At stage 3: Are we not supposed to decrement Yi to 9? I am saying this because the decision variable Pi=13 satisfying the condition that if Pi>=0, Yi +1=Yi-1 as shown below:
Question: Where am I going wrong? Please, someone explain this to me.
PS: I appreciate even more worked examples, thanks.
No,
We are not supposed to decrement value of Yi to 9 at 3rd Stage.
Let me elaborate you where you are going wrong,
Consider you are at 3rd Stage.
Now you are checking value of Pi based on previous value of itself that is from 2nd stage.
Here you will do, Pi = -1 + 4(2) + 6
The value 13 will be obtained after performing this operation at this stage.
After this value is obtained, at 4th stage, you'll check value of Pi from 3rd stage.
Now Pi will be greater than 0 and hence Yi will decrement to 9.
Hope this helps.
Related
I am working on a AI algorithm. first when program runs a random solution is generated from which ,in first iteration of the program 10 solution vectors are created, by analyzing these solutions we could give each of them a probability ( highest , second highest, third highest and so on) towards the optimal solution , for the second input of the program I want it to be a vector (possible solution) obtained from those 10 vectors previously found. But i need the vector solution to consider all the previous solutions with a different impact depending on their probability ...
i.e A=[4.7 ,5.6, 3.5,9 ] b=[-7.9 ,8 ,-2.8 ,4.6] c=[7 ,9.7 , 4,6,3.9] ......
i used mean in my program
NextPossibleSolution = mean(([A;B;C;]))
But do you think mean is the right move ? i don't think because all the solution contributes equal to Next Possible Solution (next input) regardless of their likelihood ... Please if there is a method formula or anything , Let me know that ... I really need it badly .... A Billion Thanks
It's some time I was thinking about solving this problem. I have a registration of angular data (Angle(~20000,1)) variating between 0 and 355 (a potentiometer attached to a rotary testing machine), and I wanted to convert it in an incremental form, since I want the final total angular displacement. The main issue is that between 355 and the next 0 there are no jumps but a fast decrement (with strongly negative slope in time vs angle space). I've tried 2 ways up to now:
Calculate the Angslope=diff(Angle), extract with find the indexes j1=find(Angslope>0.2 & Angslope<0.2) to avoid the negative slopes due to the inversion of angular signal, then try to apply those indexes to the original Angle(n,1), as Angle2=Angle(j1). The trouble is the n-1 length of Angslope and the fact that somehow there is not a simple shift of my indexes of one position.
For cycles and logical, wanting to exclude data if the previous one is < the current value,etc
Angle2=zeros(size(Angle,1),1);
for i=2:size(Angle,1)
if Angle(i,1)<Angle(i-1,1)
Angle2(i,1)=NaN;
else Angle2(i,1)=Angle(i,1);
end
end
Which works good, but I don't know how to "match up" the single increment steps I obtain!
Any help or simple comment would be of great help!!
You are probably looking for the unwrap function. For this you have to convert your angles into radians, but that's not a big deal.
You can get the increments in one line:
Inc = diff(unwrap(Angle*pi/180))*180/pi;
and your total angular displacement:
Tot = sum(Inc);
Best,
As I've written in the title, I'm trying to find the exact distance (dimensionless distance in this case) when two functions start to differ from each other a 5% of the Y-axis. The two functions intersect at the value of 1 in the X-axis and I need to find the described distance before the intersection, not after (i.e., it must be less than 1). I've written a Matlab code for you to see the shape of the functions and the following calculations which I'm trying to make them work but they don't, I don't know why. "Explicit solution could not be found".
I don't know if I explained it clearly. Please let me know if you need a more detailed explanation.
I hope you can throw some light in this issue.
