Skip to content
Related Articles

Related Articles

8085 program for bubble sort

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 27 Jun, 2021

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 – 

 

AddressLabelInstructionComment
2000HSTARTLXI H, 2040HLoad size of array
2003H MVI D, 00HClear D register to set up a flag
2005H MOV C, MSet C register with number of elements in list
2006H DCR CDecrement C
2007H INX HIncrement memory to access list
2008HCHECKMOV A, MRetrieve list element in Accumulator
2009H INX HIncrement memory to access next element
200AH CMP MCompare Accumulator with next element
200BH JC NEXTBYTEIf accumulator is less then jump to NEXTBYTE
200EH JZ NEXTBYTEIf accumulator is equal then jump to NEXTBYTE
2011H MOV B, MSwap the two elements
2012H MOV M, A 
2013H DCX H 
2014H MOV M, B 
2015H INX H 
2016H MVI D, 01HIf exchange occurs save 01 in D register
2018HNEXTBYTEDCR CDecrement C for next iteration
2019H JNZ CHECKJump to CHECK if C>0
201CH MOV A, DTransfer contents of D to Accumulator
201DH CPI 01HCompare accumulator contents with 01H
201FH JZ STARTJump to START if D=01H
2022H HLTHALT

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. 
     

 

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!