Towers of Hanoi: Difference between revisions

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define hanoi(n, i, j, k);
define hanoi(n, src, dst, via);
if n > 0 then
if n > 0 then
hanoi(n - 1, i, k, j);
hanoi(n - 1, src, via, dst);
printf('Move disk ' >< n >< ' from ' >< i >< ' to ' >< j >< '.\n');
printf('Move disk ' >< n >< ' from ' >< src >< ' to ' >< dst >< '.\n');
hanoi(n - 1, k, j, i);
hanoi(n - 1, via, dst, src);
endif;
endif;
enddefine;
enddefine;

Revision as of 17:46, 12 May 2007

Task
Towers of Hanoi
You are encouraged to solve this task according to the task description, using any language you may know.

In this task, the goal is to solve the Towers of Hanoi problem with recursivity.

Ada

with Ada.Text_Io; use Ada.Text_Io;

procedure Towers is
   type Pegs is (Left, Center, Right);
   procedure Hanoi (Ndisks : Natural; Start_Peg : Pegs := Left; Via_Peg : Pegs := Center; End_Peg : Pegs := Right) is
   begin
      if Ndisks > 0 then
         Hanoi(Ndisks - 1, Start_Peg, Via_Peg, End_Peg);
         Put_Line("Move disk" & Natural'Image(Ndisks) & " from " & Pegs'Image(Start_Peg) & " to " &
            Pegs'Image(End_Peg));
         Hanoi(Ndisks - 1, Via_Peg, End_Peg, Start_Peg);
      end if;
   end Hanoi;
begin
   Hanoi(4);
end Towers;


AppleScript

global moves --this is so the handler 'hanoi' can see the 'moves' variable
set moves to ""
hanoi(4, "peg A", "peg C", "peg B")

on hanoi(ndisks, fromPeg, toPeg, withPeg)
    if ndisks is greater than 0 then
        hanoi(ndisks - 1, fromPeg, withPeg, toPeg)
        set moves to moves & "Move disk " & ndisks & " from " & fromPeg & " to " & toPeg & return
        hanoi(ndisks - 1, withPeg, toPeg, fromPeg)
    end if
    return moves
end hanoi


C++

Compiler: GCC

void move(int n, int from, int to, int via) {
  if (n == 1) {
    std::cout << "Move disk from pole " << from << " to pole " << to << std::endl;
  } else {
    move(n - 1, from, via, to);
    move(1, from, to, via);
    move(n - 1, via, to, from);
  }
}


E

def move(out, n, fromPeg, toPeg, viaPeg) {
    if (n.aboveZero()) {
        move(out, n.previous(), fromPeg, viaPeg, toPeg)
        out.println(`Move disk $n from $fromPeg to $toPeg.`)
        move(out, n.previous(), viaPeg, toPeg, fromPeg)
    }
}

move(stdout, 4, def left {}, def right {}, def middle {})

Java

 public void move(int n, int from, int to, int via) {
   if (n == 1) {
     System.out.println("Move disk from pole " + from + " to pole " + to);
   } else {
     move(n - 1, from, via, to);
     move(1, from, to, via);
     move(n - 1, via, to, from);
   }
 }

Pop11

define hanoi(n, src, dst, via);
if n > 0 then
    hanoi(n - 1, src, via, dst);
    printf('Move disk ' >< n >< ' from ' >< src >< ' to ' >< dst >< '.\n');
    hanoi(n - 1, via, dst, src);
endif;
enddefine;
hanoi(4, "left", "middle", "right");

Python

def hanoi(ndisks, startPeg=1, endPeg=3):
    if ndisks:
        hanoi(ndisks-1, startPeg, 6-startPeg-endPeg)
        print "Move disk %d from peg %d to peg %d" % (ndisks, startPeg, endPeg)
        hanoi(ndisks-1, 6-startPeg-endPeg, endPeg)

hanoi(ndisks=4)

Seed7

const proc: hanoi (in integer: disk, in string: source, in string: dest, in string: via) is func
  begin
    if disk > 0 then
      hanoi(pred(disk), source, via, dest);
      writeln("Move disk " <& disk <& " from " <& source <& " to " <& dest);
      hanoi(pred(disk), via, dest, source);
    end if;
  end func;