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The only cherry picking is that the numbers need to be small enough that the decimal digit does not overflow (4 times 2 is fine but 4 times 3 is not). A more

The only cherry picking is that the numbers need to be small enough that the decimal digit does not overflow (4 times 2 is fine but 4 times 3 is not). A more 'convolved' example: 121 * 23 = 2783 If you use hex instead it also works for larger digits like 0x42 * 0x31 = 0xCA2 = [12, 10, 2] So, if you convert to a digit of a large enough base you can do any multiplication with just a convolution and no carry.

rualca 6y ago
> The only cherry picking is that the numbers need to be small enough that the decimal digit does not overflow Aren't you restating my point? > So, if you convert to a digit of a large enough base you can do any multiplication with just a convolution and no carry. There is no convolution at all. There's only a cherry-picked example of how plain old multiplication feels similar to a sliding dot product, which for some reason some people confuse with convolution. This is not helpful, and only adds to the confusion already expressed.
- antonfire 6y ago
  The first few Google results for "sliding dot product" basically describe it as a convolution with the domain of one of the functions reversed. (Note as far as whether or not to reverse one of the functions, multiplication in base-b agrees with "convolution" and not with "sliding dot product".) Elsewhere in the thread you seem to be suggesting that convolution is a way to convert between a frequency domain and a time domain, which suggests that you are confusing convolution with Fourier transforms. There's some nice relationships between these things, so they are often discussed together, but even then a single convolution happens either entirely in the time domain or entirely in the frequency domain, not as a way of going from one of those domains to the other. E.g. the pointwise product (in the frequency domain) of the Fourier transforms of two functions is the Fourier transform of the convolution (in the time domain) of those two functions. There's also a chance you are focused on the difference between functions with a discrete domain and functions with a continuous domain. The term "convolution" is often applied to both.
- AdhemarVandamme 6y ago
  You’re right that the analogy between elementary-school-multiplication and convolution only works for multiplication without carrying. That still leaves a large amount of possible multiplication instances, for the term “cherry picking” to apply here. You’re wrong to suggest that elementary-school-multiplication and (discrete) convolution are so dissimilar that the analogy is so unhelpful that it only adds to everyone’s confusion. Maybe to yours, but not to everyone’s. If I understand your objection correctly (other than the already conceded fact that the analogy does not work in case of multiplication with carrying), you already have internalized that convolution is an operator on 2 functions, resulting in a function. And since you have already internalized that, the analogy with any operator on two natural numbers resulting in a natural number confuses you — although I assume you can see the similarity in the sliding operation. For people who have not internalized what a convolution is, the visualisationg of the sliding operation, and the analogy with elementary-school-multiplication is the helpful bit. To clarify why the operations are really similar: consider a natural number as a function that takes a natural number as argument, and returns the digit (between 0 and 9 inclusive) for that decimal position. So f = 2 031 is considered to be: n → f(n) = if n = 3 then return 2 else if n = 1 then return 3 else if n = 0 then return 1 else return 0 (end if). And g = 320 is considered to be: n → g(n) = if n = 2 then return 3 else if n = 1 then return 2 else return 0 (end if). What is then the discrete convolution f ∗ g? It’s f ∗ g = n → (f ∗ g)(n) = f(3) × g(n – 3) + f(1) × g(n – 1) + f(0) × g(n) = n → if n = 5 then return 6 else if n = 4 then return 4 else if n = 3 or n = 2 then return 9 else n = 1 then return 2 else return 0 (end if). This is how we consider 649 920 = 2 031 × 320.
  - rualca 6y ago
    > You’re wrong to suggest that elementary-school-multiplication and (discrete) convolution are so dissimilar (...) They are way more than dissimilar, they are radically different concepts altogether. Convolution is a function, or an operator that outputs a function given two input functions that share the same domain. Plain old algebra over real numbers is nothing of the sort. You're trying to compare a function, with its domain and codomain, with an operation between scalars that results in a scalar. It's way more than apples to oranges. It's apples to orange tree plantations.

rualca 6y ago

> The only cherry picking is that the numbers need to be small enough that the decimal digit does not overflow Aren't you restating my point? > So, if you convert to a digit of a large enough base you can do any multiplication with just a convolution and no carry. There is no convolution at all. There's only a cherry-picked example of how plain old multiplication feels similar to a sliding dot product, which for some reason some people confuse with convolution. This is not helpful, and only adds to the confusion already expressed.

antonfire 6y ago
The first few Google results for "sliding dot product" basically describe it as a convolution with the domain of one of the functions reversed. (Note as far as whether or not to reverse one of the functions, multiplication in base-b agrees with "convolution" and not with "sliding dot product".) Elsewhere in the thread you seem to be suggesting that convolution is a way to convert between a frequency domain and a time domain, which suggests that you are confusing convolution with Fourier transforms. There's some nice relationships between these things, so they are often discussed together, but even then a single convolution happens either entirely in the time domain or entirely in the frequency domain, not as a way of going from one of those domains to the other. E.g. the pointwise product (in the frequency domain) of the Fourier transforms of two functions is the Fourier transform of the convolution (in the time domain) of those two functions. There's also a chance you are focused on the difference between functions with a discrete domain and functions with a continuous domain. The term "convolution" is often applied to both.
AdhemarVandamme 6y ago
You’re right that the analogy between elementary-school-multiplication and convolution only works for multiplication without carrying. That still leaves a large amount of possible multiplication instances, for the term “cherry picking” to apply here. You’re wrong to suggest that elementary-school-multiplication and (discrete) convolution are so dissimilar that the analogy is so unhelpful that it only adds to everyone’s confusion. Maybe to yours, but not to everyone’s. If I understand your objection correctly (other than the already conceded fact that the analogy does not work in case of multiplication with carrying), you already have internalized that convolution is an operator on 2 functions, resulting in a function. And since you have already internalized that, the analogy with any operator on two natural numbers resulting in a natural number confuses you — although I assume you can see the similarity in the sliding operation. For people who have not internalized what a convolution is, the visualisationg of the sliding operation, and the analogy with elementary-school-multiplication is the helpful bit. To clarify why the operations are really similar: consider a natural number as a function that takes a natural number as argument, and returns the digit (between 0 and 9 inclusive) for that decimal position. So f = 2 031 is considered to be: n → f(n) = if n = 3 then return 2 else if n = 1 then return 3 else if n = 0 then return 1 else return 0 (end if). And g = 320 is considered to be: n → g(n) = if n = 2 then return 3 else if n = 1 then return 2 else return 0 (end if). What is then the discrete convolution f ∗ g? It’s f ∗ g = n → (f ∗ g)(n) = f(3) × g(n – 3) + f(1) × g(n – 1) + f(0) × g(n) = n → if n = 5 then return 6 else if n = 4 then return 4 else if n = 3 or n = 2 then return 9 else n = 1 then return 2 else return 0 (end if). This is how we consider 649 920 = 2 031 × 320.
- rualca 6y ago
  > You’re wrong to suggest that elementary-school-multiplication and (discrete) convolution are so dissimilar (...) They are way more than dissimilar, they are radically different concepts altogether. Convolution is a function, or an operator that outputs a function given two input functions that share the same domain. Plain old algebra over real numbers is nothing of the sort. You're trying to compare a function, with its domain and codomain, with an operation between scalars that results in a scalar. It's way more than apples to oranges. It's apples to orange tree plantations.