[PD] tri function in [expr~]

[Claude Heiland-Allen]

Hi Derek,

On 2017-08-13 17:33, Derek Holzer via Pd-list wrote:
> Dear list,
> 
> I am transcribing some code from C to Pd expressions, and I am
> wondering how I might best get the tri function working since it's not
> native to expr.

My favourite triangle function is more expensive (division, branching) 
but has a variable shape:

[expr~ if ($v1 < $v2, $v1 / $v2, (1 - $v1) / (1 - $v2))]

all of $v1, $v2, output are in [0..1), with $v2=0.5 for symmetrical 
triangle

> I have been using this expression as my triangle function:
> 
> (min($v1, 1- $v1) - 0.25) * 4
> 
> So for this example:
> 
> wave_x = .5-triangle((phasor-9/16)/(2/16))*.5;

here you have 0.5 - (... * 0.5)

> I have tried the following, which doesn't function correctly compared
> with the original C:
> 
> [phasor~ 200]
> |
> [expr~ ($v1-9./16.)/(2./16.)]
> |
> [expr~ (0.5 - (min($v1, 1.- $v1) - 0.25) * 4.) * 0.5]

here you have (0.5 - ...) * 0.5, which isn't the same

> |
> [s~ wave_x]
> 
> I am quite sure I have made some errors in the syntax somewhere, and I
> did trial-and-error many other combinations before arriving at this
> one. Would someone more math-literate than me care to help work this
> out?
> 
> SIDENOTE: when is it necessary to force expr~ to consider a number as
> a float by using a decimal?
> 
> (9/16) doesn't work but (9./16.) does.

I think int/int -> int with truncation is the most common occasion where 
it matters in practice (except possibly overflow/wrapping with large 
ints?).  If at least one argument is a float it should promote the other 
to float too and give a float result.

> ((min($v1, 1- $v1) - 0.25) * 4) works regardless of whether the 1 and
> the 4 have a decimal point following them or not.

$v1 is a float so the 1 in 1-$v1 is promoted from int to float.
min(float, float) is a float so the 4 is promoted to float too.

> Thanks much!
> Derek

hth,


Claude
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