Re: [math-fun] When is base 2 not binary

28 Apr 2018

      * Lucas, Stephen K - lucassk <lucassk@jmu.edu> [Apr 28. 2018 09:09]:
...
Joerg,
...
Yes, you have the reference correct to Katai & Szabo. If you read
their proof carefully, you discover that they essentially use a
carry rule to represent a Gaussian integer in base -n+i, and prove
existence and uniqueness. But explicitly writing things in terms of
a carry rule makes number conversion and arithmetic much easier to
deal with.
Thanks!
...
I must admit, after looking back at your list of representations in
base 2+i using {1,-1,i,-1,0}, I’m intrigued at how you came up with
the digits. It is far from obvious, and the number of digits going
up then down is bizarre!
I am doing this in a truly brain-dead way:
testing every possible digit!

Here is my (UGLY) pari/gp code to do this:

-----------
to_rdx(n, rx, dg)=
/* convert n into radix rx expansion using digits given in dg[] */
{
    my(d, t, v, ct, i);

    i = 1;
    v = vector(30); /* vector of digits, must be long enough */

    while ( n != 0,
        for (j=1, #dg,  \\ for each digit d...
            \\ test whether denom((n-d)/rx)==1:
            d = dg[j];
            t = (n - d) / rx;
            dt = denominator(t);
            if ( dt == 1,
                v[i] = d;
                i += 1;
                n = t;
                break();
            );
            \\ no digit worked:
            if ( j==#dg, error(" radix rx and digit set dg[] are not compatible") );
        );
    );
    v = vector(i-1, k, v[k]);  /* reverse digits */
    return(v);
} /* ----- */

from_rdx(v, rx, dg)=
/* inverse of to_rdx(), for testing */
{ return( sum(j=1, #v, v[j] * rx^(j-1) ) ); }

\\ Our parameters:
rx=+2+I; dg=[0,+1,-1,+I,-I];

{ for (n=-33, 125,
    x = n; 
    v=to_rdx(x, rx, dg);
    m=from_rdx(v, rx, dg);
    print(x,": ",v);
    if ( x!=m, error("Epic fail!") );
); }
-----------

Oh, there is s reference in that file:
\\ From: rcs@xmission.com
\\ To: math-fun@mailman.xmission.com
\\ Subject: Re: [math-fun] Converting numbers with complex bases
\\ cf. Message-ID: <20110713000524.2e83jckd4cgw400g@webmail.xmission.com>

The message my code seems to refer to is this one:

-----------
...
Date: Tue, 12 Jul 2011 12:51:50 -0400
From: Andy Latto <andy.latto@pobox.com>
To: math-fun <math-fun@mailman.xmission.com>
Reply-To: math-fun <math-fun@mailman.xmission.com>
Subject: Re: [math-fun] Converting numbers with complex bases
X-Spam-Score: -3.3 (---)
On Tue, Jul 12, 2011 at 11:22 AM, Kerry Mitchell <lkmitch@gmail.com> wrote:
...
Hi,
Using the digits {0, 1} and the complex base -1+i, it's straightforward to
show that 1010 = 1+3i.  Is there an easy way to go backward?  That is, given
a complex number in the form a+bi, recover its "binary" representation?  I'm
immediately interested in the case for bases -1+i and -1-i (digits {0, 1}),
but also interested in the general case.
The standard procedure for converting a number to another base works
fine. If we want to express a number N in base B, divide N by B,
finding Q and R such that
(*) N = B.Q + R
where R is a digit. Then find the base-B representation of Q, and add
an R on the end. If (*) always has a solution, then any number can be
represented in that base with that digit set. If the solution of (*)
is always unique, the representation is unique.
Andy

Best regards,   jj
...
Steve
[...]

Re: [math-fun] When is base 2 not binary

Joerg Arndt