Counting from 1 to n without any consecutive numbers

Question

Goal

You are given an integer n (n > 1). You must output how many permutations of the integers 1 to n there are which start at 1, end at n, and don't have two consecutive integers which differ by 1.

Alternatively, if you take the complete graph K_n and remove the edges of the path 1-2-3-...-n you must count the Hamiltonian paths from 1 to n in the remaining graph.

The examples will use f(n) for a function that takes in n and outputs the number of valid permutations, but your submission can be a function or a program.

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Examples

For n = 6, a possible solution is 1-3-5-2-4-6

However, 1-3-5-2-6-4 is not a valid solution since it does not end with 6.

In fact, for n = 6, there are only 2 solutions (1-4-2-5-3-6 is the other one).

Hence f(6) = 2.

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For n = 4 the only permutations which start in 1 and end in 4 are 1-2-3-4 and 1-3-2-4. In both of them the 2 is adjacent to the 3, giving consecutive integers which differ by 1. Therefore f(4) = 0.

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Test cases

f(6) = 2
f(4) = 0
f(8) = 68
f(13) = 4462848

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Winning criterion

This is code-golf, the shortest answer wins.

Answer 1

MATL, 16 bytes

qtq:Y@0&Yc!d|qAs

Try it online!

For inputs exceeding 12 it runs out of memory.

Explanation

q      % Implicitly input n. Push n-1
tq     % Duplicate and subtract 1: pushes n-2
:      % Range [1 2 ... n-2]
Y@     % Matrix with all permutations, each in a row
0      % Push 0
&Yc    % Append n-1 and predend 0 to each row
!      % Tranpose
d      % Consecutive differences along each column
|      % Absolute value
q      % Subtract 1
A      % All: true if all values in each column are non-zero
s      % Sum. Implicitly display

Answer 2

Mathematica, 58 bytes, polynomial(n) time

Abs[Sum[(k-1)Hypergeometric2F1[k,k-#,2,2](#-k)!,{k,#}]-1]&

How it works

Rather than iterating over permutations with brute force, we use the inclusion–exclusion principle to count them combinatorially.

Let S be the set of all permutations of [1, …, n] with σ₁ = 1, σ_n = n, and let S_i be the set of permutations σ ∈ S such that |σ_i − σ_{i + 1}| = 1. Then the count we are looking for is

|S| − |S₁ ∪ ⋯ ∪ S_{n − 1}| = ∑_{2 ≤ k ≤ n + 1; 1 ≤ i2 < ⋯ < ik − 1 < n} (−1)^{k − 2}|S_i2 ∩ ⋯ ∩ S_{ik − 1}|.

Now, |S_i2 ∩ ⋯ ∩ S_{ik − 1}| only depends on k and on the number j of runs of consecutive indices in [i₁, i₂, …, i_{k − 1}, i_k] where for convenience we fix i₁ = 0 and i_k = n. Specifically,

|S_i2 ∩ ⋯ ∩ S_{ik − 1}| = 2^{j − 2}(n − k)!, for 2 ≤ j ≤ k ≤ n,
|S_i2 ∩ ⋯ ∩ S_{ik − 1}| = 1, for j = 1, k = n + 1.

The number of such index sets [i₁, i₂, …, i_{k − 1}, i_k] with j runs is

(_{k − 1}C_{j − 1})(_{n − k}C_{j − 2}), for 2 ≤ j ≤ k ≤ n,
1, for j = 1, k = n + 1.

The result is then

(−1)^{n − 1} + ∑_{2 ≤ k ≤ n} ∑_{2 ≤ j ≤ k} (−1)^{k − 2}(_{k − 1}C_{j − 1})(_{n − k}C_{j − 2})2^{j − 2}(n − k)!

The inner sum over j can be written using the hypergeometric ₂F₁ function:

(−1)^{n − 1} + ∑_{2 ≤ k ≤ n} (−1)^k(k − 1)₂F₁(2 − k, k − n; 2; 2)(n − k)!

to which we apply a Pfaff transformation that lets us golf away the powers of −1 using an absolute value:

(−1)^{n − 1} + ∑_{2 ≤ k ≤ n} (−1)ⁿ(k − 1)₂F₁(k, k − n; 2; 2)(n − k)!
= |−1 + ∑_{1 ≤ k ≤ n} (k − 1)₂F₁(k, k − n; 2; 2)(n − k)!|.

