I have been away from Mathematica for quite a while and am trying to fix some old notebooks from v4 that are no longer working under v11. I'm also a tad rusty.
I am attempting to use functional minimization to fit a polynomial of variable degree to an arbitrary function (F) given a starting guess (ao) and domain of interest (d). Note that while F is arbitrary, its nature is such that the integral of the product of F and a polynomial (or F^2) can always be evaluated algebraically.
For the sake of example, I'll use the following inputs:
ao = { 1, 2, 3, 4 }
d = { -1, 1 }
F = Sin[x]
To do so, I create an array of 'indexed' variables
polyCoeff = Array[a,Length[a],0]
Result: polycoeff = {a[0], a[1], a[2], a[3]}
I then create the polynomial itself using the following
genPoly[{},x_] := 0
genPoly[a_List,x_] := First[a] + x genPoly[Rest[a],x]
poly = genPoly[polyCoeff,x]
Result: poly = a[0] + x (a[1] + x (a[2] + x a[3]))
I then define my objective function as the integral of the square of the error of the difference between this poly and the function I am attempting to fit:
Q = Integrate[ (poly - F[x])^2, {x, d[[1]],d[[2]]} ]
result: Q = 0.545351 - 2. a[0.]^2 + 0.66667 a[1.]^2 + .....
And this is where things break down. poly looks just as I expected: a polynomial in x with coefficients that look like a[0], a[1], a[2], ... But, Q is not exactly what I expected. I expected and got a new polynomial. But not the coefficients contained a[0.], a[1.], a[2.], ...
The next step is to create the initial guess for FindMinimum
init = Transpose[{polyCoeff,ao}]
Result: {{a[0],1},{a[1],2},{a[3],3},{a[4],4}}
This looks fine.
But when I make the call to FindMinimum, I get an error because the coefficients passed in the objective (a[0.],a[1.],...) do not match those passed in the initial guess (a[0],a[1],...).
S = FindMinimum[Q,init]
So I think my question is how do I keep Integrate from changing the arguments to my coefficients? But, I am open to other approaches as well. Keep in mind though that this is "legacy" work that I really don't want to have to completely revamp.
Thanks much for any/all help.
I've tried to find a good way to speed up the code for a problem I've been working on. The basic idea of the code is very simple. There are five inputs:
Four 1xm (for some m < n, they can be different sizes) matrices (A, B, C, D) that are pairwise-disjoint subsets of {1,2,...,n} and one nxn symmetric binary matrix (M). The basic idea for the code is to check an inequality for for every combination of elements and if the inequality holds, return the values that cause it to hold, i.e.:
for a = A
for b = B
for c = C
for d = D
if M(a,c) + M(b,d) < M(a,d) + M(b,c)
result = [a b c d];
return
end
end
end
end
end
I know there has to be a better way to do this. First, since it's symmetric, I can cut down half of the items checked since M(a,b) = M(b,a). I've been researching vectorization, found several functions I'd never heard of with MATLAB (since I'm relatively new), but I can't find anything that will particularly help me with this specific problem. I've thought of other ways to approach the problem, but nothing has been perfected, and I just don't know what to do at this point.
For example, I could possibly split this into two cases:
1) The right hand side is 1: then I have to check that both terms on the left side are 0.
2) The right hand side is 2: then I have to check that at least one term on the left hand side is 0.
But, again, I won't be able to avoid nesting.
I appreciate all the help you can offer. Thank you!
You're asking two questions here: (1) is there a more efficient algorithm to perform this search, and (2) how can I vectorize this in MATLAB. The first one is very interesting to think about, but may be a little beyond the scope of this forum. The second one is easier to answer.
As pointed out in the comments below your question, you can vectorize the for loop by enumerating all of the possibilities and checking them all together, and the answers from this question can help:
[a,b,c,d] = ndgrid(A,B,C,D); % Enumerate all combos
a=a(:); b=b(:); c=c(:); d=d(:); % Reshape from 4-D matrices to vectors
ac = sub2ind(size(M),a,c); % Convert subscript pairs to linear indices
bd = sub2ind(size(M),b,d);
ad = sub2ind(size(M),a,d);
bc = sub2ind(size(M),b,c);
mask = (M(ac) + M(bd) < M(ad) + M(bc)); % Test the inequality
results = [a(mask), b(mask), c(mask), d(mask)]; % Select the ones that pass
Again, this isn't an algorithmic change: it still has the same complexity as your nested for loop. The vectorization may cause it to run faster, but it also lacks early termination, so in certain cases it may be slower.
