@simon_brooke @vashti Only if it doesn’t also need to be fast.
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@alwayscurious @vashti define fast. I can compute the factorial of 1,000 in under one millisecond on an ordinary laptop in several different #Lisp dialects, without optimisation.
Is that not fast enough for you?
https://gist.github.com/simon-brooke/fcb59705950c5ad515e18fba065510ae
Agreed. #CommonLisp on Android ( #cl_repl) is capable and plenty fast enough on a Pixel 8a running your factorial code. 😎
You can make CL faster than the C/C++ used by NumPy: https://stewart123579.github.io/blog/posts/code/calculating-a-dot-product/#python-vs-dot-r-vs-dot-lisp-timing
![Partial screenshot with the following text:
QML-USER [1]
(defun fact (n) "Compute the factorial o...
real time : 0.000 secs
run time : 0.000 secs
gc count : 1 times
consed : 654512 bytes](https://cdn.fosstodon.org/media_attachments/files/115/771/823/878/357/887/original/3374a426f3cc16d6.png)
@svw @alwayscurious @vashti from the cons space usage I'm guessing that was the recursive algorithm? If so, the more impressive. Small #Lisp imlementations tend not to have dynamic stack. But allocating and deallocating stack frames on a dynamic stack is a time cost.
@svw
Sorry, could you time the
(mytime naïve-dot-v2 "naïve loop v2"
(loop for x across aa
for y across ab
summing (* x y)))
idiom instead?
Thinking about it, I'm not sure how to write DO for this. I have been interested in DO recently.
@simon_brooke @alwayscurious @vashti
@screwlisp Right, what arguments do you want me to test?
@simon_brooke could you do something comparable to svw. but with
(loop for x fixnum across arrayx for y fixnum across arrayy summing (* x y) into z fixnum finally (return z))
and
(do ((x (pop listx) (pop listx)) (y (pop listy) (pop listy)) (z 0))
((null listx) (incf z (* x y)))
(declare (fixnum x y z))
(incf z (* x y)))
which are my swing at fast forms closely analogous to svw's fast lisp blogging. I'm also interested in how similar performance gains and losses are between lisps.
@screwlisp aye, but arguments! What arguments to these functions would it be useful to you for me to test?
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@svw
Though, based on my laptop,
(defun doot2 (a b)
(do ((z 0)
(x (pop a) (pop a))
(y (pop b) (pop b)))
((null a) (incf z (* x y)))
(declare (fixnum x y z))
(incf z (* x y))))
consistently beats
(defun loopdot (a b)
(loop :for x fixnum :in a
:for y fixnum :in b
:summing (* x y) :into z fixnum
:finally (return z)))
@simon_brooke @alwayscurious @vashti
@alwayscurious @vashti however, I'm grateful to you for raising that issue, because it prompted me to find out how fast it would run in Portable Standard #Lisp, so I've spent a happy couple of hours trying to debug bootstrapping problems in Blake McBride's version of the PSL compiler. I don't have it working yet, but I have made progress. Will it be faster than SBCL? Don't know, but I suspect so.
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