Difference between revisions of "CSC231 Analysis of the Towers of Hanoi"
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We see that the number of moves for 2 disks is 3, for 3 disks it's 7, for 4 disks it's 15, for ''n'' disks, it's 2<sup>n</sup>-1. | We see that the number of moves for 2 disks is 3, for 3 disks it's 7, for 4 disks it's 15, for ''n'' disks, it's 2<sup>n</sup>-1. | ||
<br /> | <br /> | ||
| − | =Execution time | + | =Estimating the Execution time= |
==Call Tree of MoveDisks( 4, 'A', 'B', 'C' )== | ==Call Tree of MoveDisks( 4, 'A', 'B', 'C' )== | ||
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3,A,C,B 1,A,B,C 3,C,B,A | 3,A,C,B 1,A,B,C 3,C,B,A | ||
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</pre></code> | </pre></code> | ||
Revision as of 18:19, 25 November 2012
--D. Thiebaut 16:24, 25 November 2012 (EST)
Contents
A High-Level Version in Python
# hanoi.py (Python V 2.7)
# D. Thiebaut
# implements the game of hanoi in python.
# uses recursion.
#
# to run, type
# python hanoi.py
# at the command line
def moveDisk( source, dest, extra, n ):
if n==1:
print "move disk from %s to %s" % ( source, dest )
return
# more than 1 disk...
moveDisk( source, extra, dest, n-1)
print "move disk from %s to %s" % ( source, dest )
moveDisk( extra, dest, source, n-1)
def main():
while True:
n = input( "# of disks? " )
moveDisk( "A", "B", "C", n )
main()
Top Down Approach: Version 1
Source
In this version all we do is simply set a main program that calls the function moveDisks and passes it
- the number of disk to move (in al)
- the source peg (in bl)
- the destination peg (in cl)
- and the extra or temporary peg (in dl)
The moveDisks function prints a single move, testing the printMove function that prints "X Y" on a line by itself, indicating that a disk must be move from X to Y.
;;; hanoi1.asm
;;; D.T.
;;; Towers of Hanoi, top down...
;;; assumptions:
;;; al = n Number of disk
;;; bl = Source character identifying the source
;;; cl = Dest character identifying the destination
;;; dl = Temp character identifying the extra peg
section .data
moveStr db "X --> X", 10
moveStrLen equ $-moveStr
source equ moveStr + 0
dest equ moveStr + 6
section .text
global _start
;;; ------------------------------------------------------------
;;; M A I N
;;; ------------------------------------------------------------
_start: mov al, 4 ; move 4 disks
mov bl, 'A' ; from A
mov cl, 'B' ; to B
mov dl, 'C' ; using C as extra
call moveDisks
;;; return to OS
mov eax, 1
mov ebx, 0
int 0x80
;;; ------------------------------------------------------------
;;; moveDisks:
;;; assumptions:
;;; al = n Number of disk
;;; bl = Source character identifying the source
;;; cl = Dest character identifying the destination
;;; dl = Temp character identifying the extra peg
;;; Does not return anything
;;; does not modify registers
;;; ------------------------------------------------------------
moveDisks: push ax
push bx
push cx
push dx
mov byte[source], bl
mov byte[dest], cl
call printMove
pop dx
pop cx
pop bx
pop ax
ret
;;; ------------------------------------------------------------
;;; printMove
;;; Prints the string moveStr on the screen
;;; Does not alter any register
;;; ------------------------------------------------------------
printMove: pushad
mov eax, 4
mov ebx, 1
mov ecx, moveStr
mov edx, moveStrLen
int 0x80
popad
ret
Output
[231a@beowulf ~/handout]$ nasmld hanoi1 A --> B
Top Down Version 2
In this version we fully flesh out the recursive function. It would be hard to go half way with the implementation of moveDisks.
Source
;;; hanoi2.asm
;;; D.T.
;;; Towers of Hanoi, top down...
;;; assumptions:
;;; al = n Number of disk
;;; bl = Source character identifying the source
;;; cl = Dest character identifying the destination
;;; dl = Temp character identifying the extra peg
%define N 6 ; number of disks to move
section .data
moveStr db "X X", 10
moveStrLen equ $-moveStr
source equ moveStr + 0
dest equ moveStr + 2
section .text
global _start
;;; ------------------------------------------------------------
;;; M A I N
;;; ------------------------------------------------------------
_start: mov al, N ; move 4 disks
mov bl, 'A' ; from A
mov cl, 'B' ; to B
mov dl, 'C' ; using C as extra
call moveDisks
;;; return to OS
mov eax, 1
mov ebx, 0
int 0x80
;;; ------------------------------------------------------------
;;; moveDisks:
;;; assumptions:
;;; al = n Number of disk
;;; bl = Source character identifying the source
;;; cl = Dest character identifying the destination
;;; dl = Temp character identifying the extra peg
;;; Does not return anything
;;; does not modify registers
;;; ------------------------------------------------------------
moveDisks: push ax
push bx
push cx
push dx
cmp al, 1
ja .recurse
;;; only 1 to move: move from source to dest!
