程序代写代做代考 algorithm DNA mips CSE 220: Systems Fundamentals I

CSE 220: Systems Fundamentals I
Homework #5
Spring 2017

Assignment Due: Friday, April 21, 2017 by 11:59 pm

Assignment Overview
The focus of this homework assignment is writing recursive functions and its application to string func-
tions commonly used for studies like DNA sequencing. This assignment also reinforces MIPS function
calling and register conventions.
Please read the assignment completely before implementing any of the functions.
You MUST implement all the functions in the assignment as defined. It is OK to implement additional
helper functions of your own in the hw5.asm file.
o YouMUST follow the efficient MIPS calling and register conventions. Do not brute-force save reg-
isters! YouWILL lose points.
o DoNOT rely on changes you make in the main files! We will test your functions with our own testing
mains. Functions will not be called in the same order andwill be called independent of each other.
o Do not submit a file with the functions/labels main or _start defined. You will obtain a ZERO for
the assignment if you do this.
� If you are having difficulties implementing these functions,write out the pseudocodeor implement the
functions in a higher-level language first. Once you understand the algorithm andwhat steps to perform,
then translate the logic toMIPS.
� When writing your program, try to comment as much as possible. Try to stay consistent with your
formatting. It is much easier for your TA and the professor to help you if we can figure out what your
code does quickly.

Getting started
Download hw5.zip from Piazza in the Homework section of Resources. This file contains hw5.asm
and 4 different main files (one for testing each of your functions), which you need for the assignment.

CSE 220 – Spring 2017 Homework #5 Page 1

At the top of your hw5.asm program in comments put your name and SBU ID number.

# Homework #5
# name: MY_NAME
# sbuid: MY_SBU_ID

How to Test Your Functions
To test your functions, simply open theprovided hw5.asm andoneof the main files inMARS.Assemble
the main file and run the file. MARS will take the contents of the file referenced with the .include
hw5.asm at the bottom of the file and add the contents of your file to themain file before assembling it.
Once the contents have been substituted into the file, MARSwill then assemble it as normal.
o Your assignment will not be graded using the provided main. Any modifications to main files will
not be graded. You will only submit your hw5.asm file via Sparky. Make sure that all code require for
implementing your functions ( .text and .data ) are included in the hw5.asm file!
o Make sure to initialize all of your valueswithin your functions! Never assume registers ormemorywill
hold any particular values!

Part 1 -Matching a globbedDNA pattern
DNA sequences are usually very long. The human genome has over 3 billion base pairs! Searching for ex-
act patterns in such a long sequence can be very tedious and hence sometimes searching for approximate
patterns is useful to narrow down the search space. In this part you will be writing a function check if a
sequencematches with a globbed sequence.

Globbing
A glob pattern is a pattern that includes wildcard characters. A wildcard character is a single character,
such as an asterisk (’*’), used to represent a number of characters or an empty string. It is often used in
file searches so the full name need not be typed. Youmight have seen *.txt used to specify all text files
in UNIX systems. If you have not yet, you will in CSE320 😉 For this homework you will only support the *
wildcard. It matches any number of any characters, including none.
match_glob – Recursive function
This function compares a DNA sequence to a globbed pattern and tells you if there is an exact match,
along with the length of all the globbed section(s) that wasmatched.
(boolean match, int glob_len) match_glob(char[] seq , char[] pat)

• seq : A string containing only letters representing DNA nucleotides (A,C,G or T). You can assume
CSE 220 – Spring 2017 Homework #5 Page 2

that seq will always be valid. Note, the characters are CASE INSENSITIVE!
• pat : A string specifying the globbed pattern. You can assume the string contains only letters rep-
resenting DNA nucleotides (A,C,G or T). It may contain 0 or more * characters. You can assume that
pat will always be valid, and there will be 0 or 1 exact matches of pat in seq .

• returns: match as true (1) if pat is matchwith seq , false (0) if not. glob_len is the total length
of the string(s) that was matched by the wildcard(s). glob_len is irrelevant if match is false. All
matches are case insensitive.

Examples:
Code Return Value
match_glob(“”,”*”) true (1), 0
match_glob(“ACGTTCAAGAGTACC”,””) false (0), X (don’t care)
match_glob(“acgttcaagagtacc”,”ACG*”) 1, 12
match_glob(“ACGTTCAAGAGTACC”,”AcG*AcC”) 1, 9
match_glob(“acgttcaagagtacc”,”*tACc”) 1, 11
match_glob(“ACGTTCAAGAGTACC”,”ACG”) 0, X (don’t care)
match_glob(“acgttcaagagtacc”,”*GTA*”) 1, 12
match_glob(“ACGTTCAAGAGTACC”,”*”) 1, 15
o Don’t care values will not be checkedwhen grading. Any valuemay be returned.
o Note: Wewill not test with two ormore wildcard (*) characters sequentially.
The logic for this recursive function can be confusing or difficult. We have provided the pseudo code for
the algorithm below.

