有没有直接的方法来获得所有 nCr 组合的有序集合的第 N 个组合?
示例:我有四个元素:[6, 4, 2, 1]。一次取三个的所有可能组合是:[[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]。
有没有一种算法可以在有序结果集中给我例如第三个答案 [6, 2, 1],而不枚举所有先前的答案?
有没有直接的方法来获得所有 nCr 组合的有序集合的第 N 个组合?
示例:我有四个元素:[6, 4, 2, 1]。一次取三个的所有可能组合是:[[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]。
有没有一种算法可以在有序结果集中给我例如第三个答案 [6, 2, 1],而不枚举所有先前的答案?
请注意,您可以通过递归生成包含第一个元素的所有组合,然后生成不包含的所有组合来生成序列。在这两种递归情况下,您删除第一个元素以从 n-1 个元素中获取所有组合。在 Python 中:
def combination(l, r):
if r == 0:
yield []
elif len(l) == r:
yield l
else:
for c in (combination(l[1:], r-1)):
yield l[0:1]+c
for c in (combination(l[1:], r)):
yield c
每当您通过这样的选择生成序列时,您都可以通过计算选择生成的元素数量并将计数与 k 进行比较来递归地生成第 k个元素。如果 k 小于计数,则您做出该选择。否则,减去计数并重复您当时可以做出的其他可能选择。如果总是有b
选择,您可以将其视为在 base 中生成一个数字b
。如果选择的数量不同,该技术仍然有效。在伪代码中(当所有选择始终可用时):
kth(k, choicePoints)
if choicePoints is empty
return empty list
for each choice in head of choicePoints:
if k < size of choice
return choice and kth(k, tail of choicePoints)
else
k -= size of choice
signal exception: k is out-of-bounds
这为您提供了一个基于 0 的索引。如果您想要基于 1,请将比较更改为k <= size of choice
.
棘手的部分(以及伪代码中未指定的部分)是选择的大小取决于先前的选择。请注意,伪代码可用于解决比问题更普遍的情况。
对于这个特定问题,有两个选择(b
= 2),第一个选择(即包括第一个元素)的大小由n-1 C r-1给出。这是一种实现(需要合适的nCr
):
def kthCombination(k, l, r):
if r == 0:
return []
elif len(l) == r:
return l
else:
i=nCr(len(l)-1, r-1)
if k < i:
return l[0:1] + kthCombination(k, l[1:], r-1)
else:
return kthCombination(k-i, l[1:], r)
如果你颠倒选择的顺序,你就会颠倒顺序。
def reverseKthCombination(k, l, r):
if r == 0:
return []
elif len(l) == r:
return l
else:
i=nCr(len(l)-1, r)
if k < i:
return reverseKthCombination(k, l[1:], r)
else:
return l[0:1] + reverseKthCombination(k-i, l[1:], r-1)
使用它:
>>> l = [6, 4, 2, 1]
>>> [kthCombination(k, [6, 4, 2, 1], 3) for k in range(nCr(len(l), 3)) ]
[[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]
>>> powOf2s=[2**i for i in range(4,-1,-1)]
>>> [sum(kthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[28, 26, 25, 22, 21, 19, 14, 13, 11, 7]
>>> [sum(reverseKthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[7, 11, 13, 14, 19, 21, 22, 25, 26, 28]
我在寻找方法时偶然发现了这个问题_ '在后者上找不到太多(你的问题的反面并不是那么难以捉摸)。
因为在我想我会在这里发布我的解决方案之前,我也解决了(我认为是)你的确切问题。
**
编辑:我的要求也是您的要求 - 我看到了答案并认为递归很好。好吧,经过六年的漫长岁月,你拥有了它;只需向下滚动。
**
对于您在问题中提出的要求(我认为是),这可以很好地完成工作:
def iterCombinations(n, k):
if k==1:
for i in range(n):
yield [i]
return
result = []
for a in range(k-1, n):
for e in iterCombinations(n, k-1):
if e[-1] == a:
break
yield e + [a]
然后,您可以在按降序排列的集合中查找项目(或使用一些等效的比较方法),因此对于有问题的情况:
>>> itemsDescending = [6,4,2,1]
>>> for c in iterCombinations(4, 3):
... [itemsDescending[i] for i in c]
...
[6, 4, 2]
[6, 4, 1]
[6, 2, 1]
[4, 2, 1]
这也可以在 Python 中直接使用,但是:
>>> import itertools
>>> for c in itertools.combinations(itemsDescending, 3):
... c
...
