A True Tale of Love Lost and Found in an Infinite Number of Acts...

expanded.js

//Our little runner, celebration animation and placeholder 
// for an empty set of possible maze generation steps
L="♥", 
// q = current font size, 
// z = width and generic constant, 
// h = animation loop counter
q=z=h=21, 

// All the colours we need, 
// the hsla is pretty but tricky to shorten
b="#fff3#ffb3#4df3hsla(0,100%,84%,.1)".split(3), 

// shorthand reference to a function we use more than once
o=setTimeout; 

// set the size of the canvas (z^2)
c.width=c.height=441; 

// The 'paint' function. 
// d = array index (in b above), f and g are the x and y location to draw
i=function(d,f,g){ 
	
// to save having to write out 'a.textAlign, a.Baseline' etc...
 with(a)   

// Set the properties of the text on the canvas...
  textAlign="center",  
  
// ...these are annoyingly long.
  textBaseline="middle", 
  
// Sneaky trick: When setting the font size, you must also set the font family
// but rather than set 'Arial' (5 bytes), this tries to set a font called 'a' (1 byte).
// As that isn't available, it falls back to the system default
  font=q+"px a",    
  
// Set the colour using the array and index
  fillStyle=b[d],   
// Draw the current square
  fillRect(z*f,z*g,z,z),
  
// Set the colour for the runner and the animation
  fillStyle="red",  

// Handy Evaluation Shortcut: 
// the right-hand side of the OR is only evaluated if d is falsy (in this case 0)
// i.e. "only draw the text if this is the runner"
  d||fillText(L,z*f+10,z*g+11) 
};

// Our 'Solver' function
s=function(){        
	
// Set previous step as 'visited'
 n[y][x]=3;       
 
// If this direction (South) is free
 2>n[y+1][x]?     
 
// Go here
  (n[++y][x]=1,k.push(y,x)):  
  
// Repeat for north
 2>n[y-1][x]?     
  (n[--y][x]=1,k.push(y,x)):
  
// Repeat for west
 2>n[y][x-1]?     
  (n[y][--x]=1,k.push(y,x)):
  
// Repeat for east
 2>n[y][x+1]?     
  (n[y][++x]=1,k.push(y,x)):
  
// If there is no way forward from here, backtrack
 (x=k.pop(),y=k.pop());    
 
// Move the runner here
 n[y][x]=0;     
 
// If we've found our way to the goal
 19==y&&19==x?     
 
//   Run the celebration animation
  e(h=z):       
  
// otherwise, setTimeout for our next animation/solution frame
 o(s,50);       
 
// For each row,
 for(r in n)       
 
// For each column,
  for(l in n[r])   
  
// Paint the square
   i(n[r][l],l,r)  
};

// The combination celebration and maze generation function
(e=function(){    
 
// if there are steps left to complete in the animation
 if(h--)       
 
// Set the font size to get bigger
  q=z*(z-h),   
  
// Draw the celebration heart in the middle
  i(0,10,10),     
  
// Reset the original font size (for the trails)
  q=z,      
  
// setTimeout on this
  o(e,40);     
  
// If we're not animating, generate a maze
 else{      
	 
// Empty array to keep track of where we've been
  k=[];     
  
// Loop through the existing maze and set everything back to being wall
  for(n=[],y=z;0<y--;){ 
   n[y]=[];
   for(x=z;0<x--;)
    n[y][x]=2
  }
  
// Start top left
  x=y=1;      
  
// This is us
  n[y][x]=0;     

// Mark this square as corridor then loop while there is 
// more corridor to carve
  for(k.push(y,x);k.length;) 
   (      
// Our stack of potential directions. To save space, 
// this is done as a string but in order to be able 
// to use numbers as directions, we set p to a string 
// so we can concatenate.
   p=L,      
   
// We use steps of two for the maze so we can draw chunky walls

// If we aren't at the bottom edge and we're looking at a wall, 
// South is a valid direction
   18>y&&2==n[y+2][x]&&(p+=1), 
   
// Repeat for North
   2<=y&&2==n[y-2][x]&&(p+=2), 
   
// Repeat for West
   2<=x&&2==n[y][x-2]&&(p+=3), 

// Repeat for East
   2==n[y][x+2]&&(p+=4),    
// While there are potential directions (i.e. our stack contains something)
   p!=L)?      
    (
// Pick a random direction. 
// If it is South...
    1==(W=p[~~(Math.random()*p.length)])?   
// ...clear the two squares south of here 
     n[++y][x]=n[++y][x]=1:     
	 
// Repeat for North
    2==W?           
     n[--y][x]=n[--y][x]=1:     
// Repeat for West
    3==W?           
     n[y][--x]=n[y][--x]=1:     
// Repeat for East
    4==W&&(n[y][++x]=n[y][++x]=1),    
    
