FlareBlog/content/en/posts/csci-1100/exam-3-overview/index.md

---
title: CSCI 1100 - Test 3 Overview and Practice Questions
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date: 2024-04-03T03:54:07-04:00
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description: This blog post provides an overview of Test 3 for CSCI 1100 - Computer Science 1, including important logistical instructions, topics covered, and practice questions on sets, dictionaries, classes, and file I/O in Python
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summary: This blog post provides an overview of Test 3 for CSCI 1100 - Computer Science 1, including important logistical instructions, topics covered, and practice questions on sets, dictionaries, classes, and file I/O in Python
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# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
---

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## Important Logistical Instructions:

- Test 3 will be held Thursday, April 4, 2024.
- Most students will take the exam from 6:00 - 7:30 pm (90 minutes).
- Students who provided us with an accommodation letter indicating the need for extra time or a quiet location will be given extra time beyond 7:30.
- Room assignments will be posted on Submitty by the Wednesday night before the exam, April 3rd.
- Students MUST:
  - Go to their assigned rooms.
  - Bring their IDs to the exam.
  - Sit in the correct section.
  - Put away all calculators, phones, etc. and take off/out all headphones and earbuds
  - Hand over their tests as soon as the time is up. Those found working on the test after 90 minutes will receive a zero.

Failing to do one of these may result in a 20 point penalty on the exam score. Failure to do all can cost up to 80 points.

- You cannot leave the exam room (not even for a bathroom break) until you hand over your exam.
- Similar to exam 1 and 2, a one-page crib-sheet is allowed during the test.

## Overview

- Primary coverage is Lectures 14-19, Labs 7-9, HW 5-7.
- Please review lecture notes, class exercises, labs, homework, practice programs, and tests, working through problems on your own before looking at the solutions.
- Some problems will be related to material covered in Exam 2:
  - Lists and files
  - List and string splitting; ranges
- You should review relevant material from the Exam 2 practice problems if you are not yet comfortable with these topics.
- No calculators, no textbook, no classnotes, no electronics of any kind! BUT, you may bring a one-page, double-sided, 8.5” x 11” “crib sheet” sheet with you. You may prepare this as you wish, and you may work in groups to prepare a common crib sheet. Of course, each of you must have your own copy during the test. You will need to turn in a signed copy of your crib sheet at the end of the test. If you need to keep a copy for your records, please photocopy it prior to coming to the exam.
- Please refer back to the Test 1 and Test 2 practice problems for further instructions.

## Questions

### Not Used

> Write a function called `notused` that takes a list of words as its single parameter, and returns a set containing the letters of the English alphabet that are not used by any of the words in the input list. Your function must use sets. Here is an example of how your function should work:
>
> ```python
> >>> notused([ "Dog", "pony", "elephant", "Tiger", "onyx", "Zebu" ])
> {'c', 'f', 'k', 'j', 'm', 'q', 's', 'w', 'v'}
> ```
>
> Hint: you can use the following set in your solution:
>
> ```python 
> all_letters = set("abcdefghijklmnopqrstuvwxyz")
> ```

{{< details summary="**Answer of This Question**" >}}
Here's a Python function `notused` that takes a list of words and returns a set containing the letters of the English alphabet that are not used by any of the words in the input list:

```python
def notused(words):
    all_letters = set("abcdefghijklmnopqrstuvwxyz")
    used_letters = set()

    for word in words:
        used_letters |= set(word.lower())

    return all_letters - used_letters
```

Explanation:

1. We define the function `notused` that takes a list of words as its parameter.

2. We create a set called `all_letters` that contains all the lowercase letters of the English alphabet using `set("abcdefghijklmnopqrstuvwxyz")`.

3. We initialize an empty set called `used_letters` to store the letters used by the words in the input list.

4. We start a loop that iterates over each word in the `words` list.

5. For each word, we convert it to lowercase using `word.lower()` to ensure case-insensitivity. Then, we convert the lowercase word into a set of its individual letters using `set(word.lower())`.

6. We update the `used_letters` set by taking the union of itself with the set of letters from the current word using the `|=` operator. This adds any new letters from the current word to the `used_letters` set.

