python - 大额的子集和

Question

subset sum problem 以 NP 完全性而闻名，但也有各种技巧可以稍微快速地解决问题的版本。

通常的动态规划算法需要随着目标总和增长的空间。我的问题是:我们可以减少这个空间需求吗？

我正在尝试解决具有适度元素数量但目标总和非常大的子集总和问题。元素数量对于指数时间算法(和 shortcut method )来说太大，目标和对于通常的动态规划方法来说太大。

考虑这个说明问题的玩具问题。给定集合 A = [2, 3, 6, 8] 找出总和为 target = 11 的子集数。枚举所有子集，我们看到答案是 2:(3, 8) 和 (2, 3, 6)。

动态规划解决方案给出相同的结果，当然 - ways[11] 返回 2:

def subset_sum(A, target):
    ways = [0] * (target + 1)
    ways[0] = 1
    ways_next = ways[:]
    for x in A:
        for j in range(x, target + 1):
            ways_next[j] += ways[j - x]
        ways = ways_next[:]

    return ways[target]

现在考虑将总和 target = 1100 设为 A = [200, 300, 600, 800]。显然还有 2 个解决方案:(300, 800) 和 (200, 300, 600)。但是，ways 数组增长了 100 倍。

是否可以在填写动态规划存储数组时跳过某些权重？对于我的示例问题，我可以计算输入集的最大公分母，然后将所有项目减去该常数，但这不适用于我的实际应用。

This SO question 是相关的，但这些答案并没有使用我想到的方法。 Akshay 在 this page 上的第二条评论说:

...in the cases where n is very small (eg. 6) and sum is very large (eg. 1 million) then the space complexity will be too large. To avoid large space complexity n HASHTABLES can be used.

这似乎更接近我正在寻找的东西，但我似乎无法真正实现这个想法。这真的可能吗？

编辑添加:要解决的问题的较小示例。有 1 个解决方案。

target = 5213096522073683233230240000
A = [2316931787588303659213440000,
     1303274130518420808307560000,
     834095443531789317316838400,
     579232946897075914803360000,
     425558899761116998631040000,
     325818532629605202076890000,
     257436865287589295468160000,
     208523860882947329329209600,
     172333769324749858949760000,
     144808236724268978700840000,
     123386899930738064691840000,
     106389724940279249657760000,
     92677271503532146368537600,
     81454633157401300519222500,
     72153585080604612224640000,
     64359216321897323867040000,
     57762842349846905631360000,
     52130965220736832332302400,
     47284322195679666514560000,
     43083442331187464737440000,
     39418499221729173786240000,
     36202059181067244675210000,
     33363817741271572692673536,
     30846724982684516172960000,
     28604096143065477274240000,
     26597431235069812414440000,
     24794751591313594450560000,
     23169317875883036592134400,
     21698632766175580575360000,
     20363658289350325129805625,
     19148196591638873216640000,
     18038396270151153056160000,
     17022355990444679945241600]

一个真正的问题是:

