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> This is cool I don't agree the example has anything to do with convolution. The example is explained with plain old primary school multiplication. 100101 *

by rualca 6y ago

> This is cool I don't agree the example has anything to do with convolution. The example is explained with plain old primary school multiplication. 100101 * 23 = (100000+100+1)*23 = 2300000+2300+23 There isn't any fantastic property, only cherry-picked number which works as decimal left-shifts. Convolution is not a number. Convolution is an operator that outputs a function.

Tomminn 6y ago
The parents point is that when you do the primary school multiplication algorithm, you are actually performing a discrete convolution.
- rualca 6y ago
  You really aren't. That assertion is like commenting that elementary arithmetics is discrete signal analysis because 1+1=2, which happens to match the amplitude of a resonance of a normalized vibration mode. It isn't. It's a cherry picked example. A broken clock which coincides with the current time.
  - enriquto 6y ago
    > It isn't. It's a cherry picked example. A broken clock which coincides with the current time. No. It really is. Look at the example. Nobody can convince you if you are already set up against that idea, but look at what happens when you multiply two polyomials P(z)Q(z). You obtain a polynomial whose coefficients are the convolution of those of P and Q. Now, a decimal number is "just" a polynomial evaluated on z=10. And a vibrating string is "just" a polynomial evaluated at z=exp(it). The algebra of convolutions and products is exactly the same in all three cases.
jaggirs 6y ago
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.
sriku 6y ago
(sum_i ai x^i)(sum_j bj x^j) when you expand that, the coefficient of x^k is the sum of aibj such that i+j=k which is discrete convolution. Set x=10 and you're dealing with decimal representation of numbers.
- rualca 6y ago
  > Set x=10 and (...) ...and you no longer have a function, only the image of a function at a specific point of its domain. Convolutions are functions, not functions evaluated at a specific, and cherry-picked point.

by rualca 6y ago

Tomminn 6y ago

The parents point is that when you do the primary school multiplication algorithm, you are actually performing a discrete convolution.

rualca 6y ago
You really aren't. That assertion is like commenting that elementary arithmetics is discrete signal analysis because 1+1=2, which happens to match the amplitude of a resonance of a normalized vibration mode. It isn't. It's a cherry picked example. A broken clock which coincides with the current time.
- enriquto 6y ago
  > It isn't. It's a cherry picked example. A broken clock which coincides with the current time. No. It really is. Look at the example. Nobody can convince you if you are already set up against that idea, but look at what happens when you multiply two polyomials P(z)Q(z). You obtain a polynomial whose coefficients are the convolution of those of P and Q. Now, a decimal number is "just" a polynomial evaluated on z=10. And a vibrating string is "just" a polynomial evaluated at z=exp(it). The algebra of convolutions and products is exactly the same in all three cases.

jaggirs 6y ago

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.

sriku 6y ago

(sum_i ai x^i)(sum_j bj x^j) when you expand that, the coefficient of x^k is the sum of aibj such that i+j=k which is discrete convolution. Set x=10 and you're dealing with decimal representation of numbers.

rualca 6y ago
> Set x=10 and (...) ...and you no longer have a function, only the image of a function at a specific point of its domain. Convolutions are functions, not functions evaluated at a specific, and cherry-picked point.