7. After the loop, `used_letters` will contain all the letters used by the words in the input list.

8. Finally, we return the set difference between `all_letters` and `used_letters` using `all_letters - used_letters`. This gives us the set of letters from the English alphabet that are not used by any of the words in the input list.

Here's an example of how the function works:

```python
print(notused(["Dog", "pony", "elephant", "Tiger", "onyx", "Zebu"]))
# Output: {'c', 'f', 'k', 'j', 'm', 'q', 's', 'w', 'v'}
```

In this example, the function `notused` is called with a list of words. It returns a set containing the letters 'c', 'f', 'k', 'j', 'm', 'q', 's', 'w', and 'v', which are the letters of the English alphabet that are not used by any of the words in the input list.

The solution uses sets to efficiently handle the comparison and difference operations between the sets of letters.
{{< /details >}}

### Set Intersection

> Given three sets, `s1`, `s2`, and `s3`, write a short segment of Python code to find the values that are in exactly one of the three sets. The result should be stored in a set called `s`. You may NOT use any loops or ifs.

{{< details summary="**Answer of This Question**" >}}
To find the values that are in exactly one of the three sets `s1`, `s2`, and `s3` without using any loops or conditionals, you can use set operations in Python. Here's the code segment:

```python
s = (s1 ^ s2 ^ s3) - (s1 & s2) - (s1 & s3) - (s2 & s3)
# Or this one
#s = (s1 - s2 - s3) | (s2 - s1 - s3) | (s3 - s1 - s2)
```

Explanation:

1. The `^` operator performs the symmetric difference operation between sets. It returns a new set that contains elements that are in either of the sets but not in both.
   - `s1 ^ s2 ^ s3` gives us a set containing elements that are in exactly one or all three sets.

2. To remove the elements that are in more than one set, we need to subtract the intersections of each pair of sets.
   - `s1 & s2` gives us the elements that are common to both `s1` and `s2`.
   - `s1 & s3` gives us the elements that are common to both `s1` and `s3`.
   - `s2 & s3` gives us the elements that are common to both `s2` and `s3`.

3. By subtracting the intersections `(s1 & s2)`, `(s1 & s3)`, and `(s2 & s3)` from the symmetric difference `(s1 ^ s2 ^ s3)`, we remove the elements that are in more than one set.

4. The resulting set `s` will contain only the elements that are in exactly one of the three sets `s1`, `s2`, and `s3`.

This code segment achieves the desired result without using any loops or conditionals, utilizing only set operations.
{{< /details >}}

### Words in All Strings

> Given three strings of words, with each word separated by a space, write code to output the number of words that appear in all three strings. Assume the strings are associated with the variables `w1`, `w2` and `w3`.
>
> For 
> ```python
> w1 = "the quick brown fox jumps over the lazy dog"
> w2 = "hey diddle diddle the cat and the fiddle the cow jumps over the moon"  
> w3 = "jack and jill went over the hill to fetch a pail of water"
> ```
> The output should be `2` because `the` and `over` appear in all three. No loops are allowed. You can solve this in one (long) line of code. In this case, it is acceptable to use more than one line, but make sure you understand the one-line solution when you see it.

{{< details summary="**Answer of This Question**" >}}
To find the number of words that appear in all three strings without using loops, you can use the `set` data structure and the `intersection` method. Here's the solution:

```python
print(len(set(w1.split()) & set(w2.split()) & set(w3.split())))
```

Explanation:

1. We use the `split()` method on each string (`w1`, `w2`, `w3`) to split them into lists of words. By default, `split()` splits the string on whitespace, which separates the words.

2. We convert each list of words into a set using `set()`. A set is an unordered collection of unique elements, so it eliminates any duplicate words within each string.

3. We use the `&` operator (set intersection) to find the common elements between the sets. The intersection of two sets returns a new set containing only the elements that are present in both sets.

4. We chain the `&` operator to find the intersection of all three sets, which gives us a set of words that appear in all three strings.

5. Finally, we use the `len()` function to count the number of elements in the resulting set, which represents the number of words that appear in all three strings.

The one-line solution combines all these steps into a single line of code, making it concise but potentially less readable. It's important to understand the individual steps and the concept of set intersection to comprehend the one-line solution.

