small modification to 9.1

epequeno · Dec 28, 2012 · 4f53aa5 · 4f53aa5
1 parent c67b5a9
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diff --git a/ch06/6.01.py b/ch06/6.01.py
@@ -3,12 +3,13 @@
 
 # Current Status: Complete
 
+
 def compare(x, y):
     if x > y:
         return 1
     elif x == y:
         return 0
     else:
         return -1
-        
+
 print compare(2, 2)
diff --git a/ch06/6.02.py b/ch06/6.02.py
@@ -7,9 +7,10 @@
 
 import math
 
+
 def hypotenuse(x, y):
-    return math.sqrt(x**2 + y**2)
-    
+    return math.sqrt(x ** 2 + y ** 2)
+
 print hypotenuse(3, 4)
 
 """pretty much just wrote this out with one try (coming back to this problem

diff --git a/ch06/6.03.py b/ch06/6.03.py
@@ -1,4 +1,4 @@
-# Exercise 6.3 Write a function is_between(x, y, z) that returns True 
+# Exercise 6.3 Write a function is_between(x, y, z) that returns True
 # if x <= y <= z or False otherwise.
 
 # Current Status = Complete
@@ -7,6 +7,7 @@
 y = int(raw_input('y?\n'))
 z = int(raw_input('z?\n'))
 
+
 def is_between(x, y, z):
     return x <= y and y <= z
 

diff --git a/ch06/6.04.py b/ch06/6.04.py
@@ -1,25 +1,28 @@
-# Exercise 6.4 Draw a stack diagram for the following program. What does the 
+# Exercise 6.4 Draw a stack diagram for the following program. What does the
 # program print?
 
 # Current Status: Complete
 
+
 def b(z):
     prod = a(z, z)
     print z, prod
     return prod
-
+
+
 def a(x, y):
     x = x + 1
     return x * y
 
+
 def c(x, y, z):
     sum = x + y + z
-    pow = b(sum)**2
+    pow = b(sum) ** 2
     return pow
 
 x = 1
 y = x + 1
-print c(x, y+3, x+y)
+print c(x, y + 3, x + y)
 
 # The program prints 8100. c() calls b() with the sum of 1, 5, 3 which is 9.
 # b() then calls a(9, 9). The first 9 has 1 added to it so a returns

diff --git a/ch06/6.05.py b/ch06/6.05.py
@@ -1,19 +1,20 @@
 # Exercise 6.5 The Ackermann function, A(m, n), is defined:
 #    [picture]
-# Write a function named ack that evaluates Ackerman’s function. Use your 
+# Write a function named ack that evaluates Ackerman’s function. Use your
 # function to evaluate ack(3, 4), which should be 125. What happens for larger
 # values of m and n?
 
 # Current Status = Complete
 
-m = 3 
+m = 3
 n = 4
 
+
 def ack(m, n):
-   if m == 0:
-       return n + 1
-   elif n == 0:
-       return ack(m - 1, 1)
-   else:
-       return ack(m - 1, ack(m, n - 1))
+    if m == 0:
+        return n + 1
+    elif n == 0:
+        return ack(m - 1, 1)
+    else:
+        return ack(m - 1, ack(m, n - 1))
 ack(m, n)
diff --git a/ch06/6.06.py b/ch06/6.06.py
@@ -3,17 +3,22 @@
 # if the first and last letters are the same and the middle is a palindrome.
 # The following are functions that take a string argument and return the
 # first, last, and middle letters:
-
+
+
 def first(word):
     return word[0]
+
+
 def last(word):
     return word[-1]
+
+
 def middle(word):
     return word[1:-1]
-    
+
 # We'all see how they work in Chapter 8.
 # 1. Type these functions into a file named palindrome.py and test them out.
-# What happens if you call middle with a string with two letters? One letter? 
+# What happens if you call middle with a string with two letters? One letter?
 # What about the empty string, which is written '' and contains no letters?
 # 2. Write a function called is_palindrome that takes a string argument and
 # returns True if it is a palindrome and False otherwise. Remember that you
@@ -23,12 +28,13 @@ def middle(word):
 
 word = str(raw_input('Want to see if it is a palindrome?\n'))
 
