What degree is this palindrome?

Question

Your task is to determine the palindromic degree of a given non-empty string.

To do this, loop until one of the following conditions are met:

• The string is not a palindrome.
• You encounter a string that has already been encountered.

And do the following:

• If it is a palindrome, add one to the degree.
• Depalindromize.

For example, if I had the string babababab, the calculation would go:

babababab     1
babab         2
bab           3
ba

For the string zzzzz, it would go:

zzzzz         1
zzz           2
zz            3
z             4
z

Test cases

abc 0
cenea 0
pappap 2
aa 2
aba 1
aaa 3
cccc 3
zzzzz 4

Explanation:

abc : 0
cenea : 0
pappap -> pap -> pa : 2
aa -> a -> a : 2
aba -> ab : 1
aaa -> aa -> a -> a: 3
cccc -> cc -> c -> c: 3
zzzzz -> zzz -> zz -> z -> z : 4

Remember, this is code-golf, so the code with the fewest bytes wins.

Answer 1

Perl, 53 bytes

52 bytes of code + 1 byte for -p.

$\++while s/^((.+).?)(??{reverse$2})$/$1/||s/^.$//}{

To run it :

perl -pE '$\++while s/^((.+).?)(??{reverse$2})$/$1/||s/^.$//}{' <<< "zzzzz"

Answer 2

Jelly, 11 9 bytes

ŒHḢ$ƬUƑƤS

Try it online!

-2 bytes thanks to Unrelated String!

How it works

ŒHḢ$ƬUƑƤS - Main link. Takes a string S on the left
   $Ƭ     - Until reaching a previously encountered string:
ŒH        -   Split into halves
  Ḣ       -   Take the first one
       Ƥ  - Over each prefix:
     U    -   Reverse each string in the prefix
      Ƒ   -   Does that equal the prefix?
        S - Sum

Answer 3

JavaScript (ES6), 85 bytes

f=s=>s[1]?(t=s.slice(0,l=-~s.length/2))==[...s.slice(-l)].reverse().join``?1+f(t):0:1

Answer 4

Haskell, 58 bytes

p[a]=1
p a|reverse a==a=1+p(take(div(1+length a)2)a)
p _=0

Answer 5

05AB1E, 14 bytes

[ÐÂÊ#¼g#2ä¬])\

Uses the CP-1252 encoding. Try it online!

Answer 6

Scala, 56 bytes

s=>if(s.reverse==s&&s.size>1)1+h(s.take(s.size/2))else 2

requires an assignmentt to a variable with type declaration:

val f:(String=>Int)=...

Answer 7

Pyth - 21 bytes

The case of palindromes never running out really cost me, will look for a better way.

.xxK.uhc2NQ)hf!_ITKlK

Test Suite.

Answer 8

Java 7,147 bytes

int c(String s,int t){int l=s.length();return l<2?t+1:s.equals(new StringBuilder(s).reverse().toString())?c(s.substring(0,l%2>0?l/2+1:l/2),++t):t;}

Answer 9

Python 2, 94 87 77 bytes

The recursive approach... this currently assumes that empty & single letter strings get a score of 1.

def P(x,c=0):a=len(x);return c+(a<2)if(x[::-1]!=x)+(a<2)else P(x[:-~a/2],-~c)

Try it online!

Answer 10

Pyth, 17 bytes

a!_IQl.u?_IJhc2NJ

Test suite

The previous Pyth answer uses an older version of the language and a significantly different approach.

Explanation:

a!_IQl.u?_IJhc2NJ   | Full program
a!_IQl.u?_IJhc2NJNQ | with implicit variables filled
--------------------+-------------------------------------------------------------------------------------------------------------------------
      .u          Q | Repeat the following until a repeated value is found, collecting the intermediate results, with N starting as Q (input):
           Jhc2N    |  J = the first half of N (longer half is necessary)
        ?_I     JN  |  if J is invariant over reversal (i.e. is a palindrome), N = J, else N = N
     l              | take the length of the resulting list
a                   | subtract
 !_IQ               |  0 if Q is a palindrome, 1 otherwise

Answer 11

Python 3, 53 bytes

f=lambda x:2>>len(x)or-~f(x[len(x)//2:])*(x==x[::-1])

Try it online!

How it works:

• 2>>len(x) will give us 1 if len(x)==1 and 0 if len(x)>1 (we don't have to treat the case where x is the empty string)

• if the previous gives 0 then we recurse with x[len(x)//2:] which is the second part of our palindrome (including the center), -~ we add 1 and multiply by (x==x[::-1]) to return 0 if it is not a palidrome