1999 1:03 AM
Subject: Mersenne: These go to 11 (WAS: blahblah...)
> > >or, more concisely, (1+1+1)^(1+1) + 1.
> > >Can anyone represent that number in fewer than (1+1+1)! ones?
>
> This all depends on what operators and notations are accepted and
> without specifying that,
> >or, more concisely, (1+1+1)^(1+1) + 1.
> >Can anyone represent that number in fewer than (1+1+1)! ones?
This all depends on what operators and notations are accepted and
without specifying that, the whole question is useless.
What about without any ones at all: (With C++ operators)
((0++)++)
> takes two.
>
> - Original Message -
> From: Ernst W. Mayer <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Tuesday, June 08, 1999 11:47 AM
> Subject: Mersenne: These go to 11 (WAS: blahblah...)
>
> >
> > Paul Leyland <[EMAIL PROTECTE
> Ground rules are critical, but how about
>
> /.1
>
> where "/" represents the APL-style monadic divide or
> multiplicative inverse.
>
> 1/.1
>
> takes two.
Indeed, but ".1" represents 1 / radix, so 1/.1 is just radix.
The whole point of my tongue-in-cheek posting was to indicate how an
im
7 AM
Subject: Mersenne: These go to 11 (WAS: blahblah...)
>
> Paul Leyland <[EMAIL PROTECTED]> wrote:
>
> >The radix is always 10.
> {snip}
> >or, more concisely, (1+1+1)^(1+1) + 1.
> >
> >Can anyone represent that number in fewer than (1+1+1)! ones?
>
&
Paul Leyland <[EMAIL PROTECTED]> wrote:
>The radix is always 10.
{snip}
>or, more concisely, (1+1+1)^(1+1) + 1.
>
>Can anyone represent that number in fewer than (1+1+1)! ones?
How about
1 << 1,
where the shift is, of course, decimal.
Your shifty friend,
-Ernst
