8085 program for bubble sort

Prerequisite – Bubble Sort
Problem – Write an assembly language program in 8085 microprocessor to sort a given list of n numbers using Bubble Sort. 

Example –

Assumption – Size of list is stored at 2040H and list of numbers from 2041H onwards. 

Algorithm –

1. Load size of list in C register and set D register to be 0
2. Decrement C as for n elements n-1 comparisons occur
3. Load the starting element of the list in Accumulator
4. Compare Accumulator and next element
5. If accumulator is less than or equal to the next element jump to step 8
6. Swap the two elements
7. Set D register to 1
8. Decrement C
9. If C>0 take next element in Accumulator and go to point 4
10. If D=0, this means in the iteration, no exchange takes place consequently we know that it won’t take place in further iterations so the loop in exited and program is stopped
11. Jump to step 1 for further iterations

Program –

Address	Label	Instruction	Comment
2000H	START	LXI H, 2040H	Load size of array
2003H		MVI D, 00H	Clear D register to set up a flag
2005H		MOV C, M	Set C register with number of elements in list
2006H		DCR C	Decrement C
2007H		INX H	Increment memory to access list
2008H	CHECK	MOV A, M	Retrieve list element in Accumulator
2009H		INX H	Increment memory to access next element
200AH		CMP M	Compare Accumulator with next element
200BH		JC NEXTBYTE	If accumulator is less then jump to NEXTBYTE
200EH		JZ NEXTBYTE	If accumulator is equal then jump to NEXTBYTE
2011H		MOV B, M	Swap the two elements
2012H		MOV M, A
2013H		DCX H
2014H		MOV M, B
2015H		INX H
2016H		MVI D, 01H	If exchange occurs save 01 in D register
2018H	NEXTBYTE	DCR C	Decrement C for next iteration
2019H		JNZ CHECK	Jump to CHECK if C>0
201CH		MOV A, D	Transfer contents of D to Accumulator
201DH		CPI 01H	Compare accumulator contents with 01H
201FH		JZ START	Jump to START if D=01H
2022H		HLT	HALT

Explanation-

• Retrieve an element in accumulator.
• Compare it with next element, if it is greater then swap otherwise move to next index.
• If in one entire loop there has been no exchange, halt otherwise start the whole iteration again.
• The following approach has two loops, one nested inside other so- 

  Worst and Average Case Time Complexity: O(n*n). Worst case occurs when array is reverse sorted. 
  Best Case Time Complexity: O(n). Best case occurs when array is already sorted. 
  

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