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-

  • Retrive 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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