Due by 11:59pm on Wednesday, 2/20
Submission. See the online submission instructions. We have provided a starter file for the questions below.
Readings. Section 1.7 of the online lecture notes.
Q1. In the lecture notes, we defined a pig_latin function to compute the Pig Latin equivalent of an English word. The code that we used is as follows:
def pig_latin_original(w): """Return the Pig Latin equivalent of a lowercase English word w.""" if starts_with_a_vowel(w): return w + 'ay' return pig_latin_original(rest(w) + first(w)) def first(s): """Returns the first character of a string.""" return s[0] def rest(s): """Returns all but the first character of a string.""" return s[1:] def starts_with_a_vowel(w): """Return whether w begins with a vowel.""" c = first(w) return c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u'
This code repeatedly moves a letter from the beginning of a word to the end, until the first letter is a vowel, at which point it adds on 'ay' to the end. However, this code fails when the original word has no vowels in the set {a, e, i, o, u}, such as the word "sphynx." Write a new version of the pig_latin function that just adds 'ay' to the original word if it does not contain a vowel in this set. Use only the first, rest, and starts_with_a_vowel functions to access the contents of a word, and use the built-in len function to determine its length. Do not use any loops.
def pig_latin(w): """Return the Pig Latin equivalent of a lowercase English word w. >>> pig_latin('pun') 'unpay' >>> pig_latin('sphynx') 'sphynxay' """ "*** YOUR CODE HERE ***"
Q2. A classic puzzle called the Towers of Hanoi is a game that consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.
The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:
Complete the definition of towers_of_hanoi which prints out the steps to solve this puzzle for any number of n disks starting from the start rod and moving them to the end rod:
def towers_of_hanoi(n, start, end): """Print the moves required to solve the towers of hanoi game if we start with n disks on the start pole and want to move them all to the end pole. The game is to assumed to have 3 poles (which is traditional). >>> towers_of_hanoi(1, 1, 3) Move 1 disk from rod 1 to rod 3 >>> towers_of_hanoi(2, 1, 3) Move 1 disk from rod 1 to rod 2 Move 1 disk from rod 1 to rod 3 Move 1 disk from rod 2 to rod 3 >>> towers_of_hanoi(3, 1, 3) Move 1 disk from rod 1 to rod 3 Move 1 disk from rod 1 to rod 2 Move 1 disk from rod 3 to rod 2 Move 1 disk from rod 1 to rod 3 Move 1 disk from rod 2 to rod 1 Move 1 disk from rod 2 to rod 3 Move 1 disk from rod 1 to rod 3 """ "*** YOUR CODE HERE ***" def move_disk(start, end): print("Move 1 disk from rod", start, "to rod", end)
Q3. The number of partitions of a positive integer n is the number of ways in which n can be expressed as the sum of positive integers in increasing order. For example, the number 5 has 7 partitions.
5 = 5
5 = 1 + 4
5 = 2 + 3
5 = 1 + 1 + 3
5 = 1 + 2 + 2
5 = 1 + 1 + 1 + 2
5 = 1 + 1 + 1 + 1 + 1
Write a tree-recursive function part(n) that returns the number of partitions of n.
Hint: Introduce a helper function that computes partitions of n using only a subset of the integers less than or equal to n. Once you have done so, you can use very similar logic to the count_change function from the lecture notes:
def part(n): """Return the number of partitions of positive integer n. >>> part(5) 7 >>> part(10) 42 >>> part(15) 176 >>> part(20) 627 """ "*** YOUR CODE HERE ***"
Q4. (Extra for experts) Recall that the summation function computes the sum of a sequence of terms from 1 to n:
def summation(n, term): """Return the sum of the first n terms of a sequence. >>> summation(5, lambda x: pow(x, 3)) 225 """ total, k = 0, 1 while k <= n: total, k = total + term(k), k + 1 return total
Write a function interleaved_sum that similarly computes the sum of a sequence of terms from 1 to n, but uses different function to compute the terms for odd and even numbers. Do so without using any loops or testing in any way if a number is odd or even. (You may test if a number is equal to 0, 1, or n.)
def interleaved_sum(n, odd_term, even_term): """Compute the sum odd_term(1) + even_term(2) + odd_term(3) + ..., up to n. >>> # 1 + 2^2 + 3 + 4^2 + 5 ... interleaved_sum(5, lambda x: x, lambda x: x*x) 29 """ "*** YOUR CODE HERE ***"