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Create a Program To Search String In MIPS Assembly Language (Using MARS Simulator).


Instructions

Objective
Write program to search DNA sequence in C and MIPS assembly language.

Requirements and Specifications

DNA Search: This project explores pattern matching techniques to find a pattern in a DNA sequence containing letters in the DNA alphabet {A, C, G, T}. For example, suppose we have a DNA sequence as follows:
ATGACGATCTACGTATGGCAGCCACGCTTTTGATGTTAAGTCACACAGCCAAGTCAACAAGGGCGACTTCATGATCTTTCCGCTCCGTTGGTGTAGGCCCGTGTTCAAATTCAATGGCGATTGGAATTACCTTTGAAATACTCCAACCGACCGCCACGGCCAGGGTCCCGCTCGCTCTCTGTGGCCCTCCCACAAAACTCCGGTGAAAGTTGATTTGGACACGGACCCAAAGCAGCGTAGATTATTCGAGCGTATTCGGTAGTCATTGAGGCCCCAA
The pattern “GCTTTT” can be found at index 27 (where the first letter of the sequence is at index 0). Note that overlapping matches are treated individually. For example, if the sequence is ‘AAAAAA’ and the pattern is ‘AAA’, there are 4 occurrences of the pattern: at indices 0, 1, 2, and 3. You will write a C program and a MIPS program to find the indices at which a given pattern occurs in a given DNA sequence.
Strategy: Unlike many “function only” programming tasks where a solution can be quickly envisioned and implemented, this task requires a different strategy.
  1. Before writing any code, reflect on the task requirements and constraints. Mentally explore different approaches and algorithms, considering their potential performance and costs. The metrics of merit are static code length, dynamic execution time, and storage requirements. There are often trade-offs between these parameters.
  2. Once a promising approach is chosen, a high level language (HLL) implementation (e.g., in C) can deepen its understanding. The HLL implementation is more flexible and convenient for exploring the solution space and should be written before constructing the assembly version where design changes are more costly and difficult to make. For P1-1, you will write a C assignment implementation of the program.
  3. Once a working C version is created, it’s time to “be the compiler” and see how it translates to MIPS assembly. This is an opportunity to see how HLL constructs are supported on a machine platform (at the ISA level). This level requires the greatest programming effort; but it also uncovers many new opportunities to increase performance and efficiency. You will write the assembly version for P1-2.
Screenshots of output
String search in MIPS assembly language
String search in MIPS assembly language 1
String search in MIPS assembly language 2
String search in MIPS assembly language 3
String search in MIPS assembly language 4

Source Code

/*

ECE 2035 Project P1-1

Please fill in the following

Student name:

Current date:

This is the only file that should be modified for the C implementation

of Project 1.

This program initializes a DNA sequence of 10,240 random characters and

a pattern of 3 to 7 random characters, all characters are from the DNA

alphabet {A, T, G, C}.

For Project P1-1, you will implement the function Match, called by main.

The function is defined at the end of this file.

*/

#include

#include

#include

// DO NOT include any additional libraries.

#define DEBUG 0 // RESET THIS TO 0 BEFORE SUBMITTING YOUR CODE

#define SEQLEN 10240

#define MAXPATLEN 10

#define NUMCHAR 4

char Alphabet[] = "ACTG";

int MatchIndices[SEQLEN];

int main(int argc, char *argv[]) {

   char Seq[SEQLEN], Pat[MAXPATLEN];

   int I, PatLen;

   void Print_Seq(char *Seq, int SeqLen);

   void Print_Pat(char *Pat, int PatLen);

   void Match(char *Pat, int PatLen, char *Seq, int SeqLen);

   if (DEBUG){

     printf("Sample debugging print statement.\n");

   }

   srand((unsigned int) time(NULL)); // seed random number generator

   PatLen = (rand() % MAXPATLEN) + 1; // compute pattern length

   for (I = 0; I < SEQLEN; I++)

      Seq[I] = Alphabet[rand() % NUMCHAR]; // create sequence

   for (I = 0; I < PatLen; I++)

      Pat[I] = Alphabet[rand() % NUMCHAR]; // create pattern

   Print_Pat(Pat, PatLen); // print pattern

   Print_Seq(Seq, SEQLEN); // print sequence

   Match(Pat, PatLen, Seq, SEQLEN); // match pattern in sequence

   printf("Pattern detected at the following locations:\n");

