Updated Palindrome-Partitioning w/ README

Palindrome-Partitioning/README.md

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+# Lab 4: Coding Palindrome Partitioning with Backtracking
+
+## Description
+
+In this lab you will develop a recursive algorithm for the palindrome partition
+counting problem.
+
+A **palindrome** is a string of letters that is the same backwards as forwards.
+For example, "racecar", "eevee", and "12321" are all palindromes. A string with
+a single character is trivially a palindrome.
+
+A **palindromic partition** of a string is a partition of the string into
+substrings whose concatenation is equal to the original string, but such that
+every substring is itself a palindrome.
+
+For example, the string, "seeks" can be broken up into the palindrome partition
+["s", "ee", "k", "s"] or as ["s", "e", "e", "k", "s"]. Your task is to design
+and implement a recursive algorithm that counts the number of palindromic
+partitions of a given string.
+
+## Input
+
+Your input will begin with a single line containing a nonnegative integer n
+followed by exactly n lines each of which contains an input string.
+
+## Output
+
+You should output **exactly** n lines, one for each input string with each
+ending in a newline character. The value of each line should be the number of
+unique palindromic partitions that can be made with the input string.
+
+<table>
+    <tr>
+        <td>Sample Input</td>
+        <td>Sample Output</td>
+    </tr>
+    <tr>
+        <td><pre>3<br>abc<br>bcccb<br>seeks</pre></td>
+        <td><pre>1<br>5<br>2</pre></td>
+    </tr>
+</table>
+
+## Turning it in
+
+**Submit your Python code as cs412_palindrome_count_bt.py to Gradescope.**
+
+## Hints
+
+- As always, when devising a recursive backtracking algorithm, think of your
+  output as a series of choices. Almost always you can get to the recursive
+  solution by having each recursive call brute-force make all possible next
+  choice and use the recursion fairy to handle the remaining choices.
+  - What are the choices here? Think about what you need to do to the string to
+    turn it into a palindromic partition--this will tell you exactly what a
+    "choice" is in terms of producing palindromic partitions.
+- Don't worry about speed here. You are almost certainly going to have a rather
+  bad runtime. We'll fix this in a few weeks when we talk about dynamic
+  programming. You probably want to write a tiny helper predicate
+  isPalindrome(s) that returns true if s is a palindrome and false otherwise.
+- How can you easily have a string evaluate to the reverse of itself? Recall
+  that slicing takes 3 values (start pos, end pos, step size). See this website
+  for details:
+  https://www.digitalocean.com/community/tutorials/how-to-index-and-slice-strings-in-python-3