From: Niklas Holsti <niklas.holsti@tidorum.invalid>
Subject: Re: Inefficient algorithms
Date: Sat, 11 Sep 2010 14:20:09 +0300
Date: 2010-09-11T14:20:09+03:00 [thread overview]
Message-ID: <8f16vaF2q1U1@mid.individual.net> (raw)
In-Reply-To: <87lj78ab6m.fsf@ludovic-brenta.org>
Ludovic Brenta wrote:
> Niklas Holsti <niklas.holsti@tidorum.invalid> writes:
>> Niklas Holsti wrote:
>>> Rick wrote:
>>>> I am working on a class in algorithmic efficiency and, to make a
>>>> point, need an algorithm that runs O(2^n) - a factoring algorithm
>>>> perhaps. All I can find is 'C' code I can't decipher. Can someone
>>>> please help with an Ada example.
>>> Perhaps the canonical example is this: you are given a Boolean
>>> formula containing n Boolean variables. Find a valuation (a value,
>>> True or False, for each variable) that makes the formula True.
>> Replying to myself, since I realize that I probably misunderstood your
>> need: what you want is an inefficient O(2^n) solution to an easy
>> (non-exponential) problem. The problem I suggested is not, of course,
>> an easy one. But it has easy variants. For example, require that the
>> Boolean formula uses at most 10 of the n variables, where this limit
>> (10) is considered a fixed number independent of the total number (n)
>> of variables.
>
> Actually I do think that the problem is trivial and that you can solve
> it without recursion. If you see your array of Booleans as a number in
> binary representation:
>
> procedure Inefficient (N : in Natural) is
>
> subtype Valuation is Natural range 1 .. 2**N;
That will of course fail to compile when N grows large enough, as it
will when we are talking of order-of-complexity questions. (Also, I
think you really wanted the range 0 .. 2**N - 1, giving the N-bit binary
numbers).
> -- We assume the solution is Valuation'Last but it could be any value.
> Solution : constant Valuation := Valuation'Last;
> begin
> for K in Valuation'Range loop
> exit when K = Solution;
Here you are assuming that the Boolean formula has the simple form
Var1 = K1 and Var2 = K2 and Var3 = K3 and ... Var<N> = K<N>
where K1..K<N> are the Boolean constants that correspond to the constant
bits in your Solution. I had in mind a more general formula where, for
example, we can compare two variables (Var1 = Var4), negate variables,
and so on.
In the terms of my first post, this algorithm would be:
-- The formula for which a solution is sought:
function Formula (Input : Valuation) return Boolean
is begin
return <a complex Boolean formula depending on the N bits
in the Input parameter>;
end Formula;
-- The search for the solution:
for K in Valuation loop
exit when Formula (K);
end loop;
A trivial amount of code, indeed, but since Valuation has 2**N values,
the loop will iterate 2**N times, so this algorithm is also O(2**N), as
for the recursive one.
This example is just the Boolean satisfiability problem (SAT), a
well-known difficult problem, which is not trivial in the general case.
--
Niklas Holsti
Tidorum Ltd
niklas holsti tidorum fi
. @ .
next prev parent reply other threads:[~2010-09-11 11:20 UTC|newest]
Thread overview: 23+ messages / expand[flat|nested] mbox.gz Atom feed top
2010-09-11 4:24 Inefficient algorithms Rick
2010-09-11 6:51 ` J-P. Rosen
2010-09-13 3:45 ` Robert A Duff
2010-09-11 6:54 ` Niklas Holsti
2010-09-11 7:07 ` Niklas Holsti
2010-09-11 9:07 ` Rick
2010-09-11 15:05 ` Niklas Holsti
2010-09-17 5:26 ` Rick
2010-09-11 9:20 ` Ludovic Brenta
2010-09-11 9:23 ` Ludovic Brenta
2010-09-11 11:20 ` Niklas Holsti [this message]
2010-09-11 18:29 ` Peter C. Chapin
2010-09-11 14:28 ` stefan-lucks
2010-09-12 1:04 ` Wilson
2010-09-12 1:53 ` Rick
2010-09-12 8:35 ` Georg Bauhaus
2010-09-12 11:56 ` stefan-lucks
2010-09-15 1:11 ` BrianG
-- strict thread matches above, loose matches on Subject: below --
2010-09-15 8:51 Rick
2010-09-15 21:45 ` John B. Matthews
2010-09-16 12:05 ` Chad R. Meiners
2010-09-16 20:19 ` John B. Matthews
2010-09-17 5:24 ` Rick
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