(6, 4, 2)
(6, 4, 1)
(6, 2, 1)
(4, 2, 1)
这是我为我的要求(实际上是为你的要求!)所做的,它不创建或遍历任一方向的有序列表,而是使用简单但有效的n C r非递归实现,选择(n,k):
def choose(n, k):
'''Returns the number of ways to choose k items from n items'''
reflect = n - k
if k > reflect:
if k > n:
return 0
k = reflect
if k == 0:
return 1
for nMinusIPlus1, i in zip(range(n - 1, n - k, -1), range(2, k + 1)):
n = n * nMinusIPlus1 // i
return n
要在前向排序列表中的某个(从零开始的)索引处获取组合:
def iterCombination(index, n, k):
'''Yields the items of the single combination that would be at the provided
(0-based) index in a lexicographically sorted list of combinations of choices
of k items from n items [0,n), given the combinations were sorted in
descending order. Yields in descending order.
'''
if index < 0 or index >= choose(n, k):
return
n -= 1
for i in range(k):
while choose(n, k) > index:
n -= 1
yield n
index -= choose(n, k)
n -= 1
k -= 1
要获取某个组合将驻留在反向排序列表中的(从零开始的)索引:
def indexOfCombination(combination):
'''Returns the (0-based) index the given combination would have if it were in
a reverse-lexicographically sorted list of combinations of choices of
len(combination) items from any possible number of items (given the
combination's length and maximum value)
- combination must already be in descending order,
and it's items drawn from the set [0,n).
'''
result = 0
for i, a in enumerate(combination):
result += choose(a, i + 1)
return result
对于您的示例来说,这太过分了(但我现在意识到这只是一个示例);这就是每个索引依次进行的方式:
def exampleUseCase(itemsDescending=[6,4,2,1], k=3):
n = len(itemsDescending)
print("index -> combination -> and back again:")
for i in range(choose(n, k)):
c = [itemsDescending[j] for j in iterCombination(i, n, k)][-1::-1]
index = indexOfCombination([itemsDescending.index(v) for v in c])
print("{0} -> {1} -> {2}".format(i, c, index))
>>> exampleUseCase()
index -> combination -> and back again:
0 -> [6, 4, 2] -> 0
1 -> [6, 4, 1] -> 1
2 -> [6, 2, 1] -> 2
3 -> [4, 2, 1] -> 3
这可以找到给定一些长列表的索引,或者在眨眼之间返回某个天文索引的组合,例如:
>>> choose(2016, 37)
9617597205504126094112265433349923026485628526002095715212972063686138242753600
>>> list(iterCombination(_-1, 2016, 37))
[2015, 2014, 2013, 2012, 2011, 2010, 2009, 2008, 2007, 2006, 2005, 2004, 2003,
2002, 2001, 2000, 1999, 1998, 1997, 1996, 1995, 1994, 1993, 1992, 1991, 1990, 1989,
1988, 1987, 1986, 1985, 1984, 1983, 1982, 1981, 1980, 1979]
或者,因为这是最后一个,并且由于选择(n,k)中的反射可能很快,所以这是一个从中间开始的,它似乎同样快......
>>> choose(2016, 37)//2
4808798602752063047056132716674961513242814263001047857606486031843069121376800
>>> list(iterCombination(_, 2016, 37))
[1978, 1973, 1921, 1908, 1825, 1775, 1747, 1635, 1613, 1598, 1529, 1528, 1521,
1445, 1393, 1251, 1247, 1229, 1204, 1198, 922, 901, 794, 699, 685, 633, 619, 598,
469, 456, 374, 368, 357, 219, 149, 93, 71]
最后一个例子停下来思考片刻,但你不会吗?
