@Mike
Yes, it does help, as
it is a sort of eye-opener, rephrasing the fourth factor (s) as (s-0), and
it answers my (yet un-uttered) question about "m14".
-M
At 2018-09-04 13:46, 'Mike Day' via Programming wrote:
Does this help?
Each line is a small amendment to the preceding one...
(-:@(+/))3 4 5 NB. Semiperimeter, s
(-:@(+/)-0&,)3 4 5 NB. s - 0, a, b, c
6 3 2 1
(-:@(+/)*/@:-0&,)3 4 5 NB. s * (s - a) * ...
36
(-:@(+/)%:@(*/)@:-0&,)3 4 5 NB. Heronâs formula applied to 3 4 5
(-:@(+/)%:@(*/)@:-0&,) NB. Let interpreter remove unnecessary brackets...
-:@(+/) %:@(*/)@:- 0&,
So the semiperimeter is calculated just the
once. It relies on converting the triplet a,b,c
to the quadruplet 0, a, b, c, rather than doing particularly smart bracketing.
I donât often use [: but if you prefer
it, the following arises from a similar
building process using [: rather than @ and @:
([:-:(+/))3 4 5
(([:-:(+/)) - 0&,)3 4 5
6 3 2 1
(([:-:(+/))( [: */ - )0&,)3 4 5
36
(([:-:(+/))( [: %: [: */ - )0&,)3 4 5
(([:-:(+/))( [: %: [: */ - )0&,) NB. Get rid of extra brackets
([: -: +/) ([: %: [: */ -) 0&,
Cheers,
Mike
Sent from my iPad
> On 4 Sep 2018, at 12:50, Martin Kreuzer <[email protected]> wrote:
>
> Hi all -
>
> To calculate the area of a flat triangle, using Heron's formula,
> A(a,b,c)= sqrt( s2*(s2-a)*(s2-b)*(s2-c) )
> I wrote a simple function doing this:
>
> * get the three sides (as list input y)
> * compute the half perimeter s2
> * build the differences s2-y
> * build product
> * take square root
>
> My explicit solution looks like this
>
> taher=: 13 : '%: s2 * */ s2-y [ s2=. -: +/ y'
>
> and works
>
> taher 3 4 5
> 6
>
> Suggested tacit version looks like this (and works too)
>
> tahert=: [: %: ([: -: +/) * [: */ ([: -: +/) - ]
>
> Q: Is there a way to reference the
intermediate result of ([: -: +/) the half perimeter s2
> within the tacit expression, as has been done in the explicit..?
> [Guess the interpreter takes care of this
anyway; my question aims at whether a shorter formulation could be reached.]
>
> Thanks
> -M
>
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