Barbanegra era un pirata Inglés de principios del 18 ^º siglo. Aunque era conocido por saquear y tomar barcos, comandó sus naves con el permiso de sus tripulaciones. No hay informes de que él haya dañado o asesinado a sus cautivos.

Este desafío es en honor al infame Barbanegra e inspirado en el Día Internacional de la Charla como un pirata (19 de septiembre). También es lo contrario de este desafío de Pyrrha .

El reto

Cree un programa que tome un mapa del tesoro como entrada (compuesto por los caracteres que se enumeran a continuación) y muestre sus instrucciones.

Entrada

Todas las entradas consistirán en v, >, <, ^, espacio en blanco, y una sola X.

Puede asumir lo siguiente:

• el mapa nunca se enrollará o se cruzará

• la flecha de inicio siempre será el carácter inferior en la columna de la izquierda

• siempre habrá un tesoro ( X)

Una entrada de muestra se muestra a continuación.

  >>v   >>>>>>v
  ^ v   ^     v
  ^ v   ^   v<<
  ^ v   ^   v
  ^ >>>>^   >>X
  ^
>>^

Salida

La salida debe ser una ", "cadena de direcciones delimitada. A continuación se muestra la salida correcta del mapa anterior.

E2, N6, E2, S4, E4, N4, E6, S2, W2, S2, E2

Se permite una nueva línea o espacio final.

Ejemplos

In:
>>>>>>>>>>>>>>>v
               v
               v
               >>>>X

Out:
E15, S3, E4

In:
>>>>>>v
^     v
^     >>>>X

Out:
N2, E6, S2, E4

In:
X
^
^
^

Out:
N3

In:
>>>>>>v
^     v
^     v
      v
      >>>>>>X

Out:
N2, E6, S4, E6

In:
 X
 ^
 ^
>^

Out:
E1, N3

In:
>X

Out:
E1

In:
v<<<<<
vX<<<^
>>>>^^
>>>>>^

Out:
E5, N3, W5, S2, E4, N1, W3

¡Feliz charla internacional como un día pirata!

CJam, 78 bytes

qN/_:,$W=:Tf{Se]}s:U,T-{_U="><^vX"#"1+'E1-'WT-'NT+'S0"4/=~_@\}g;;]e`{(+}%", "*

Pruébalo en línea .

Explicación

La idea principal aquí es encontrar la línea más larga (llamaremos a esta longitud T), luego rellenar todas las líneas a la misma longitud y concatenarlas (esta nueva cadena es U). De esta manera, solo se necesita un contador para moverse en el mapa. Sumar / restar 1significa moverse hacia la derecha / izquierda en la misma línea, sumar / restar Tsignifica moverse hacia abajo / arriba una línea.

qN/    e# Split the input on newlines
_:,    e# Push a list of the line lengths
$W=:T  e# Grab the maximum length and assign to T
f{Se]} e# Right-pad each line with spaces to length T
s:U    e# Concatenate lines and assign to U

Ahora es el momento de configurar el ciclo.

,T-    e# Push len(U) - T
       e# i.e. position of first char of the last line
{...}g e# Do-while loop
       e# Pops condition at the end of each iteration

El cuerpo del bucle utiliza una tabla de búsqueda y evalúa para elegir qué hacer. Al comienzo de cada iteración, el elemento de la pila superior es la posición actual. Debajo hay todas las direcciones NSWE procesadas. Al final de la iteración, la nueva dirección se coloca debajo de la posición y se usa una copia de la misma como condición para el bucle. Los caracteres distintos de cero son verdaderos. Cuando se encuentra una X, se empuja 0 como la dirección, terminando el ciclo.

_U=      e# Push the character in the current position
"><^vx"# e# Find the index in "><^Vx"
"..."4/  e# Push the string and split every 4 chars
         e# This pushes the following list:
         e# [0] (index '>'): "1+'E" pos + 1, push 'E'
         e# [1] (index '<'): "1-'W" pos - 1, push 'W'
         e# [2] (index '^'): "T-'N" pos - T, push 'N'
         e# [3] (index 'v'): "T+'S" pos + T, push 'S'
         e# [4] (index 'X'): "0"    push 0
=~       e# Get element at index and eval
_@\      e# From stack: [old_directions position new_direction]
         e# To stack: [old_directions new_direction position new_direction]
         e# (You could also use \1$)
         e# new_direction becomes the while condition and is popped off

Ahora las miradas se encaja como esto: [directions 0 position]. Vamos a generar la salida.

;;    e# Pop position and 0 off the stack
]     e# Wrap directions in a list
e`    e# Run length encode directions
      e# Each element is [num_repetitions character]
{     e# For each element:
 (+   e#   Swap num_repetitions and character
}%    e# End of map (wraps in list)
", "* e# Join by comma and space

CJam, 86 bytes

qN/_,{1$=cS-},W=0{_3$3$==_'X-}{"^v<>"#_"NSWE"=L\+:L;"\(\ \)\ ( )"S/=~}w];Le`{(+}%", "*

Pruébalo en línea

Explicación:

qN/     Get input and split into rows.
_,      Calculate number of rows.
{       Loop over row indices.
  1$=     Get row at the index.
  c       Get first character.
  S-      Compare with space.
},      End of filter. The result is a list of row indices that do not start with space.
W=      Get last one. This is the row index of the start character.
0       Column number of start position. Ready to start tracing now.
{       Start of condition in main tracing loop.
  _3$3$   Copy map and current position.
  ==      Extract character at current position.
  _'X-    Check if it's the end character `X.
}       End of loop condition.
{       Start of loop body. Move to next character.
  "^v<>"  List of directions.
  #       Find character at current position in list of directions.
  _       Copy direction index.
  "NSWE"  Matching direction letters.
  =       Look up direction letter.
  L\+:L;  Append it to directions stored in variable L.
  "\(\ \)\ ( )"
          Space separated list of commands needed to move to next position for each of
          the 4 possible directions.
  S/      Split it at spaces.
  =       Extract the commands for the current direction.
  ~       Evaluate it.
}w      End of while loop for tracking.
];      Discard stack content. The path was stored in variable L.
Le`     Get list of directions in variable L, and RLE it.
{       Loop over the RLE entries.
  (+      Swap from [length character] to [character length].
}%      End of loop over RLE entries.
", "*   Join them with commas.

