Re: convolution + basic image processing question


"Rune Allnor" <allnor@xxxxxxxxxxxx> wrote in message
news:1174206588.310906.68140@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 18 Mar, 00:31, Martin Jørgensen <hotmail_s...@xxxxxxxxxxx> wrote:
Rune Allnor wrote:
On 17 Mar, 20:31, Martin Jørgensen <hotmail_s...@xxxxxxxxxxx> wrote:
Rune Allnor wrote:
On 17 Mar, 05:21, Martin Jørgensen <hotmail_s...@xxxxxxxxxxx> wrote:
Hi,
I didn't know anything about convolution before, but the
documentation
says: "Algebraically, convolution is the same operation as
multiplying
the polynomials whose coefficients are the elements of u and v."
That's true for 1D data. Don't try to convert this analogy to
2D signals, it is likely to confuse far more than it helps.
Back to the example:
....
The solution is: 2*x2+4*x+2. Is it just the sum of the 2 other? No, I
don't get it...

Because there is nothing to get. This is one of those cases

Please don't reply if you have nothing to contribute with.

I am contributing the best I can. I have seen all too
often that apparently intelligent people get bogged down
with a completely irrelevant detail in a useless analogy,
just as you are about getting bogged down.

where a "simple", "intuitive" analogy does more damage and
confusion than you would experience without that same analogy.

It must be because you don't understand it yourself.

I understand convoluton, in 1D, 2D and 3D. I don't
understand polynomial multiplication in other dimensions
than 1D. Which is why I can't see why your belowed
analogy of plynomials is interesting at all, in 2D.

Don't spend your enery, in vain, on understanding an irrelevant

IT IS NOT IRRELEVANT....

That's your claim. Why don't you explain to me what
the matrixes

A = [1 1; 1 1];
B = [1 1 1; 1 1 1; 1 1 1];

represent in terms of plynomials?

You could treat A(i, j) as the coefficient of x^(i-1)*y^(j-1) in the
polynomial: [Note: this code requires the Symbolic Math Toolbox to
execute.]


syms x y

A = [1 2; 3 4];
xv = x.^(0:1);
yv = y.^(0:1);
Ap = expand(xv*A*yv.')

B = [1 2 3;4 5 6;7 8 9];
xv = x.^(0:2);
yv = y.^(0:2);
Bp = expand(xv*B*yv.')

C = conv2(A, B);
xv = x.^(0:3);
yv = y.^(0:3);
Cp = expand(xv*C*yv.')
Delta_polynomial = simplify(Ap*Bp-Cp)


The equation in DOC CONV2 looks familiar ... kind of like what you'd use if
you were multiplying polynomials in this manner. In fact, the 1-D case is a
special case of this

[1 2 3] -> x^0*[1 2 3]*y.^((0:2)')

*snip*

--
Steve Lord
slord@xxxxxxxxxxxxx


.

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