(boolean, int) match_glob(char[] seq, char[] pat) {
/* CHECK BASE CASES */
// Wildcard is the only character left
if(pat.equals(“*”))

return (true, seq.length());

// Seq and pat are identical
if(seq.equalsIgnoreCase(pat))

return (true, 0);

// If seq or pat is empty and the other is not, then no match possible.
// LOGICAL XOR, not bitwise. Equivalently, ((seq.length() != 0) XOR
// (pat.length() != 0))
if((seq.length() XOR pat.length()) != 0)

CSE 220 – Spring 2017 Homework #5 Page 3

return (false, 0);

/* RECURSIVE CALLS for remaining sequence */
// Check if the characters are equal
if(seq[0].toLower() == pat[0].toLower())

return match_glob(seq.substring(1), pat.substring(1));

//the current char is a glob match
if(pat[0] == ‘*’) {

// recursively call on the remaining pattern
(match1, glob_len1) = match_glob(seq, pat.substring(1));

if (match1) {
// match is found!
return (match1, glob_len1);

} else {
// no match found. Continue searching from the next character
(match2, glob_len2) = match_glob(seq.substring(1), pat);
return (match2, glob_len2 + 1);

}
}
return (false, 0);

}

Part 2 – DNA Sequence Permutation Generation
A nucleotide is the basic structural unit of DNA. There are 4 different kinds of nucleotides in DNA: Ade-
nine (A), Guanine (G), Cytosine (C) and Thymine (T). A strand of DNA has a double-helix structure, which
looks like a spiraling ladder. To form such a structure, the nucleotides are paired using complementary
base pairing. Without delving into the details, this means that A can only connect with T (and vice versa)
and C can only connect to G (and vice versa).
In this part, you will be writing functions to generate and display all permutations of a DNA sequence of
desired length.
save_perm – Helper function
This function saves seq in character pairs separated by a hyphen (-) to the address specified by dst . A
newline character is saved at the end of the permutation.
char[] save_perm(char[] dst, char[] seq)

CSE 220 – Spring 2017 Homework #5 Page 4

• seq : A string containing only letters representing DNA nucleotides (A,C,G or T). You can assume
that seq will always be valid, be of even length, and contain only CAPITAL letters.

• dst : The starting destination address of where to save the permutation characters with hyphens.
• returns: The address of the next byte after the newline char.

Examples:
Code
save_perm(“@@@@@@@@@@@@1234”, “ATGCCGTA”)
dstAfter Call Return Value
“AT-GC-CG-TA
1234” Address of byte after

Code
save_perm(“@@@@@@@@@@@@1234”, “AT”)
dstAfter Call Return Value
“AT
@@@@@@@@@1234” Address of byte after

construct_candidates – Helper function
This function constructs the possible candidates for the next character in the permutation. If the next
character is the start of a newpair it can be any nucleotide, however, if it is a continuation of the same pair
then the next character has to contain the complementary base pair.
int construct_candidates(char[] cand, char[] seq, int n)

• cand : Space to store the candidates for the next character of the permutation. This space is de-
clared in the caller function.

• seq : A character array representing the current state of the permutation. This will only contain
valid DNA nucleotides but will not be the completed permutation. You can assume that seq will
always be valid but not necessarily null-terminated.

• n : A number describing the location of the character-to-be-filled in the seq array.
• returns: Returns the number of candidates that are present (Maximum is 4).

