Bicubic interpolation direct interpolation formula Matlab source code

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Gobert
Gobert 2018-6-15
评论： Rik 2021-2-23
Can anyone help by sharing the source code of the bicubic image interpolation algorithm using/involving 'direct interpolation formula'? See the figure below?
Also, here's an incomplete (and possibly erroneous) example showing the 'direct interpolation formula' -- V(m',n'):
a = y2-y;
b = x2-x;
d1 = -a*((1-a)^2)*(-b*((1-b)^2))*img(x0,y0); %%(m-1,n-1)
d2 = (1-2*(b^2) + (b^3))*img(x0,y1); %%(m,n-1)
d3 = b*(1+b-(b^2))*img(x0,y2); %%(m+1,n-1)
d4 = -(b^2)*(1-b)*img(x0,y3); %%(m+2,n-1)
d5 = (1-2*(a^2)+(a^3))*(-b*((1-b)^2))*img(x1,y0); %%(m-1,n)
d6 = (1-2*(b^2)+(b^3))*img(x1,y1); %%(m,n)
d7 = b*(1+b-(b^2))*img(x1,y2); %%(m+1,n)
d8 = -(b^2)*(1-b)*img(x1,y3); %%(m+2,n)
d9 = a*(1+a-(a^2))*(-b*((1-b)^2))*img(x2,y0); %%(m-1,n+1)
d10 =(1-2*(b^2)+(b^3))*img(x2,y1); %%(m,n+1)
d11 = b*(1+b-(b^2))*img(x2,y2); %%(m+1,n+1)
d12 = -(b^2)*(1-b)*img(x2,y3); %%(m+2,n+1)
d13 = -(a^2)*(1-a)*(-b*((1-b)^2))*img(x3,y0); %%(m-1,n+2)
d14 =(1-2*(b^2) + (b^3))*img(x3,y1); %%(m,n+2)
d15 = b*(1+b-(b^2))*img(x3,y2); %%(m+1,n+2)
d16 = -(b^2)*(1-b)*img(x3,y3); %%(m+2,n+2)
V(m',n') = (d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9 + d10 + d11 + d12 + d13 + d14 + d15 + d16);
PS: I don't want the source code shown here: https://thilinasameera.wordpress.com/2010/12/24/digital-image-zooming-sample-codes-on-matlab/. Also, I don't want to use 'interp2' or any other shortcut.
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Shubha B.
Shubha B. 2021-2-23
To understand the theory concept correctly. For above written code I am not getting the theory for that.
Rik
Rik 2021-2-23
If your main point is understanding the theory, does the exact implementation matter?

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回答（1 个）

Steven Lord
Steven Lord 2018-6-15
Why not just use interp2? The help for the 'cubic' method states:
'cubic' - bicubic interpolation as long as the data is
uniformly spaced, otherwise the same as 'spline'
If you're not allowed to use interp2 because this is homework, tell us what's blocking you from completing the implementation and we may be able to offer guidance.
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Gobert
Gobert 2018-6-15
I didn't want to use 'interp2' or any other short-cut. What is intriguing to me is simply the 'geometric transformation' that will enable accessing (x0, y0), etc., at the edges of an image, without generating errors (even if it requires padding that image, first).

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