   I = 0;

   while (MatchIndices[I] != -1)

     printf("Base pair %d in the sequence\n", MatchIndices[I++]);

   return 0;

}

/* Print Sequence

This routine prints the sequence. */

void Print_Seq(char *Seq, int SeqLen) {

   int I;

   printf("The sequence is ...\n");

   for (I = 0; I < SeqLen; I++) {

      putchar(Seq[I]);

      if (I % 80 == 79)

         printf("\n");

   }

}

/* Print Pattern

This routine prints the match pattern. */

void Print_Pat(char *Pat, int PatLen) {

   int I;

   printf("The pattern is ... \"");

   for (I = 0; I < PatLen; I++)

      putchar(Pat[I]);

   printf("\"\n");

}

/* Match

This routine find indices of occurrences of a variable length DNA

pattern in a DNA sequence and stores them in the global array MatchIndices.

It stores a -1 at MatchIndices[n] (where n = number of occurrences of the pattern) to indicate the end of the sequence of indices. */

void Match(char Pat[], int PatLen, char Seq[], int SeqLen) {

  // insert your code here

  int indSeq; // index in dna sequence

  int indPat; // index in pattern

  int indMat; // index in matched indices

  int matched; // 1 if pattern is matched, 0 otherwise

  indMat = 0; // start at initial position in match indices array;

  for (indSeq = 0; indSeq <= SeqLen - PatLen; indSeq++) // search pattern in all sequence positions

  {

    matched = 1; // assume pattern matches at start

    for (indPat = 0; indPat < PatLen && matched; indPat++) // match all characters in pattern, continue while chars are equal

    {

      if (Seq[indSeq + indPat] != Pat[indPat]) // if not equal,

        matched = 0; // pattern is not matched

    }

    if (matched) // if all pattern characters were matched

    {

      MatchIndices[indMat++] = indSeq; // save current sequence index as a match and increment matches

    }

  }

  MatchIndices[indMat] = -1; // save ending -1

  return;

}

Assembly Language

# DNA Search

#

# This program computes the indices of all (possibly overlapping) occurrences

# of a pattern string in an input DNA sequence, given in the array starting

# at the label Input. It stores the indices in the output array starting at

# label Indices.

#

# Please fill in the following

# Student name:

# Date:

.data

Input: .alloc 600

Indices: .alloc 4800

.text

DNAsearch: addi $1, $0, Input # point to input base

  swi 548 # create DNA search

                # your code goes here

  srl $3, $2, 16 # get length from returned value

  andi $2, $2, 0xFFFF # clear upper word to leave only the pattern

  addi $4, $0, 1 # load a 1

  sll $5, $3, 1 # length *2

  sllv $4, $4, $5 # calculate 2 ^ length*2

  addi $4, $4, -1 # calculate 2 ^ length*2 - 1, which is the bit mask for the given length

  lw $5, 0($1) # load first word from input

  lw $6, 4($1) # load second word from input

  addi $1, $1, 8 # advance to third word

  sll $6, $6, 16 # move second word to upper part

  or $5, $5, $6 # combine both words into a single 32 bit one

  addi $6, $0, 4800 # number of input characters

  sub $6, $6, $3 # num - length

  addi $6, $6, 1 # num - length + 1 = maximum index

  addi $7, $0, Indices # point to start of indices

  addi $8, $0, -1 # load -1

  sw $8, 0($7) # store initial index as -1

  addi $8, $0, 0 # start in position 0 in input

searchLoop: addi $9, $0, 8 # start check count

matchLoop: and $10, $4, $5 # leave only pattern length bits of input

  bne $10, $2, nextMatch # if they are not equal, is not a match

  sw $8, 0($7) # else, is a match, store index

  addi $7, $7, 4 # move to next position in indices

nextMatch: srl $5, $5, 2 # shift right twice to advance to next character

  addi $8, $8, 1 # increment position in input

  beq $8, $6, endSearch # if i = max index end the search

  addi $9, $9, -1 # decrement count

  bne $9, $0, matchLoop # if not zero, do another match

  lw $10, 0($1) # load next word from input

  sll $10, $10, 16 # move second word to upper part

  or $5, $5, $10 # combine both words into a single 32 bit one

  addi $1, $1, 4 # advance to next word

  j searchLoop # repeat the loop to search more matches

endSearch: addi $10, $0, -1 # load -1

  sw $10, 0($7) # store last index as -1

                addi $2, $0, Indices # store the output

  swi 580 #

  jr $31 # return to caller