Demo

In[1]:= Table[Abs[Sum[(k-1)Hypergeometric2F1[k,k-#,2,2](#-k)!,{k,#}]-1]&[n],{n,50}]

Out[1]= {1, 0, 0, 0, 0, 2, 10, 68, 500, 4174, 38774, 397584, 4462848, 

>    54455754, 717909202, 10171232060, 154142811052, 2488421201446, 

>    42636471916622, 772807552752712, 14774586965277816, 297138592463202402, 

>    6271277634164008170, 138596853553771517492, 3200958202120445923684, 

>    77114612783976599209598, 1934583996316791634828454, 

>    50460687385591722097602304, 1366482059862153751146376304, 

>    38366771565392871446940748410, 1115482364570332601576605376898, 

>    33544252621178275692411892779180, 1042188051349139920383738392594332, 

>    33419576037745472521641814354312790, 

>    1105004411146009553865786545464526206, 

>    37639281863619947475378460886135133496, 

>    1319658179153254337635342434408766065896, 

>    47585390139805782930448514259179162696722, 

>    1763380871412273296449902785237054760438426, 

>    67106516021125545469475040472412706780911268, 

>    2620784212531087457316728120883870079549134420, 

>    104969402113244439880057492782663678669089779118, 

>    4309132147486627708154774750891684285077633835734, 

>    181199144276064794296827392186304334716629346180848, 

>    7800407552443042507640613928796820288452902805286368, 

>    343589595090843265591418718266306051705639884996218154, 

>    15477521503994968035062094274002250590013877419466108978, 

>    712669883315580566495978374316773450341097231239406211100, 

>    33527174671849317156037438120623503416356879769273672584588, 

>    1610762789255012501855846297689494046193178343355755998487686}

Answer 3

Jelly, 17 16 bytes

ṖḊŒ!ð1;;⁹IỊṀðÐḟL

A monadic link.

Try it online!

How?

ṖḊŒ!ð1;;⁹IỊṀðÐḟL - Link: number n
Ṗ                - pop (implicit range build) -> [1,n-1]
 Ḋ               - dequeue -> [2,n-1]
  Œ!             - all permutations of [2,n-1]
    ð       ðÐḟ  - filter discard those entries for which this is truthy:
     1;          -   1 concatenated with the entry
       ;⁹        -   ...concatenated with right (n)
         I       -   incremental differences
          Ị      -   is insignificant (absolute value <=1)
           Ṁ     -   maximum
               L - length (the number of valid arrangements)

Answer 4

05AB1E, 17 bytes

L¦¨œʒ¹1Š)˜¥Ä1å_}g

Try it online!

Answer 5

Japt, 19 18 bytes

o2 á è_pU äÉ m²e>1

Test it online! I would not recommend testing on anything larger than 10.

Explanation

o2 á è_  pU äÉ  m²  e>1
o2 á èZ{ZpU ä-1 mp2 e>1}
                          : Implicit: U = input integer
o2                        : Create the range [2..U-1].
   á                      : Generate all permutations of this range.
     èZ{               }  : Check how many permutations Z return a truthy value:
        ZpU               :   Push U to the end of Z.
            ä-1           :   Push 1 to the beginning of Z, then take the difference
                          :   of each pair of items.
                m         :   Map each item X to
                 p2       :     X ** 2. This gives a number greater than 1 unless the
                          :     item is 1 or -1.
                    e>1   :   Return whether every item in this list is greater than 1.
                          :   This returns `true` iff the permutation contains no
                          :   consecutive pairs of numbers.
                          : Implicit: output result of last expression

Answer 6

Haskell, 76 65 bytes

Saved 11 bytes thanks to @xnor.

Using the result for Q_rec on page 7 of @ChristiaanWesterbeek's find, we get

f 1=1
f n|n<6=0
f n=sum$zipWith((*).f)[n-5..][n-4,1,10-2*n,4,n-2]

I don't understand how their next result ha relates to this, but after speeding up (first by memoization, see earlier versions, then as below) I get their numbers.

While the above is okay for n=20, it is essentialy an example how not to do recursion. Here is a faster version (only for n>=6) that also would only need constant memory - if only the numbers didn't keep increasing...

f n=last$foldl(#)[1,0,0,0,0][6..n]
l#n=tail l++[sum$zipWith(*)l[n-4,1,10-2*n,4,n-2]]

That gives

Prelude> f 50
1610762789255012501855846297689494046193178343355755998487686
Prelude> f 500
659178618863924802757920269977240274180092211041657762693634630044383805576666007245903670780603497370173231423527767109899936008034229541700392144282505597945561328426013937966521561345817045884498867592832897938083071843810602104434376305964577943025310184523643816782047883794585616331928324460394146825636085453532404319881264974005968087265587062691285454120911586459406436421191277596121471930913837355151842093002557978076653884610826296845041929616496533544124347765641367732716560025553179112645454078955409181466212732427071306363820080109636358537270466838558068527692374178581063316309789026101221004745226182671038004326069705775312654329754698423385241664984156235692539255677944294995403233446243315371404887473868003155621849544566385172835597260848972758443874423271017007843907015007416644383573987606586308556317833384896267539628278571497402655322562624217658332870157802254043614726316296058329670971054977099155788604175817828380564156329839201579006169173002756295957371639199917376529472990059986681882194726437566769717959443857298155265292535858523609764515938314672724480762724541633037484152303637096

It's no problem to also get f 5000 but I don't want to paste the result...

