8085 program to find minimum value of digit in the 8 bit number

Problem – Write an assembly language program in 8085 microprocessor to find minimum value of digit in the 8 bit number. Example – Assume 8 bit number is stored at memory location 2050, and minimum value digit is stored at memory location 3050. Algorithm –

1. Load the content of memory location 2050 in accumulator A.
2. Move the content of A in register B.
3. Perform AND operation of content of A with 0F and store the result in A.
4. Move the content of A in register C.
5. Move the content of B in A.
6. Reverse the content of A by using RLC instruction 4 times.
7. Perform AND operation of content of A with 0F and store the result in A.
8. Compare the contents of A and C by help of CMP C instruction.
9. Check if carry flag is set then jump to memory location 2013 otherwise move the content of C in A. Go to memory location 2013.
10. Store the value of A in memory location 3050.

Note – CMP C instruction compares the value of A and C. If A>C then carry flag becomes reset otherwise set. Program –

MEMORY ADDRESS	MNEMONICS	COMMENT
2000	LDA 2050	A <- M[2050]
2003	MOV B, A	B <- A
2004	ANI 0F	A <- A (AND) 0F
2006	MOV C, A	C <- A
2007	MOV A, B	A <- B
2008	RLC	Rotate content of accumulator right by 1 bit without carry
2009	RLC	Rotate content of accumulator right by 1 bit without carry
200A	RLC	Rotate content of accumulator right by 1 bit without carry
200B	RLC	Rotate content of accumulator right by 1 bit without carry
200C	ANI 0F	A <- A (AND) 0F
200E	CMP C	A – C
200F	JC 2013	Jump if CY = 1
2012	MOV A, C	A <- C
2013	STA 3050	M[3050] <- A
2016	HLT	END

Explanation – Registers A, B, C are used for general purpose.

1. LDA 2050: loads the contents of memory location 2050 in A.
2. MOV B, A: moves the content of A in B.
3. ANI 0F: perform AND operation between contents of A and value 0F.
4. MOV C, A: moves the content of A in C.
5. MOV A, B: moves the content of B in A.
6. RLC: shift the content of A left by 1 bit without carry. Use this instruction 4 times to reverse the content of A.
7. ANI 0F: perform AND operation between contents of A and value 0F.
8. CMP C: compare contents of A, C and update the value of carry flag accordingly.
9. JC 2013: jump to memory location 2013 if CY = 1.
10. MOV A, C: moves the content of C in A.
11. STA 3050: stores the content of A in memory location 3050.
12. HLT: stops executing the program and halts any further execution.

Advantages:

1. Simple and easy to understand logic
   
2. Efficient use of the accumulator and registers
   
3. Can be easily modified to find the maximum value of a digit
   

Disadvantages:

1. Only works for 8-bit numbers
   
2. May not be the most efficient method for finding the minimum digit, especially for larger numbers
   
3. Requires a counter variable to keep track of the number of digits, which adds complexity to the program.

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