Practice Program 5.7:

! A palindrome is a sequence of characters that reads the same backward and forward.

! Punctuation marks and spaces are ignored. No distinction is made betwen

! lowercase and uppercase letters. An example palindrome is the following string:

! "Name no one man".

! Ask the user to enter a string and determine if this string is a palindrome.

! test for Hannah, George, "Able was I ere I saw Elba", "Name no one man"

! Pseudocode:

! 1. remove all characters that are not letters

! 2. make all letters capital

! 3. make a reverse version

! 4. compare the original with the reverse

! 5. if not equal print "The string is not a palindrome"

! 6. if equal print: "The string is a palindrome"

! PP5.7.1 clean and capitalize the string:

!LET ccaps$ = ucase$("Name no one man")

!PRINT "the capitalized string is ..."; ccaps$

!LET word$ = ""

!LET index = len(ccaps$)

!LET j=0

!DO until j= index

! LET j=j+1

! LET B$=ccaps$[j:j]

! PRINT B$

! IF ord(B$) <90 and ord(B$)>64 then

! LET word$ =word$&B$

! ELSE

! LET B$=""

! PRINT word$

! END IF

!LOOP

!PRINT word$

!END

! Practice program PP5.7.2 to make a reverse version of cleanword$

! LET word$ = "SAMENOONEMAN"

! LET revword$=""

! FOR n = len(word$) to 1 step -1

! LET revword$= revword$&word$[n:n]

! NEXT n

! PRINT "the original word is...", word$

! PRINT "the reversed word is...", revword$

! END

! The main program

INPUT prompt "Please enter a phrase...":word$

!LET word$= "Name no one man"

! 2. make all letters capital

LET capword$= UCASE$(word$)

PRINT "capword$ is ...",capword$

! 1. remove all characters that are not letters

LET cleanword$=""

LET j=0 ! initialize control variable for do loop

LET index = len(capword$) ! the range of the do loop control variable

DO until j= index

LET j=j+1

LET B$=capword$[j:j] !test each letter in capword$

IF ord(B$) <90 and ord(B$)>64 then !test for capital letters

LET cleanword$ =cleanword$&B$ ! build up cleanword$

ELSE

LET B$="" ! throw away noncaps in capword$

PRINT cleanword$ !see how it stands as the loop works

END IF

LOOP

PRINT "cleanword$ is...", cleanword$

LET revword$="" ! initialize

FOR n = len(cleanword$) to 1 step -1

LET revword$= revword$&cleanword$[n:n]

NEXT n

PRINT "the original phrase is...", word$

PRINT "the reversed phrase is...", revword$

IF revword$=cleanword$ then

PRINT "the phrase is a palindrome"

ELSE

PRINT "the phrase is not a palindrome"

END IF

END

Sample output:

Please enter a phrase...Able was I ere I saw Elba

capword$ is ... ABLE WAS I ERE I SAW ELBA

ABLE

ABLEWAS

ABLEWASI

ABLEWASIERE

ABLEWASIEREI

ABLEWASIEREISAW

cleanword$ is... ABLEWASIEREISAWELBA

the original phrase is... Able was I ere I saw Elba

the reversed phrase is... ABLEWASIEREISAWELBA

the phrase is a palindrome