+
 def is_palindrome(word):
-   if len(word) <= 2 and word[0] == word[-1]:
-       print 'True'
-   elif word[0] == word[-1]:
-       is_palindrome(word[1:-1])
-   else:
-       print 'False'
+    if len(word) <= 2 and word[0] == word[-1]:
+        print 'True'
+    elif word[0] == word[-1]:
+        is_palindrome(word[1:-1])
+    else:
+        print 'False'
 
 is_palindrome(word)
diff --git a/ch06/6.07.py b/ch06/6.07.py
@@ -5,9 +5,10 @@
 # Current Status: Complete
 
 # I really couldn't figure this one out for the life of me so this solution
-# is stolen directly from 
+# is stolen directly from
 # http://stackoverflow.com/questions/4429462/python-recursion-exercise
 
+
 def is_power(a,b):
     if(a%b != 0):
         return False

diff --git a/ch06/6.08.py b/ch06/6.08.py
@@ -1,5 +1,5 @@
 # Exercise 6.8 The greatest common divisor (GCD) of a and b is the largest
-# number that divides both of them with no remainder 
+# number that divides both of them with no remainder
 # One way to find the GCD of two numbers is Euclid's algorithm, which
 # is based on the observation that if r is the remainder when a is divided by
 # b, then gcd(a, b) = gcd(b, r). As a base case, we can consider gcd(a, 0) = a.
@@ -9,12 +9,13 @@
 
 # Current Status: Incomplete
 
+
 def gcd(a, b):
     if a == 0:
         return b
     elif b == 0:
         return a
     else:
         return gcd(b, a % b)
-        
+
 print gcd(1989, 867)
diff --git a/ch07/7.01.py b/ch07/7.01.py
@@ -17,6 +17,7 @@
 
 # Current Status: Complete
 
+
 def print_n(s, n):
     i = 0
     while i < n:

diff --git a/ch07/7.02.py b/ch07/7.02.py
@@ -6,14 +6,15 @@
 
 n = raw_input('Square root of what?\n')
 
+
 def square_root(n):
-   n = float(n)   
-   x = n / 2
-   i = 0
-   while i < 10:
-       y = (x + n / x) / 2
-       x = y
-       i += 1
-   return x
+    n = float(n)
+    x = n / 2
+    i = 0
+    while i < 10:
+        y = (x + n / x) / 2
+        x = y
+        i += 1
+    return x
 
-print square_root(n)
+print square_root(n)
diff --git a/ch07/7.03.py b/ch07/7.03.py
@@ -11,33 +11,37 @@
 
 import math
 
+
 def newtons(n):
-   n = float(n) # convert input to float so printout() doesn't have to
-   x = n / 2 # rough estimate
-   i = 0
-   while i < 10:
-       y = (x + n / x) / 2 # newtons method
-       x = y
-       i += 1
-   return y
+    n = float(n)  # convert input to float so printout() doesn't have to
+    x = n / 2  # rough estimate
+    i = 0
+    while i < 10:
+        y = (x + n / x) / 2  # newtons method
+        x = y
+        i += 1
+    return y
+
 
 def libmath(n):
-   n = float(n) 
-   return math.sqrt(n)
+    n = float(n)
+    return math.sqrt(n)
 
-# this function has a mix of int, str and float so there is a bit of 
+# this function has a mix of int, str and float so there is a bit of
 # conversion going on.
 
