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MathGroup Archive 2007

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Re: Re: Working with factors of triangular numbers.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80039] Re: [mg79864] Re: Working with factors of triangular numbers.
  • From: "Oleksandr Pavlyk" <pavlyk at gmail.com>
  • Date: Sat, 11 Aug 2007 02:08:03 -0400 (EDT)
  • References: <200707030923.FAA17995@smc.vnet.net>

Hi,

First of all, very nice work.

It threw me off initially that lld5 did not work for my example:

In[70]:= lld5[chi = 2^84*5^18*7^6]

Out[70]= 14

In[71]:= lldaux5[FactorInteger[chi][[All, 2]] ]

Out[71]= 26

But then I realize that factorCounts is supposed to work only for
triangular numbers,
as is implied by the link given in the mail, and I would like to
explicitly state for the record:

In[72]:= factorCounts[chi]

Out[72]= {1,1,1,1,1,1,1,1,1,1,1,1,5}

In[73]:= FactorInteger[chi]

Out[73]= {{2,84},{5,18},{7,6}}

Oleksandr Pavlyk

On 8/10/07, DrMajorBob <drmajorbob at bigfoot.com> wrote:
> Here's a significantly faster code (for n=12,13 anyway).
>
> Clear[lldaux5, lld5, t5, factorCounts]
> factorCounts[1] = factorCounts[2] = factorCounts[3] = {1};
> factorCounts[4] = {2};
> factorCounts[7] = {3};
> factorCounts[k_Integer] :=
>   If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
>    factorCounts[(k - 1)/2, k + 2]]
> factorCounts[j_Integer, k_Integer] :=
>   Switch[GCD[j, k],
>      1,
>      Join[FactorInteger@j, FactorInteger@k],
>      3, Replace[
>       Join[FactorInteger[j/3],
>        FactorInteger[k/3]], {3, s_} :> {3, s + 2}, {1}]][[All, -1]] //
>    Sort
> lldaux5[{}] := 0
> lldaux5[x_List] /; MemberQ[x, 1] :=
>   Count[x, 1] + lldaux5@DeleteCases[x, 1]
> lldaux5[Lpow_] :=
>   lldaux5[Lpow] =
>    With[{Mtup = Tuples[Range[0, #] & /@ Lpow]},
>     Total[Quiet@
>        LinearProgramming[ConstantArray[-1, Length[Mtup]],
>         Transpose@Mtup, Thread@{Lpow, 0},
>         Array[{0, 1} &, Length[Mtup]], Integers]] - 1]
> lld5[n_Integer] := lldaux5@factorCounts@n
> t5[n_] := # (# + 1)/2 &@NestWhile[# + 1 &, 2, lld5[#] < n &]
>
> T2c /@ Range[10] // Timing
> t5 /@ Range[10] // Timing
>
> {0.422, {3, 15, 55, 253, 1081, 13861, 115921, 665281, 18280081,
>    75479041}}
>
> {1.046, {3, 15, 55, 325, 1081, 18145, 226801, 665281, 18280081,
>    75479041}}
>
> T2c[11] // Timing
> t5[11] // Timing
>
> {1.625, 2080995841}
>
> {3.063, 2080995841}
>
> Timing differences aren't reliable either way for n<=11, but beyond that
> it's a different story:
>
> T2c[12] // Timing
> t5[12] // Timing
>
> {27., 68302634401}
>
> {17.187, 68302634401}
>
> T2c[13] // Timing
> t5[13] // Timing
>
> {150.172, 924972048001}
>
> {64.032, 924972048001}
>
> t5[14] // Timing
>
> {516.546, 48318396825601}
>
> The trick (already used by someone else, I think)
>
> lldaux5[x_List] /; MemberQ[x, 1] :=
>   Count[x, 1] + lldaux5@DeleteCases[x, 1]
>
> means that if the factor counts includes n ones, the LP problem used by
> T2c has 2^n times as many variables as the one used by t5. Yet this
> yielded only a 10-15% improvement, by itself.
>
> A bigger gain was achieved in factorCounts, based on one of Carl Woll's
> ideas in
>
> http://forums.wolfram.com/mathgroup/archive/2007/Jul/msg00411.html
>
> factorCounts[k_Integer] :=
>   If[EvenQ@k, factorCounts[(k + 2)/2, k - 1],
