understand how s and s11 shorten one character and then grow one character at a time because there is no loop.

def toChars(s):
import string
#s = string.lower(s)
s = s.lower()
ans = ''
for c in s:
#if c in string.lowercase:
if c in s:
ans = ans + c
return ans

def isPal(s):
if len(s) <= 1:
return True
else:
s11=s[1:-1]
equation=(s[0] == s[-1])
returnequation=equation
returnispal=isPal(s[1:-1])
returnboth=returnequation and returnispal
return returnboth

def isPalindrome(s):
"""Returns True if s is a palindrome and False otherwise"""
mychars=toChars(s)
returnispal=isPal(mychars)
return returnispal

print (isPalindrome('Guttag'))
print (isPalindrome('Guttug'))
print (isPalindrome('Able was I ere I saw Elba'))
print (isPalindrome('Are we not drawn onward, we few, drawn onward to new era?'))

I created variables so I can see changes in a debugger ide

in def is Palindrome(s)
I do not understand how s and s11 shorten one character and then grow one character at a time because there is no loop.
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Commented:
the ispal function is being called recursively.
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Author Commented:
line
19
26
or both

and how
0
Commented:
isPalindrome is the initial invocation of the isPal function.  The recursion only happens once you are inside the isPal function.  So, I think the answer to your question is 19.
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Author Commented:
thanks
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