World: r3wp

It might be a good idea to run a peephole optimizer on the patterns 
before compiling them, to convert ones like "a*a" to "aa*".

It looks like repeat doesn't bind its argument, so it is definitely 
faster in this case.

; Version with support for decreasing ranges
regset: func [expression /local out negate? b e x] [
    negate?: false
    out: make bitset! []
    parse/all expression [
        opt ["~" (negate?: true)]
        some [
            "-" (insert out #"-") |
            b: skip "-" e: skip (
                b: first b  e: first e
                either b > e [
                    insert out e
                    repeat x b - e [insert out e + x]
                ] [
                    insert out b
                    repeat x e - b [insert out b + x]
                ]
            ) |
            x: skip (insert out first x)
        ]
    ]
    if negate? [out: complement out]
    out
]

; Version without support for decreasing ranges
regset: func [expression /local out negate? b e x] [
    negate?: false
    out: make bitset! []
    parse/all expression [
        opt ["~" (negate?: true)]
        some [
            "-" (insert out #"-") |
            b: skip "-" e: skip (
                b: first b  e: first e
                either b > e [
                    insert insert insert out b #"-" e
                ] [
                    insert out b
                    repeat x e - b [insert out b + x]
                ]
            ) |
            x: skip (insert out first x)
        ]
    ]
    if negate? [out: complement out]
    out
]

; Slightly faster
bitset-to-string: func [b [bitset!] /local s x] [
    s: copy either find b 0 ["^(00)"] [""]
    repeat x 255 [
        if find b x [append s to-char x]
    ]
    s
]

I'll look more at this tomorrow.

Hi Brian, thanks for support, I was out for a sleep :)

BrianH, what ? Repeat does bind its argument, doesn't it ?
>> repeat n 4 []
>> n
** Script Error: n has no value
** Near: n

Yeah, so it does. I wonder why the docs don't say (will be local) 
like it does for foreach. It still ends up faster than loop when 
you have to keep track of an index or a counter.

Brian, you've stated that  "It is faster for parse to match a one-character 
string than a character value." It seems to me that the opposite 
statement is true. (matching a char! is faster than matching a on-character 
string!)

It seems to me that the opposite _should_ be true, but parse converts 
the character to a string before matching it - no conversion is performed 
for string values. It's just one of those weird things.

This is one of those things that I actually profiled to determine 
for sure. Strange, but true.

my measurements show:

>> time-block [parse "a" ["a"]] 0.05
== 3.83615493774414E-7
>> time-block [parse "a" [#"a"]] 0.05
== 3.61204147338867E-7

, i.e. the opposite

Which version? Nevermind, my timing differences may just be a multitasking 
artifact.

>> rebol/version
== 1.3.2.3.1

Strange. When I tested before the difference turned out to be about 
the same as the overhead of string conversion. Quite the cooincidence. 
I guess I should have repeated my timings more :(

Too small a sample for a busy computer.

use can use time-block to make sure (it's on my site)

you can use...

nice to see char is faster than string, so I'm not going to rewrite 
this ;) Anyway, new version with BrianH's version of regex and bitset-to-string 
and with endless loop in tail-parse fixed is posted on http://bolek.techno.cz/reb/regex.r

That's funny, I thought we had determined that string was faster 
than char.

It's probably close enough to ignore either way.

Rebolek, I gather you made the parse go in reverse to handle rules 
like "a+a" better. How does your reverse code handle "aa+", or "aa+a" 
- same problem?

Here's another benchmark:

>> data: head insert/dup make string! 10'000'000 #"a" 10'000'000

>> t0: now/time/precise loop 10 [parse data [some "a"]] now/time/precise 
- t0
== 0:00:06.078

>> t0: now/time/precise loop 10 [parse data [some #"a"]] now/time/precise 
- t0
== 0:00:04.296


Running this test several times shows that char! matching is, in 
average, 30 % faster than string! matching.

Well there you go. That's different numbers than last time, but more 
dramatic. It's just a #, easy fix :)

Thanks for answering that question though.

Just tried your benchmark on my machine.....I get similar % faster 
for char! matching.
Though the elapse times vary depending on the version of REBOL.

Didn't want to sound "dramatic", but just wanted to provide a more 
accurate measure. Sure whatever datatype is used (char! or string!) 
in regex.r, that won't  change much the overall speed. ;-)

I'm more concerned about the first and follow sets - doing that wrong 
could mean a slowdown of orders of magnitude.