>    factorCounts[(k - 1)/2, k + 2]]
>
> Several fine adjustments were required to get any improvement at all.
> Defining factorCounts for 1,2,3,4, and 7 removed steps from the code for
> other cases, and it was crucial to take full advantage of the fact that
> the GCD of the two factors could only be 1 or 3.
>
> Bobby
>
> On Tue, 07 Aug 2007 00:31:22 -0500, sashap <pavlyk at gmail.com> wrote:
>
> > On Aug 5, 3:58 am, Carl Woll <ca... at wolfram.com> wrote:
> >> sashap wrote:
> >> >Carl's code is very good on numbers not involving
> >> >products of large powers of primes. Otherwise it
> >> >it is memory intensive and not as efficient:
> >>
> >> >In[109]:= Table[Timing[LargestPartition[{k, k}]], {k, 7, 15}]
> >>
> >> >Out[109]=
> >> >{{0.031,7},{0.141,7},{0.172,8},{0.187,9},{1.531,9},{1.875,10},
> >> >{12.016,10},{16.234,11},{141.391,11}}
> >>
> >> >One can use LinearProgramming to improve on the
> >> >situation.
> >>
> >> >Additional improvement comes from the following
> >> >observation. The longest factorization can be built
> >> >by taking the longest sequence of divisors of degree 1,
> >> >then of degree 2 and so on. The degree of the divisor
> >> >is defined as
> >>
> >> >  deg[n_] := Total[ FactorInteger[n][[All,2]] ]
> >>
> >> >In other words, let n == p1^e1 *...* pn^en.
> >> >First take {p1, p2,.., pn}, then the longest
> >> >sequence of pairs with repetitions and so on.
> >>
> >> Nice work! However, I think there is a problem with your algorithm.
> >> There are many possible ways to a longest sequence of (distinct) pairs
> >> with repetitions. It is possible that some of these sequences will not
> >> produce the longest factorization. For instance, consider 2^10 * 3^3 *
> >> 5^2. After removing the degree 1 factors we are left with 2^9 * 3^2 * 5.
> >> Two possible length 3 sequences of pairs are:
> >>
> >> a) 2^2, 2*3, 2*5 leaving 2^5*3
> >>
> >> and
> >>
> >> b) 2^2, 2*3, 3*5 leaving 2^6
> >>
> >> Now, for case a) we can create two more unique factors, 2^3 and 2^2*3,
> >> while for case b) we can't create two more unique factors.
> >>
> >> So, I think one more thing is needed in your algorithm, a way of knowing
> >> which longest sequence of pairs to pick.
> >>
> >> Now, it turns out that the above example is handled correctly by your
> >> algorithm, that is, your algorithm by design or chance picks a longest
> >> sequence that leads to a longest factorization. I played around with a
> >> bunch of different examples, and came up with one example where your
> >> algorithm picks a longest sequence that doesn't lead to a longest
> >> factorization. This wasn't easy because my algorithm is soo slow (but I
> >> think, correct).
> >>
> >> In[594]:= counterexample =  {2,3,4,5,6,7,8,10,12,14,15,35};
> >>
> >> In[595]:= Times @@ counterexample
> >>
> >> Out[595]= 35562240000
> >>
> >> In[596]:= Length[counterexample]
> >>
> >> Out[596]= 12
> >>
> >> In[597]:= LengthOfLongestDecomposition[Times @@ counterexample]
> >>
> >> Out[597]= 11
> >>
> >> My guess is that a simple heuristic probably exists which will allow you
> >> to pick a good longest sequence instead of a bad longest sequence.
> >> Perhaps something like choosing the longest sequence that keeps as many
> >> different types of prime factors available for the next stage? If so,
> >> then the biggest bottleneck will be just factoring the triangular
