Re: Re: Working with factors of triangular numbers.

*To*: mathgroup at smc.vnet.net*Subject*: [mg80019] Re: [mg79864] Re: Working with factors of triangular numbers.*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Fri, 10 Aug 2007 06:44:09 -0400 (EDT)*References*: <200707030923.FAA17995@smc.vnet.net> <24165820.1186478071750.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

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 ma ny >> 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, Tota l[list]]]; >> >> 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 factor= s. >> >> >>Partitionable can be improved, but another bottleneck for larger ca= ses >> >>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 instea= d >> >>somehow combine FactorInteger[n+2] and FactorInteger[n-1]. >> >> >>At any rate, it takes a bit longer, but with enough patience one ca= n >> >>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 fixe= d >> >>>it, I hope for the final time. Here is the new FF: >> >> >>>FFF[n_] := Module[{u = FactorInteger[n], s, k, partialQ, final= Q, 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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**Re: Working with factors of triangular numbers.**

**Re: Re: Working with factors of triangular numbers.**