背景

一个前增加的一组序列的顺序被定义为整数集的序列s ^ 1，S ^ 2，⋯ ，小号Ñ其满足以下的：$N$$S_1,S_2,\cdots,S_n$

• 每个是一个非空的子集{ 1 ，2 ，⋯ ，Ñ }。$S_i$$\{1,2,\cdots,N\}$
• 对于，小号我 ∩ 小号我+ 1 = ∅，即任何两个连续集没有共同的元素。$1\le i<n$$S_i \cap S_{i+1} = \varnothing$
• 对于，平均的（平均值）小号我比的严格更少小号我+ 1。$1\le i<n$$S_i$$S_{i+1}$

挑战

给定一个正整数N，输出order的最长递增集合序列的长度N。

测试用例

这些是基于Project Euler用户thundre的结果。

1 => 1 // {1}
2 => 2 // {1} {2}
3 => 3 // {1} {2} {3}
4 => 5 // {1} {2} {1,4} {3} {4}
5 => 7 // {1} {2} {1,4} {3} {2,5} {4} {5}
6 => 10 // {1} {2} {1,4} {3} {1,4,5} {2,3,6} {4} {3,6} {5} {6}
7 => 15 // {1} {2} {1,4} {3} {1,2,7} {3,4} {1,2,5,7} {4} {1,3,6,7} {4,5} {1,6,7} {5} {4,7} {6} {7}
8 => 21
9 => 29
10 => 39
11 => 49
12 => 63
13 => 79
14 => 99
15 => 121
16 => 145
17 => 171
18 => 203
19 => 237
20 => 277
21 => 321
22 => 369
23 => 419
24 => 477
25 => 537

规则

适用标准代码高尔夫球规则。有效提交的最短字节数为准。

赏金

大约4年前在Project Euler论坛上已经在此问题上进行了讨论，但是我们未能提出可证明的多项式时间算法（用表示N）。因此，我将向首次提交的+200赏金给予奖励，或者证明其不可能。

Brachylog，28个字节

⟦₁⊇ᶠk⊇pSs₂ᶠ{c≠&⟨+/l⟩ᵐ<ᵈ}ᵐ∧Sl

在线尝试！

这真是太慢了。大约需要30秒N = 3，而12分钟后并没有完成N = 4。

说明

⟦₁                             Take the range [1, …, Input]
  ⊇ᶠk                          Find all ordered subsets of that range, minus the empty set
     ⊇                         Take an ordered subset of these subsets
      pS                       Take a permutation of that subset and call it S
       Ss₂ᶠ                    Find all substrings of 2 consecutive elements in S
           {           }ᵐ      Map for each of these substrings:
            c≠                   All elements from both sets must be different
              &⟨+/l⟩ᵐ            And the average of both sets (⟨xyz⟩ is a fork like in APL)
                     <ᵈ          Must be in strictly increasing order
                         ∧Sl   If all of this succeeds, the output is the length of L.

更快的版本，39字节

⟦₁⊇ᶠk⊇{⟨+/l⟩/₁/₁}ᵒSs₂ᶠ{c≠&⟨+/l⟩ᵐ<₁}ᵐ∧Sl

在计算机上，此过程大约需要50秒N = 4。

这是同一程序，除了我们通过平均对子集的子集进行排序，而不是进行随机排列。因此，我们使用{⟨+/l⟩/₁/₁}ᵒ代替p。

{         }ᵒ     Order by:
 ⟨+/l⟩             Average (fork of sum-divide-length)
      /₁/₁         Invert the average twice; this is used to get a float average

我们需要得到一个浮点平均值，因为我刚刚发现了一个荒谬的错误，其中的浮点数和整数不按值进行比较，而是按类型与排序谓词进行比较（这也是为什么我使用<ᵈ而不是<₁比较两个平均值；后者会要求双重反转技巧即可工作）。

