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Re^2: Non-recursive Ackermann

by BrowserUk (Pope)
on Feb 04, 2011 at 09:52 UTC ( #886168=note: print w/ replies, xml ) Need Help??


in reply to Re: Non-recursive Ackermann
in thread [OT]: threading recursive subroutines.

I was able to construct a non-recursive implementation of the Ackermann function.

Impressive work!

Maybe you'll have better luck with that.

I couldn't see much opportunity for parallisation in your first version.

Update: Cleaned up code a bit.

That's certainly easier to follow. (*)

The 'obvious' thing would be to async the inner for loop. Except that with maximum realistic values of m being 5, starting a thread for 5 iterations is nonsense.

So then my attention moves to the outer loop, with the thought that you can often split the range of a loop into subranges and run them in parallel.

But, firstly we don't know the range. Secondly, even iteratively, the dependency of each subsequent iteration upon the previous would require lock-stepping the threading--either through complex locking or queuing--and with the minimal effort required for each step of each iteration versus the cost of locking and context switching, again make that a non-starter.

Upshot: I think you called it earlier with your z = F(x); return F(z); comment that went over my head at the time. In the light of that, I've modified my first tentative rule from a post above to be:

For a recursive algorithm--that is not tail recursive--to be a suitable candidate for parallelisation, it requires that it has multiple, independant, recursive calls at each level.

The interesting thing now is: how many recursive algorithms fit that pattern?

m < 4 can be special-cased. Known formula exist for A(0,n), A(1,n), A(2,n) and A(3,n).

Indeed. The following returns all the tractable cases within the machine capability in a blink of an eye:

sub ackermann_c { use constant INF => exp(~0 >> 1); my( $m, $n ) = @_; --$m, $n=1 if $n == 0; return $n + 1 if $m == 0; return $n + 2 if $m == 1; return 2*( $n+3)-3 if $m == 2; return 2**($n+3)-3 if $m == 3; return INF unless $m == 4; my( $p, $r ) = ( $n+3, 2 ); $r = 2**$r while --$p; return $r - 3; }

But that now means I need to find a more suitable testcase for my experiments.

An in-place, parallised qsort maybe?

(*)Though wonder why you do a loop initialisation rather than my @n = (-1 ) x $m; my @A = ( 1 ) x $m; given that m will realistically max out at 5. And whilst I'm on unimportant, purely subjective matters, why for(;;) not while( 1 )?


Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.


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Re^3: Non-recursive Ackermann
by ikegami (Pope) on Feb 04, 2011 at 16:24 UTC

    [ Just a few rushed little comments ]

    Thanks!

    we don't know the range.

    would require lock-stepping the threading-

    with the minimal effort required for each step

    I reached the same conclusions.

    it requires that it has multiple, independant, recursive calls at each level.

    That's why I used fib at first and called Ackermann a whole other ball game.

    The interesting thing now is: how many recursive algorithms fit that pattern?

    Any divide and conquer algorithm, for starters. That's where the dependency graph came in. I wanted to see if there was a split in it that could be exploited.

    wonder why you do a loop initialisation rather than

    I didn't know what I was going to end up with when I started. The code got refactored multiple times. I didn't micro-optimise.

    (Upd: Put differently, the snippet is a theoretical implementation, not a practical one. Work still needs to be done to make it practical. Well, as practical as Ackermann can be. )

    why for(;;) not while( 1 )?

    It's what the code I learned from used. I've never used while (1). I like for (;;) cause I read it as "for ever". Trivia: Since the entire bound check of while (CONSTANT) gets optimised away, and since "while" and C-style "for" use the same opcodes, while (1) and for (;;) are practically identical internally.

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