Re: [flint-devel] Function Interface Updates

[<D.C...@wa...>]

Hi,

> Wow, that's detailed. Whoever goes to implement this will find it quite
>easy.
>
> One thing that worries me is we decided to represent a p-adic as a
>multiprecision integer, but if I am reading what you've written
> correctly, you are thinking of a padic as a power series in number 6
>below. Or am I just misunderstanding what you mean?

I think you may be misunderstanding me, all the arithmetic I've planned is
by representing a p-adic as an integer. My reasoning for (6) was in case a
user wished to input a p-adic as you would by hand, say 3^-1 + 1 + 2*3 +
3^2 + O(3^3) (or in the notation below; 1 . 1 2 1) instead of inputting it
more directly as an integer. This function would allow us to convert the
input into a format that we would do arithmetic with. All the arithmetic
I've planned is done with integers that represent the p-adic, theres no
arithmetic planned using the power series representation.

I was thinking as to what would be the best way to deal with say 16/5
which lies in Z_3. I was proposing that we store two integers to represent
such a p-adic as (16,5) instead of storing (16/5)%(3^n); However will
storing (16/5)%3^n be faster than storing two integers? The actual
manipulation of the integers will take longer if we store two integers. We
have to check if a power of p divides the denominator straight away
otherwise the inverse modulo p^n wont exist. For example if we wanted
16/15 +O(3^5), we would have to take out 3^-1 straight away to get 16/5
and then find 16/5%3^6. We need to check the numerator for powers of p
also since if we perform a division we could end up with a power of p on
the denominator, in which case the reductions modulo p^k will be
impossible. This was one of the things I was trying to take care of in my
last email outlining division.

In my previous email I was working on the basis that we store two integers
for each p-adic, hence the a,b,c,d, but things will be simpler if we store
a single integer. I'll have a look at this, re-write it and send it again
later.


> By the way, I think you asked a few days ago about how to represent
>fractions in Qp. I think it will be enough to represent them as an
>element of Zp multiplied by the appropriate power of p (where all we
store is >the exponent of the power), e.g:

Yes, all the information about the structure of Q_p we need is stored in
Z_p so we can reduce arithmetic of Q_p to Z_p in this way which is how
I've been thinking of doing things. So for example if we had 17/6 +
O(3^5), we would write this as (3^-1)*(17/2) + O(3^6).



> 2*3^-2+3^-1+1+2+O(3^2) would be represented as:
>
> -2, 2+3+3^2+2*3^3+O(3^4) which will really get stored as the quadruple
of numbers:
>
> -2, 68, 3, 4 i.e. /3^-2)*(68 mod 3^4)
>
> Does that look like it works?
>
> Bill.
>


If x = 2*3^-2 +3^-1 + 1 +2*3 + O(3^2)
then I would write x*3^2 = 2 + 3^1 + 3^2 + 2*3^3 + O(3^4)

So we would need to store -2, 68, 3 and 4 as you say.


Daniel Ellam