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Create a 7x7 convolution kernel which has an equivalent effect to three passes with a 3x3 mean filter.

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Emma_m 2021 年 1 月 5 日
コメント済み: Rik 2021 年 1 月 11 日
How do I create a 7x7 convolution kernel that would have the have the same effect as 3 passes of a 3x3 mean filter. Assume I have a picture thats 7x7, just trying to quicken up the process but unsure of the kernel values.
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Rik 2021 年 1 月 11 日
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回答 (1 件)

Steve Eddins
Steve Eddins 2021 年 1 月 11 日
Convolution is associative:
Associativity of convolution
So, your 7x7 convolution kernel would be the convolution of the mean filters:
Use the 'full' option of conv2 to do this.
Note: I wouldn't necessarily expect convolving once with 7x7 to be faster than convolving three times with 3x3 filters. Convolving with a 7x7 filter is 49 multiplications and adds per pixel, whereas convolving three times with 3x3 filters is 3 times 9 = 27 multiplications and adds per pixel.
There is some overhead (such as memory allocation and input processing) associated with each convolution operation, so execution time is not just about counting floating-point operations. Generally, though, decomposing a larger filter into a sequence of smaller filters is a technique used to speed up operations, not slow them down.

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