Single moves

The board is an infinite 2 dimensional square grid, like a limitless chess board. A piece with value N (an N-mover) can move to any square that is a distance of exactly the square root of N from its current square (Euclidean distance measured centre to centre).

For example:

• A 1-mover can move to any square that is horizontally or vertically adjacent


• A 2-mover can move to any square that is diagonally adjacent


• A 5-mover moves like a chess knight

Note that not all N-movers can move. A 3-mover can never leave its current square because none of the squares on the board are a distance of exactly root 3 from the current square.

Multiple moves

If allowed to move repeatedly, some pieces can reach any square on the board. For example, a 1-mover and a 5-mover can both do this. A 2-mover can only move diagonally and can only reach half of the squares. A piece that cannot move, like a 3-mover, cannot reach any of the squares (the starting square is not counted as "reached" if no movement occurs).

The images show which squares can be reached. More details on hover. Click for larger image.

• Squares reachable in 1 or more moves are marked in black


• Squares reachable in exactly 1 move are shown by red pieces
  (apart from the 3-mover, which cannot move)
  

What proportion of the board can a given N-mover reach?

Input

• A positive integer N

Output

• The proportion of the board that an N-mover can reach


• This is a number from 0 to 1 (both inclusive)


• For this challenge, output as a fraction in lowest terms, like 1/4, is allowed

So for input 8, both 1/8 and 0.125 are acceptable outputs.

Scoring

This is code golf. The score is the length of the code in bytes. For each language, the shortest code wins.

Test cases

In the format input : output as fraction : output as decimal

  1 : 1     : 1

  2 : 1/2   : 0.5

  3 : 0     : 0

  4 : 1/4   : 0.25

  5 : 1     : 1

  6 : 0     : 0

  7 : 0     : 0

  8 : 1/8   : 0.125

  9 : 1/9   : 0.1111111111111111111111111111

 10 : 1/2   : 0.5

 13 : 1     : 1

 16 : 1/16  : 0.0625

 18 : 1/18  : 0.05555555555555555555555555556

 20 : 1/4   : 0.25

 25 : 1     : 1

 26 : 1/2   : 0.5

 64 : 1/64  : 0.015625

 65 : 1     : 1

 72 : 1/72  : 0.01388888888888888888888888889

 73 : 1     : 1

 74 : 1/2   : 0.5

 80 : 1/16  : 0.0625

 81 : 1/81  : 0.01234567901234567901234567901

 82 : 1/2   : 0.5

144 : 1/144 : 0.006944444444444444444444444444

145 : 1     : 1

146 : 1/2   : 0.5

148 : 1/4   : 0.25

153 : 1/9   : 0.1111111111111111111111111111

160 : 1/32  : 0.03125

161 : 0     : 0

162 : 1/162 : 0.006172839506172839506172839506

163 : 0     : 0

164 : 1/4   : 0.25

241 : 1     : 1

242 : 1/242 : 0.004132231404958677685950413223

244 : 1/4   : 0.25

245 : 1/49  : 0.02040816326530612244897959184

260 : 1/4   : 0.25

261 : 1/9   : 0.1111111111111111111111111111

288 : 1/288 : 0.003472222222222222222222222222

290 : 1/2   : 0.5

292 : 1/4   : 0.25

293 : 1     : 1

324 : 1/324 : 0.003086419753086419753086419753

325 : 1     : 1

326 : 0     : 0

360 : 1/72  : 0.01388888888888888888888888889

361 : 1/361 : 0.002770083102493074792243767313

362 : 1/2   : 0.5

369 : 1/9   : 0.1111111111111111111111111111

370 : 1/2   : 0.5

449 : 1     : 1

450 : 1/18  : 0.05555555555555555555555555556

488 : 1/8   : 0.125

489 : 0     : 0

490 : 1/98  : 0.01020408163265306122448979592

520 : 1/8   : 0.125

521 : 1     : 1

522 : 1/18  : 0.05555555555555555555555555556

544 : 1/32  : 0.03125

548 : 1/4   : 0.25

549 : 1/9   : 0.1111111111111111111111111111

584 : 1/8   : 0.125

585 : 1/9   : 0.1111111111111111111111111111

586 : 1/2   : 0.5

592 : 1/16  : 0.0625

593 : 1     : 1

596 : 1/4   : 0.25

605 : 1/121 : 0.008264462809917355371900826446

610 : 1/2   : 0.5

611 : 0     : 0

612 : 1/36  : 0.02777777777777777777777777778

613 : 1     : 1

624 : 0     : 0

625 : 1     : 1

Clean, 189 185 172 171 bytes

import StdEnv

$n#r=[~n..n]

#p=[[x,y]\x<-r,y<-r|x^2+y^2==n]

=sum[1.0\_<-iter n(q=removeDup[k\[a,b]<-[[0,0]:p],[u,v]<-q,k<-[[a+u,b+v]]|all(e=n>=e&&e>0)k])p]/toReal(n^2)

Try it online!

Finds every position reachable in the n-side-length square cornered on the origin in the first quadrant, then divides by n^2 to get the portion of all cells reachable.

JavaScript (Node.js), 144 bytes

a=>Object.keys((w=F=(x,n=2)=>n*n>x?w[x]=-~w[x]:x%n?F(x,n+1):w[F(x/n,n),n]=-~w[n])(a)&&w).map(y=>z*=[,1,.5**w[y],w[y]%2?0:1/y**w[y]][y%4],z=1)&&z

Try it online!

Inspired by the video Pi hiding in prime regularities by 3Blue1Brown.

For each prime factor $p^n$ in the factorization of the number, calculate $f(p^n)$:

• If $n$ is odd and $pequiv 3text{ (mod 4)}$ - $f(p^n)=0$. Because there is no place to go.


• If $n$ is even and $pequiv 3text{ (mod 4)}$ - $f(p^n)=frac{1}{p^n}$.


• If $p=2$ - $f(2^n)=frac{1}{2^n}$.


• If $pequiv 1text{ (mod 4)}$ - $f(p^n)=1$.

Multiply all those function values, there we are.

Mathematica, 80 bytes

d[n_]:=If[#=={},0,1/Det@LatticeReduce@#]&@Select[Tuples[Range[-n,n],2],#.#==n&];

This code is mostly reliant on a mathematical theorem. The basic idea is that the code asks for the density of a lattice given some generating set.

More precisely, we are given some collection of vectors - namely, those whose length squared is N - and asked to compute the density of the set of possible sums of these vectors, compared to all integer vectors. The math at play is that we can always find two vectors (and their opposite) that "generate" (i.e. whose sums are) the same set as the original collection. LatticeReduce does exactly that.

If you have just two vectors, you can imagine drawing an identical parallelogram centered at each reachable point, but whose edge lengths are the given vectors, such that the plane is completely tiled by these parallelograms. (Imagine, for instance, a lattice of "diamond" shapes for n=2). The area of each parallelogram is the determinant of the two generating vectors. The desired proportion of the plane is the reciprocal of this area, since each parallelogram has just one reachable point in it.

The code is a fairly straightforward implementation: Generate the vectors, use LatticeReduce, take the determinant, then take the reciprocal. (It can probably be better golfed, though)