Follow the path

JavaScript (ES6), 204 211 210

Edit 1 Bug fix - output '*' for void input
Edit 2 Simpler decoding of direction to x and y diff

Here is my answer to The treasure map, revised to fulfill the specs.

F=m=>(m.replace(/\D(\d+)/g,(d,z)=>{for(;z--;r=[...r],r[x]=m,p[y]=r.join``)for(d<'>'?--x:d<'^'?++x:d<'v'?--y:++y,p=~x?~y?p:[y=0,...p]:p.map(r=>' '+r,x=0),r=p[y]||'';!r[x];)r+=' '},x=y=0,p=[m='*']),p.join`
`)

Less golfed and explained more or less

f=m=>(
  x=y=0, // starting position
  p=['*'], // output string array (initialized with minimum output)
  m.replace( /\D(\d+)/g, 
  (d,z) => // execute the following for each group direction/length. Length in z, direction in d[0]
  {
    while( z--)  // repeat for the len
    {
      // check d to change values of x and y
      // all the comparison are with > and <, not equal
      // so that they work with the whole d, not just d[0]
      d<'>'?--x:d<'^'?++x:d<'v'?--y:++y,
      // now if x or y are < 0 then p must be adjusted  
      p = ~x 
        ? ~y
          ? p // both x and y are >= 0, p is not changed
          : [y = 0, ...p] // y < 0, shift p by on adding a 0 element and set y to 0
        : p.map(r=> ' ' + r, x = 0); // x < 0, add a space to the left for each row in p and set x to 0
      r = p[y] || ''; // get current row in r
      for( ; !r[x]; ) // if the current row is empty or too short
        r += ' '; // ... add spaces up to position x
      // set character in x position
      r = [...r], // the shorter way is converting to array ...
      r[x] = '*', // setting the element
      p[y] = r.join`` // and the back to string using join
    }
  }),
  p.join`\n` // return output array as a newline separated string
}

Test

F=m=>(m.replace(/\D(\d+)/g,(d,z)=>{for(;z--;r=[...r],r[x]='*',p[y]=r.join``)for(d<'>'?--x:d<'^'?++x:d<'v'?--y:++y,p=~x?~y?p:[y=0,...p]:p.map(r=>' '+r,x=0),r=p[y]||'';!r[x];)r+=' '},x=y=0,p=['*']),p.join`
`)

// TEST
console.log = x => O.textContent += x + '\n';

console.log(F('')+'\n')

console.log(F('v6>4^3<7')+'\n')

console.log(F('^2v2>3<3v3>4^5v5>3^5>4v2<4v3>4^3v3>3^5>4v2<4v3>7^5>4v2<4v3>9^3<2^2v2>4^2v2<2v3>8^5>2v4>2^4v5<3>6^5>5<5v2>5<5v2>5<4v1>8^3<1^2v2>1v2>2^3v3>2^2>1^2v2<1v3<3>11^3<2^2v2>4^2v2<2v3>5^5>5<5v2>5<5v2>5<4v1>7^5>4v2<4v3>4^3v3>3^5>4v2<3v1<1v2>3^1>1v1'))
<pre id=O></pre>


MATL, 71 bytes

1thj'.\d+'XX"@Z)XK6L)U:"K1)XK118=K94=-K62=K60=-hv]]YstY)X<1--lqg10*32+c

Uses current release (6.0.0) of the language/compiler. Works in Matlab and in Octave.

EDIT (June 21, 2016): due to changes in the language, the code requires a few modifications to run in current release (16.0.0). You can try it online including the needed modifications.

Examples

>> matl
 > 1thj'.\d+'XX"@Z)XK6L)U:"K1)XK118=K94=-K62=K60=-hv]]YstY)X<1--lqg10*32+c
 > 
> ^4>3v2<1
    ****
    *  *
    * **
    *   
    *  

>> matl
 > 1thj'.\d+'XX"@Z)XK6L)U:"K1)XK118=K94=-K62=K60=-hv]]YstY)X<1--lqg10*32+c
 > 
> ^2v2>3<3v3>4^5v5>3^5>4v2<4v3>4^3v3>3^5>4v2<4v3>7^5>4v2<4v3>9^3<2^2v2>4^2v2<2v3>8^5>2v4>2^4v5<3>6^5>5<5v2>5<5v2>5<4v1>8^3<1^2v2>1v2>2^3v3>2^2>1^2v2<1v3<3>11^3<2^2v2>4^2v2<2v3>5^5>5<5v2>5<5v2>5<4v1>7^5>4v2<4v3>4^3v3>3^5>4v2<3v1<1v2>3^1>1v1
  *   *  *****  *****  *****  *   *     *** *  ******  *     *    *   *  ******  *****  *****
  *   *  *   *  *   *  *   *  *   *     * * *  *       *  *  *    *   *  *       *   *  *   *
  *****  *****  *****  *****  *****     * * *  ******  ** * **    *****  ******  *****  *****
  *   *  *   *  *      *        *       * * *  *        * * *       *    *       *   *  **   
  *   *  *   *  *      *        *       * ***  ******   *****       *    ******  *   *  *  **
  *******************************************************************************************

Explanation

The program has four main steps:

  1. Read input string and split into its components.
  2. Build a 2-column matrix where each row describes a unit displacement in the appropriate direction. For example, [0 -1] indicates one step to the left. The first row is the origin of the path, [1 1]
  3. Compute the cumulative sum of that matrix along the first dimension . Now each row describes the coordinates of a *. Normalize to minumum value 1
  4. Create a new matrix that contains 1 at the coordinates indicated by the matrix from step 3, and 0 otherwise. This is then transformed into a char matrix.

Code:

1th                         % row vector [1 1]. Initiallize matrix of step 2
j                           % (step 1) read input string 
'.\d+'XX                    % split into components. Creates cell array of substrings
"                           % (step 2) for each component
   @Z)XK                    % unbox to obtain substring and copy
   6L)U:                    % obtain number and build vector of that size
   "                        % repeat as many times as that number
      K1)                   % paste substring. Get first character: '^', 'v', '>', '<'
      XK118=K94=-           % vertical component of unit displacement: -1, 0 or 1
      K62=K60=-             % horizontal component of unit displacement: -1, 0 or 1
      h                     % concatenate horizontally
      v                     % append vertically to existing matrix
   ]                        % end
]                           % end
Ys                          % (step 3) cumulative sum along first dimension
tY)X<1--                    % normalize to minimum value 1
lqg                         % (step 4) build matrix with 0/1
10*32+c                     % replace 0 by space and 1 by asterisk

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