Is the path correct?

Question

Is the path correct ?

Given a string, your program must determine whether or not Alice can reach Bob, following the signs on the map. You must return a truthy or falsy value.

The input string will contain the following characters :

• (space) and \n. Only used for padding.
• >, <, ^, v, are the "signs". They show the direction.
• &. Ignore next char, whatever it is (including spaces and Bob)
• A and B respectively Alice and Bob.

The map is a torus. If Alice reaches the end of a row, then she is teleported back to the beginning of that row. Same applies for columns.

Examples

Given those examples as input, your program must return a truthy value (each input is separated with a - line):

A>B
-----------------
<BA
-----------------
A B
-----------------
BA
-----------------
Av >B
 > ^
-----------------
Av
 B
-----------------
A>v
 >B
-----------------
A>v> v>B
 v< > &<^
 > ^
-----------------
A>v> v
 v< >B&<
 > ^
-----------------
A&<B
-----------------
A&B<
-----------------
<A &B

For the following, your program must output a falsy value or nothing :

AvB
-----------------
A&B
-----------------
A><B
-----------------
A>v
 ^<B

Path viewer

You can have a visual representation of the input here. Thanks to arnemart for this.

Some rules

• The input may be padded with spaces to form a rectangle
• Initially, Alice goes straight forward (from above, she goes to the right)
• Your program should handle infinite loop cases (see Hint)
• A is not always the first character, as you may have seen, however, you may write a program that handles only input with A being the first char, but 20 bytes will be added to your score.
• A map (the input) is considered invalid in case there are other characters than those specified above. Your program won't have to handle them.
• This is code-golf, so shortest code in bytes wins.

Hint

Your program will most likely be a loop, unless you find a more mathematical solution to this challenge. To handle infinite path, I recommend using a finite loop (e.g for) instead of a potentially infinite one (e.g while).

Therefore, the maximum iteration of the loop I can think of (I did not do any maths, any correction is welcome) is (length of input)*4

• AdmBorkBork: Closely related -- codegolf.stackexchange.com/q/57952/42963
• user81655: Can the input be padded with spaces so that it forms a rectangle and each line is the same length?
• THC: @user81655 Yes it can
• mbomb007: If only there was a way to replace a few chars then exec in ><>...
• THC: @mbomb007 Sorry :D

Answer 1

C, 250 bytes

#define h x=(j+x-1+d%3)%j,i=(j+i+(d=='v')-(d==94))%j
char a[999],b[999];x,i,d=2,k,f;main(j){for(;gets(a+j*i);)j=strlen(a+j*i++);for(x=strchr(a,65)-a,i=x/j,x=x%j;!k*d;h)k|=b[f=j*i+x]&1<<d%5,b[f]|=1<<d%5,(f=a[f])&16?d=f:f&4?h:f&2?d=0:0;putchar(48+!k);}

This also requires that the input is padded with spaces.

Ungolfed:

#define h x=(j+x-1+d%3)%j,i=(j+i+(d=='v')-(d==94))%j 
char a[999],b[999];x,i,d=2,k,f;
main(j){
 for(;gets(a+j*i);)j=strlen(a+j*i++); //read lines and determine line length
 for(x=strchr(a,65)-a,i=x/j,x=x%j; //the current position is 'A'
 !k*d; //as long as we have not arrived at 'B' 
 //and not travelled through the same cell and in the same direction twice
 h) //advance one in the current direction
 k|=b[f=j*i+x]&1<<d%5,b[f]|=1<<d%5, //if we have already travelled 
 //through this cell in this direction
 (f=a[f])&16?d=f //change direction if this cell is one of <>^v
 :f&4?h //if this cell is '&', advance one in the current direction
 :f&2?d=0:0; //if we have reached 'B'
 putchar(48+!k);
}

Answer 2

JavaScript (ES6), 176 bytes

m=>[...m,q=~m.length,l=~m.search`
`||q,a=m.search`A`,p=d=r=1].map(_=>(a+=d,a-=d*((a-l+1)%l?m[a]?0:q/l:~l),c=m[a],p=="&"?0:d={">":1,"<":-1,"^":l,v:-l}[c=="B"?r=0:c]||d,p=c))&&!r

Explanation

Requires the input to be space-padded to form a rectangle (I hope this is OK).

var solution =
m=>[
 ...m, // iterate m.length times
 q=~m.length, // q = total length of m
 l=~m.search`\n`||q, // l = line length
 a=m.search`A`, // a = index of Alice
 p= // p = previous character
 d= // d = direction offset
 r=1 // r = result
 ].map(_=>( // loop
 a+=d, // move Alice
 a-=d*((a-l+1)%l?m[a]?0:q/l:~l), // wrap
 c=m[a],
 p=="&"?0: // do nothing after &
 d={">":1,"<":-1,"^":l,v:-l} // set the new direction
 [c=="B"?r=0:c] // if on Bob, r = 0
 ||d,
 p=c
 ))&&!r // return result
var testCasesTrue =
`A>B
-----------------
<BA
-----------------
A B
-----------------
BA
-----------------
Av >B
 > ^ 
-----------------
Av
 B
-----------------
A>v 
 >B
-----------------
A>v> v>B
 v< > &<^ 
 > ^ 
-----------------
A>v> v
 v< >B&<
 > ^ 
-----------------
A&<B
-----------------
A&B<
-----------------
<A &B`
.split("\n-----------------\n");
var testCasesFalse =
`AvB
-----------------
A&B
-----------------
A><B
-----------------
A>v 
 ^<B`
.split("\n-----------------\n");
var test = (cases, expected) => cases.map((t,i)=>++i+": "+(solution(t)==expected?"Pass":"Fail")).join`\n`;
document.write(`<pre>${test(testCasesTrue,true)}\n\n${test(testCasesFalse,false)}</pre>`);

• THC: Congratulations !