• Same as usual:
  • This quiz is SOLO.
  • This portion is worth 80 pts.  The written portion tomorrow will be worth 20 pts.
  • No notes, books, web sites (except reading this quiz and accessing Autolab).
  • IDLE hint:  python -m idlelib.idle &
  • You must submit your quiz5.py file to Autolab before leaving the lab!  You will not have access to this file after you leave the lab!
  • To start, download this one file: quiz5.py
• Different from usual:
  • The private autograder you have is "very nearly" the final version.  We will keep the same tests worth the same points, but the constants will be changed when we autograde after the quiz so you cannot just hardcode the answers.
  • You may use up to 5 submissions during the quiz, though (as always) your last submission is the only one that counts.  Obviously, you may not submit nor resubmit after you leave the lab (do not attempt to do so, even if Autolab for some reason allows it).
• Good luck! :-)

1. leastFrequentLetters [40 pts]
2. getCaesarShift [40 pts]

1. leastFrequentLetters [40 pts]
   Write the function leastFrequentLetters(s) that takes a string s which may contain arbitrary characters and returns a lowercase string containing all the alphabetic letters that occur least frequently in s, in alphabetic order.  The test is case insensitive.  If s contains no letters, return the empty string.  To further help you understand the task, here is a sample test function (also in quiz5.py):

   def testLeastFrequentLetters():
       print "Testing leastFrequentLetters()...",
       assert(leastFrequentLetters("abc") == "abc")
       assert(leastFrequentLetters("cba") == "abc")
       assert(leastFrequentLetters("abcab") == "c")
       assert(leastFrequentLetters("abca") == "bc")
       assert(leastFrequentLetters("aBcA") == "bc")
       assert(leastFrequentLetters("") == "")
       assert(leastFrequentLetters("^%#$*&!@#*&#!@#") == "")
       print "Passed!"

2. getCaesarShift [40 pts]
   Background:  in a Caesar cipher, we start with a "shift", which here we will assume is an integer between 0 and 25 inclusive, and each letter in a word is shifted by that amount.  For example, if the shift is 2, then "abc" is shifted to become "cde" ("a" is shifted to "c", "b" is shifted to "d", and "c" is shifted to "e").  The shift wraps around as needed, so if the shift is 2, then "xyz" becomes "zab".
   
   With this in mind, write the function getCaesarShift(word1, word2) that takes two words, both guaranteed to only contain lowercase letters, and returns the shift required to change word1 into word2, or -1 if this is not possible.  If the words are the same, return 0.  To further help you understand the task, here is a sample test function (also in quiz5.py):

   def testGetCaesarShift():
       print "Testing getCaesarShift()...",
       assert(getCaesarShift("abc", "abc") == 0)
       assert(getCaesarShift("abc", "bcd") == 1)
       assert(getCaesarShift("abc", "zab") == 25)
       assert(getCaesarShift("zab", "abc") == 1)
       assert(getCaesarShift("zab", "abd") == -1)
       assert(getCaesarShift("zab", "abcd") == -1)
       assert(getCaesarShift("", "") == 0)
       print "Passed!"

15-112 Fall 2012 Quiz 5b (non-autograded)

* 10 Minutes.  No calculators, no notes, no books, no computers.
* 20 pts (4 problems, 5 pts each; 5 pts bonus).  (The other 80 points are on the autograded quiz5a)

# What will each of the following print?
def f1(s):
    t = s.replace("a","")
    for c in s:
        if (c in t):
            t += c.upper()
    return t
print f1("abAB")

def f2(x):
    s = ""
    for n in xrange(15,15+x):
        s += str(n)
    return s[3:6]*2
print f2(5)

def f3(s, t):
    u = ""
    for i in xrange(min(len(s), len(t))):
        if (s[i] == t[len(t)-1-i]):
            u += s[i]
    return u
print f3("a1bc2de", "ad2ab3e")

def f4(m, n):
    s = ""
    while (m > 0):
        for x in xrange(n-m):
            s += "."
        s += "\n"
        m -= 1
    return s
print f4(3, 5)

def f5bonus():
    x = count = 0
    while (count < 7):
        (c,d,e) = (0,x*x**2*x,str(x*x**2*x))
        while (d): (c,d) = (c+1,d/10)
        (x,count) = (x+1,count+(c == 5))
    while (str(c)*len(e) != e): e = str(int(e)-1)
    return e
print f5bonus()
 

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