diff --git a/palindrome checker 4 methods.py b/palindrome checker 4 methods.py
new file mode 100644
index 0000000..35fba25
--- /dev/null
+++ b/palindrome checker 4 methods.py	
@@ -0,0 +1,85 @@
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Apr  8 18:10:50 2018
+
+@author: Administrator
+"""
+#check if a string is a palindrome
+#reversing a string is extremely easy
+#at first i was thinking about using pop function
+#however, for an empty string, pop function is still functional
+#it would not stop itself
+#it would show errors of popping from empty list
+#so i have to use the classic way, slicing
+#each time we slice the final letter
+#to remove the unnecessary marks, we have to use regular expression
+
+import re
+
+def f(n):
+    #this is the base case, when string is empty
+    #we just return empty
+    if n=='':
+        return n
+    else:
+        return n[-1]+f(n[:-1])
+    
+def check(n):
+    #this part is the regular expression to get only characters
+    #and format all of em into lower cases
+    c1=re.findall('[a-zA-Z0-9]',n.lower())
+    c2=re.findall('[a-zA-Z0-9]',(f(n)).lower())
+    if c1==c2:
+        return True
+    else:
+        return False
+    
+    
+    
+#alternatively, we can use deque
+#we pop from both sides to see if they are equal
+#if not return false
+#note that the length of string could be an odd number
+#we wanna make sure the pop should take action while length of deque is larger than 1
+
+from collections import deque
+
+def g(n):
+    
+    c=re.findall('[a-zA-Z0-9]',n.lower())
+    de=deque(c)
+    while len(de) >=1:
+        if de.pop()!=de.popleft():
+            return False
+    return True
+
+
+#or we can use non recursive slicing
+def h(n):
+    c=re.findall('[a-zA-Z0-9]',n.lower())
+    if c[::-1]==c:
+        return True
+    else:
+        return False
+    
+#or creating a new list
+#using loop to append new list from old list s popped item
+def i(n):
+    c=re.findall('[a-zA-Z0-9]',n.lower())
+    d=[]
+    for i in range(len(c)):
+        d.append(c.pop())
+    c=re.findall('[a-zA-Z0-9]',n.lower())
+
+    if d==c:
+        return True
+    else:
+        return False
+    
+
+
+print(check('Evil is a deed as I live!'))
+print(g('Evil is a deed as I live!'))
+print(h('Evil is a deed as I live!'))
+print(i('Evil is a deed as I live!'))
+#for time and space complexity, python non recursive slicing wins
\ No newline at end of file