Loop at 8:15

Overshoot at 10:38

It is doable, there are a lot of background math and other items from CS that would require you to study, but others have done it.

I am working on a desktop version - stay posted.

@@REAGLIN Can you please tell any alternate till the desktop version comes?

Probably, it is pretty powerful.

Yep - it can also be configured in settings.

var num = 0; var Node = function(_content) { this.next = null; this.last = null; this.content = _content; } var Stack = function() { this.bottom = null; this.top = null; this.push = function(_content) { // No head - create one if (this.bottom == null) { this.bottom = new Node(_content); this.top = this.bottom; return this; } var addedNode = new Node(_content); addedNode.last = this.top; // pointer to previous node this.top.next = addedNode; // current top points to new this.top = addedNode; // which becomes new top return this; } this.pop = function() { if (this.bottom == null) { alert("The Stack is Empty"); return null; } // Case of one node if (this.bottom == this.top) { var n = this.top; this.top = null; this.bottom = null; return n; } // Now remove top Node var a = this.top; // hold value for return this.top = this.top.last; this.top.next = null; return a; } this.toString = function() { var str = ""; var node = this.bottom; while (node != null) { str += node.content + ":"; node = node.next; } return str; } } function run() { var stackA = new Stack(); var stackB = new Stack(); var stackC = new Stack(); // Fill Original Stack var i = parseInt(document.getElementById('uInput').value); for (var t = i; t > 0; t--) { stackA.push(t); } document.getElementById('output1').innerHTML = '<br/>Stack A Before: ' + stackA.toString(); document.getElementById('output1').innerHTML += '<br/>Stack B Before: ' + stackB.toString(); runHanoi(i, stackA, stackB, stackC); document.getElementById('output1').innerHTML += '<br/>Stack A After: ' + stackA.toString(); document.getElementById('output1').innerHTML += '<br/>Stack B After: ' + stackB.toString(); document.getElementById('output1').innerHTML += '<br/><br/>Congratulations! You solved the Towers of Hanoi in ' + num + ' moves.'; } function runHanoi(z, A, B, C) { // All recursive functions MUST HAVE a return condition if (z == 1) { move(A, B); num++; } else { // Recursive Call runHanoi(z - 1, A, C, B); // Where all the action occurs move(A, B); num++; runHanoi(z - 1, C, B, A); } } function move(A, B) { B.push(A.pop().content); }

@@REAGLIN thanks