You can test the code with the provided example:

```python
w1 = "the quick brown fox jumps over the lazy dog"
w2 = "hey diddle diddle the cat and the fiddle the cow jumps over the moon"  
w3 = "jack and jill went over the hill to fetch a pail of water"

print(len(set(w1.split()) & set(w2.split()) & set(w3.split())))  # Output: 2
```

The output will be `2` because the words "the" and "over" appear in all three strings.
{{< /details >}}

### Set Operations Output

> What is the output when the following code is run by Python? For sets, do not worry about getting the exact order of the output correct.
> ```python
> s1 = set([7, 9, 12, 7, 9] )
> s2 = set(['abc', 12, 'b', 'car', 7, 10, 12 ])
> s3 = set([12, 14, 12, 'ab'])
> print(s1 & s2)
> print(s1 | s2)
> print('b' in s2)
> print('ab' in s2)
> print('ab' in s3)
> s2.discard(12)
> print((s1 & s2) ^ s3)
> ```
> Of course, you can make up many other questions about set operations.

{{< details summary="**Answer of This Question**" >}}
The output of the given code will be:

```text
{12, 7}
{'b', 7, 9, 10, 12, 'car', 'abc'}
True
False
True
{12, 7, 14, 'ab'}
```

Explanation:

1. `s1 = set([7, 9, 12, 7, 9])`: This creates a set `s1` with elements 7, 9, and 12. Duplicates are automatically removed.

2. `s2 = set(['abc', 12, 'b', 'car', 7, 10, 12])`: This creates a set `s2` with elements 'abc', 12, 'b', 'car', 7, and 10. Again, duplicates are removed.

3. `s3 = set([12, 14, 12, 'ab'])`: This creates a set `s3` with elements 12, 14, and 'ab'.

4. `print(s1 & s2)`: This performs the intersection operation between sets `s1` and `s2`, which returns a new set containing the common elements. The output will be `{12, 7}`.

5. `print(s1 | s2)`: This performs the union operation between sets `s1` and `s2`, which returns a new set containing all the elements from both sets. The output will be `{'b', 7, 9, 10, 12, 'car', 'abc'}`.

6. `print('b' in s2)`: This checks if the element 'b' is present in set `s2`. It will output `True` since 'b' is in `s2`.

7. `print('ab' in s2)`: This checks if the element 'ab' is present in set `s2`. It will output `False` since 'ab' is not in `s2`.

8. `print('ab' in s3)`: This checks if the element 'ab' is present in set `s3`. It will output `True` since 'ab' is in `s3`.

9. `s2.discard(12)`: This removes the element 12 from set `s2` if it exists. After this operation, `s2` will be `{'abc', 'b', 'car', 7, 10}`.

10. `print((s1 & s2) ^ s3)`: This performs the following operations:
    - `s1 & s2` calculates the intersection of sets `s1` and `s2`, which is `{7}`.
    - `(s1 & s2) ^ s3` performs the symmetric difference operation between the result of `s1 & s2` and set `s3`. It returns a new set containing elements that are in either `(s1 & s2)` or `s3`, but not in both. The output will be `{12, 7, 14, 'ab'}`.

The order of elements in the output sets may vary since sets are unordered collections. The actual output you provided matches the expected output.
{{< /details >}}

### Restaurant Reviews

> You are given a dictionary containing reviews of restaurants. Each key is the name of the restaurant. Each item in the dictionary is a list of reviews. Each review is a single string. See the example below.
> ```python
> rest_reviews = {"DeFazio's":["Great pizza", "Best in upstate"], \
>   "I Love NY Pizza":["Great delivery service"], \
>   "Greasy Cheese": [ "Awful stuff", "Everything was terrible" ] }
> ```
> Assuming `rest_reviews` has already been created, solve the following problems.
>
> (a) Write code to find all restaurants where the review contains at least one of the following words: awful, terrible, dump. For each restaurant found, output the name of the restaurant and the number of reviews that have at least one of the words. Be careful about capitalization. 'Awful' and 'awful' should match.
>
> (b) Write code to find and print the name of the restaurant with the highest number of reviews. If there is more than one restaurant with the same number of reviews, print the names of each of these restaurants.
>  
> (c) Write a function that takes as arguments the review dictionary, a new review, and the name of a restaurant. The function should add the review to the dictionary. If the restaurant is already in the dictionary, the function should add the review to the existing list of reviews for that restaurant. If the restaurant is not in the dictionary, the function should add a new item to the dictionary. Your function should be called by `add_review(rest_reviews, new_review, rest_name)`
>
> (d) Write a function that takes the same arguments as `add_review`, but deletes the given review. Specifically, if the review is in the dictionary associated with the restaurant, the function should delete the review and return True. Otherwise, the function should return False. If the given restaurant is not in the dictionary, the function should also return False. The function should be called by `del_review(rest_reviews, old_review, rest_name)`