> >> numbers.
> >>
> >> Carl
> >>
> >
> > Carl,
> >
> > good points ! Indeed:
> >
> > In[98]:= ffaux /@ Permutations[FactorInteger[chi][[All, 2]] ]
> >
> > Out[98]= {12,12,11,12,12,12,12,12,12,11,11,11}
> >
> > The heuristics you mention might be different sorting of exponents,
> > even though we can not guarantee it, that is
> > changing LengthOfLongestDecomposition to the following:
> >
> > LengthOfLongestDecomposition[nn_Integer] :=
> >   ffaux[Reverse@Sort@FactorInteger[nn][[All, 2]]]
> >
> > Another approach is to write n == d1^s1 * ... * dk^sk and just use
> > LinearProgramming once. Here
> > s1, ..., sk are either 0 or 1, and d1, ..., dk are all distinct
> > divisors of the given integer n.
> >
> > Clear[lldaux, lld];
> > lldaux[Lpow_] := (lldaux[Lpow] =
> >    With[{Mtup = Tuples[ Range[0, #] & /@ Lpow]},
> >     Total[
> >       Quiet@LinearProgramming[ConstantArray[-1, Length[Mtup]],
> >         Transpose@Mtup, Thread@{Lpow, 0},
> >         Array[{0, 1} &, Length[Mtup]], Integers]] - 1])
> >
> > lld[n_Integer] := lldaux[FactorInteger[n][[All, 2]] // Sort]
> >
> > T2c[n_] := # (# + 1)/2 &@
> >   NestWhile[# + 1 &, 2, lld[# (# + 1)/2 - 1] < n &]
> >
> > The performance of T2c is competitive to Carl's code for small values
> > of the argument
> > and has an edge over it for large ones:
> >
> > In[168]:= {Timing@T2[11], Timing@T2c[11]}
> >
> > Out[168]= {{3.829,2080995841},{3.921,2080995841}}
> >
> > In[169]:= {Timing@T2[12], Timing@T2c[12]}
> >
> > Out[169]= {{26.532,68302634401},{22.797,68302634401}}
> >
> > In[170]:= {Timing@T2[13], Timing@T2c[13]}
> >
> > Out[170]= {{144.406,924972048001},{123.875,924972048001}}
> >
> > Oleksandr Pavlyk and Maxim Rytin
> >
> >> >In order to choose the longest sequence of factors of a given degree,
> >> >we set this up as an integer linear programming problem. We illustrate
> >> >this for degree 2. Let p1*p1, p1*p2, and so on be candidate factors of
> >> >degree 2. Then
> >>
> >> >   (p1*p1)^s1 (p1*p2)^s2 ...
> >>
> >> >must divide the original number. Additionally, each sk is either 0 or
> >> >1.
> >> >This gives us the constraints for variables s1, s2, ...   and
> >> >the objective function being maximized is s1+s2+....
> >>
> >> >The code implementing those ideas:
> >>
> >> >CombinationsWithRepetitions =
> >> >  Compile[{{n, _Integer}, {k, _Integer}},
> >> >   Module[{ans, i = 1, j = 1, l = 1},
> >> >    ans = Array[0 &, {Binomial[n + k - 1, k], k}];
> >> >    While[True, While[j <= k, ans[[i, j++]] = l];
> >> >     If[i == Length@ans, Break[]];
> >> >     While[ans[[i, --j]] == n,];
> >> >     l = ans[[i++, j]] + 1;
> >> >     ans[[i]] = ans[[i - 1]];];
> >> >    ans]];
> >>
> >> >Clear[LengthOfLongestDecomposition, ffaux]
> >> >LengthOfLongestDecomposition[nn_] :=
> >> > ffaux[Sort@FactorInteger[nn][[All, 2]]]
> >> >ffaux[$Lpow_] :=
> >> > ffaux[$Lpow] =
> >> >  Module[{Lpow = $Lpow, Ldiv, n, m, k = 1, Lind, ans = 0},
> >> >   n = Length@Lpow;
> >> >   While[Total@Lpow >= k,
> >> >    Ldiv = CombinationsWithRepetitions[n, k++];
> >> >    m = Length@Ldiv;
> >> >    Lind =
> >> >     Quiet@LinearProgramming[ConstantArray[-1, m],
> >> >       Total[1 - Unitize[Ldiv - #], {2}] & /@ Range@n,
> >> >       Thread@{Lpow, -1}, Array[{0, 1} &, m], Integers];
> >> >    ans += Total@Lind;
> >> >    Lpow -= BinCounts[Flatten@Pick[Ldiv, Lind, 1], {Range[n + 1]}]];
> >> >   ans]
> >>
> >> >Compare the timing given earlier with the following
> >>
> >> >In[132]:= Table[Timing[ffaux[{k, k}]], {k, 7, 15}]
> >>
> >> >Out[132]= \
> >> >{{0.016,7},{1.78746*10^-14,7},{1.78746*10^-14,8},{0.015,9},{3.15303*\
> >> >10^-14,9},{0.016,10},{0.,10},{0.015,11},{0.,11}}
> >>
> >> >Clearly, the use of heavy machinery of LinearProgramming comes at the
> >> >cost of
> >> >noticeable overhead which is why the performance of this algorithm is
> >> >comparable
> >> >to that of Carl's code when measured over a sequence of triangular
> >> >numbers because
> >> >problematic numbers are rare.
> >>
> >> >T2b[n_] := # (# + 1)/2 &@
> >> >  NestWhile[# + 1 &, 1,
> >> >   LengthOfLongestDecomposition[# (# + 1)/2 - 1] < n &]
> >>
> >> >In[66]:= {Timing[T2[11]], Timing[T2b[11]]}
> >>
> >> >Out[66]= {{3.172,2080995841},{2.328,2080995841}}
> >>
> >> >In[67]:= {Timing[T2[12]], Timing[T2b[12]]}
> >>
> >> >Out[67]= {{26.312,68302634401},{18.235,68302634401}}
> >>
> >> >In[68]:= {Timing[T2[13]], Timing[T2b[13]]}
> >>
> >> >Out[68]= {{142.765,924972048001},{111.422,924972048001}}
> >>
> >> >Oleksandr Pavlyk and Maxim Rytin
> >>
> >> >On Jul 10, 5:37 am, Carl Woll <ca... at wolfram.com> wrote:
> >>
> >> >>Andrzej and Diana,
> >>
> >> >>Here is a faster algorithm. First, an outline:
> >>
> >> >>For a given number, the easy first step in writing  it as a
> >> >>factorization of the most distinct factors is to use each prime as a
> >> >>factor. Then, the hard part is to take the remaining factors and
> >> >>partition them in a way to get the most distinct factors. However,
> >> note
> >> >>that this step only depends on the counts of each prime factor, and
> >> not
> >> >>on the values of the primes themselves. Hence, we can memoize the
> >> number
> >> >>of distinct factors possible for a set of counts. I do this as
> >> follows:
> >>
> >> >>Let f be the list of remaining factors, of length l.
> >>
> >> >>1. Determine possible integer partitions of l, with the smallest
> >> >>partition allowed being 2. We also need to order these partitions by
> >> >>length, from largest to smallest:
> >>
> >> >>partitions = IntegerPartitions[l, All, Range[2, l]];
> >> >>partitions = Reverse@partitions[[ Ordering[Length/@partitions] ]];
> >>
> >> >>2. Now, we need to test each partition to see if it's possible to fill
> >> >>in the partitions with the factors such that each partition is unique.
> >> >>Once we find such a partition, we are done, and we know how many
> >> >>distinct factors that number can be written as. This step I do by
> >> >>recursion, i.e., take the first member of a partition and fill it in
> >> >>with one of the possible subsets of factors, and then repeat with the
> >> >>remaining members of a partition. I do this with the function
> >> Partition=
> > able.
> >>
> >> >>Here is the code:
> >>
> >> >>LargestPartition[list : {1 ..}] := Length[list]
> >> >>LargestPartition[list_List] :=  Length[list] +
> >> >>LargestTwoPartition[Reverse@Sort@list - 1 /. {a__, 0 ..} -> {a}]
> >>
> >> >>Clear[LargestTwoPartition]
> >> >>LargestTwoPartition[list_] :=  LargestTwoPartition[list] = Module[{=
> > set,
> >> >>partitions, res},
> >> >>   set = CountToSet[list];
> >> >>   partitions = IntegerPartitions[Total[list], All, Range[2, Total[li=
> > st]]];