Since M is binary, we can think about this as a graph problem. i,j in {1..n} correspond to nodes, and M(i,j) indicates whether there is an undirected edge connecting them.
Since A,B,C,D are disjoint, that simplifies the problem a bit. We can approach the problem in stages:
Find all (c,d) for which there exists a such that M(a,c) < M(a,d). Let's call this set CD_lt_a, (the subset of C*D such that the "less than" inequality holds for some a).
Find all (c,d) for which there exists a such that M(a,c) <= M(a,d), and call this set CD_le_a.
Repeat for b, forming CD_lt_b for M(b,d) < M(b,c) and CD_le_b for M(b,d)<=M(b,c).
One way to satisfy the overall inequality is for M(a,c) < M(a,d) and M(b,d) <= M(b,c), so we can look at the intersection of CD_lt_a and CD_le_b.
The other way is if M(a,c) <= M(a,d) and M(b,d) < M(b,c), so look at the intersection of CD_le_a and CD_lt_b.
With (c,d) known, we can go back and find the (a,b).
And so my implementation is:
% 0. Some preliminaries
% Get the size of each set
mA = numel(A); mB = numel(B); mC = numel(C); mD = numel(D);
% 1. Find all (c,d) for which there exists a such that M(a,c) < M(a,d)
CA_linked = M(C,A);
AD_linked = M(A,D);
CA_not_linked = ~CA_linked;
% Multiplying these matrices tells us, for each (c,d), how many nodes
% in A satisfy this M(a,c)<M(a,d) inequality
% Ugh, we need to cast to double to use the matrix multiplication
CD_lt_a = (CA_not_linked * double(AD_linked)) > 0;
% 2. For M(a,c) <= M(a,d), check that the converse is false for some a
AD_not_linked = ~AD_linked;
CD_le_a = (CA_linked * double(AD_not_linked)) < mA;
% 3. Repeat for b
CB_linked = M(C,B);
BD_linked = M(B,D);
CD_lt_b = (CB_linked * double(~BD_linked)) > 0;
CD_le_b = (~CB_linked * double(BD_linked)) < mB;
% 4. Find the intersection of CD_lt_a and CD_le_b - this is one way
% to satisfy the inequality M(a,c)+M(b,d) < M(a,d)+M(b,c)
CD_satisfy_ineq_1 = CD_lt_a & CD_le_b;
% 5. The other way to satisfy the inequality is CD_le_a & CD_lt_b
CD_satisfy_ineq_2 = CD_le_a & CD_lt_b;
inequality_feasible = any(CD_satisfy_ineq_1(:) | CD_satisfy_ineq_2(:));
Note that you can stop here if feasibility is your only concern. The complexity is A*C*D + B*C*D, which is better than the worst-case A*B*C*D complexity of the for loop. However, early termination means your nested for loops may still be faster in certain cases.
The next block of code enumerates all the a,b,c,d that satisfy the inequality. It's not very well optimized (it appends to a matrix from within a loop), so it can be pretty slow if there are many results.
% 6. With (c,d) known, find a and b
% We can define these functions to help us search
find_a_lt = #(c,d) find(CA_not_linked(c,:)' & AD_linked(:,d));
find_a_le = #(c,d) find(CA_not_linked(c,:)' | AD_linked(:,d));
find_b_lt = #(c,d) find(CB_linked(c,:)' & ~BD_linked(:,d));
find_b_le = #(c,d) find(CB_linked(c,:)' | ~BD_linked(:,d));
% I'm gonna assume there aren't too many results, so I will be appending
% to an array inside of a for loop. Bad for performance, but maybe a bit
% more readable for a StackOverflow answer.
