6 ms·

Woha This is cool. It literally is the same, if you use base-infinity digits (not binary, or decimal, but infinity-ary?). How do you multiply two polynomials

Woha This is cool. It literally is the same, if you use base-infinity digits (not binary, or decimal, but infinity-ary?). How do you multiply two polynomials with a single variable?

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
- 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.
sriku 6y ago
Ref reply below - https://news.ycombinator.com/item?id=25195046 https://news.ycombinator.com/item?id=25195046
antonfire 6y ago
A polynomial with a single variable is more or less what you would mean by "use base-infinity digits" in the first place. Roughly, in base 10, the number represented by the sequence of digits d_0, d_1, ..., d_n is d_0 + d_1 * 10 + d_2 * 10^2 + ... + d_n * 10^n. In "base infinity", the "number" represented by the sequence of digits d_0, d_1, ..., d_n is the polynomial d_0 + d_1 * x + d_2 * x^2 + ... + d_n * x^n. Multiplying these "numbers" (polynomials) is like decimal multiplication, except the digits can be arbitrarily high and you do no carrying at all.

Woha This is cool. It literally is the same, if you use base-infinity digits (not binary, or decimal, but infinity-ary?). How do you multiply two polynomials with a single variable?

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
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.

sriku 6y ago

Ref reply below - https://news.ycombinator.com/item?id=25195046 https://news.ycombinator.com/item?id=25195046

antonfire 6y ago

A polynomial with a single variable is more or less what you would mean by "use base-infinity digits" in the first place. Roughly, in base 10, the number represented by the sequence of digits d_0, d_1, ..., d_n is d_0 + d_1 * 10 + d_2 * 10^2 + ... + d_n * 10^n. In "base infinity", the "number" represented by the sequence of digits d_0, d_1, ..., d_n is the polynomial d_0 + d_1 * x + d_2 * x^2 + ... + d_n * x^n. Multiplying these "numbers" (polynomials) is like decimal multiplication, except the digits can be arbitrarily high and you do no carrying at all.