+
 def printout():
-   for i in range(1, 10): 
-       n = str(newtons(i)) # newtowns() gets int and returns float. change to str.
-       l = str(libmath(i))  # same here
-       ab = abs(newtons(i) - libmath(i)) # out as int in as float, no str
-       if (len(n) or len(l)) == 3:
-           print i, n, '         ', l, '          ', ab
-       elif len(n) == 12:
-           print i, n, '', l, ' ', ab
-       else:
-           print i, n, l, '', ab
+    for i in range(1, 10):
+        n = str(newtons(i))  # newtowns() gets int and returns float. change
+        # to str.
+        l = str(libmath(i))  # same here
+        ab = abs(newtons(i) - libmath(i))  # out as int in as float, no str
+        if (len(n) or len(l)) == 3:
+            print i, n, '         ', l, '          ', ab
+        elif len(n) == 12:
+            print i, n, '', l, ' ', ab
+        else:
+            print i, n, l, '', ab
 
 printout()
diff --git a/ch07/7.04.py b/ch07/7.04.py
@@ -5,7 +5,7 @@
 @author: steven
 """
 
-# The built-in function eval takes a string and evaluates it using the 
+# The built-in function eval takes a string and evaluates it using the
 # Python interpreter.
 # For example:
 # >>> eval('1 + 2 * 3')
@@ -23,6 +23,7 @@
 
 # Current Status: Complete
 
+
 def eval_loop():
     while True:
         n = raw_input('Input?\n:: ')
@@ -32,5 +33,5 @@ def eval_loop():
             result = eval(n)
             print result
     print result
-        
+
 eval_loop()
diff --git a/ch07/7.05.py b/ch07/7.05.py
@@ -5,27 +5,28 @@
 @author: steven
 """
 
-# The brilliant mathematician Srinivasa Ramanujan found an infinite series 
+# The brilliant mathematician Srinivasa Ramanujan found an infinite series
 # that can be used to generate a numerical approximation of pi:
 # http://en.wikipedia.org/wiki/Srinivasa_Ramanujan#Mathematical_achievements
 #
-# Write a function called estimate_pi that uses this formula to compute and 
+# Write a function called estimate_pi that uses this formula to compute and
 # return an estimate of pi. It should use a while loop to compute terms of
-# the summation until the last term is smaller than 1e-15 (which is Python 
+# the summation until the last term is smaller than 1e-15 (which is Python
 # notation for 10^−15 ). You can check the result by comparing it to math.pi.
 # You can see my solution at thinkpython.com/code/pi.py.
 
 # Current Status: Complete
 
 import math
 
+
 def estimate_pi():
     k = 0.0
     last_term = 1.0
     sigma = 0
     while last_term > 1e-15:
         last_term = ((math.factorial(4.0 * k)) * (1103.0 + 26390.0 * (k))) \
-        / ((math.factorial(k)**4.0) * (396.0**(4.0 * k)))
+        / ((math.factorial(k) ** 4.0) * (396.0 ** (4.0 * k)))
         k += 1.0
         sigma += last_term
     result = ((2 * math.sqrt(2)) / 9801) * sigma

diff --git a/ch08/8.05.py b/ch08/8.05.py
@@ -10,14 +10,15 @@
 
 # Current Status: Complete
 
-word = 'banana'
-target = 'a'
+myword = 'banana'
+mytarget = 'a'
+
 
 def count(word, target):
-    count = 0
+    i = 0
     for letter in word:
         if letter == target:
-            count += 1
-    return count
+            i += 1
+    return i
 
-print count(word, target)
+print count(myword, mytarget)
diff --git a/ch09/9.01.py b/ch09/9.01.py
@@ -1,10 +1,11 @@
-# Write a program that reads words.txt and prints only the words with 
+# Write a program that reads words.txt and prints only the words with
 # more than 20 characters (not counting whitespace).
 
 # Current Status: Complete
 
 wordList = open('words.txt')
 
+
 def words(word):
     wordCount = 0
     lineCount = 0
@@ -13,7 +14,7 @@ def words(word):
             print word.rstrip('\r\n')
             wordCount += 1
         lineCount += 1
-    percent  = (float(wordCount) / float(lineCount)) * 100.0
-    print "%0.3f" % percent + "%"
+    percent = (float(wordCount) / float(lineCount)) * 100.0
+    print "%0.3f%%" % percent
 
-words(wordList)
+words(wordList)
-Original file line number
+Diff line change
@@ Expand Up / @@ -17,6 +17,7 @@ @@
     # Current Status: Complete
     def print_n(s, n):
         i = 0
         while i < n:
@@ Expand Down @@