Re: Working with factors of triangular numbers.
- To: mathgroup at smc.vnet.net
- Subject: [mg78758] Re: [mg78490] Working with factors of triangular numbers.
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 8 Jul 2007 06:19:52 -0400 (EDT)
- References: <200707030923.FAA17995@smc.vnet.net> <6D2F6E70-2462-4B69-A617-4DEC322D69BF@mimuw.edu.pl> <93E8C621-E93F-4761-A0FF-205BF84249CD@mimuw.edu.pl> <a851af150707070640g1f6a22e9ue072d4eba5c9262a@mail.gmail.com>
You must have made some mistake.
Let me do it all form the beginning, with Mathematica 6.0 for Mac and
fresh kernel:
<< Combinatorica`
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] ||
!MemberQ[Most[l], Last[l]], Thread[Total[l] <= s - 1]}];
finalQ[l_List] := And @@ Flatten[{Last[l] == Array[0 & , k] ||
!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]]]
T[n_] := Block[{k = 1, $Messages},
While[k++; FF[k*((k + 1)/2) - 1] < n, Null]; k*((k + 1)/2)]
Map[T, Range[7]]
{3, 15, 55, 253, 1081, 13861, 115921}
which is what you expected?
Andrzej Kozlowski
On 7 Jul 2007, at 22:40, Diana Mecum wrote:
> Andrzej,
>
> I appreciate all of the work you have done with my question.
>
> I tested this latest update, and got
>
> Map[T, Range[6]]
> {3,15,91, 1431, 10011, 218791}
>
> The expect that the first 6 terms would be:
>
> {3,15,55,253,1081,13861}
>
> I used the first's e-mail code for T, and the second code for FF.
>
> Thanks,
>
> Diana
>
> On 7/7/07, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: *This
> message was transferred with a trial version of CommuniGate(tm) Pro*
> 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:
>
> > *This message was transferred with a trial version of CommuniGate
> > (tm) Pro*
> > 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
>
>
>
>
> --
> "God made the integers, all else is the work of man."
> L. Kronecker, Jahresber. DMV 2, S. 19.
- References:
- Working with factors of triangular numbers.
- From: Diana <diana.mecum@gmail.com>
- Working with factors of triangular numbers.