Where is that snake going?
Wolfram Language (Mathematica), 16+83=99 bytes
Library import statement (16 bytes):
<<Combinatorica`
Actual function body (83 bytes):
Length@HamiltonianCycle[MakeGraph[#~Position~1~Join~{1>0},##||Norm[#-#2]==1&],All]&
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Note that the question just ask for the number of Hamiltonian path in the graph.
However, (for some reason) the HamiltonianPath function doesn't really work with directed graph (example). So, I used the workaround described in this Mathematica.SE question:
- Add an vertex (called
True) that is connected to all other vertices. - Count the number of Hamiltonian cycle on the resulting graph.
The graph is constructed using MakeGraph (annoyingly there are no directly equivalent built-in), using the boolean function ##||Norm[#-#2]==1&, which returns True if and only if one of the arguments is True or the distance between the two vertices are 1.
Tr[1^x] cannot be used instead of Length@x, and <2 cannot be used instead of ==1.
HamiltonianPath can be used if the graph is undirected, with the function body takes 84 bytes (exactly 1 byte more than the current submission):
Length@HamiltonianPath[MakeGraph[#~Position~1,Norm[#-#2]==1&,Type->Undirected],All]&
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Jelly, 12 11 bytes
ŒṪŒ!ạƝ€§ÐṂL
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Explanation.
ŒṪ Positions of snake blocks.
Œ! All permutations.
For each permutation:
ạƝ€ Calculate the absolute difference for each neighbor pair
§ Vectorized sum.
Now we have a list of Manhattan distance between snake
blocks. Each one is at least 1.
ÐṂL Count the number of minimum values.
Because it's guaranteed that there exists a valid snake,
the minimum value is [1,1,1,...,1].
JavaScript (ES6), 154 134 bytes
m=>m.map((r,Y)=>r.map(g=(_,x,y,r=m[y=1/y?y:Y])=>r&&r[x]&&[-1,0,1,2].map(d=>r[r[x]=0,/1/.test(m)?g(_,x+d%2,y+~-d%2):++n,x]=1)),n=0)|n/4
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How?
Method
Starting from each possible cell, we flood-fill the matrix, clearing all cells on our way. Whenever the matrix contains no more 1's, we increment the number n of possible paths.
Each valid path is counted 4 times because of the direction chosen on the last cell, which actually doesn't matter. Therefore, the final result is n / 4.
Recursive function
Instead of calling the recursive function g() from the callback of the second map() like this...
m=>m.map((r,y)=>r.map((_,x)=>(g=(x,y,r=m[y])=>...g(x+dx,y+dy)...)(x,y)))
...we define the recursive function g() directly as the callback of map():
m=>m.map((r,Y)=>r.map(g=(_,x,y,r=m[y=1/y?y:Y])=>...g(_,x+dx,y+dy)...))
Despite the rather long formula y=1/y?y:Y which is needed to set the initial value of y, this saves 2 bytes overall.
Commented code
m => // given the input matrix m[][]
m.map((r, Y) => // for each row r[] at position Y in m[][]:
r.map(g = ( // for each entry in r[], use g() taking:
_, // - the value of the cell (ignored)
x, // - the x coord. of this cell
y, // - either the y coord. or an array (1st iteration),
// in which case we'll set y to Y instead
r = m[y = 1 / y ? y : Y] // - r = the row we're currently located in
) => // (and update y if necessary)
r && r[x] && // do nothing if this cell doesn't exist or is 0
[-1, 0, 1, 2].map(d => // otherwise, for each direction d,
r[ // with -1 = West, 0 = North, 1 = East, 2 = South:
r[x] = 0, // clear the current cell
/1/.test(m) ? // if the matrix still contains at least one '1':
g( // do a recursive call to g() with:
_, // a dummy first parameter (ignored)
x + d % 2, // the new value of x
y + ~-d % 2 // the new value of y
) // end of recursive call
: // else (we've found a valid path):
++n, // increment n
x // \_ either way,
] = 1 // / do r[x] = 1 to restore the current cell to 1
) // end of map() over directions
), // end of map() over the cells of the current row
n = 0 // start with n = 0
) | n / 4 // end of map() over the rows; return n / 4