target = 262988806539946324131984661067039976436265064677212251086885351040000
A = [116883914017753921836437627140906656193895584300983222705282378240000,
     65747201634986581032996165266759994109066266169303062771721337760000,
     42078209046391411861117545770726396229802410348353960173901656166400,
     29220978504438480459109406785226664048473896075245805676320594560000,
     21468474003260924418937523352411426647858372626711204170357987840000,
     16436800408746645258249041316689998527266566542325765692930334440000,
     12987101557528213537381958571211850688210620477887024745031375360000,
     10519552261597852965279386442681599057450602587088490043475414041600,
     8693844844295746252297013588993057072273225278585528961549928960000,
     7305244626109620114777351696306666012118474018811451419080148640000,
     6224587137040149683597270084426981690799173128454727836375984640000,
     5367118500815231104734380838102856661964593156677801042589496960000,
     4675356560710156873457505085636266247755823372039328908211295129600,
     4109200102186661314562260329172499631816641635581441423232583610000,
     3639983481521748430892521260443459881470796742937193786669693440000,
     3246775389382053384345489642802962672052655119471756186257843840000,
     2914003396564502206448583502127866774917064428556368433095682560000,
     2629888065399463241319846610670399764362650646772122510868853510400,
     2385386000362324935437502594712380738650930291856800463373109760000,
     2173461211073936563074253397248264268068306319646382240387482240000,
     1988573206351200938616141104476672789688204647842814753019927040000,
     1826311156527405028694337924076666503029618504702862854770037160000,
     1683128361855656474444701830829055849192096413934158406956066246656,
     1556146784260037420899317521106745422699793282113681959093996160000,
     1443011284169801504153550952356872298690068941987447193892375040000,
     1341779625203807776183595209525714165491148289169450260647374240000,
     1250838556670374906691960338012080744048823137584838292922165760000,
     1168839140177539218364376271409066561938955843009832227052823782400,
     1094646437211014876720019400903392201607763016346356924399106560000,
     1027300025546665328640565082293124907954160408895360355808145902500,
     965982760477305139144112620999228563585913919842836551283325440000,
     909995870380437107723130315110864970367699185734298446667423360000,
     858738960130436976757500934096457065914334905068448166814319513600,
     811693847345513346086372410700740668013163779867939046564460960000,
     768411414287644482489363509326632509674989232073666182868912640000,
     728500849141125551612145875531966693729266107139092108273920640000,
     691620793004461075955252231602997965644352569828303092930664960000,
     657472016349865810329961652667599941090662661693030627717213377600,
     625791330255672395317036671188673352614551016483550865168079360000,
     596346500090581233859375648678095184662732572964200115843277440000,
     568931977371436071675467087219123799753953628290345594563299840000,
     543365302768484140768563349312066067017076579911595560096870560000,
     519484062301128541495278342848474027528424819115480989801255014400,
     497143301587800234654035276119168197422051161960703688254981760000,
     476213321032044045508347054897310957784092466595223632570186240000,
     456577789131851257173584481019166625757404626175715713692509290000,
     438132122515529069774235170457376054037925971973698044293020160000,
     420782090463914118611175457707263962298024103483539601739016561664,
     404442609057972047876946806715939986830088526993021531852188160000,
     389036696065009355224829380276686355674948320528420489773499040000,
     374494562534633427030238036407319297168052779889230688624970240000,
     360752821042450376038387738089218074672517235496861798473093760000,
     347753793771829850091880543559722282890929011143421158461997158400,
     335444906300951944045898802381428541372787072292362565161843560000,
     323778155173833578494287055791985197213007158728485381455075840000,
     312709639167593726672990084503020186012205784396209573230541440000,
     302199145693704480473409550206308504954053507241841138853071360000,
     292209785044384804591094067852266640484738960752458056763205945600,
     282707666261699891568916593460940582033071824431295083135592960000,
     273661609302753719180004850225848050401940754086589231099776640000,
     265042888929147215048611399412486748738992254650755607041456640000,
     256825006386666332160141270573281226988540102223840088952036475625,
     248983485481605987343890803377079267631966925138189113455039385600,
     241495690119326284786028155249807140896478479960709137820831360000,
     234340660761814501342824380545368657996226388663143017230461440000,
     227498967595109276930782578777716242591924796433574611666855840000,
     220952578483466770957349011608519198854244960871423861446658560000,
     214684740032609244189375233524114266478583726267112041703579878400,
     208679870295533683104133831435857945991878646837700655494453760000,
     202923461836378336521593102675185167003290944966984761641115240000,
     197401994025105141026072179446079922264038329650750423033879040000,
     192102853571911120622340877331658127418747308018416545717228160000,
     187014262428406274938300203425450649910232934881573156328451805184,
     182125212285281387903036468882991673432316526784773027068480160000,
     177425404985627474536673746714144021883127046501745489011223040000,
     172905198251115268988813057900749491411088142457075773232666240000,
     168555556186474170249629649778586749838977769381324948621621760000,
     164368004087466452582490413166899985272665665423257656929303344400]

Answer 1

在您链接到的特定评论中，建议使用哈希表仅存储实际上作为某个子集的总和出现的值。在最坏的情况下，这在元素数量上呈指数增长，因此它基本上等同于您已经提到并排除的蛮力方法。