mov byte[source], bl
mov byte[dest], cl
call printMove
jmp .return
.recurse:
;;; move n-1 disk from source to temp
dec al ; al <-- n-1
xchg cl, dl ; swap dest and temp
call moveDisks
xchg cl, dl ; swap back dest and temp
;;; move 1 from source to dest
push ax
mov al, 1
mov byte[source], bl
mov byte[dest], cl
call printMove
pop ax
;;; move n-1 disk from temp to dest
xchg bl, dl ; make temp the source
call moveDisks
.return: pop dx
pop cx
pop bx
pop ax
ret
;;; ------------------------------------------------------------
;;; printMove
;;; Prints the string moveStr on the screen
;;; Does not alter any of the registers
;;; ------------------------------------------------------------
printMove: pushad
mov eax, 4
mov ebx, 1
mov ecx, moveStr
mov edx, moveStrLen
;;; int 0x80
popad
ret
Note here that we are using a macro, N, defined by the %define statement. This method of defining a constant is interesting as it allows us to modify the constant at assembly time.
Here's how.
- comment out the %define statement in the program (this will result in an undefined variable if we assemble as usual)
- when assembling the program, provide a value for the macro N as follows:
nasm -f elf -F stabs -dN=4 hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2
- The result will be that your program will be assembled just as if you had included the line
%define N 4
- inside your program.
The section below shows how to make the program run for several values of N:
Output
[231a@beowulf ~/handout]$ nasm -f elf -dN=2 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2
A C
A B
C B
[231a@beowulf ~/handout]$ nasm -f elf -dN=3 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2
A B
A C
B C
A B
C A
C B
A B
[231a@beowulf ~/handout]$ nasm -f elf -dN=4 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2
A C
A B
C B
A C
B A
B C
A C
A B
C B
C A
B A
C B
A C
A B
C B
Trick
If we want to count the number of moves for a given number of disk, we can pipe the output of the ./hanoi2 program in the command wc, which stands for word-count. wc outputs the number of lines, number of words, and number of characters in the text given to it.
Below are the same commands as above, but filtered by wc:
[231a@beowulf ~/handout]$ nasm -f elf -dN=2 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2 | wc
3 6 12
[231a@beowulf ~/handout]$ nasm -f elf -dN=3 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2 | wc
7 14 28
[231a@beowulf ~/handout]$ nasm -f elf -dN=4 -F stabs hanoi2.asm ; ld -melf_i386 -o hanoi2 hanoi2.o ; ./hanoi2 | wc
15 30 60
We see that the number of moves for 2 disks is 3, for 3 disks it's 7, for 4 disks it's 15, for n disks, it's 2n-1.
Estimating the Execution time
Call Tree of MoveDisks( 4, 'A', 'B', 'C' )
N, Source, Dest, Extra
4,A,B,C
|
+---------------------------+-------------------------------+
| | |
3,A,C,B 1,A,B,C 3,C,B,A
| |
+-------------+------------+ +------------+-------------+
| | | | | |
2,A,B,C 1,A,C,B 2,B,C,A 2,C,A,B 1,C,B,A 2,A,B,C
| | | |
+--------+--------+ +--------+--------+ +--------+--------+ +--------+--------+
| | | | | | | | | | | |
1,A,C,B 1,A,B,C 1,C,B,A 1,B,A,C 1,B,C,A 1,A,C,B 1,C,B,A 1,C,A,B 1,B,A,C 1,A,C,B 1,A,B,C 1,C,B,A
| | | | | | | |
1,A,C,B 1,C,B,A 1,B,A,C 1,A,C,B 1,C,B,A 1,B,A,C 1,A,C,B 1,C,B,A
Let's look at the top two levels of the tree:
4,A,B,C
|
+---------------------------+-------------------------------+
| | |
3,A,C,B 1,A,B,C 3,C,B,A
| |
We read it as follows.
- MoveDisks is called to move 4 disks from A to B, using C as an extra peg. To do so it calls itself once to move 3 disks from A to C, to put them "out of the way", and uses B as a temporary peg to do so. Then it moves 1 disk from A to B. Then it has to move the 3 disks that were on C back to B where the belong, and it uses A as an extra peg in the process.
Let's figure out how many instructions are executed by the different parts:
| Tree part | Number of instructions | |
|---|---|---|
4,A,B,C |
4 pushes + 1 cmp + 1 jump = 6 |
