Before you turn this problem in, make sure everything runs as expected. First, restart the kernel (in the menubar, select Kernel$\rightarrow$Restart) and then run all cells (in the menubar, select Cell$\rightarrow$Run All).
Make sure you fill in any place that says YOUR CODE HERE
or "YOUR ANSWER HERE", as well as your name and collaborators below:
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NAME = "Alyssa P. Hacker"
COLLABORATORS = "Ben Bitdiddle"
For this problem set, we'll be using the Jupyter notebook:
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def squares(n):
"""Compute the squares of numbers from 1 to n, such that the
ith element of the returned list equals i^2.
"""
if n < 1:
raise ValueError
return [i ** 2 for i in range(1, n + 1)]
Your function should print [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
for $n=10$. Check that it does:
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squares(10)
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"""Check that squares returns the correct output for several inputs"""
from nose.tools import assert_equal
assert_equal(squares(1), [1])
assert_equal(squares(2), [1, 4])
assert_equal(squares(10), [1, 4, 9, 16, 25, 36, 49, 64, 81, 100])
assert_equal(squares(11), [1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121])
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"""Check that squares raises an error for invalid inputs"""
from nose.tools import assert_raises
assert_raises(ValueError, squares, 0)
assert_raises(ValueError, squares, -4)
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def sum_of_squares(n):
"""Compute the sum of the squares of numbers from 1 to n."""
return sum(squares(n))
The sum of squares from 1 to 10 should be 385. Verify that this is the answer you get:
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sum_of_squares(10)
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"""Check that sum_of_squares returns the correct answer for various inputs."""
assert_equal(sum_of_squares(1), 1)
assert_equal(sum_of_squares(2), 5)
assert_equal(sum_of_squares(10), 385)
assert_equal(sum_of_squares(11), 506)
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"""Check that sum_of_squares relies on squares."""
orig_squares = squares
del squares
try:
assert_raises(NameError, sum_of_squares, 1)
except AssertionError:
raise AssertionError("sum_of_squares does not use squares")
finally:
squares = orig_squares
$\sum_{i=1}^n i^2$
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import math
def hypotenuse(n):
"""Finds the hypotenuse of a right triangle with one side of length n and
the other side of length n-1."""
# find (n-1)**2 + n**2
if (n < 2):
raise ValueError("n must be >= 2")
elif n == 2:
sum1 = 5
sum2 = 0
else:
sum1 = sum_of_squares(n)
sum2 = sum_of_squares(n-2)
return math.sqrt(sum1 - sum2)
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print(hypotenuse(2))
print(math.sqrt(2**2 + 1**2))
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print(hypotenuse(10))
print(math.sqrt(10**2 + 9**2))