> >> >>   partitions = Reverse@partitions[[Ordering[Length /@ partitions]]];
> >> >>   res = Cases[partitions, x_ /; Partitionable[{}, set, x], 1, 1];
> >> >>   If[res === {},
> >> >>      0,
> >> >>      Length@First@res
> >> >>   ]
> >> >>]
> >>
> >> >>Partitionable[used_, unused_, part_] := Module[{first, rest},
> >> >>  first = Complement[DistinctSubsets[unused, First@part], used];
> >> >>  If[first === {}, Return[False]];
> >> >>  rest = CountToSet /@ Transpose[
> >> >>    SetToCount[unused, Max[unused]] -
> >> >>    Transpose[SetToCount[#, Max[unused]] & /@ first]
> >> >>  ];
> >> >>  Block[{Partitionable},
> >> >>    Or @@ MapThread[
> >> >>      Partitionable[Join[used, {#1}], #2, Rest[part]] &,
> >> >>      {first, rest}
> >> >>    ]
> >> >>  ]
> >> >>]
> >>
> >> >>Partitionable[used_, rest_, {last_}] := ! MemberQ[used, rest]
> >>
> >> >>CountToSet[list_] := Flatten@MapIndexed[Table[#2, {#1}] &, list]
> >> >>SetToCount[list_, max_] := BinCounts[list, {1, max + 1}]
> >>
> >> >>DistinctSubsets[list_, len_] := Union[Subsets[list, {len}]]
> >>
> >> >>T2[n_] := Module[{k=1},
> >> >>  While[k++; LargestPartition[FactorInteger[k*((k + 1)/2) - 1][[All,
> >> >>2]]] < n];
> >> >>  k*((k + 1)/2)
> >> >>]
> >>
> >> >>Then:
> >>
> >> >>In[11]:= T2[6] // Timing
> >> >>Out[11]= {0.047,13861}
> >>
> >> >>In[12]:= T2[7] // Timing
> >> >>Out[12]= {0.031,115921}
> >>
> >> >>In[13]:= T2[8] // Timing
> >> >>Out[13]= {0.11,665281}
> >>
> >> >>In[14]:= T2[9] // Timing
> >> >>Out[14]= {0.625,18280081}
> >>
> >> >>In[15]:= T2[10] // Timing
> >> >>Out[15]= {1.157,75479041}
> >>
> >> >>In[16]:= T2[11] // Timing
> >> >>Out[16]= {5.875,2080995841}
> >>
> >> >>In[17]:= T2[12] // Timing
> >> >>Out[17]= {48.703,68302634401}
> >>
> >> >>For Diana, note that
> >>
> >> >>In[20]:= 481 482/2 - 1 == 115921 - 1 == 2 3 4 5 6 7 23
> >> >>Out[20]= True
> >>
> >> >>so 115921-1 can indeed be written as a product of 7 distinct factors.
> >>
> >> >>Partitionable can be improved, but another bottleneck for larger cases
> >> >>is evaluating FactorInteger. One possibility for improving
> >> FactorInteger
> >> >>speed is to note that
> >>
> >> >>In[21]:= n (n + 1)/2 - 1 == (n + 2) (n - 1)/2 // Expand
> >>
> >> >>Out[21]= True
> >>
> >> >>So, rather than applying FactorInteger to n(n+1)/2-1 you can instead
> >> >>somehow combine FactorInteger[n+2] and FactorInteger[n-1].
> >>
> >> >>At any rate, it takes a bit longer, but with enough patience one can
> >> >>also find:
> >>
> >> >>In[53]:= 1360126 1360127/2 - 1 == 924972048001 - 1 ==  2 3 4 =
> > 5 6 7 10 11
> >> >>12 13 15 23 31
> >> >>Out[53]= True
> >>
> >> >>and
> >>
> >> >>In[56]:= 9830401 9830402/2 - 1 == 48318396825601 - 1 ==  2 3 =
> > 4 5 6 8 9
> >> >>10 11 12 17 22 32 59
> >> >>Out[56]= True
> >>
> >> >>Carl Woll
> >> >>Wolfram Research
> >>
> >> >>Andrzej Kozlowski wrote:
> >>
> >> >>>Well, I stayed up longer than I wanted and I think I have now fixed
> >> >>>it, I hope for the final time. Here is the new FF:
> >>
> >> >>>FFF[n_] := Module[{u = FactorInteger[n], s, k, partialQ, finalQ, s=
> > pace},
> >> >>>   s = u[[All,2]]; k = Length[u]; 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]], Total[l] == s - 1}];
> >> >>>    space =
> >>
> >> ...
> >>
> >> read more =BB
> >
> >
> >
> >
>
>
>
> --
> DrMajorBob at bigfoot.com
>


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