Determine which of N points is not on sin(ax+b), where a and b are unknown.

Suppose N points ((x1,y1),(x2,y2),...(xN,yN)) are given from a curve y=sin(ax+b) where a,b values are unknown. Before giving these N points to you, y coordinate of one point is randomly tampered so that it does not lie on the curve. Write a program to determine which point among N points is NOT on the sinusoidal curve, whose a, b values are unknown.

Any logic on how to approach this question, would be highly appreciate would be very thankful! Please dont suggest inbuilt functions and toolbox (for e.g. NonLinearModel.fit and curve fitting toolbox)

[Walter Roberson]

C and C++ code are mostly off topic for this forum, except in connection with Mex or loadlibrary or Polyspace.

Answer 1

[James Tursa]

A modification of your posted approach:

Calculate the a and b using two points at a time, then use that to generate an associated y vector and compare it to the original y vector to find the outlier. Do this for several pairs of points to make sure several of the pairs do not contain the outlier (it is OK if a couple of them do). Then pick off the outlier that most of the pairs agree on. E.g., an outline:

x=(0:0.01:2*pi);   % sample data
y=sin(2*x + 3);    % sample data
y(100) = something_else; % pick some arbitrary spot to be the outlier
m = zeros(numel(x)-1,1); % calculated outlier index array, one element for each pair of points
for i=1:numel(x)-1  % loop through a bunch of point pairs
    % you insert code here to find the a and b values for the i and i+1 pair of points using posted P\R method
    % you insert code here to calculate a y vector for the entire x vector using the a and b you just calculated
    % you insert code here to find the index of the max abs difference between original y and just calculated y
    % save that index in the m(i) spot
end
outlier = mode(m);  % the most frequent value in m is the index of the outlier

So, you simply need to fill in the code inside the loop using the techniques that are already discussed in this thread. I could have given you this code, but thought I would leave it to you to figure out (it is just the appropriate pieces from what has already been posted in this thread). I picked the looping above to match the looping that you already had in your code, but it doesn't have to be this. You could pick the pairs randomly if you wanted to ... you just need to generate enough different pairs to guarantee that most of them will not contain the outlier. The mode(m) stuff will get rid of any pair results that may have been corrupted by an outlier.

[sangeet sagar]

You just made my day! I have been stuck on this since last 2 days....and the feeling I am in!!!!!!!.

It ran smoothly. Your idea of mode(m) is really appreciable. Thanks a lot.

[sangeet sagar]

Hello, I would rather suggest you to go through each comments discussed for this question. Actually everything has been written, just need to go through carefully and use some of yours. Try once on your own. Write me back if you face issues.

Answer 2

[Walter Roberson]

Loop N times
  take data without sample #N
  fit arcsin(y) = a*x+b as a linear fit to get a and b
  project yp = sin(a*x+b)
  calculate residue(N) between y and yp
end

Lowest residue matches the case where the tampered point was excluded.

[sangeet sagar]

Suppose I have two vectors x and y (both x and y satisfy the equation) of a finite length. Now

   fit arcsin(y) = a*x+b as a linear fit to get a and b

How do I do this ?

[Walter Roberson]

polyfit() or the \ operator.

[sangeet sagar]

The point where we get the largest residue between y and yp will be my point which lies outside the curve, right?

[sangeet sagar]

Hello Sir, I am still finding it difficult to write the code for this line:

fit arcsin(y) = a*x+b as a linear fit to get a and b

I don't understand how the value of a and b calculated in the single run of the loop.

[Walter Roberson]

ab = polyfit(x, arcsin(y), 1);
a = ab(1); b = ab(2);

[Walter Roberson]

"The point where we get the largest residue between y and yp will be my point which lies outside the curve, right?"

No, the one with the smallest residue will be the one where the point that was omitted was the tweaked point, because the remaining points would all fit very nicely to the curve.

[sangeet sagar]

Sir, for example y=sin(2x + 3) , a=2 and b=3 (general form y=sin(a*x + b)). When I run the following code why don't I get Q=[2 3] . For every iteration I get different value of a and b, not even close to 2 and 3 respectively.

   x=(0:0.01:2*pi);
    y=sin(2*x + 3); 
     for i=1:62
         P=[x(i)   1;
            x(i+1) 1];
         R=[asin(y(i));
           asin(y(i+1))];
         Q=R\P    % %Q=[a b];
      end

[James Tursa]

See my reply to this post:

https://www.mathworks.com/matlabcentral/answers/385752-when-i-run-the-below-code-why-dont-i-get-value-of-q-2-3

Bottom line is that this method is fundamentally flawed.

[sangeet sagar]

Hello Sir, I am really sorry for troubling you, but I am stuck with the following

I ran the code below: (all coordinates x and y lie on the curve sin(2x + 3). )

x=(0:0.1:2*pi);
y=sin(2*x + 3);

n=length(x);

for i=1:n
    ab = polyfit(x(i), asin(y(i)), 1);
    a(i) = ab(1); b(i) = ab(2);
    yp = sin(a(i)*x+b(i));
    [r,p,k] =residue (y,yp);
end

Now how do I conclude after looking at the residue value that all the coordinates lie on the curve and that none of them lies outside the curve.

If there is any error in this code, kindly help me correct it.

[sangeet sagar]

Hello Sir, @WalterRoberson PLease help me with the above problem. I am really having a hard time solving this

https://in.mathworks.com/matlabcentral/answers/385668-determine-which-of-n-points-is-not-on-sin-ax-b-where-a-and-b-are-unknown#comment_540994

Answer 3

[Srikanth KS]

How about trying using a different approach. I assume that a and b ae unknown given the coordinates x and y I will substitute x and y and solve the simultaneous equations to get a and b. I will solve it for a couple of points to prove that my a and b are correct it was mentioned that only 1 point was messed up so once I know a and b I will iterate and find out the difference between y and to typically y -yp should tend to 0 if I get a point where y -yp is not close to 0 I will pick that point as manipulated point.

[sangeet sagar]

Hello, I tried exactly same. So according to you arcsin(y) = a*x + b

and you will run this in a loop. Now the problem arises, arcsin is a multi valued function. SO each time you run the loop, you will get different values of a and b. Here goes the code:

x=(0:0.01:2*pi);
    y=sin(2*x + 3); 
     for i=1:62
         P=[x(i)   1;
            x(i+1) 1];
         R=[asin(y(i));
           asin(y(i+1))];
         Q=P\R    % %Q=[a b];
      end