Thank you so much in advanced.
r=0:0.001:1.2;
ro=0.335;
rt=r./ro;
De=0.3534;
k=2.8552;
B=(2*k/De)^0.5;
Fm=2.*De.*B.*ro.*[1-exp(B.*ro.*(1-rt))].*exp(B.*ro.*(1-rt));
A=5;
b=2.2347;
C=167.4692;
Ftt=(C.*(exp(-b.*rt).*((b.^6.*rt.^5)./120 + (b.^5.*rt.^4)./24 + (b.^4.*rt.^3)./6 + (b.^3.*rt.^2)./2 + b.^2.*rt + b) - b.*exp(-b.*rt).*((b.^6.*rt.^6)./720 + (b.^5.*rt.^5)./120 + (b.^4.*rt.^4)./24 + (b.^3.*rt.^3)./6 + (b.^2.*rt.^2)./2 + b.*rt + 1)))./rt.^6 - (6.*C.*(exp(-b.*rt).*((b.^6.*rt.^6)./720 + (b.^5.*rt.^5)./120 + (b.^4.*rt.^4)./24 + (b.^3.*rt.^3)./6 + (b.^2.*rt.^2)./2 + b.*rt + 1) - 1))./rt.^7 - A.*b.*exp(-b.*rt);
plot(rt,-Fm,'red')
axis([0 2 -1 3])
xlabel('Dimensionless distance')
ylabel('Force, -dU/dr')
hold on
plot(rt,-Ftt,'green')
clear rt
syms rt
%assume(0<rt<1)
r1=solve((Fm-Ftt)/Ftt==0.05,rt)
r2=solve((Ftt-Fm)/Fm==0.05,rt)
Welcome to the crux of floating point data. The reason why is because for the values of r that you are providing, the exact solution of 0.05 may be in between two of the values in your r array and so you won't be able to get an exact solution. Also, FWIW, your equation may never generate a solution of 0.05, which is why you're getting that error too. Either way, doing that explicit solve on floating point data is never recommended, unless you know very well how your data are shaped and what values you expect for the output of the function you're applying the data to.
As such, it's always recommended that you find the nearest value that satisfies your condition. As such, you should do something like this:
[~,ind] = min(abs((Fm-Ftt)./Ftt - 0.05));
r1 = r(ind);
The first line will find the nearest location in your r array that satisfies the 5% criterion. The next line of code will then give you the value that is in your r array that satisfies this. You can do the same with r2 by:
[~,ind2] = min(abs((Ftt-Fm)./Fm - 0.05));
r2 = r(ind2);
What the above code is basically doing is that it is trying to find at what point in your array would the difference between your data and 5% be 0. In other words, which point in your r array would be close enough to make the above relation equal to 0, or essentially when it is as close to 5% as possible.
If you want to improve this, you can always change the step size of r... perhaps make it 0.00001 or something. However, the smaller the step size, the larger your array and you'll eventually run out of memory!
I have a vector that contains a long list of data (time series). I would like to find the maximum of every 60 elements without going through manually C=[max(B(1:60)), etc... ] because it is a rather large data set. Is there a clean way of doing this? Thanks for any ideas! I appreciate it.
Oli's suggestion deserves to be made into a formal answer. Try this:
C = max(reshape(B,60,[]));
As another option you can look at blkproc.
A= randn(600,1);
blkproc( A, [60,1], 'max');
blkproc is being phased out, so you will also have to look at blockproc.
Though, reshaping and taking the max will probably be more efficient as was mentioned in the comments.
max( reshape(A, [60, 10] ) )
[update]
As a note... don't use blkproc :-). Using a very large array (A), blkproc is 100x slower than the max,reshape.
You can also use 'buffer' function.
A= randn(600,1);
max(buffer(A,60));
This solution works even when the length of the vector is not exact multiple of 60 and is faster in comparison to 'reshape' function.
A quick question because i fear there may already be an answer (although i cant find it)
i am getting the error: Matrix dimensions must agree.
because i am useing '<'
now with all the other operators there is a way around this either by putting '.' infront or by using a different formula. So what do people do about the less than operator????
i don't see why the greater than or equal to (>=) works but yet less than does not!?
am i being stupid and missed something really obvious??
code snippet
matrix 1 represents an array of 16 numbers
matrix 2 can represents anywhere between 10 and 20 numbers
idx = (matrix2 >= matrix1 * 0.1 & matrix2 < matrix1 * 1.5);
any help guidance or advice on the topic would be much appreciated! thank you!
EDIT
i know the matrices are different sizes but is there a way to use less then with different size arrays? as im not bothered about the size of the array but the numbers within
If you want to compare parts of matrices, like M(1:3,10:12)>A(5:7,1:3), you, probably, have to use the function squeeze():
squeeze(M(1:3,10:12))>squeeze(A(5:7,1:3))
This function remotes singleton dimensions and everything works fine after.