{{< details summary="**Answer of Part A**" >}}
```python
for restaurant, reviews in rest_reviews.items():
    count = 0
    for review in reviews:
        if any(word in review.lower() for word in ["awful", "terrible", "dump"]):
            count += 1
    if count > 0:
        print(f"{restaurant}: {count} review(s) containing the specified words")
```

Explanation:
- We iterate over each restaurant and its reviews in the `rest_reviews` dictionary.
- For each restaurant, we initialize a `count` variable to keep track of the number of reviews containing the specified words.
- We iterate over each review and check if any of the specified words ("awful", "terrible", "dump") are present in the review (case-insensitive).
- If a review contains any of the specified words, we increment the `count`.
- After checking all reviews for a restaurant, if the `count` is greater than 0, we print the restaurant name and the count of reviews containing the specified words.
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
```python
max_reviews = 0
restaurants_with_max_reviews = []

for restaurant, reviews in rest_reviews.items():
    num_reviews = len(reviews)
    if num_reviews > max_reviews:
        max_reviews = num_reviews
        restaurants_with_max_reviews = [restaurant]
    elif num_reviews == max_reviews:
        restaurants_with_max_reviews.append(restaurant)

print("Restaurant(s) with the highest number of reviews:")
for restaurant in restaurants_with_max_reviews:
    print(restaurant)
```

Explanation:
- We initialize `max_reviews` to keep track of the highest number of reviews and `restaurants_with_max_reviews` to store the restaurant(s) with the highest number of reviews.
- We iterate over each restaurant and its reviews in the `rest_reviews` dictionary.
- For each restaurant, we calculate the number of reviews using `len(reviews)`.
- If the number of reviews is greater than the current `max_reviews`, we update `max_reviews` and set `restaurants_with_max_reviews` to a list containing only the current restaurant.
- If the number of reviews is equal to `max_reviews`, we append the current restaurant to `restaurants_with_max_reviews`.
- Finally, we print the restaurant(s) with the highest number of reviews.
{{< /details >}}

{{< details summary="**Answer of Part C**" >}}
```python
def add_review(rest_reviews, new_review, rest_name):
    if rest_name in rest_reviews:
        rest_reviews[rest_name].append(new_review)
    else:
        rest_reviews[rest_name] = [new_review]
```

Explanation:
- The `add_review` function takes the `rest_reviews` dictionary, a `new_review`, and the `rest_name` as arguments.
- If the `rest_name` is already present in the `rest_reviews` dictionary, we append the `new_review` to the existing list of reviews for that restaurant.
- If the `rest_name` is not in the dictionary, we create a new entry in the dictionary with the `rest_name` as the key and a list containing the `new_review` as the value.
{{< /details >}}

{{< details summary="**Answer of Part D**" >}}
```python
def del_review(rest_reviews, old_review, rest_name):
    if rest_name in rest_reviews:
        if old_review in rest_reviews[rest_name]:
            rest_reviews[rest_name].remove(old_review)
            return True
    return False
```

Explanation:
- The `del_review` function takes the `rest_reviews` dictionary, an `old_review`, and the `rest_name` as arguments.
- If the `rest_name` is present in the `rest_reviews` dictionary, we check if the `old_review` is in the list of reviews for that restaurant.
- If the `old_review` is found, we remove it from the list using the `remove` method and return `True` to indicate successful deletion.
- If the `rest_name` is not in the dictionary or the `old_review` is not found, we return `False` to indicate that the deletion was not successful.
{{< /details >}}

### Python Output

> For each of the following sections of code, write the output that Python would generate:
>
> