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 >