>>> import random
>>> rSet = set(random.randint(0, 10000000) for i in range(900))
>>> len(rSet)
900
>>> rList = sorted(rSet, reverse=True)
>>> combinations.indexOfCombination(rList)
61536587905102303838316048492163850175478325236595592744487336325506086930974887
88085020093159925576117511028315621934208381981476407812702689774826510322023536
58905845549371069786639595263444239118366962232872361362581506476113967993096033
00541202874946853699568596881200225925266331936183173583581021914595163799417151
30442624813775945054888304722079206982972852037480516813527237183254850056012217
59834465303543702263588008387352235149083914737690225710105023486226582087736870
38383323140972279867697434315252036074490127510158752080225274972225311906715033
86851377357968649982293794242170046400174118714525559851836064661141086690326842
25236658978135989907667078625869419802333512020715700514133380517628637151215549
05922388534567108671308819960483147825031620798631811671493891643972220604919591
22785587505280326638477135315176731640100473359830821781905546117103137944239120
34912084544221250309244925308316352643060056100719194985568284049903555621750881
39419639825279398618630525081169688672242833238889454445237928356800414839702024
66807635358129606994342005075585962080795273287472139515994244684088406544976674
84183671032002497594936116837768233617073949894918741875863985858049825755901232
89317507965160689287607868119414903299382093412911433254998227245783454244894604
83654290108678890682359278892580855226717964180806265176337132759167920384512456
91624558534942279041452960272707049107641475225516294235268581475735143470692000
78400891862852130481822509803019636619427631175355448729708451565341764545325720
79277290914349746541071731127111532099038538549697091038496002102703737347343739
96398832832674081286904287066696046621691978697914823322322650123025472624927566
99891468668052668317066769517155581261265629289158798073055495539590686279250097
27295943276536772955923599217742543093669565147228386873469711200278811335649924
13587219640724942441913695193417732608127949738209466313175361161142601108707568
19470026889319648128790363676253707359290547393198350533094409863254710237344552
47692325209744353688541868412075798500629908908768438513508959321262250985142709
19794478379412756202638771417821781240327337108495689300616872374578607430951230
96908870723878513999404242546015617238957825116802801618973562178005776911079790
22026655573872019955677676783191505879571719659770550759779880002320421606755826
75809722478174545846409923210824885805972611279030267270741509747224602604003738
30411365119180944456819762167312738395140461035991994771968906979578667047734952
21981545694935313345331923300019842406900689401417602004228459137311983483386802
30352489602769346000257761959413965109940729263098747702427952104316612809425394
85037536245288888254374135695390839718978818689595231708490351927063849922772653
26064826999661128817511630298712833048667406916285156973335575847429111697259113
53969532522640227276562651123634766230804871160471143157687290382053412295542343
14022687833967461351170188107671919648640149202504369991478703293224727284508796
06843631262345918398240286430644564444566815901074110609701319038586170760771099
41252989796265436701638358088345892387619172572763571929093224171759199798290520
71975442996399826830220944004118266689537930602427572308646745061258472912222347
18088442198837834539211242627770833874751143136048704550494404981971932449150098
52555927020553995188323691320225317096340687798498057634440618188905647503384292
79493920419695886724506109053220167190536026635080266763647744881063220423654648
36855624855494077960732944499038847158715263413026604773216510801253044020991845
89652657529729792772055725210165026891724511953666038764273616212464901231675592
46950937136633665320781952510620087284589083139308516989522633786063418913473703
96532777760440118656525488729217328376766171004246127636983612583177565603918697
15557602015171235214344399010185766876727226408494760175957535995025356361689144
85181975631986409708533731043231896096597038345028523539733981468056497208027899
6245509252811753667386001506195
然而,从该索引返回到 900-choose-10,000,000 与先前实现的组合将非常慢(因为它只是在每次迭代时从 n 中减去一个)。
对于如此大的组合列表,我们可以改为对空间进行二进制搜索,而我们添加的开销意味着对于小的组合列表只会慢一点:
def iterCombination(index, n, k):
'''Yields the items of the single combination that would be at the provided
(0-based) index in a lexicographically sorted list of combinations of choices
of k items from n items [0,n), given the combinations were sorted in
descending order. Yields in descending order.
'''
if index < 0 or n < k or n < 1 or k < 1 or choose(n, k) <= index:
return
for i in range(k, 0, -1):
d = (n - i) // 2 or 1
n -= d
while 1:
nCi = choose(n, i)
while nCi > index:
d = d // 2 or 1
n -= d
nCi = choose(n, i)
if d == 1:
break
n += d
d //= 2
n -= d
yield n
index -= nCi
从此人们可能会注意到,所有要求choose
取消条款的电话,如果我们取消所有内容,我们最终会得到更快的实施,我认为是什么......
def iterCombination(index, n, k):
'''Yields the items of the single combination that would be at the provided
(0-based) index in a lexicographically sorted list of combinations of choices
of k items from n items [0,n), given the combinations were sorted in
descending order. Yields in descending order.