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BTW, it's possible to not use fancy math and still not use (ultra) brute force. First, instead of looking at all permutations, look at partial permutations and only extend them when they are not already invalid. It's no use to look at all permutations starting with 1 6 5. Second, some partial permutations like 1 3 5 7 and 1 5 3 7 have exactly the same valid continuations, so handle them together. Using these ideas, I could compute the values up to n=16 in 0.3s.

Answer 7

Python, 125 bytes

from itertools import*
lambda n:sum(p[-1]-p[0]==n-1and all(~-abs(x-y)for x,y in zip(p,p[1:]))for p in permutations(range(n)))

Answer 8

Mathematica, 66 bytes

Count[Permutations@Range@#,x:{1,__,#}/;FreeQ[Differences@x,1|-1]]&

Explanation

Function with first argument #.

Count[                                                             (* Count the number of *)
      Permutations@                                                (* permutations of *)
                   Range@#,                                        (* the list {1, ..., #} *)
                           x:{1,__,#}                              (* of the form {1, __, #} *)
                                     /;                            (* such that *)
                                             Differences@x,        (* the list of differences of consecutive elements *)
                                       FreeQ[                      (* is free of elements of the form *)
                                                           1|-1    (* 1 or -1 *)
                                                               ]]&

Answer 9

Javascript (ES6), 100 74 72 60 bytes

f=n=>n--<6?!n|0:f(n)*--n+4*f(n--)-2*f(n--)*--n+f(n)*++n+f(n)

Below is the version before the golf-mastery of @PeterTaylor

f=n=>n<6?n==1|0:(n-4)*f(n-5)+f(n-4)-2*(n-5)*f(n-3)+4*f(n-2)+(n-2)*f(n-1)

Thanks to the answer from @ChristianSievers that managed to draft a Haskell solution from a paper that I found after googling '0, 2, 10, 68, 500, 4174, 38774, 397584', here's a Javascript version that does not permutate too.

Usage

for (i=1; i<=20; i++) {
  console.log(i, f(i))
}

1 1 
2 0 
3 0 
4 0 
5 0 
6 2 
7 10 
8 68 
9 500 
10 4174 
11 38774 
12 397584 
13 4462848 
14 54455754 
15 717909202 
16 10171232060 
17 154142811052 
18 2488421201446 
19 42636471916622 
20 772807552752712

Answer 10

Python 3, 109 107 102 bytes

q=lambda s,x,n:sum(q(s-{v},v,n)for v in s if(v-x)**2>1)if s else x<n;f=lambda n:q({*range(2,n)},1,n-1)

Try it online!

Removed four bytes by not trying to one-line the function (as suggested by @shooqie) and another byte by replacing abs with a square. (Requires Python 3.5+)

Answer 11

Brachylog, 26 bytes

{⟦₁pLh1&~tLs₂ᶠ{-ȧ>1}ᵐ}ᶜ|∧0

Try it online!

Explanation

{                    }ᶜ       Output = count the number of outputs of:
 ⟦₁pL                           L is a permutation of [1, …, Input]
    Lh1                         The head of L is 1
       &~tL                     The tail of L is the Input
          Ls₂ᶠ                  Find all sublists of length 2 of L
              {    }ᵐ           Map on each sublist:
               -ȧ>1               The elements are separated by strictly more than 1
                       |      Else (no outputs to the count)
                        ∧0    Output = 0

Answer 12

Python 2, 136 bytes

-10 bytes thanks to @ovs.

lambda n,r=range:sum(x[0]<1and~-n==x[-1]and 2+~any(abs(x[i]-x[i+1])<2for i in r(n-1))for x in permutations(r(n)))
from itertools import*

Try it online!

Answer 13

Mathematica, 134 bytes

(s=Permutations@Range[2,#-1];g=Table[Join[Prepend[s[[i]],1],{#}],{i,Length@s}];Length@Select[Union@*Abs@*Differences/@g,FreeQ[#,1]&])&

test cases n: 2 to 12

{0, 0, 0, 0, 2, 10, 68, 500, 4174, 38774, 397584}

Answer 14

Python 2, 105 bytes

lambda n:reduce(lambda a,i:a+[i*a[-5]+a[-4]+2*(1-i)*a[-3]+4*a[-2]+(i+2)*a[-1]],range(2,n),[0,1]+4*[0])[n]

Try it online!

This is based on Philippe Flajolet's paper discovered by @Christiaan Westerbeek; it's much faster and two bytes shorter than my Python 3 solution which enumerates the possible permutations. (In Python 3, reduce has annoyingly been moved to functools.)

There is a much shorter version using numpy's dot product, but that overflows quite rapidly and requires numpy to have been imported. But for what it's worth:

lambda n:reduce(lambda a,i:a+[dot([i,1,2-2*i,4,i+2],a[-5:])],range(2,n),[0,1]+4*[0])[n]