Knight's Shortest Path on Chessboard

Question

I\'ve been practicing for an upcoming programming competition and I have stumbled across a question that I am just completely bewildered at. However, I feel as though it\'s

Answer 1

Very interesting problem which I was encountered recently. After looking some solutions I was tried to recover analytic formula (O(1) time and space complexity) given on SACO 2007 Day 1 solutions.

First of all I want to appreciate Graeme Pyle for very nice visualization which helped me to fix formula.

For some reason (maybe for simplification or beauty or just a mistake) they moved minus sign into floor operator, as a result they have got wrong formula floor(-a) != -floor(a) for any a.

Here is the correct analytic formula:

var delta = x-y;
if (y > delta) {
    return delta - 2*Math.floor((delta-y)/3);
} else {
    return delta - 2*Math.floor((delta-y)/4);
}

The formula works for all (x,y) pairs (after applying axes and diagonal symmetry) except (1,0) and (2,2) corner cases, which are not satisfy to pattern and hardcoded in the following snippet:

function distance(x,y){
     // axes symmetry 
     x = Math.abs(x);
     y = Math.abs(y);
     // diagonal symmetry 
     if (x < y) {
        t = x;x = y; y = t;
     }
     // 2 corner cases
     if(x==1 && y == 0){
        return 3;
     }
     if(x==2 && y == 2){
        return 4;
     }
    
    // main formula
    var delta = x-y;
		if(y>delta){
  		return delta - 2*Math.floor((delta-y)/3);
  	}
  	else{
  		return delta - 2*Math.floor((delta-y)/4);
  	}
}


$body = $("body");
var html = "";
for (var y = 20; y >= 0; y--){
	html += '';
	for (var x = 0; x <= 20; x++){
  	html += ''+distance(x,y)+'';
  }
  html += '';
}

html = ''+html+'';
$body.append(html);

Note: The jQuery used for only illustration, for code see distance function.