CJam（81字节）

{YY@#(#{{2bW%ee{)*~}%}:Z~{)Z__1b\,d/\a+}%$}%{_,1>{2ew{z~~&!\~=>}%0&!}{,}?},:,:e>}

在线演示。它应该4在合理的时间内为输入执行，但是我不会尝试使用更高的输入。

解剖

{                 e# Declare a block (anonymous function)
  YY@#(#          e# There are 2^N subsets of [0, N), but the empty subset is problematic
                  e# so we calculate 2^(2^N - 1) subsets of the non-empty subsets
  {               e# Map integer to subset of non-empty subsets:
    {             e#   Define a block to map an bitset to its set indices; e.g. 9 => [0 3]
      2bW%ee      e#     Convert to base 2, reverse, and index
      {)*~}%      e#     If the bit was set, keep the index
    }:Z           e#   Assign the block to variable Z
    ~             e#   Evaluate it
    {             e#   Map those indices to non-empty subsets of [0, N):
      )Z          e#     Increment (to skip the empty set) and apply Z
      __1b\,d/    e#     Sum one copy, take length of another, divide for average
      \a+         e#     Wrap the subset and prepend its average value
    }%
    $             e#   Sort (lexicographically, so by average value)
  }%
  {               e# Filter out subsets of subsets with conflicts:
    _,1>{         e#   If the length is greater than 1
      2ew         e#     Take each consecutive pair of subsets
      {           e#     Map:
        z~        e#       Zip and expand to get [score1 score2] [subset1 subset2]
        ~&!\      e#       No element in common => 1
        ~=        e#       Different scores => 0
        >         e#       1 iff both constraints are met
      }%
      0&!         e#     1 iff no consecutive pair failed the test
    }{
      ,           e#   Otherwise filter to those of length 1
    }?
  },
  :,:e>           e# Map to size of subset and take the greatest
}

JavaScript（ES6），175个字节

天真且缓慢的递归搜索。大约需要15秒来计算TIO的前7个项。

n=>(a=[...Array(n)].reduce(a=>[...a,...a.map(y=>[x,...y],x=n--)],[[]]),g=(p,b=a,n)=>a.map(a=>(m=a.map(n=>s+=++k*b.includes(n)?g:n,s=k=0)&&s/k)>p&&g(m,a,-~n),r=r>n?r:n))(r=0)|r

在线尝试！

或测试此修改版本以输出最长的递增设置序列。

怎么样？

$\{1,2,\dots,n\}$$a$

a = [...Array(n)].reduce(a =>
  [...a, ...a.map(y => [x, ...y], x = n--)],
  [[]]
)

递归部分：

g = (                         // g = recursive function taking:
  p,                          //   p = previous mean average
  b = a,                      //   b = previous set
  n                           //   n = sequence length
) =>                          //
  a.map(a =>                  // for each set a[] in a[]:
    (m = a.map(n =>           //   for each value n in a[]:
      s +=                    //     update s:
        ++k * b.includes(n) ? //       increment k; if n exists in b[]:
          g                   //         invalidate the result (string / integer --> NaN)
        :                     //       else:
          n,                  //         add n to s
      s = k = 0)              //     start with s = k = 0; end of inner map()
      && s / k                //   m = s / k = new mean average
    ) > p                     //   if it's greater than the previous one,
    && g(m, a, -~n),          //   do a recursive call with (m, a, n + 1)
    r = r > n ? r : n         //   keep track of the greatest length in r = max(r, n)
  )                           // end of outer map()

Python 3中，205个 197 184 182字节

• 由于ovs，节省了八个二十一字节。
• 多亏了ceilingcat，节省了两个字节。

f=lambda N,c=[[1]]:max([len(c)]+[f(N,c+[n])for k in range(N)for n in combinations(range(1,N+1),k+1)if not{*c[-1]}&{*n}and sum(c[-1])/len(c[-1])<sum(n)/len(n)]);from itertools import*

在线尝试！