// Add this to the stack of places we've been
    k.push(y,x)        
    ):  
// Once we've got gone as far as we can, work our way back to continue generating
     (x=k.pop(),y=k.pop());     
// Put the goal at the bottom right
     n[19][19]=0;      
// Start solving
     s(y=x=1)       
 }
})() // Kick it all off by running the animation as an immediately executing function

minified-in-shim.html

<!doctype html>
<html>
	<head>
		<title>JS1k, 1k demo submission [ID]</title>
		<meta charset="utf-8" />
	</head>
	<body>
		<canvas id="c"></canvas>
		<script>
			var b = document.body;
			var c = document.getElementsByTagName('canvas')[0];
			var a = c.getContext('2d');
			document.body.clientWidth; // fix bug in webkit: http://qfox.nl/weblog/218
		</script>
		<script>
// start of submission //
L="♥",q=z=h=21,b="#fff3#ffb3#4df3hsla(0,100%,84%,.1)".split(3),o=setTimeout;c.width=c.height=441;i=function(d,f,g){with(a)textAlign="center",textBaseline="middle",font=q+"px a",fillStyle=b[d],fillRect(z*f,z*g,z,z),fillStyle="red",d||fillText(L,z*f+10,z*g+11)};s=function(){n[y][x]=3;2>n[y+1][x]?(n[++y][x]=1,k.push(y,x)):2>n[y-1][x]?(n[--y][x]=1,k.push(y,x)):2>n[y][x-1]?(n[y][--x]=1,k.push(y,x)):2>n[y][x+1]?(n[y][++x]=1,k.push(y,x)):(x=k.pop(),y=k.pop());n[y][x]=0;19==y&&19==x?e(h=z):o(s,50);for(r in n)for(l in n[r])i(n[r][l],l,r)};(e=function(){if(h--)q=z*(z-h),i(0,10,10),q=z,o(e,40);else{k=[];for(n=[],y=z;0<y--;){n[y]=[];for(x=z;0<x--;)n[y][x]=2}x=y=1;n[y][x]=0;for(k.push(y,x);k.length;)(p=L,18>y&&2==n[y+2][x]&&(p+=1),2<=y&&2==n[y-2][x]&&(p+=2),2<=x&&2==n[y][x-2]&&(p+=3),2==n[y][x+2]&&(p+=4),p!=L)?(1==(W=p[~~(Math.random()*p.length)])?n[++y][x]=n[++y][x]=1:2==W?n[--y][x]=n[--y][x]=1:3==W?n[y][--x]=n[y][--x]=1:4==W&&(n[y][++x]=n[y][++x]=1),k.push(y,x)):(x=k.pop(),y=k.pop());n[19][19]=0;s(y=x=1)}})()
// end of submission //
		</script>
	</body>
</html>

minified.js

L="♥",q=z=h=21,b="#fff3#ffb3#4df3hsla(0,100%,84%,.1)".split(3),o=setTimeout;c.width=c.height=441;i=function(d,f,g){with(a)textAlign="center",textBaseline="middle",font=q+"px a",fillStyle=b[d],fillRect(z*f,z*g,z,z),fillStyle="red",d||fillText(L,z*f+10,z*g+11)};s=function(){n[y][x]=3;2>n[y+1][x]?(n[++y][x]=1,k.push(y,x)):2>n[y-1][x]?(n[--y][x]=1,k.push(y,x)):2>n[y][x-1]?(n[y][--x]=1,k.push(y,x)):2>n[y][x+1]?(n[y][++x]=1,k.push(y,x)):(x=k.pop(),y=k.pop());n[y][x]=0;19==y&&19==x?e(h=z):o(s,50);for(r in n)for(l in n[r])i(n[r][l],l,r)};(e=function(){if(h--)q=z*(z-h),i(0,10,10),q=z,o(e,40);else{k=[];for(n=[],y=z;0<y--;){n[y]=[];for(x=z;0<x--;)n[y][x]=2}x=y=1;n[y][x]=0;for(k.push(y,x);k.length;)(p=L,18>y&&2==n[y+2][x]&&(p+=1),2<=y&&2==n[y-2][x]&&(p+=2),2<=x&&2==n[y][x-2]&&(p+=3),2==n[y][x+2]&&(p+=4),p!=L)?(1==(W=p[~~(Math.random()*p.length)])?n[++y][x]=n[++y][x]=1:2==W?n[--y][x]=n[--y][x]=1:3==W?n[y][--x]=n[y][--x]=1:4==W&&(n[y][++x]=n[y][++x]=1),k.push(y,x)):(x=k.pop(),y=k.pop());n[19][19]=0;s(y=x=1)}})()