Count the number of shortest paths to n

Question

This code challenge will have you compute the number of ways to reach \$n\$ starting from \$2\$ using maps of the form \$x \mapsto x + x^j\$ (with \$j\$ a non-negative integer), and doing so in the minimum number of steps.

(Note, this is related to OEIS sequence A307092.)

Example

So for example, \$f(13) = 2\$ because three maps are required, and there are two distinct sequences of three maps that will send \$2\$ to \$13\$:

$$\begin{array}{c} x \mapsto x + x^0 \\ x \mapsto x + x^2 \\ x \mapsto x + x^0\end{array} \qquad \textrm{or} \qquad \begin{array}{c}x \mapsto x + x^2 \\ x \mapsto x + x^1 \\ x \mapsto x + x^0\end{array}$$

Resulting in \$2 \to 3 \to 12 \to 13\$ or \$2 \to 6 \to 12 \to 13\$.

Example values

f(2)   = 1 (via [])
f(3)   = 1 (via [0])
f(4)   = 1 (via [1])
f(5)   = 1 (via [1,0])
f(12)  = 2 (via [0,2] or [2,1])
f(13)  = 2 (via [0,2,0] or [2,1,0], shown above)
f(19)  = 1 (via [4,0])
f(20)  = 2 (via [1,2] or [3,1])
f(226) = 3 (via [2,0,2,1,0,1], [3,2,0,0,0,1], or [2,3,0,0,0,0])
f(372) = 4 (via [3,0,1,0,1,1,0,1,1], [1,1,0,2,0,0,0,1,1], [0,2,0,2,0,0,0,0,1], or [2,1,0,2,0,0,0,0,1])

Challenge

The challenge is to produce a program which takes an integer \$n \ge 2\$ as an input, and outputs the number of distinct paths from \$2\$ to \$n\$ via a minimal number of maps of the form \$x \mapsto x + x^j\$.

This is code-golf, so fewest bytes wins.

Answer 1

Jelly, 16 bytes

2+*¥þ³Ḷ¤F$n³Ạ$¿ċ

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A full program taking as its argument \$n\$ and returning the number of ways to reach \$n\$ using the minimal path length. Inefficient for larger \$n\$.

Answer 2

JavaScript (ES6),  111 ... 84  80 bytes

Returns true rather than \$1\$ for \$n=2\$.

f=(n,j)=>(g=(i,x,e=1)=>i?e>n?g(i-1,x):g(i-1,x+e)+g(i,x,e*x):x==n)(j,2)||f(n,-~j)

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Commented

f = (                     // f is the main recursive function taking:
  n,                      //   n = input
  j                       //   j = maximum number of steps
) => (                    //
  g = (                   // g is another recursive function taking:
    i,                    //   i = number of remaining steps
    x,                    //   x = current sum
    e = 1                 //   e = current exponentiated part
  ) =>                    //
    i ?                   // if there's still at least one step to go:
      e > n ?             //   if e is greater than n:
                          //     add the result of a recursive call with:
        g(i - 1, x)       //       i - 1, x unchanged and e = 1
      :                   //   else:
                          //     add the sum of recursive calls with:
        g(i - 1, x + e) + //       i - 1, x + e and e = 1
        g(i, x, e * x)    //       i unchanged, x unchanged and e = e * x
    :                     // else:
      x == n              //   stop recursion; return 1 if x = n
)(j, 2)                   // initial call to g with i = j and x = 2
|| f(n, -~j)              // if it fails, try again with j + 1

Answer 3

Haskell, 78 75 bytes

This implementation uses a breath-first-search in the "tree" of iteratively all necessary mappings x -> x + x^j.

j#x=x+x^j
f n=[sum[1|x<-l,x==n]|l<-iterate((#)<$>[0..n]<*>)[2],n`elem`l]!!0

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Explanation

-- computes the mapping x -> x + x^j
j#x=x+x^j                          
--iteratively apply this function for all exponents [0,1,...,n] (for all previous values, starting with the only value [2])
                            iterate((#)<$>[0..n]<*>)[2] 
-- find each iteration where our target number occurs
    [                   |l<-...........................,n`elem`l] 
-- find how many times it occurs
     sum   [1|x<-l,x==n] 
-- pick the first entry
f n=.............................................................!!0

Answer 4

05AB1E, 17 bytes

2¸[˜DIå#εIÝmy+]I¢

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Answer 5

Python 2, 72 bytes

f=lambda n,l=[2]:l.count(n)or f(n,[x+x**j for x in l for j in range(n)])

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Answer 6

Perl 5 (-lp), 79 bytes

$e=$_;@,=(2);@,=map{$x=$_;map$x+$x**$_,0..log($e)/log$x}@,until$_=grep$_==$e,@,

TIO

Answer 7

CJam (27 bytes)

qi2a{{_W$,f#f+~2}%_W$&!}ge=

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Warning: this gets very memory-intensive very fast.

Dissection:

qi            e# Read input and parse to int n (accessed from the bottom of the stack as W$)
2a            e# Start with [2]
{             e# Loop
  {           e#   Map each integer x in the current list
    _W$,f#f+~ e#     to x+x^i for 0 <= i < n
    2         e#   and add a bonus 2 for the special case
  }%          e#   Gather these in the new list
  _W$&!       e#   Until the list contains an n
}g
e=            e# Count occurrences

The bonus 2s (to handle the special case of input 2, because while loops are more expensive than do-while loops) mean that the size of the list grows very fast, and the use of exponents up to n-1 means that the values of the larger numbers in the list grow very fast.

Answer 8

Haskell, 65 bytes

([2]%)
l%n|elem n l=sum[1|x<-l,x==n]|1>0=[x+x^k|x<-l,k<-[0..n]]%n

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Golfing flawr's breadth-first-search. I also tried going backwards from n, but it was longer:

73 bytes

q.pure
q l|elem 2l=sum[1|2<-l]|1>0=q[x|n<-l,x<-[0..n],i<-[0..n],x+x^i==n]

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Answer 9

R, 78 77 bytes

function(n,x=2){while(!{a=sum(x==n)})x=rep(D<-x[x<n],n+1)+outer(D,0:n,'^')
a}

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Using a simplified Breadth-first Search

Unrolled code with explanation :

function(n){                              # function taking the target value n

  x=2                                     # initialize vector of x's with 2

  while(!(a<-sum(x==n))) {                # count how many x's are equal to n and store in a
                                          # loop while a == 0

    x=rep(D<-x[x<n],n+1)+outer(D,0:n,'^') # recreate the vector of x's 
                                          # with the next values: x + x^0:n
  }
a                                         # return a
}   

---

Shorter version with huge memory allocation (failing for bigger cases) :

R, 70 69 bytes

function(n,x=2){while(!{a=sum(x==n)})x=rep(x,n+1)+outer(x,0:n,'^')
a}

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-1 byte thanks to @RobinRyder

Answer 10

Pyth, 24 bytes

VQIJ/mu+G^GHd2^U.lQ2NQJB

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This should produce the correct output, but is very slow (the 372 test case times out on TIO). I could make it shorter by replacing .lQ2 with Q, but this would make the runtime horrendous.

Note: Produces no output for unreachable numbers \$(n \leq 1)\$

Explanation

VQ                        # for N in range(Q (=input)):
   J                      #   J =
     m                    #     map(lambda d:
      u                   #       reduce(lambda G,H:
       +G^GH              #         G + G^H,
            d2            #         d (list), 2 (starting value) ),
              ^U.lQ2N     #       cartesian_product(range(log(Q, 2)), N)
    /                Q    #     .count(Q)
  IJ                  JB  #   if J: print(J); break