New answers tagged grid
6
Python 3 + SymPy, 480 455 bytes
lambda w,h:len({min(I(D(H(V(x)))for x in o)for H,V,D in Q((lambda p:(w-1-p[0],p[1]),I),(lambda p:(p[0],h-1-p[1]),I),(lambda p:p[::-1],I)))for o in permutations(Q(range(w),range(h)))for s in[[*zip(o,o[1:])]]if any(x+X-w+1|y+Y-h+1for(x,y),(X,Y)in zip(o,o[::-1]))+any(gcd(x-X,y-Y)>1for(x,y),(X,Y)in s)+any((len({*u,*v})>3)*...
1
Julia, 254 bytes
Not very golfed, just to keep it going;
l=length
function f(M,r,s)
l(r)<7&&return l(s)>6&&g(M,r,s)
h,j=rand([f,g],2)
a,c=extrema(r)
b=rand(a+1:c-1)
M[b,s].=2
h(M,b+1:c,s)
j(M,a:b-1,s)
end
g(M,r,s)=f(M',s,r)
n=80
M=fill(1,n,n)
f(M,1:n,1:n)
for i∈1:n
print.(getindex.(" #",M[i,:]))
println()
end
Verification, try it ...
3
Javascript (ES6), 208 bytes
Recursive, depth-first search. A breadth-first approach would be probably faster, but less golfy.
Input: a multi line string, using 1 for ghosts, 4 for solid objects, 6 for Jimmy and 2 for empty space.
f=(s,l=-~s.search`
`,j=s.search(6),g=[...s],x=j%l)=>j<l|!g[j+l]|!x|x>l-3||(g[j]=6,![...g].some((a,i)=>a&1&&...
1
Jelly, 94 89 bytes
Ø.,U$;N$+®ŒṬ€×8+®ŒṬ¤+&7$Ç€Ẹ
&Ɱ8,2ŒṪ€Ḣ©_¥/Ṡ01¦Ạ¡€+ƲṪŒṬḤ+&13$ÑÇFṀ>8Ʋ?
|Ø.¦4Z$⁺FṀ=12ƲÇFṀ>8Ʋ?
Try it online!
A full program that takes an integer matrix as its argument, with 0 as space, 1 as wall, 2 as ghost and 8 as Jimmy. Returns 1 for escape and 0 for no escape.
6
Python 2, 199 bytes
def g(I,G,W,w,h):
u,v=I.real,I.imag;R=1-(w-1>u>0<v<h-1);H=[z+cmp(u,z.real)+cmp(v,z.imag)*(u==z.real)*1jfor z in G]
for d in(R<1>(I in H+W))*range(4):J=I+1j**d;R|=(J in W)<g(J,H,W+[I],w,h)
return R
Try it online!
Rewrite of Chas Brown's solution using complex numbers.
All the coordinates are represented as a ...
8
Python 2, 236 235 227 212 bytes
def f((x,y),G,M,U=[]):
R=1-(len(M[0])-1>x>0<y<len(M)-1);H=[(u+cmp(x,u),v+cmp(y,v)*(u==x))for u,v in G];d,e=0,1
for _ in' '*4*(R<1>((x,y)in H)):J=z,w=x+d,y+e;d,e=-e,d;R|=M[w][z]>(J in U)<f(J,H,M,U+[J])
return R
Try it online!
-15 bytes thx to Bubbler
As input, takes a tuple (x,y) as jimmy's ...
1
J, 125 bytes
p=.((,-)#:1 2)|.!.0]
f=.([:(''-:2 0~.@-.~[:,]*+/@p)' '~:])*1-1 e.[:([:,@(,]+1=(2|#\)*/4|#\@|:)([:>./]*"2 p)^:_)_(<0 0)}' '~:]
Try it online!
Just a first draft, but all test cases are passing. Will attempt to golf and add explanation this week.
5
Jelly, 34 bytes
Ø1©ȧn⁶µŒṪạ®SỊƊƇḢ©Wḟ@ƲƬṖṪ<3Ȧȧm2m€4Ȧ
A monadic Link accepting a list of lists of characters (the lines) which yields 1 for a Hamiltonian or 0 otherwise. Uses the layout in the question (any non-space chars as path parts will work).
Try it online! Or see the test-suite.
How?
Starts at the top-left, [1,1], and finds a neighbouring non-...
3
Jelly, 54 bytes
n⁶,Z$µØ0jm2N,¹ƇɗƝ)m2)Z×LƊ€2¦Ẏ€;"/µØ-ḟN}Ḣɗ,+¥ƭƒịṪ¥¥ƬL=L
Try it online!
A monadic link which takes a list of strings as input and returns a boolean. The strings use the format of the question but omitting every other character (so half the width). Full explanation to follow.
5
Charcoal, 63 55 46 bytes
ＡＭ↗ Ｗ⁼№ＫＶ#¹«✳⊗⁻¹⌕ＫＶ#¦ ¿¬∨﹪ⅈ⁴﹪ⅉ²⊞υω»¿‹ＬυＸＮ²⎚«⎚¹
Try it online! Link is to verbose version of code. Edit: Saved 8 bytes by using a more cumbersome but still acceptable input format. Saved 9 bytes by requiring the path only uses #s. Explanation:
Ａ
Copy the path to the canvas.
Ｍ↗
Print a space one square in from the bottom-left ...
4
JavaScript (ES7), 126 124 122 bytes
Takes input as (n)(matrix). Returns a Boolean value.
n=>g=(m,X=k=0,Y=0,d=0)=>m.map((r,y)=>r.map((c,x)=>++c|(x-X)**2+(y-Y)**2-1||g(m,x,r[++d,x%4|y%2||k++,x]=y)))|d>1?k=g:k==n*n
Try it online!
How?
Starting at \$(0,0)\$, we flood-fill the path and increment a counter \$k\$ each time we go ...
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