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%matplotlib inline
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import matplotlib.pyplot as plt
The following exercises let you practice Python syntax. Do not use any packages not in the standard library except for matplotlib.pyplot which has been imported for you.
If you have not done much programming, these exercises will be challenging. Don't give up! For this first exercise, solutions are provided, but try not to refer to them unless you are desperate.
1. Grading (20 points)
A = [90 - 100]
B = [80 - 90)
C = [65 - 80)
D = [0, 65)where square brackets indicate inclusive boundaries and parentheses indicate exclusive boundaries. However, studens whose attendance is 12 days or fewer get their grade reduced by one (A to B, B to C, C to D, and D stays D). The function should take a score and an attendance as an argument and return A, B, C or D as appropriate.(10 points)
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scores = [ 84, 76, 67, 23, 83, 23, 50, 100, 32, 84, 22, 41, 27,
29, 71, 85, 47, 77, 39, 25, 85, 69, 22, 66, 100, 92,
97, 46, 81, 88, 67, 20, 52, 62, 39, 36, 79, 54, 74,
64, 33, 68, 85, 69, 84, 30, 68, 100, 71, 33, 21, 95,
92, 72, 53, 50, 31, 82, 53, 68, 49, 37, 40, 21, 94,
30, 54, 58, 92, 95, 73, 80, 81, 56, 44, 22, 69, 70,
25, 50, 59, 32, 65, 79, 27, 62, 27, 31, 78, 88, 68,
53, 79, 69, 89, 38, 80, 55, 92, 55]
attendances = [17, 19, 21, 14, 10, 20, 14, 9, 6, 21, 5, 23, 21, 4, 5, 21, 20,
2, 14, 14, 21, 22, 3, 0, 11, 0, 0, 4, 20, 14, 23, 16, 24, 5,
12, 11, 22, 20, 15, 23, 0, 20, 20, 6, 4, 14, 6, 18, 17, 0, 18,
6, 3, 19, 24, 7, 9, 15, 18, 10, 2, 15, 21, 2, 9, 21, 20, 11,
24, 23, 14, 22, 4, 12, 7, 19, 6, 18, 23, 6, 14, 6, 1, 12, 7,
11, 22, 21, 7, 22, 24, 4, 10, 17, 21, 15, 0, 20, 3, 20]
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# Your answer here
def grade(score, attendance):
"""Function that returns grade based on score and attendance."""
if attendance > 12:
if score >= 90:
return 'A'
elif score >= 80:
return 'B'
elif score >= 65:
return 'C'
else:
return 'D'
else:
if score >= 90:
return 'B'
elif score >= 80:
return 'C'
else:
return 'D'
counts = {}
for score, attendance in zip(scores, attendances):
g = grade(score, attendance)
counts[g] = counts.get(g, 0) + 1
for g in 'ABCD':
print(g, counts[g])
2. The Henon map and chaos. (25 points)
The Henon map takes a pont $(x_n, y_n)$ in the plane and maps it to $$ x_{n+1} = 1 - a x_n^2 + y_n \\ y_{n+1} = b x_n $$
plt.scatter function with s=1 to make the plot. (10 points)
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# Your answer here
def henon(x, y, a=1.4, b=0.3):
"""Henon map."""
return (1 - a*x**2 + y, b*x)
henon(1, 1)
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n = 1000
n_store = 50
aa = [i/100 for i in range(100, 141)]
xxs = []
for a in aa:
xs = []
x, y = 1, 1
for i in range(n - n_store):
x, y = henon(x, y, a=a)
for i in range(n_store):
x, y = henon(x, y, a=a)
xs.append(x)
xxs.append(xs)
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%matplotlib inline
import matplotlib.pyplot as plt
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for a, xs in zip(aa, xxs):
plt.scatter([a]*n_store, xs, s=1)
3. Collatz numbers - Euler project problem 14. (25 points)
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
NOTE: Once the chain starts the terms are allowed to go above one million.
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# Your answer here
def collatz(n):
"""Returns Collatz sequence starting with n."""
seq = [n]
while n != 1:
if n % 2 == 0:
n = n // 2
else:
n = 3*n + 1
seq.append(n)
return seq
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def collatz_count(n):
"""Returns Collatz sequence starting with n."""
count = 1
while n != 1:
if n % 2 == 0:
n = n // 2
else:
n = 3*n + 1
count += 1
return count
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%%time
best_n = 1
best_length = 1
for n in range(2, 1000000):
length = len(collatz(n))
if length > best_length:
best_length = length
best_n = n
print(best_n, best_length)
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%%time
best_n = 1
best_length = 1
seen = set([])
for n in range(2, 1000000):
if n in seen:
continue
seq = collatz(n)
seen.update(seq)
length = len(seq)
if length > best_length:
best_length = length
best_n = n
print(best_n, best_length)
4. Reading Ulysses. (30 points)
text, discarding the header information (i.e. text should start with \n\n*** START OF THIS PROJECT GUTENBERG EBOOK ULYSSES ***\n\n\n\n\n). Also remove the footer information (i.e. text should not include anything from End of the Project Gutenberg EBook of Ulysses, by James Joyce). (10 points)text. For simplicity, a word is defined here to be any sequence of characters with no space between the characters (i.e. a word may include punctuation or numbers, just not spaces). If there are ties, report the starting index and length of the last word found. For example, in "the quick brow fox jumps over the lazy dog." the longest word is jumps which starts at index 19 and has length 5, and 'dog.' would be considered a 4-letter word (15 points).
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# Your answer here
import urllib.request
response = urllib.request.urlopen('http://www.gutenberg.org/files/4300/4300-0.txt')
text = response.read().decode()
# Alternative version using requests library
# Although not officially part of the standard libaray,
# it is so widely used that the standard docs point to it
# "The Requests package is recommended for a higher-level HTTP client interface."
import requests
url = 'http://www.gutenberg.org/files/4300/4300-0.txt'
text = requests.get(url).text
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with open('ulysses.txt', 'w') as f:
f.write(text)
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with open('ulysses.txt') as f:
text = f.read()
start_string = '\n\n*** START OF THIS PROJECT GUTENBERG EBOOK ULYSSES ***\n\n\n\n\n'
stop_string = 'End of the Project Gutenberg EBook of Ulysses, by James Joyce'
start_idx = text.find(start_string)
stop_idx = text.find(stop_string)
text = text[(start_idx + len(start_string)):stop_idx]
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best_len = 0
best_word = ''
for word in set(text.split()):
if len(word) > best_len:
best_len = len(word)
best_word = word
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best_word
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idx = text.rfind(best_word,)
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idx, best_len
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text[idx:(idx+best_len)]
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