Solutions to Test 1 Practice CompSci 101 Spring 2013
Feb 13, 2013

Problem 1:

27/15, int, 1
32%11, int, 10
sum(range(6)), int, 15
len("dog") < len("catsup"), boolean, true
"1,234,567".split(",") list, ["1", "234", "567"]
"platform"[-2], string, "r"
"beat"+"nick", string, "beatnick"
"snapdragon"[3:8] string, "pdrag"
0.23*10, float 2.3
[8,7,6,7,6,5,4,3,2,1][5:7] list [5,4]
2**4 int 16

Part B:
def rorhrer(weight,height):
    return 1.0*weight*2768/(height**3)

Part C:
def triangle_area(a,b,c):
    s = (a+b+c)/2.0
    area = math.sqrt(s*(s-a)*(s-b)*(s-c))
    return area

Problem 2: Part A
def divisor_sum(n):
    return sum([k for k in range(1,n) if n % k == 0])

or
    s = 0
    for k in range(n):
        if n % k == 0:
            s = s + k
    return s

PART B:
def is_amicable(x,y):
    return divisor_sum(x) == y and divisor_sum(y) == x

PART C:
def nearby(city,clist,apart):
    ret = []
    for elt in clist:
         if distance(elt[1],elt[2],city[1],city[2]) <= apart:
             ret.append(elt[0])
    return ret

OR

return [elt[0] for elt in clist if 
        distance(elt[1],elt[2],city[1],city[2]) <= apart]

*******
Problem 3

Part A

The one test that doesn't pass is the third, when "big" is b then
the expression 

b in ["bad","news", "is", "trouble"]

evaluates to true so "clean" is returned, the word "bad" is never
checked.

Move the else: outside the loop, e.g.,

def censored(bad,headline):
    words = headline.split()
    for b in bad:
        if b in bad:
            return "censored"
    return "clean"

******
Part C:

index+2 skips the comma and space to get the first name
the slice uses index to get everything up to and NOT including the comma


Part D:

.find() returns -1 when the string isn't found, so -1+2 = 1 and the
slice gets "ob jones". The value of the slice "bob jones"[:-1] is
everything upto and not include the last character or "bob jone"

******

Problem 4 Part B  

B. def howManyInRange(values, start, end):
       agelist = getAges(values)
       count = 0
       for age in agelist:
          if start <= age <= end:
               count += 1
       return count

ALTERNATE SOLUTION:
def howManyInRange(values, start, end):    
    agelist = getAges(values)
    ages = []
    for age in agelist:
        if start <= age <= end:
            ages.append(age)
    return len(ages)

ALTERNATE SOLUTION:
def howManyInRange(values, start, end):
    values = getAges(values)
    return sum([1 for num in values if num >= start and num <= end])

ALTERNATE SOLUTION:
def howManyInRange(values, start, end):
    return len([num for num in getAges(values) if num >= start and num <= end])

*******
Problem 5

5. def convertWord(word):
       if word.lower() == "hello":
           return "Ahoy"
       pos = word.find("ar")
       if (pos > 0):
           return word[:pos] + 'arrr' + word[pos+2:]
       if (len(word)> 7) and pos == -1:
           ans = ""
           for ch in word:
               if ch != 'o' and ch != 'u':
                   ans += ch
           return ans   
       return word

ALTERNATE SOLUTION:
def convertWord(word):
    if word == "hello" or word == 'Hello':
        return "Ahoy"
    if word.startsWith("ar")
        return word
    pos = word.find("ar")
    if (pos > 0):
        return word[:pos+2] + 'rr' + word[pos+2:]
    if (len(word)> 7):
        word = word.replace("o","")
        word = word.replace("u","")
        return word
    return word

******

Problem 6

Part A

def isPerfect(num):
    sum = 0
    for div in range(1,num):
        if num % div == 0:
            sum += div
    return sum == num

alternatively, replace the last line/return statement with

    if sum == num:
        return True
    return False

----
Alternatively use a list comprehension

  def isPerfect(num):
      return sum([div for div in range(1,num) if num % div == 0]) == num

---
PART B

def combine(phrase):
   st = ""
   for w in phrase:
       st += w[:firstVowelIndex(w)+2]
   return st
---

********
Problem 7
Part A

def isSquareFree(num):

    for x in range(1,num):
        sq = x*x
        if num % sq == 0:
            return False
    return True


Part B

def wordCount(words,sub):
    """
    return the number of strings in string list words
    that contain string sub
    """
    c = 0
    for w in words:
        if sub in w:
            c = c + 1
    return c

--
or 

def wordCount(words,sub):
    return sum([1 for w in words if w.find(sub) >= 0])


******
Problem 8

Question 1:
'dog', no ie and no ei found

Question 2:
line 8 since eii isn't 0 and 'c' occurs before the 'ei'

Question 3:
line 6 for similar reasons

Question 4:

control falls through to the end since all the if expressions are false
so False is returned as no other return statements execute

Question 5:

Start code with:

if iei == 0: return False
if eii == 0: return True