'''
nCk = 1
for nMinusI, iPlus1 in zip(range(n, n - k, -1), range(1, k + 1)):
nCk *= nMinusI
nCk //= iPlus1
curIndex = nCk
for k in range(k, 0, -1):
nCk *= k
nCk //= n
while curIndex - nCk > index:
curIndex -= nCk
nCk *= (n - k)
nCk -= nCk % k
n -= 1
nCk //= n
n -= 1
yield n
最后提醒一下,对于问题的用例,可以执行以下操作:
def combinationAt(index, itemsDescending, k):
return [itemsDescending[i] for i in
list(iterCombination(index, len(itemsDescending), k))[-1::-1]]
>>> itemsDescending = [6,4,2,1]
>>> numberOfItemsBeingChosen = 3
>>> zeroBasedIndexWanted = 1
>>> combinationAt(zeroBasedIndexWanted, itemsDescending, numberOfItemsBeingChosen)
[6, 4, 1]
一种方法是使用位的属性。这仍然需要一些枚举,但您不必枚举每个集合。
对于您的示例,您的集合中有 4 个数字。因此,如果您要生成 4 个数字的所有可能组合,则可以按如下方式枚举它们:
{6、4、2、1} 0000 - {(集合中没有数字)} 0001 - {1} 0010 - {2} 0011 - {2, 1} ... 1111 - {6, 4, 2, 1}
看看每个“位”如何对应于“那个数字是否在你的集合中”?我们在这里看到有 16 种可能性 (2^4)。
因此,现在我们可以遍历并找到仅打开 3 位的所有可能性。这将告诉我们所有存在的“3”组合:
0111 - {4, 2, 1} 1011 - {6, 2, 1} 1101 - {6, 4, 1} 1110 - {6, 4, 2}
让我们将每个二进制值重写为十进制值:
0111 = 7 1011 = 11 1101 = 13 1110 = 14
现在我们已经做到了——好吧,你说你想要“第三个”枚举。所以让我们看看第三大数字:11。它的位模式为 1011。对应于... {6, 2, 1}
凉爽的!
基本上,您可以对任何集合使用相同的概念。所以现在我们所做的就是将问题从“枚举所有集合”转换为“枚举所有整数”。对于您的问题,这可能会容易得多。
从 Python 3.6 itertools 食谱:
def nth_combination(iterable, r, index):
'Equivalent to list(combinations(iterable, r))[index]'
pool = tuple(iterable)
n = len(pool)
if r < 0 or r > n:
raise ValueError
c = 1
k = min(r, n-r)
for i in range(1, k+1):
c = c * (n - k + i) // i
if index < 0:
index += c
if index < 0 or index >= c:
raise IndexError
result = []
while r:
c, n, r = c*r//n, n-1, r-1
while index >= c:
index -= c
c, n = c*(n-r)//n, n-1
result.append(pool[-1-n])
return tuple(result)
在实践中:
iterable, r, index = [6, 4, 2, 1], 3, 2
nth_combination(iterable, r, index)
# (6, 2, 1)
或者,如文档字符串中所述:
import itertools as it
list(it.combinations(iterable, r))[index]
# (6, 2, 1)
另请参阅- 为您实现此配方more_itertools
的第三方库。通过以下方式安装:
> pip install more_itertools
只是一个粗略的草图:将您的数字排列成元组的上三角矩阵:
A(n-1,n-1)
Aij = [i+1, j-1]
如果您首先遍历矩阵行,您将获得两个元素按升序排列的组合。要概括为三个元素,请将矩阵行视为另一个三角矩阵,而不是向量。它有点像创建立方体的一个角。
至少这是我解决问题的方式
让我澄清一下,您不必存储矩阵,您需要计算索引。让我研究一下维度示例,您原则上可以扩展到 20 个维度(簿记可能很糟糕)。
ij = (i*i + i)/2 + j // ij is also the combination number
(i,j) = decompose(ij) // from ij one can recover i,j components
I = i // actual first index
J = j + 1 // actual second index
这个二维示例适用于任何数字 n,您不必将排列制成表格。
是的,有一种直接的方法可以获得所有 nCr 组合的有序集合的第 N 个组合吗?假设您需要生成给定集合的第 0、第 3、第 6 .. 组合。您可以直接生成它,而无需在使用 JNuberTools 之间生成组合。您甚至可以生成下一个十亿组合(如果您的集合大小很大)这是代码示例:
JNumberTools.combinationsOf(list)
.uniqueNth(8,1000_000_000) //skip to billionth combination of size 8
.forEach(System.out::println);
JNumberTools 的 Maven 依赖项是:
<dependency>
<groupId>io.github.deepeshpatel</groupId>
<artifactId>jnumbertools</artifactId>
<version>1.0.0</version>
</dependency>