Test 1 Solutions CompSci 6 Fall 2011
1. A                  1. B.
22                    o
22                    os
2                     try
3.33333333333         chickory
True                  ['chickory', 'parsley']

2A. def areaEquilateralTriangle(side):
        return (math.sqrt(3)/4.0) * (side**2)

ALTERNATE SOLUTION:
def areaEquilateralTriangle(side):
    return (3**(0.5)/4.0) * (side*side)

2B. def whoseNightToCook(day, month):
        if day %2 == 0:  # even day
            return "Ellen"
        if day == 31 and month % 2 == 0:
            return "Ellen"
        return "Oscar"

ALTERNATE SOLUTION:
    whoseNightToCook(day, month):
    if day%2 == 0 or (day == 31 and month %2 == 0):
        return "Ellen"
    else 
        return "Oscar"

3. A1. [3, 5, 5, 7, 4]
A2. 7

A3. ['chicken']

A4. 'chicken'

B1. 'rhino'

B2. Mystery returns the longest string. If there are multiple strings of
that length, it returns the first such.

B3. amount = max([len(w) for w in animals])

B4. Mystery still returns the longest string. If there are multiple strings
of that length, now it returns the last such. 

4.A.def getAges(data):
       answer = []
       for w in data:
           splitLine = w.split(':')
           numberStr = splitLine[-1]
           answer.append(int(numberStr))
       return answer

ALTERNATE SOLUTION:
def getAges(data):
    answer = []
    # each string in data is replaced by a list of three strings
    for (index, value) in enumerate(data):
        data[index] = info.split(":")
    for info in data:  # for each list in data
        answer.append(int(info[2]))
    return answer

ALTERNATE SOLUTION:
def getAges(data):
    return [int(w.split(':')[-1]) for w in data]

ALTERNATE SOLUTION:
def getAges(data):
    templist = []
    for w in data:
        templist.append(data[(w.find(':')+1):])
    ages = []
    for v in ages:
        ages.append(int(templist[(v.find(':')+1):]))
    return ages

ALTERNATE SOLUTION: (same as above but with list comprehensions)
def getAges(data):
    templist = [data[(w.find(':')+1):] for w in data]
    ages = [int(templist[(v.find(':')+1):] for v in templist]
    return ages

B. def howManyInRange(values, start, end):
       agelist = getAges(values)
       int 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])

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