results = zeros(0,4);
% Find those that satisfy it the first way
[c_list,d_list] = find(CD_satisfy_ineq_1);
for ii = 1:numel(c_list)
c = c_list(ii); d = d_list(ii);
a = find_a_lt(c,d);
b = find_b_le(c,d);
% a,b might be vectors, in which case all combos are valid
% Many ways to find all combos, gonna use ndgrid()
[a,b] = ndgrid(a,b);
% Append these to the growing list of results
abcd = [a(:), b(:), repmat([c d],[numel(a),1])];
results = [results; abcd];
end
% Repeat for the second way
[c_list,d_list] = find(CD_satisfy_ineq_2);
for ii = 1:numel(c_list)
c = c_list(ii); d = d_list(ii);
a = find_a_le(c,d);
b = find_b_lt(c,d);
% a,b might be vectors, in which case all combos are valid
% Many ways to find all combos, gonna use ndgrid()
[a,b] = ndgrid(a,b);
% Append these to the growing list of results
abcd = [a(:), b(:), repmat([c d],[numel(a),1])];
results = [results; abcd];
end
% Remove duplicates
results = unique(results, 'rows');
% And actually these a,b,c,d will be indices into A,B,C,D because they
% were obtained from calling find() on submatrices of M.
if ~isempty(results)
results(:,1) = A(results(:,1));
results(:,2) = B(results(:,2));
results(:,3) = C(results(:,3));
results(:,4) = D(results(:,4));
end
I tested this on the following test case:
m = 1000;
A = (1:m); B = A(end)+(1:m); C = B(end)+(1:m); D = C(end)+(1:m);
M = rand(D(end),D(end)) < 1e-6; M = M | M';
I like to think that first part (see if the inequality is feasible for any a,b,c,d) worked pretty well. The other vectorized answers (that use ndgrid or combvec to enumerate all combinations of a,b,c,d) would require 8 terabytes of memory for a problem of this size!
But I would not recommend running the second part (enumerating all of the results) when there are more than a few hundred c,d that satisfy the inequality, because it will be pretty damn slow.
P.S. I know I answered already, but that answer was about vectorizing such loops in general, and is less specific to your particular problem.
P.P.S. This kinda reminds me of the stable marriage problem. Perhaps some of those references would contain algorithms relevant to your problem as well. I suspect that a true graph-based algorithm could probably achieve the worst-case complexity as this while additionally offering early termination. But I think it would be difficult to implement a graph-based algorithm efficiently in MATLAB.
P.P.P.S. If you only want one of the feasible solutions, you can simplify step 6 to only return a single value, e.g.
find_a_lt = #(c,d) find(CA_not_linked(c,:)' & AD_linked(:,d), 1, 'first');
find_a_le = #(c,d) find(CA_not_linked(c,:)' | AD_linked(:,d), 1, 'first');
find_b_lt = #(c,d) find(CB_linked(c,:)' & ~BD_linked(:,d), 1, 'first');
find_b_le = #(c,d) find(CB_linked(c,:)' | ~BD_linked(:,d), 1, 'first');
if any(CD_satisfy_ineq_1)
[c,d] = find(CD_satisfy_ineq_1, 1, 'first');
a = find_a_lt(c,d);
b = find_a_le(c,d);
result = [A(a), B(b), C(c), D(d)];
elseif any(CD_satisfy_ineq_2)
[c,d] = find(CD_satisfy_ineq_2, 1, 'first');
a = find_a_le(c,d);
b = find_a_lt(c,d);
result = [A(a), B(b), C(c), D(d)];
else
result = zeros(0,4);
end
If you have access to the Neural Network Toolbox, combvec could be helpful here.
running allCombs = combvec(A,B,C,D) will give you a (4 by m1*m2*m3*m4) matrix that looks like:
[...
a1, a1, a1, a1, a1 ... a1... a2... am1;
b1, b1, b1, b1, b1 ... b2... b1... bm2;
c1, c1, c1, c1, c2 ... c1... c1... cm3;
d1, d2, d3, d4, d1 ... d1... d1... dm4]
You can then use sub2ind and Matrix Indexing to setup the two values you need for your inequality:
indices = [sub2ind(size(M),allCombs(1,:),allCombs(3,:));
sub2ind(size(M),allCombs(2,:),allCombs(4,:));
sub2ind(size(M),allCombs(1,:),allCombs(4,:));
sub2ind(size(M),allCombs(2,:),allCombs(3,:))];
testValues = M(indices);
testValues(5,:) = (testValues(1,:) + testValues(2,:) < testValues(3,:) + testValues(4,:))
Your final a,b,c,d indices could be retrieved by saying
allCombs(:,find(testValues(5,:)))
Which would print a matrix with all columns which the inequality was true.
