Re: Working with factors of triangular numbers.
- To: mathgroup at smc.vnet.net
- Subject: [mg78750] Re: [mg78490] Working with factors of triangular numbers.
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 8 Jul 2007 06:15:42 -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> <6F7B3048-4C96-48BC-AADF-2A62837DE6D3@mimuw.edu.pl> <a851af150707072125o69edf471o4b5be8932a924a32@mail.gmail.com>
You are quite right. I have checked it and can now confirm that my
code works correctly with Mathematica 6 but not with Mathematica 5.2.
Indeed, with exactly the same definitions I get:
$Version
"6.0 for Mac OS X PowerPC (32-bit) (April 20, 2007)"
T[5]
1081
but
$Version
5.2 for Mac OS X (February 24, 2006)
T[5]
10011
which is completely wrong. There seems to be some very fundamental
difference between Mathematica 5.2 and Mathematica 6. I don't think
the diffeence is between the Backtrack functions in the Combinatrica
packages, because I used the old one a number of times in the past
and it worked fine. I have to admit that at the moment I am
completely baffled. Although I can't right now devote much more time
to this problem, I think finding out what it is that makes the
difference here is important. I can now see that the difference lies
in the function FF. For example, in Mathematica 6:
FF[576]
4
but in Mathematica 5.2:
FF[576]
"Nonatomic expression expected at position `1` Thread[False]
2
It suggests that the difference may lie in error handling. I hope
someone reading this will see where the problem lies and let us know.
If not, I will try to discover it myself, but it won't be very soon.
I don't think this alone is quite good enough reason to upgrade to
version 6, but there are plenty of better ones and that would solve
your immediate problem ;-)
Andrzej Kozlowski
On 8 Jul 2007, at 13:25, Diana Mecum wrote:
> Andrzej,
>
> I copied your code verbatim, except for DiscreteMath Combinatorica.
>
> With Mathematica 5.2, a pc and a fresh kernel, I get:
>
> <<DiscreteMath`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]]
> {3,15,91,1431,10011,218791,8378371}
>
> Diana M.
>
> On 7/7/07, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: *This
> message was transferred with a trial version of CommuniGate(tm) Pro*
> 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.
>
>
>
>
> --
> "God made the integers, all else is the work of man."
> L. Kronecker, Jahresber. DMV 2, S. 19.
- Follow-Ups:
- Re: Re: Working with factors of triangular numbers.
- From: János <janos.lobb@yale.edu>
- Re: Re: Working with factors of triangular numbers.
- References:
- Working with factors of triangular numbers.
- From: Diana <diana.mecum@gmail.com>
- Working with factors of triangular numbers.