Javascript (ES6), 239 bytes

a=>(b=a.split`
`,b.reverse().some((c,d)=>c[0]!=' '&&((e=d)||1)),j=[],eval("for(g=b[e][f=0];b[e][f]!='X';g=b[e][f],j.push('NSEW'['^v><'.indexOf(i)]+h))for(h=0;g==b[e][f];e+=((i=b[e][f])=='^')-(i=='v'),f+=(i=='>')-(i=='<'),h++);j.join`, `"))

Explicación:

a=>(
    b = a.split('\n'),
    // loops through list from bottom to find arrow
    b.reverse().some(
        (c, d)=>
            // if the leftmost character is not a space, saves the index and exit
            // the loop
            // in case d == 0, the ||1 makes sure the loop is exited
            c[0] != ' ' && ((e = d) || 1)
    ),
    j = [], // array that will hold the instructions
    eval("  // uses eval to allow a for loop in a lambda without 'return' and {}

        // loops through all sequences of the same character
        // e is the first coordinate of the current character being analyzed
        // f is the second coordinate
        // defines g as the character repeated in the sequence
        // operates on reversed b to avoid using a second reverse
        // flips ^ and v to compensate

        for(g = b[e][f = 0];
            b[e][f] != 'X'; // keep finding sequences until it finds the X
            g = b[e][f],    // update the sequence character when it hits the start of a
                            // new sequence
            j.push('NSEW'['^v><'.indexOf(i)] + h)) // find the direction the sequence is
                                                   // pointing to and add the
                                                   // instruction to j

            // loops through a single sequence until it hits the next one
            // counts the length in h
            for(h = 0;
                g == b[e][f]; // loops until there is a character that isn't part of
                              // the sequence
                // updates e and f based on which direction the sequence is pointing
                // sets them so that b[e][f] is now the character being pointed toward
                e += ((i = b[e][f]) == '^') - (i == 'v'),
                f += (i == '>') - (i == '<'),
                // increments the length counter h for each character of the sequence
                h++);

            // return a comma separated string of the instructions
            j.join`, `
    ")
)

JavaScript (ES6), 189

Pruebe a ejecutar el fragmento a continuación en un navegador compatible con EcmaScript 6.

f=m=>{(m=m.split`
`).map((r,i)=>r[0]>' '?y=i:0);for(x=o=l=k=p=0;p!='X';++l,y+=(k==1)-(k<1),x+=(k>2)-(k==2))c=m[y][x],c!=p?(o+=', '+'NSWE'[k]+l,l=0):0,k='^v<>'.search(p=c);alert(o.slice(7))}

// Testable version, no output but return (same size)
f=m=>{(m=m.split`
`).map((r,i)=>r[0]>' '?y=i:0);for(x=o=l=k=p=0;p!='X';++l,y+=(k==1)-(k<1),x+=(k>2)-(k==2))c=m[y][x],c!=p?(o+=', '+'NSWE'[k]+l,l=0):0,k='^v<>'.search(p=c);return o.slice(7)}


// TEST
out = x => O.innerHTML += x+'\n';

test = 
[[`>>>>>>>>>>>>>>>v
               v
               v
               >>>>X`,'E15, S3, E4']
,[`>>>>>>v
^     v
^     >>>>X`,'N2, E6, S2, E4']
,[`X
^
^
^`,'N3']
,[`>>>>>>v
^     v
^     v
      v
      >>>>>>X`,'N2, E6, S4, E6']
,[` X
 ^
 ^
>^`,'E1, N3']
,[`>X`,'E1']
,[`v<<<<<
vX<<<^
>>>>^^
>>>>>^`,'E5, N3, W5, S2, E4, N1, W3']];

test.forEach(t=>{
  var k = t[1];
  var r = f(t[0]);
  out('Test ' + (k==r ? 'OK' : 'Fail')
      +'\n'+t[0]+'\nResult: '+r
      +'\nCheck:  '+k+'\n');
})

<pre id=O></pre>

Menos golf

f=m=>{
  m=m.split('\n'); // split in rows

  x = 0; // Starting column is 0
  m.forEach( (r,i) => r[0]>' '? y=i : 0); // find starting row

  o = 0; // output string
  p = 0; // preceding character
  l = 0; //  sequence length (this starting value is useless as will be cutted at last step)
  k = 0; //  direction (this starting value is useless as will be cutted at last step)
  while(p != 'X') // loop until X found
  {
    c = m[y][x]; // current character in c
    if (c != p) // changing direction
    {  
      o +=', '+'NSWE'[k]+l; // add current direction and length to output
      l = 0 // reset length
    );
    p = c;
    k = '^v<>'.search(c); // get new current direction
    // (the special character ^ is purposedly in first position)
    ++l; // increase sequence length
    y += (k==1)-(k<1); // change y depending on direction
    x += (k>2)-(k==2); // change x depending on direction
  }
  alert(o.slice(7)); // output o, cutting the useless first part
}