[Solved] CS061Lab 6- identification of a palindrome

The purpose of this lab is to break down the identification of a palindrome into its mostbasic steps, and implement a palindrome checker in LC3.2 Our Objectives for This Week1. Exercise 01 ~ Capture a string of text and store it2. Exercise 02 ~ Check to see if its a palindrome3. Exercise 03 ~ Refine your subroutine with Case conversion3.1 ExercisesWhat is a Palindrome?In case you didnt already know, a palindrome is a word or phrase that is spelled the sameforwards as backwards. Such words include: racecar madam deified tacocatPhrases can be palindromes too (see Exercise 03)! For example, the following are allpalindromes (with the assumption that anything except alphabetic characters are ignored) live not on evil So many dynamos Are we not drawn onward, we few, drawn onward to new eraExercise 01Write the following subroutine, which captures a user-entered string at run-time, storing it as anull-terminated character array (just like the .STRINGZ pseudo-op, with the big difference thatthat .STRINGZ requires a hard-coded character array at compile-time) .;-; Subroutine: SUB_GET_STRING; Parameter (R1): The starting address of the character array; Postcondition: The subroutine has prompted the user to input a string,; terminated by the [ENTER] key (the sentinel), and has stored; the received characters in an array of characters starting at (R1).; the array is NULL-terminated; the sentinel character is NOT stored.; Return Value (R5): The number of non-sentinel characters read from the user.; R1 contains the starting address of the array unchanged.;This subroutine should prompt the user to enter in a string of text, ending with the [ENTER] key.The string of text will be stored starting at whatever address is specified by (R1) and will beNULL-terminated (i.e. the subroutine will store zero (x0000 = #0 =) at the end of the array).The entry sentinel value itself (i.e. the newline) must not be stored in the array!The subroutine returns in R5 the number of non-sentinel characters entered.REMEMBER: no ghost writing! Echo each character as it is received from the user!Example:If the user enters: This is really hard!, followed by [ENTER], then the array will look like this:T h i s i s r e a l l y h a r d ! 0Where the initial T is stored in the address specified by R1 (this value will not be changed bythe subroutine); R5 will hold the value x14 = #20 (count the characters to confirm!)Test Harness:Now write a test harness (i.e. a program that tests your subroutine to make sure it works) thatdoes the following:1. R1 <- The address at which to store the array.Hard code this address, and reserve space there using .BLKWMake sure you have enough free memory starting from this address to store themaximum number of characters likely to be entered e.g. #100.2. Calls the subroutine3. Calls the BIOS routine PUTS (aka Trap x22) to print the string.Remember that PUTS needs the starting address of the string in R0 , not R1!Exercise 02Now add a subroutine that decides whether the newly entered string is a palindrome or not.;; Subroutine: SUB_IS_PALINDROME; Parameter (R1): The starting address of a null-terminated string; Parameter (R5): The number of characters in the array.; Postcondition: The subroutine has determined whether the string at (R1) is; a palindrome or not, and returned a flag to that effect.; Return Value: R4 {1 if the string is a palindrome, 0 otherwise};The algorithm:Hints: You already know the starting address of the array. You already know how many characters are in the array. Thus, you can calculate the address of the last character of the array If the array has n characters, compare1. array[0] with array[n-1]2. array[1] with array[n-2]3. array[2] with array[n-3]4. At what point can you decide that the string IS a palindrome?At what point can you decide that the string is NOT a palindrome?Hint: in NEITHER case is the answer after n comparisonsTest Harness:Add the following steps to your original test harness:1. Call the palindrome-checking subroutine2. Use the return value of the subroutine to report to the user whether the string was apalindrome or not in other words, report the result in the test harness , NOT in thesubroutine itself. The subroutine just sets a flag (this is a standard technique) .Your main should print out eitherThe string <print the input string here> IS a palindromeorThe string <print the input string here> IS NOT a palindromeExercise 03:The subroutine from Exercise 02 would not recognize a phrase such as Madam, Im Adam as apalindrome. It would be fairly simple to rework our subroutine to ignore whitespace andpunctuation, but for now we will just handle character case, so that your subroutine could atleast recognize as a palindrome the string MadamImAdam . We will do this by converting theentire phrase to the same case before doing the actual palindrome check.Write the following subroutine:;; Subroutine: SUB_TO_UPPER; Parameter (R1): Starting address of a null-terminated string; Postcondition: The subroutine has converted the string to upper-case in-place; i.e. the upper-case string has replaced the original string; No return value, no output (but R1 still contains the array address, unchanged).;Hints: Check the ASCII table to see how uppercase and lowercase letters differ in binary/hex Use bit-masking , not arithmetic: the conversion of a letter to uppercase can be donewith a total of two lines of LC3 code.Test Harness:Instead of writing a separate test harness for this subroutine, you can just add a call to it insideyour is_palindrome subroutine from exercise 2 , and test it with a palindrome like Racecar.This is one of the few cases where you will create nested subroutines: it is safe to do so herebecause a) the nested routine has no output and no return value; and b) you properly back up &restore only the necessary registers in all subroutines, dont you ? 🙂