How to resolve the algorithm Factors of a Mersenne number step by step in the 8086 Assembly programming language
How to resolve the algorithm Factors of a Mersenne number step by step in the 8086 Assembly programming language
Table of Contents
Problem Statement
A Mersenne number is a number in the form of 2P-1. If P is prime, the Mersenne number may be a Mersenne prime (if P is not prime, the Mersenne number is also not prime). In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test. There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1). Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar). The following is how to implement this modPow yourself: For example, let's compute 223 mod 47. Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it. Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47. Use the result of the modulo from the last step as the initial value of square in the next step: Since 223 mod 47 = 1, 47 is a factor of 2P-1. (To see this, subtract 1 from both sides: 223-1 = 0 mod 47.) Since we've shown that 47 is a factor, 223-1 is not prime. Further properties of Mersenne numbers allow us to refine the process even more. Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8. Finally any potential factor q must be prime. As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N). These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Using the above method find a factor of 2929-1 (aka M929)
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Factors of a Mersenne number step by step in the 8086 Assembly programming language
Source code in the 8086 programming language
P: equ 929 ; P for 2^P-1
cpu 8086
bits 16
org 100h
section .text
mov ax,P ; Is P prime?
call prime
mov dx,notprm
jc msg ; If not, say so and stop.
xor bp,bp ; Let BP hold k
test_k: inc bp ; k += 1
mov ax,P ; Calculate 2kP + 1
mul bp ; AX = kP
shl ax,1 ; AX = 2kP
inc ax ; AX = 2kP + 1
mov dx,ovfl ; If AX overflows (16 bits), say so and stop
jc msg
mov bx,ax ; What is 2^P mod (2kP+1)?
mov cx,P
call modpow
dec ax ; If it is 1, we're done
jnz test_k ; If not, increment K and try again
mov dx,factor ; If so, we found a factor
call msg
prbx: mov ax,10 ; The factor is still in BX
xchg bx,ax ; Put factor in AX and divisor (10) in BX
mov di,number ; Generate ASCII representation of number
digit: xor dx,dx
div bx ; Divide current number by 10,
add dl,'0' ; add '0' to remainder,
dec di ; move pointer back,
mov [di],dl ; store digit,
test ax,ax ; and if there are more digits,
jnz digit ; find the next digit.
mov dx,di ; Finally, print the number.
jmp msg
;;; Calculate 2^CX mod BX
;;; Output: AX
;;; Destroyed: CX, DX
modpow: shl cx,1 ; Shift CX left until top bit in high bit
jnc modpow ; Keep shifting while carry zero
rcr cx,1 ; Put the top bit back into CX
mov ax,1 ; Start with square = 1
.step: mul ax ; Square (result is 32-bit, goes in DX:AX)
shl cx,1 ; Grab a bit from CX
jnc .nodbl ; If zero, don't multiply by two
shl ax,1 ; Shift DX:AX left (mul by two)
rcl dx,1
.nodbl: div bx ; Divide DX:AX by BX (putting modulus in DX)
mov ax,dx ; Continue with modulus
jcxz .done ; When CX reaches 0, we're done
jmp .step ; Otherwise, do the next step
.done: ret
;;; Is AX prime?
;;; Output: carry clear if prime, set if not prime.
;;; Destroyed: AX, BX, CX, DX, SI, DI, BP
prime: mov cx,[prcnt] ; See if AX is already in the list of primes
mov di,primes
repne scasw ; If so, return
je modpow.done ; Reuse the RET just above here (carry clear)
mov bp,ax ; Move AX out of the way
mov bx,[di-2] ; Start generating new primes
.sieve: inc bx ; BX = last prime + 2
inc bx
cmp bp,bx ; If BX higher than number to test,
jb modpow.done ; stop, number is not prime. (carry set)
mov cx,[prcnt] ; CX = amount of current primes
mov si,primes ; SI = start of primes
.try: mov ax,bx ; BX divisible by current prime?
xor dx,dx
div word [si]
test dx,dx ; If so, BX is not prime.
jz .sieve
inc si
inc si
loop .try ; Otherwise, try next prime.
mov ax,bx ; If we get here, BX _is_ prime
stosw ; So add it to the list
inc word [prcnt] ; We have one more prime
cmp ax,bp ; Is it the prime we are looking for?
jne .sieve ; If not, try next prime
ret
;;; Print message in DX
msg: mov ah,9
int 21h
ret
section .data
db "*****" ; Placeholder for number
number: db "$"
notprm: db "P is not prime.$"
ovfl: db "Range of factor exceeded (max 16 bits)."
factor: db "Found factor: $"
prcnt: dw 2 ; Amount of primes currently in list
primes: dw 2, 3 ; List of primes to be extended
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