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Engineering LibreTexts

1.15: Exercises

  • Page ID
    17020
  • Exercise \(\PageIndex{1}\)

    Write a function called is_sorted that takes a list as a parameter and returns True if the list is sorted in ascending order and False otherwise. You can assume (as a precondition) that the elements of the list can be compared with the relational operators <, >, etc.

    For example, is_sorted([1,2,2]) should return True and is_sorted(['b','a']) should return False.

    Exercise \(\PageIndex{2}\)

    Two words are anagrams if you can rearrange the letters from one to spell the other. Write a function called is_anagram that takes two strings and returns True if they are anagrams.

    Exercise \(\PageIndex{3}\)

    The (so-called) Birthday Paradox:

    1. Write a function called has_duplicates that takes a list and returns True if there is any element that appears more than once. It should not modify the original list.
    2. If there are 23 students in your class, what are the chances that two of you have the same birthday? You can estimate this probability by generating random samples of 23 birthdays and checking for matches. Hint: you can generate random birthdays with the randint function in the random module.

    You can read about this problem at http://en.Wikipedia.org/wiki/Birthday_paradox.

    Solution:

    http://thinkpython.com/code/birthday.py

    Exercise \(\PageIndex{4}\)

    Write a function called remove_duplicates that takes a list and returns a new list with only the unique elements from the original. Hint: they don’t have to be in the same order.

    Exercise \(\PageIndex{5}\)

    Write a function that reads the file words.txt and builds a list with one element per word. Write two versions of this function, one using the append method and the other using the idiom t = t + [x]. Which one takes longer to run? Why?

    Hint: use the time module to measure elapsed time.

    Solution:

    http://thinkpython.com/code/wordlist.py

    Exercise \(\PageIndex{6}\)

    To check whether a word is in the word list, you could use the in operator, but it would be slow because it searches through the words in order.

    Because the words are in alphabetical order, we can speed things up with a bisection search (also known as binary search), which is similar to what you do when you look a word up in the dictionary. You start in the middle and check to see whether the word you are looking for comes before the word in the middle of the list. If so, then you search the first half of the list the same way. Otherwise you search the second half.

    Either way, you cut the remaining search space in half. If the word list has 113,809 words, it will take about 17 steps to find the word or conclude that it’s not there.

    Write a function called bisect that takes a sorted list and a target value and returns the index of the value in the list, if it’s there, or None if it’s not.

    Or you could read the documentation of the bisect module and use that!

    Solution:

    http://thinkpython.com/code/inlist.py

    Exercise \(\PageIndex{7}\)

    Two words are a “reverse pair” if each is the reverse of the other. Write a program that finds all the reverse pairs in the word list.

    Solution:

    http://thinkpython.com/code/reverse_pair.py

    Exercise \(\PageIndex{8}\)

    Two words “interlock” if taking alternating letters from each forms a new word. For example, “shoe” and “cold” interlock to form “schooled.”

    1. Write a program that finds all pairs of words that interlock. Hint: don’t enumerate all pairs!
    2. Can you find any words that are three-way interlocked; that is, every third letter forms a word, starting from the first, second or third?

    Credit: This exercise is inspired by an example at http://puzzlers.org.

    Solution:

    http://thinkpython.com/code/interlock.py