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