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Prime or not prime assemble code 8051

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Greital

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hi,
I'm working with a multiple phases project in 8051 micro-controller
and phase one involved with check if a given number is prime or not

the way I've trying to do this code is :
1. assuming 2 numbers one is prime and other is not and save them in R1, and R2
2. create a counter from 99 and save it in R7 --->> 99 because in the following phases we have to check the prime number or not from 1 to 99 (a number that user enter using keypad)
3. making a loop and divide the desire number with 99 first then check if prime or not, then decrements and repeat the loop again until we finish
4. if its prime a led will flash using delay with counter R6=3
5. if its not prime another led will flash using the same delay and same counter


but I'm really stuck with the code, something wrong but I don't now what is ?
any help is appreciable

----->>> I'm using Edsim51D1 to check my code
Code:
----------Begining---------
START:
	MOV R1,#9
	MOV R2,#31
	MOV R7,#99
	MOV R6,#3
LOOP:
	MOV A,R1
	MOV B,R7
	DIV AB
	MOV R3,#0
	MOV R3,B
	CJNE R3,#0,NOTPRIME
	DJNZ R7,LOOP

PRIME:
	ACALL DELAY
	CPL P1.3
	DJNZ R6,PRIME
AJMP DONE	


NOTPRIME:
	ACALL DELAY
	CPL P1.0
	DJNZ R6,NOTPRIME
DONE:


DELAY:
	MOV R5,#20
X3:	MOV R4,#200
X2:	MOV R3,#250
X1:	DJNZ R3,X1
	DJNZ R4,X2
	DJNZ R5,X3
RET

END
-------------End------------
 
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I thought the standard way of checking for prime is take the number (A) and divide it starting with the divisor of 2 checking for zero remainder. Increment the divisor and check again for zero remainder keep repeating until you reach half of the A value.

If you find a zero remainder the number is not prime.
If you reach a divisor of A/2 and have not found a zero remainder then the number is prime.
 

Greital

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will try your way
thanks for trying to help :) appreciate it
 

xenos

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The remain of the division is checked with factors up to SQRT(n), not n/2.

An ansient algorithm, like: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
(with a simple modification, you do not have to store anything in memory), could be helpful.

You can see a modern and efficient algorithm here: https://en.wikipedia.org/wiki/Primality_test

Code:
function is_prime(n : integer)
    if n ≤ 1
        return false
    else if n ≤ 3
        return true
    else if n mod 2 = 0 or n mod 3 = 0
        return false
    let i ← 5
    while i×i ≤ n
        if n mod i = 0 or n mod (i + 2) = 0
            return false
        i ← i + 6
    return true

Of course if you need primes up to a relatively small number (eg 256), then you could precalculate the array of primaries and store it in non volatile memory.
 

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The remain of the division is checked with factors up to SQRT(n), not n/2.
Thanks for pointing my error out. I totally forgot that it was supposed to be SQRT(n) and not n/2. The question is though how to determine SQRT(n) for any given n using that 8051? Or are we just going to assume that with the given 8-bit registers used that the maximum is 255 so is automatically <16?

You can see a modern and efficient algorithm here:
Correct me if I'm wrong but it also seems to me that mod isn't necessarily the simplest thing to compute on a 8051 in assembly.
 

KlausST

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Hi,

The question is though how to determine SQRT(n) for any given n using that 8051?
Actually square root is difficult...

Usually you test starting from 2. Then the next prime: 3, then 5, then 7... until the result of the devision is smaller than the prime number.

If the numbers to test are values from 0 ..99,
Then the prime numbers used for testing are 2, 3, 5, 7, 11, not more
99 / 7 the result is 14, larger than 7
99/11 the result is 9, smaller than 11...this indicates the end of test
Automatically this is at the square root of 99.

Klaus
 

xenos

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The question is though how to determine SQRT(n) for any given n using that 8051?
in stead of calculating sqrt(N), we use:
Code:
if(i*i>N)
 break;


Correct me if I'm wrong but it also seems to me that mod isn't necessarily
the simplest thing to compute on a 8051 in assembly.
if time is not critical, we could use a style of dynamic programming algorithm:
Code:
//check if N is divided by K
TMP = 0;
for(i=1;TMP<=N;i++)
    TMP += K;
if(TMP == K)
   the number is divided completely by K (N %K == 0)
i.e. without any multiplication or division or modulo, we check if a modulo is zero.
We should take extra care to check STATUS flags, if overflow in addition occur, then TMP got larger than N and we should terminate the loop.
 

Greital

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thanks to all of you :)
you all helped me to configure the ideas to do this code :)
really appreciate your work
 

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