Re: Working with factors of triangular numbers.
- To: mathgroup at smc.vnet.net
- Subject: [mg78726] Re: [mg78490] Working with factors of triangular numbers.
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 8 Jul 2007 06:03:16 -0400 (EDT)
- References: <200707030923.FAA17995@smc.vnet.net> <6D2F6E70-2462-4B69-A617-4DEC322D69BF@mimuw.edu.pl>
I did not intend to work any more on this code, because, as I wrote
earlier, it is certiany possible to implement the same idea much
faster wihtout using the Combinatorica package, but then I noticed a
very blatant inefficiency in the posted code and felt obliged to
correect it. Here is the corrected version of the function FF. The
fucntion T is unchanged.
Andrzej Kozlowski
FF[n_] :=
Module[{u = FactorInteger[n], s, k, partialQ, finalQ, space, sp},
s = u[[All, 2]]; k = Length[u]; sp[m_] := Tuples[Range[0, m], {k}];
partialQ[l_List] :=
And @@
Flatten[{Last[l] == Array[0 &, k] || Not[MemberQ[Most[l], Last
[l]]],
Thread[Total[l] <= s - 1]}];
finalQ[l_List] :=
And @@
Flatten[{Last[l] == Array[0 &, k] || Not[MemberQ[Most[l], Last
[l]]],
Thread[Total[l] == s - 1]}];
space =
DeleteCases[ sp /@ (s - 1), Alternatives @@ (IdentityMatrix[k]),
Infinity];
k + Max[0,
Length /@
DeleteCases[Backtrack[space, partialQ, finalQ, All], Array[0 &,
k],
Infinity]]]
On 6 Jul 2007, at 20:26, Andrzej Kozlowski wrote:
> On 3 Jul 2007, at 18:23, Diana wrote:
>
>> Math folks,
>>
>> I first generate a list of triangular numbers:
>>
>> 1, 3, 6, 10, 15, 21, ...
>>
>> and then subtract one from each as:
>>
>> 0, 2, 5, 9, 14, 20, ...
>>
>> I am trying to find the smallest triangular number (minus one) which
>> can be written as a product of "n" distinct factors, each factor > 1.
>>
>> For example:
>>
>> a(2) = 15, because 2*7 + 1 = 15.
>> a(3) = 55, because 2*3*9 + 1 = 55.
>>
>> I have worked with Divisors and FactorInteger, but am getting bogged
>> down with repeated terms. Can someone think of a neat trick to work
>> this problem?
>>
>> Diana M.
>>
>>
>
> I wil start with a grumble. Unfortunately your problem is not, in
> my judgment, solvable by means of any nice mathematics, because you
> do not require the factors to be mutually prime, that is, not to be
> divisible by the same prime. Without that one can't make use of
> uniqueness of prime decomposition and that in this kind of problems
> generally means that brute force has to be used. (I have a much
> nicer solution of the same problem when the factors are required to
> be mutually prime)
>
> So now I will present a 'brute force" argument, whose main virtue
> is that it can be much improved (but I will not do so). I will use
> the combinatorica package, which, in Mathematica 6.0 is loaded like
> this:
>
> << Combinatorica`
>
> I want to make use of the Backtrack function in this package. (This
> is the main weakness of this approach and the point which can be
> greatly improved). Here is an auxiliary function, which uses
> backtracking:
>
> FF[n_] := Module[{u = FactorInteger[n], s, k, partialQ, finalQ,
> space, sp},
> s = u[[All,2]]; k = Length[u]; sp[m_] := Tuples[Range[0, m], {k}];
> partialQ[l_List] := And @@ Flatten[
> { !MemberQ[IdentityMatrix[k], Last[l]], Last[l] == Array[0
> & , k] ||
> !MemberQ[Most[l], Last[l]], Thread[Total[l] <= s - 1]}];
> finalQ[l_List] := And @@ Flatten[{ !MemberQ[IdentityMatrix[k],
> Last[l]],
> Last[l] == Array[0 & , k] || !MemberQ[Most[l], Last[l]],
> Thread[Total[l] == s - 1]}]; space = sp /@ (s - 1);
> k + Max[0, Length /@ DeleteCases[Backtrack[space, partialQ,
> finalQ, All],
> Array[0 & , k], Infinity]]]
>
> For any positive integer n this computes the length of the largest
> factorization of n into distinct factors. For example:
>
> FF[2*3*9*11]
> 4
>
> which is obviously right. There is some minor problem in the code
> that causes a Part error message to appear sometimes, without
> however affecting the result:
>
> FF[3]
> Part::partw:Part 2 of {1} does not exist. >>
> Part::partw:Part 2 of ( {
> {{0}}
> } ) does not exist. >>
> Set::partw:Part 2 of {1} does not exist. >>
> 1
>
> However, I don't to spend time on trying to find out the cause of
> this message so on my main program I will simply suppress all
> messages:
>
> So now here is the main function T:
>
> T[n_] := Block[{k = 1, $Messages}, While[k++; FF[k*((k + 1)/2) - 1]
> < n,
> Null]; k*((k + 1)/2)]
>
> which for a given n looks for the smallest triangular number with n-
> distinct factors:
>
> Map[T, Range[8]]
> {3, 15, 55, 253, 1081, 13861, 115921, 1413721}
>
> I can't say that this is really fast, but the good news is that it
> certainly could be greatly improved. The Combinatorica general-
> purpose Backtrack function is very slow, and if someone writes a
> custom-made backtracking version suited to the problem at hand and
> compiles it, it will certainly become orders of magnitude faster.
> This has been done on this list in various situations several
> times. Unfortunately I can't spare the time necessary to do
> this. . Writing backtracking programs requires careful procedural
> programming and I am really out of practice in procedural
> programming, but there are several excellent examples in the
> archives written by Fred Simons and Maxim Rytin, and if this is
> important for you, you should either learn to do it yourself by
> studying these programs or persuade one of them to do it for you ;-)
>
> Andrzej Kozlowski
>
>
> ------------------------------------------
> Your proposition may be good
> But let's have one thing understood --
> Whatever it is, I'm against it!
> And even when you've changed it or condensed it,
> I'm against it.
>
> Professor Quincy Adams Wagstaff
> President of Huxley College
- References:
- Working with factors of triangular numbers.
- From: Diana <diana.mecum@gmail.com>
- Working with factors of triangular numbers.