Towers of Hanoi: eager moves

Suppose that one wants to return a list of the 10 first moves for a Towers of Hanoi problem with 5 disks (31 moves total). Note: The problem is more general and the numbers chosen are for example purposes only.

As a starting point, below is a solution to the Towers of Hanoi problem for 5 disks.

Here is the JavaScript code.

function move(n1, a1, b1, c1) {
	if (n1 != 0) {
		move(n1-1,a1,c1,b1);
		console.log("move "+n1+" from "+a1+" to "+b1+"\n");
		move(n1-1,c1,b1,a1);
		}
	}
move(5,"a","b","c");

Here is the output of the JavaScript code.

move 1 from a to b

move 2 from a to c

move 1 from b to c

move 3 from a to b

move 1 from c to a

move 2 from c to b

move 1 from a to b

move 4 from a to c

move 1 from b to c

move 2 from b to a

move 1 from c to a

move 3 from b to c

move 1 from a to b

move 2 from a to c

move 1 from b to c

move 5 from a to b

move 1 from c to a

move 2 from c to b

move 1 from a to b

move 3 from c to a

move 1 from b to c

move 2 from b to a

move 1 from c to a

move 4 from c to b

move 1 from a to b

move 2 from a to c

move 1 from b to c

move 3 from a to b

move 1 from c to a

move 2 from c to b

move 1 from a to b

One eager approach is to generate a list of all moves and pick out the first 10. Here is the JavaScript code.

var moveLen1 = 10;
var moveList1 = [];

function move(n1, a1, b1, c1) {
	if (n1 != 0) {
		move(n1-1,a1,c1,b1);
		var line1 = "move "+n1+" from "+a1+" to "+b1;
		moveList1.push(line1)
		move(n1-1,c1,b1,a1);
		}
	}

move(5,"a","b","c");
console.log("begin "+moveLen1+" moves\n");
for (movePos1=0; movePos1 <= moveLen1-1; movePos1++) {
	var line1 = moveList1[movePos1];
	console.log("\t"+movePos1+". "+line1+"\n");
	}
console.log("end "+moveLen1+" moves\n");

Here is the output of the JavaScript code.

begin 10 moves

	0. move 1 from a to b

	1. move 2 from a to c

	2. move 1 from b to c

	3. move 3 from a to b

	4. move 1 from c to a

	5. move 2 from c to b

	6. move 1 from a to b

	7. move 4 from a to c

	8. move 1 from b to c

	9. move 2 from b to a

end 10 moves

A lazy approach is to use a (global )variable stop when the limit of 10 has been reached.

Note that since any recursive call could cause the list to grow to the required size, there are three checks to insure that more moves are needed. Here is the JavaScript code.

var moveLen1 = 10;
var movePos1 = 0;
var moveList1 = [];

function move(n1, a1, b1, c1) {
	if (n1 != 0) {
		if (movePos1 < moveLen1) {
			move(n1-1,a1,c1,b1);
			}
		if (movePos1 < moveLen1) {
			movePos1 += 1;
			var line1 = "move "+n1+" from "+a1+" to "+b1;
			moveList1.push(line1)
			}
		if (movePos1 < moveLen1) {
			move(n1-1,c1,b1,a1);
			}
		}
	}

move(5,"a","b","c",moveLen1);
console.log("begin "+moveLen1+" moves\n");
for (movePos1=0; movePos1 <= moveLen1-1; movePos1++) {
	var line1 = moveList1[movePos1];
	console.log("\t"+movePos1+". "+line1+"\n");
	}
console.log("end "+moveLen1+" moves\n");

Here is the output of the JavaScript code.

begin 10 moves

	0. move 1 from a to b

	1. move 2 from a to c

	2. move 1 from b to c

	3. move 3 from a to b

	4. move 1 from c to a

	5. move 2 from c to b

	6. move 1 from a to b

	7. move 4 from a to c

	8. move 1 from b to c

	9. move 2 from b to a

end 10 moves