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Bytes Lang Time Link
1171Python3240522T202439ZAjax1234
185JavaScript ES7190802T181815ZArnauld

Python3, 1171 bytes

Long, but fast.

P=[[(1,3)],[(2,4)],[(1,2)],[(2,3)],[(3,4)],[(4,1)],[(1,2),(3,4)],[(2,3),(4,1)]]
M={1:(3,0,-1),3:(1,0,1),2:(4,-1,0),4:(2,1,0)}
E=enumerate
def G(d):
 q=[i for i in d if d[i]]
 while q:
  v=q.pop(0)
  Q,S=[(*v,0,[])],[v]
  for x,y,H,p in Q:
   for j in P[d[(x,y)]-1]:
    if not H or H in j:
     for J in j:
      e,X,Y=M[J];U=(x+X,y+Y)
      if d.get(U,-1)>0 and([]==p or U!=p[-1]):
       if p and U==p[0]:return p+[(x,y)]
       Q+=[(*U,e,p+[(x,y)])];S+=[U]
  q=[*{*q}-{*S}]
def V(d):
 for x,y in d:
  if d[(x,y)]:
   for j in P[d[(x,y)]-1]:
    for J in j:
     e,X,Y=M[J];U=(x+X,y+Y)
     if U not in d:return
     if d[U]==0:continue
     if not any(e in u for u in P[d[U]-1]):return
 return 1
def f(b):
 q=[{(x,y):v for x,r in E(b)for y,v in E(r)}]
 for d in q:
  if(L:=G(d)):
   if len({*L})!=len(d):continue
   return d
  D={}
  for x,y in d:
   if d[(x,y)]==0:
    for I,A in E(P,1):
     T=eval(str(d));T[(x,y)]=I
     if V(T):D[(x,y)]=D.get((x,y),[])+[I]
  if D:
   t=min(map(len,D.values()))
   k=[i for i in D if len(D[i])==t]
   if t==1:
    T=eval(str(d))
    for i in k:T[i]=D[i][0]
    q+=[T]
   else:
    for I in D[k[0]]:T=eval(str(d));T[k[0]]=I;q+=[T]

Try it online!

JavaScript (ES7),  236 ... 193  185 bytes

Outputs by modifying the input matrix.

m=>(g=(d,x,y,v,r=m[y],h=_=>++r[x]<9?g(d,x,y,v)||h():r[x]=0)=>r&&1/(n=r[x])?x|y|!v?n?g(d='21100--13203-32-21030321'[n*28+d*3+7&31],x+--d%2,y+--d%2,v+=n<7||.5):h():!m[v**.5|0]:0)(0,0,0,0)

Try it online!

(includes some post-processing code to print the result both as a matrix and as a flat list compatible with the visualization tool provided by the OP)

Results

How?

Variables

\$g\$ is a recursive function taking the current direction \$d\$, the current coordinates \$(x,y)\$ and the number of visited cells \$v\$.

The following variables are also defined in the scope of \$g\$:

Initial checks

We first make sure that our current location is valid and we load the value of the current cell into \$n\$:

r && 1 / (n = r[x]) ? ... ok ... : ... failed ...

We test whether we're back to our starting position, i.e. we're located at \$(0,0)\$ and we've visited at least a few cells (\$v>0\$):

x | y | !v ? ... no ... : ... yes ...

For now, let's assume that we're not back to the starting point.

Looking for a path

If \$n\$ is equal to \$0\$, we invoke \$h\$ to try all possible values for this tile.

If \$n\$ is not equal to \$0\$, we try to move to the next tile.

The tile connections are encoded in a lookup table, whose index is computed with \$n\$ and \$d\$, and whose valid entries represent the new values of \$d\$.

d = '21100--13203-32-21030321'[n * 28 + d * 3 + 7 & 31]

The last 8 entries are invalid and omitted. The other 4 invalid entries are explicitly marked with hyphens.

For reference, below are the decoded table, the compass and the tile-set provided in the challenge:

   | 1 2 3 4 5 6 7 8
---+-----------------
 0 | 0 - - 1 3 - 3 1          1
 1 | - 1 - - 2 0 2 0        0 + 2
 2 | 2 - 1 - - 3 1 3          3
 3 | - 3 0 2 - - 0 2

We do a recursive call to \$g\$ with the new direction and the new coordinates. We add \$1/2\$ to \$v\$ if we were on a tile of type \$7\$ or \$8\$, or \$1\$ otherwise (see the next paragraph).

g(d, x + --d % 2, y + --d % 2, v += n < 7 || .5)

If \$d\$ is invalid, \$x\$ and \$y\$ will be set to NaN, forcing the next iteration to fail immediately.

Validating the path

Finally, when we're back to \$(0,0)\$ with \$v>0\$, it doesn't necessarily mean that we've found a valid path, as we may have taken a shortcut. We need to check if we've visited the correct number of cells.

Each tile must be visited once, except tiles \$7\$ and \$8\$ that must be visited twice. This is why we add only \$1/2\$ to \$v\$ when such a tile is visited.

In the end, we must have \$v = N^2\$. But it's also worth noting that we can't possibly have \$v > N^2\$. So, it's enough to test that we don't have \$v < N^2\$, or that the \$k\$th row of the matrix (0-indexed) does not exist, where \$k=\lfloor\sqrt{v}\rfloor\$.

Hence the JS code:

!m[v ** .5 | 0]

Formatted source

m => (
  g = (
    d,
    x, y,
    v,
    r = m[y],
    h = _ => ++r[x] < 9 ? g(d, x, y, v) || h() : r[x] = 0
  ) =>
    r && 1 / (n = r[x]) ?
      x | y | !v ?
        n ?
          g(
            d = '21100--13203-32-21030321'[n * 28 + d * 3 + 7 & 31],
            x + --d % 2,
            y + --d % 2,
            v += n < 7 || .5
          )
        :
          h()
      :
        !m[v ** .5 | 0]
    :
      0
)(0, 0, 0, 0)
GOLFSCORE/189164