Eertree

Eertree is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

An eertree is a data structure designed for efficient processing of certain palindrome tasks, for instance counting the number of sub-palindromes in an input string.[1] The data structure has commonalities to both suffix tries and suffix trees.

Task

Construct an eertree for the string "eertree", then output all sub-palindromes by traversing the tree.

Python

<lang python>#!/bin/python from __future__ import print_function

class Node(object): def __init__(self): self.edges = {} # edges (or forward links) self.link = None # suffix link (backward links) self.len = 0 # the length of the node

class Eertree(object): def __init__(self): self.nodes = [] # two initial root nodes self.rto = Node() #odd length root node, or node -1 self.rte = Node() #even length root node, or node 0

# Initialize empty tree self.rto.link = self.rte.link = self.rto; self.rto.len = -1 self.rte.len = 0 self.S = [0] # accumulated input string, T=S[1..i] self.maxSufT = self.rte # maximum suffix of tree T

def get_max_suffix_pal(self, startNode, a): # We traverse the suffix-palindromes of T in the order of decreasing length. # For each palindrome we read its length k and compare T[i-k] against a # until we get an equality or arrive at the -1 node. u = startNode i = len(self.S) k = u.len while id(u) != id(self.rto) and self.S[i - k - 1] != a: assert id(u) != id(u.link) #Prevent infinte loop u = u.link k = u.len

return u

def add(self, a):

# We need to find the maximum suffix-palindrome P of Ta # Start by finding maximum suffix-palindrome Q of T. # To do this, we traverse the suffix-palindromes of T # in the order of decreasing length, starting with maxSuf(T) Q = self.get_max_suffix_pal(self.maxSufT, a)

# We check Q to see whether it has an outgoing edge labeled by a. createANewNode = not a in Q.edges

if createANewNode: # We create the node P of length Q+2 P = Node() self.nodes.append(P) P.len = Q.len + 2 if P.len == 1: # if P = a, create the suffix link (P,0) P.link = self.rte else: # It remains to create the suffix link from P if |P|>1. Just # continue traversing suffix-palindromes of T starting with the suffix # link of Q. P.link = self.get_max_suffix_pal(Q.link, a).edges[a]

# create the edge (Q,P) Q.edges[a] = P

#P becomes the new maxSufT self.maxSufT = Q.edges[a]

#Store accumulated input string self.S.append(a)

return createANewNode

def get_sub_palindromes(self, nd, nodesToHere, charsToHere, result): #Each node represents a palindrome, which can be reconstructed #by the path from the root node to each non-root node.

#Traverse all edges, since they represent other palindromes for lnkName in nd.edges: nd2 = nd.edges[lnkName] #The lnkName is the character used for this edge self.get_sub_palindromes(nd2, nodesToHere+[nd2], charsToHere+[lnkName], result)

#Reconstruct based on charsToHere characters. if id(nd) != id(self.rto) and id(nd) != id(self.rte): #Don't print for root nodes tmp = "".join(charsToHere) if id(nodesToHere[0]) == id(self.rte): #Even string assembled = tmp[::-1] + tmp else: #Odd string assembled = tmp[::-1] + tmp[1:] result.append(assembled)

if __name__=="__main__": st = "eertree" print ("Processing string", st) eertree = Eertree() for ch in st: eertree.add(ch)

print ("Number of sub-palindromes:", len(eertree.nodes))

#Traverse tree to find sub-palindromes result = [] eertree.get_sub_palindromes(eertree.rto, [eertree.rto], [], result) #Odd length words eertree.get_sub_palindromes(eertree.rte, [eertree.rte], [], result) #Even length words print ("Sub-palindromes:", result)</lang>

Output:

Processing string eertree
Number of sub-palindromes: 7
Sub-palindromes: ['r', 'e', 'eertree', 'ertre', 'rtr', 't', 'ee']