Hi Victor,
> On 13 August 2012 15:05, Ralph Corderoy <ra...@inputplus.co.uk> wrote:
> > $ by3() {
> > > sed -rn '
> > > h; s/[^147258]//g; y/147258/aaabbb/
> > > :l; s/ab//g; s/ba//g; tl
> > > G; s/^\n//p
> > > ' "$@"
> > > }
> >
>
> I must have missed this conversation.
Yes, me and Tim before your arrival, I think.
> I looked at your script and thought "divisibke by 3? That's true if
> sum of digits is divisible by three, isn't it?"
Right, that's the one where if the
http://en.wikipedia.org/wiki/Digital_root of the number is 0, 3, 6, or 9
then it's a multiple of 3 and is the first one of the two listed at
http://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_rules_for_numbers_1.E2.80.9320.
I was using the second one which I read as if the number of digits that
are 1, 4, or 7 is equal to the number that are 2, 5, or 8, then it's a
multiple of 3. But on going back for this email I realise that's wrong.
It's if the difference between the number of 1/4/7s and 2/5/8s is a
*multiple* of 3 then so is the original.
It happens that the 1/4/7s and 2/5/8s balance for all multiples up to
100 so the output in my email was correct but it breaks for 111; 3 - 0
= 3. The fix is to the last line of sed; I've widened the test too.
:-)
$ by3() {
> sed -rn '
> h; s/[^147258]//g; y/147258/aaabbb/
> :l; s/ab//g; s/ba//g; tl
> G; s/^(...)*\n//p
> ' "$@"
> }
$
$ diff <(seq -99 3 100000) <(seq -100 100000 | by3)
$
Cheers, Ralph.
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