This article might be of some use.
I am just a beginner of programming, and sorry in advance for bothering you by a (presumably) basic question.
I would like to perform the following task:
(I apologize for inconvenience; I don't know how to input a TeX-y formula in Stack Overflow ). I am primarily considering an implementation on MATLAB or Scilab, but language does not matter so much.
The most naive approach to perform this, I think, is to form an n-nested for loop, that is (the case n=2 on MATLAB is shown for example),
n=2;
x=[x1,x2];
for u=0:1
y(1)=u;
if x(1)>0 then
y(1)=1;
end
for v=0:1
y(2)=v;
if x(2)>0 then
y(2)=1;
end
z=Function(y);
end
end
However, this implementation is too laborious for large n, and more importantly, it causes 2^n-2^k abundant evaluations of the function, where k is a number of negative elements in x. Also, naively forming a k-nested for loop with knowledge of which element in x is negative, e.g.
n=2;
x=[-1,2];
y=[1,1];
for u=0:1
y(1)=u;
z=Function(y);
end
doesn't seem to be a good way; if we want to perform the task for different x, we have to rewrite a code.
I would be grateful if you provide an idea to implement a code such that (a) evaluates the function only 2^k times (possible minimum number of evaluations) and (b) we don't have to rewrite a code even if we change x.
You can evaluate Function on y in Ax easily using recursion
function eval(Function, x, y, i, n) {
if(i == n) {
// break condition, evaluate Function
Function(y);
} else {
// always evaluate y(i) == 1
y(i) = 1;
eval(Function, x, y, i + 1, n);
// eval y(i) == 0 only if x(i) <= 0
if(x(i) <= 0) {
y(i) = 0;
eval(Function, x, y, i + 1, n);
}
}
}
Turning that into efficient Matlab code is another problem.
As you've stated the number of evaluations is 2^k. Let's sort x so that only the last k elements are non-positive. To evaluate Function index y using the reverse of the permutation of the sort of x: Function(y(perm)). Even better the same method allows us to build Ax directly using dec2bin:
// every column of the resulting matrix is a member of Ax: y_i = Ax(:,i)
function Ax = getAx(x)
n = length(x);
// find the k indices of non-positives in x
is = find(x <= 0);
k = length(is);
// construct Y (last k rows are all possible combinations of [0 1])
Y = [ones(n - k, 2 ^ k); (dec2bin(0:2^k-1)' - '0')];
// re-order the rows in Y to get Ax according to the permutation is (inverse is)
perm([setdiff(1:n, is) is]) = 1:n;
Ax = Y(perm, :);
end
Now rewrite Function to accept a matrix or iterate over the columns in Ax = getAx(x); to evaluate all Function(y).
I have been given an assignment in which I am supposed to write an algorithm which performs polynomial interpolation by the barycentric formula. The formulas states that:
p(x) = (SIGMA_(j=0 to n) w(j)*f(j)/(x - x(j)))/(SIGMA_(j=0 to n) w(j)/(x - x(j)))
I have written an algorithm which works just fine, and I get the polynomial output I desire. However, this requires the use of some quite long loops, and for a large grid number, lots of nastly loop operations will have to be done. Thus, I would appreciate it greatly if anyone has any hints as to how I may improve this, so that I will avoid all these loops.
In the algorithm, x and f stand for the given points we are supposed to interpolate. w stands for the barycentric weights, which have been calculated before running the algorithm. And grid is the linspace over which the interpolation should take place:
function p = barycentric_formula(x,f,w,grid)
%Assert x-vectors and f-vectors have same length.
if length(x) ~= length(f)
sprintf('Not equal amounts of x- and y-values. Function is terminated.')
return;
end
n = length(x);
m = length(grid);
p = zeros(1,m);
% Loops for finding polynomial values at grid points. All values are
% calculated by the barycentric formula.