Examples:
Code
construct_candidates(“@@@@!!!!”, “AT”, 2)
candAfter Call Return Value
“ACGT!!!!” 4
CSE 220 – Spring 2017 Homework #5 Page 5

Code
construct_candidates(“@@@@!!!!”, “ATC”, 3)
candAfter Call Return Value
“G@@@!!!!” 1
Code
construct_candidates(“@@@@!!!!”, “ATCGG”, 5)
candAfter Call Return Value
“C@@@!!!!” 1
Code
construct_candidates(“@@@@!!!!”, “ATCGGCT”, 7)
candAfter Call Return Value
“A@@@!!!!” 1
Code
construct_candidates(“@@@@!!!!”, “ATCGGCTAA”, 9)
candAfter Call Return Value
“T@@@!!!!” 1
Code
construct_candidates(“@@@@!!!!”, “ATCGGCTAAB@!PQYRSHR.,@TAAG”, 9)
candAfter Call Return Value
“T@@@!!!!” 1
We have provided the pseudo code for the algorithm below.

int construct_candidates(char[] candidates, char[] seq, int n) {
//If the next candidate is part of a pair, pick complement
if(n%2 != 0) {

if(seq[n-1] == ‘A’) {
candidates[0] = ‘T’;
return 1;

} else if(seq[n-1] == ‘T’) {
candidates[0] = ‘A’;
return 1;

} else if(seq[n-1] == ‘C’) {
candidates[0] = ‘G’;
return 1;

} else {
candidates[0] = ‘C’;
return 1;

CSE 220 – Spring 2017 Homework #5 Page 6

}
} else {

candidates[0] = ‘A’;
candidates[1] = ‘C’;
candidates[2] = ‘G’;
candidates[3] = ‘T’;
return 4;

}
}

permutations – Recursive function
This is the recursive function that will generate all the possible permutations of desired length and store
them intomemory using the save_perm function.
(int rv, char[] next) permutations(char[] seq, int n, char[] res, int length)

• seq : Buffer character array of size at least length+1 , which is not necessarily initialized, to build
up the permutation.

• n : Continue creating permutation starting at the n th character ( seq[n-1] ).
• res : Pointer of location to store the result when the permutation is complete (size == length ).
The value of this pointer will be updated as permutations are stored.

• length : The total number of characters for each permutation.
• returns: Error (-1,0) if length is invalid, otherwise return (0, pointer), where pointer is the
address of the next location for storing the next permutation (i.e the address that is returned by
save ).

Return (-1, 0) in any of the following cases:
• length is less than or equal to 0
• length is an odd number

Examples:
Code Return Value
permutations(buf, 0, 0x400, 3) -1, 0
permutations(buf, 0, 0x400, 0) -1, 0
permutations(buf, 0, 0x400, 4) 0, 0x460
permutations(buf, 3, 0x400, 6) 0, 0x424
CSE 220 – Spring 2017 Homework #5 Page 7

o 0x400 is an arbitrary address for res . This address will changewhen each full permutation is com-
pleted and is decided by the value returned by save_perm .
Here is what is stored for permutations(buf, 0, 0x400, 4)
AT-AT
AT-CG
AT-GC
AT-TA
CG-AT
CG-CG
CG-GC
CG-TA
GC-AT
GC-CG
GC-GC
GC-TA

TA-AT
TA-CG
TA-GC
TA-TA

� The last example is an example of a partial recursive call. In this recursive call the first 3 elements of
buf are part of a solution and has been set (e.g: ATC@@@). A full example of such a call is below.
Here is what is stored for a partial permutation where buf already contains “ATC”
permutations(buf, 3, 0x400, 6)

AT-CG-AT
AT-CG-CG
AT-CG-GC
AT-CG-TA

The logic for this recursive function can be confusing or difficult. We have provided the pseudo code for
the algorithm below.
Note: add_null function added to code below.
(length%2 == 0) corrected to (length%2 == 1) .
Function arguments updated for correct ordering.

(int, char[]) permutations(char[] seq, int n, char[] res, int length) {

// Validation: length cannot be 0 and it has to be an even number
if(length%2 == 1 || length == 0) {

return (-1, 0);
}

// A Permutation of the final length has been reached. Save it to res.
if (n == length) {

add_null(seq, length+1)
char[] next = save_perm(res, seq);
return (0, next);

}
else {

// Create candidates for the next character and repeat
// Declare space on the stack and you can use the frame pointer
// for reference.
char[] candidates = new char[4];

int ncand = construct_candidates(candidates, seq, n);

CSE 220 – Spring 2017 Homework #5 Page 8

// Iterate through the candidates, creating permutations starting
// with each candidate
for(int i = 0; i < ncand; i++) { seq[n] = candidates[i]; // Recursive call to select candidates for next character (ret, res) = permutations(seq, n+1, res, length); } } return (0, res); } Hand-in Instructions See Sparky Submission Instructions on Piazza for hand-in instructions. V There is no tolerance for homework submission via email. Work must be submitted through Sparky. Please do not wait until the last minute to submit your homework. If you are struggling, stop by office hours for additional help. CSE 220 – Spring 2017 Homework #5 Page 9