Actually, big O slowdowns.

BrianH: "How does your reverse code handle "aa+", or "aa+a"" - damn... 
:)

Anyway, today I was thinking about different way how to do it, had 
no time to code it yet.
It's slow, I know...

Are you familiar with the theories behind parser generators? That 
is what you are doing. I studied the theories in college - hence 
the talks about first and follow sets.

I've been thinking about ways to do this too - you got me going. 
I was also thinking of a way to cache charsets for later reuse.

didn't you guys try using a charset?  I would guess its an order 
of magnitude faster still... no?

well, checking with one letter didn't give any changes (bitset and 
char seem the same) but just change that with an OR of two letter 
and the char goes to 4 times slower and the bitset stays exactly 
the same... so do use bitsets as often as you can.

>> t0: now/time/precise loop 10 [parse data [some #"a"]] now/time/precise 
- t0
== 0:00:06.389


>> t0: now/time/precise loop 10 [parse data [some [#"b" | #"a"]]] 
now/time/precise - t0
== 0:00:25.496

>> b: charset "ab"

>> t0: now/time/precise loop 10 [parse data [some b]] now/time/precise 
- t0
== 0:00:06.379

I'm not sure how can I test for end in case like this:
r: ["a" "b" "c" end] parse "abc" [some [r/1 (r: next r)]]
anybody?

r: ["a" "b" "c" break] parse "abc" [some [r/1 (r: next r)]]

great, thanks

Here are C sources for Regex engine used in MySQL5.0.9 http://leithal.cool-tools.co.uk/sourcedoc/mysql509/html/dir_000079.html

So I left tail-parse becuase of the "aa+" problem mentioned by BrianH. 
I wrote 'lazy-parse that works different and it seems to handle both 
cases "a+a" and "aa+" well.

>> lazy-parse "aaa" [[#"a"][some #"a"][#"a"]]
== true
>> lazy-parse "aaa" [[#"a"][some #"a"]]
== true
>> lazy-parse "aaa" [[some #"a"][#"a"]]

lazy-parse: func [string rules /local parsed?][
;does not support ANY yet (wrong results)
	parsed?: false
	rules: copy rules
	parse string [
		some [
			b: rules/1 e:
			(
				parsed?: true
				rules: next rules
				if empty? e [
					either empty? rules [
						parsed?: true
						rules: copy [break]
					] [
						e: next b
						if empty? e [parsed?: false]
					]
				]
			)
			:e
		]
	]
	parsed?
]

Warning: this does not suport ANY rule yet.

Grrr, I should make more test before posting...forget last four posts

'regset is now on rebol.org, rest of parser is still in works.

Rebolek...You posted the magic prime 797'th Script.  Well Done!  
:)

Hm thanks, but I was aiming for 800th one... ;)

Yep...I've been waiting...  :)

just a quick idea:

FORALL is implemented as mezzanine function. It calls FORSKIP which 
is mezzanine also. As you can see, it's probably not the fastest 
method. So here's the idea. Cannnot be FORALL rewritten to make it 
faster and is it possible to use PARSE to do this?
So I tried and came up with this simple function:

parall: func [
	'word body
	/local data
][
	data: get :word
	parse data compose/deep [

  some [(to set-word! word) any-type! (to paren! [do bind body :word])]
	]
]

(parall is just a shortcut for parse version of forall).


this is very simple function written in five minutes and not very 
well checked, it needs some additional work (eg. it does not return 
same value as 'forall etc).

So let's do some test (using Ladislav's %timblk.r):
>> n: copy [] repeat i 100 [append n i]

== [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 4...
>> time-block [forall n [b: n/1 + 1]] 0.05
== 3.623046875E-4
>> time-block [parall n [b: n/1 + 1]] 0.05
== 3.814697265625E-6
>> 3.62e-4 / 3.81e-6
== 95.0131233595801

95x faster? whooo....

and what about bigger block?

>> n: copy [] repeat i 10000 [append n i]

== [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 4...
>> time-block [forall n [b: n/1 + 1]] 0.05
== 3.540625E-2
>> time-block [parall n [b: n/1 + 1]] 0.05
== 3.7994384765625E-6
>> 3.54e-2 / 3.8e-6
== 9315.78947368421

9000x ? omg...

comments?

the devil is in the details... you need to handle break :)