for i = 1:m
var = 0;
sum1 = 0;
sum2 = 0;
for j = 1:n
if grid(i) == x(j)
p(i) = f(j);
var = 1;
else
sum1 = sum1 + (w(j)*f(j))/(grid(i) - x(j));
sum2 = sum2 + (w(j)/(grid(i) - x(j)));
end
end
if var == 0
p(i) = sum1/sum2;
end
end
This is a classical case for matlab 'vectorization'. I would say - just remove the loops. It is almost that simple. First, have a look at this code:
function p = bf2(x, f, w, grid)
m = length(grid);
p = zeros(1,m);
for i = 1:m
var = grid(i)==x;
if any(var)
p(i) = f(var);
else
sum1 = sum((w.*f)./(grid(i) - x));
sum2 = sum(w./(grid(i) - x));
p(i) = sum1/sum2;
end
end
end
I have removed the inner loop over j. All I did here was in fact removing the (j) indexing and changing the arithmetic operators from / to ./ and from * to .* - the same, but with a dot in front to signify that the operation is performed on element by element basis. This is called array operators in contrast to ordinary matrix operators. Also note that treating the special case where the grid points fall onto x is very similar to what you had in the original implementation, only using a vector var such that x(var)==grid(i).
Now, you can also remove the outermost loop. This is a bit more tricky and there are two major approaches how you can do that in MATLAB. I will do it the simpler way, which can be less efficient, but more clear to read - using repmat:
function p = bf3(x, f, w, grid)
% Find grid points that coincide with x.
% The below compares all grid values with all x values
% and returns a matrix of 0/1. 1 is in the (row,col)
% for which grid(row)==x(col)
var = bsxfun(#eq, grid', x);
% find the logical indexes of those x entries
varx = sum(var, 1)~=0;
% and of those grid entries
varp = sum(var, 2)~=0;
% Outer-most loop removal - use repmat to
% replicate the vectors into matrices.
% Thus, instead of having a loop over j
% you have matrices of values that would be
% referenced in the loop
ww = repmat(w, numel(grid), 1);
ff = repmat(f, numel(grid), 1);
xx = repmat(x, numel(grid), 1);
gg = repmat(grid', 1, numel(x));
% perform the calculations element-wise on the matrices
sum1 = sum((ww.*ff)./(gg - xx),2);
sum2 = sum(ww./(gg - xx),2);
p = sum1./sum2;
% fix the case where grid==x and return
p(varp) = f(varx);
end
The fully vectorized version can be implemented with bsxfun rather than repmat. This can potentially be a bit faster, since the matrices are not explicitly formed. However, the speed difference may not be large for small system sizes.
Also, the first solution with one loop is also not too bad performance-wise. I suggest you test those and see, what is better. Maybe it is not worth it to fully vectorize? The first code looks a bit more readable..
I want to write some functions as follows
y = f(x) and another function,
x = g(y) that acts as a reversible, where
y = f(g(y)) and where x and y are permutated integers.
For very simple example in the range of integers in 0 to 10 it would look like this:
0->1
1->2
2->3
...
9->10
10->0
but this is the simplest method by adding 1 and reversing by subtracting 1.
I want to have a more sofisticated algorithm that can do the following,
234927773->4299
34->33928830
850033->23234243423
but the reverse can be obtained by conversion
The solution could be obtained with a huge table storing pairs of unique integers but this will not be correct. This must be a function.
You could just XOR.
y = x XOR p
x = y XOR p
Though not my area of expertise, I think that cryptography should provide some valuable answers to your question.
If the domain of your permutation is a power of 2, you can use any block cipher: 'f' is encryption with a specific key, and 'g' is decryption with the same key. If your domain is not a power of 2, you can probably still use a block cipher: see this article.
You could use polynomial interpolation methods to interpolate a function one way, then do reverse interpolation to find the inverse function.
Here is some example code in MATLAB:
function [a] = Coef(x, y)
n = length(x);
a = y;
for j = 2:n
for i = n:-1:j
a(i) = (a(i) - a(i-1)) / (x(i) - x(i-j+1));
end
end
end
function [val] = Eval(x, a, t)
n = length(x);
val = a(n);
for i = n-1:-1:1
val = a(i) + val*(t-x(i));
end
end
It builds a Divided Difference table and evaluates a function based on Newtons Interpolation.
Then if your sets of points are x, and y (as vectors of the same length, where x(i) matches to y(i), your forward interpolation function at value n would be Eval(x, Coef(x, y), n) and the reverse interpolation function would be Eval(y, Coef(y, x), n).
Depending on your language, there are probably much cleaner ways to do this, but this gets down and dirty with the maths.
Here is an excerpt from the Text Book which is used in my Numerical Methods class: Google Book Link