一般来说，这个问题有两个参数——集合中元素的数量和目标总和的大小。朴素的蛮力在第一个是指数级的，而标准动态规划解决方案在第二个中是指数级的。这在其中一个参数较小时效果很好，但您已经指出这两个参数对于指数解来说都太大了。因此，您陷入了问题的“困难”一般情况。

大多数 NP 完全问题都有一些潜在的图形，无论是隐式的还是显式的。使用图划分和DP，它可以在图的树宽上以指数形式求解，但在树宽保持不变的情况下，在图的大小上只能以多项式形式求解。当然，如果无法访问您的数据，就不可能说出底层图可能是什么样子，或者它是否属于具有有界树宽的图类之一，因此可以有效地求解。

编辑:我刚刚写了下面的代码来展示我减少它模小数的意思。下面的代码在不到一秒的时间内解决了你的第一个问题，但它对更大的问题不起作用(尽管它确实将它减少到 n=57, log(t)=68)。

target = 5213096522073683233230240000
A = [2316931787588303659213440000,
     1303274130518420808307560000,
     834095443531789317316838400,
     579232946897075914803360000,
     425558899761116998631040000,
     325818532629605202076890000,
     257436865287589295468160000,
     208523860882947329329209600,
     172333769324749858949760000,
     144808236724268978700840000,
     123386899930738064691840000,
     106389724940279249657760000,
     92677271503532146368537600,
     81454633157401300519222500,
     72153585080604612224640000,
     64359216321897323867040000,
     57762842349846905631360000,
     52130965220736832332302400,
     47284322195679666514560000,
     43083442331187464737440000,
     39418499221729173786240000,
     36202059181067244675210000,
     33363817741271572692673536,
     30846724982684516172960000,
     28604096143065477274240000,
     26597431235069812414440000,
     24794751591313594450560000,
     23169317875883036592134400,
     21698632766175580575360000,
     20363658289350325129805625,
     19148196591638873216640000,
     18038396270151153056160000,
     17022355990444679945241600]

import itertools, time
from fractions import gcd

def gcd_r(seq):
     return reduce(gcd, seq)

def miniSolve(t, vals):
     vals = [x for x in vals if x and x <= t]
     for k in range(len(vals)):
          for sub in itertools.combinations(vals, k):
               if sum(sub) == t:
                    return sub
     return None

def tryMod(n, state, answer):
     t, vals, mult = state
     mods = [x%n for x in vals if x%n]
     if (t%n or mods) and sum(mods) < n:
          print 'Filtering with', n
          print t.bit_length(), len(vals)
     else:
          return state

     newvals = list(vals)
     tmod = t%n
     if not tmod:
          for x in vals:
               if x%n:
                    newvals.remove(x)
     else:
          if len(set(mods)) != len(mods):
               #don't want to deal with the complexity of multisets for now
               print 'skipping', n
          else:
               mini = miniSolve(tmod, mods)
               if mini is None:
                    return None
               mini = set(mini)
               for x in vals:
                    mod = x%n
                    if mod:
                         if mod in mini:
                              t -= x
                              answer.add(x*mult)
                         newvals.remove(x)
     g = gcd_r(newvals + [t])
     t = t//g
     newvals = [x//g for x in newvals]
     mult *= g
     return (t, newvals, mult)

def solve(t, vals):
     answer = set()
     mult = 1
     for d in itertools.count(2):
          if not t:
               return answer
          elif not vals or t < min(vals):
               return None #no solution'
          res = tryMod(d, (t, vals, mult), answer)
          if res is None:
               return None
          t, vals, mult = res
          if len(vals) < 23:
               break

          if (d % 10000) == 0:
               print 'd', d

     #don't want to deal with the complexity of multisets for now
     assert(len(set(vals)) == len(vals))
     rest = miniSolve(t, vals)
     if rest is None:
          return None
     answer.update(x*mult for x in rest)
     return answer

start_t = time.time()
answer = solve(target, A)
assert(answer <= set(A) and sum(answer) == target)
print answer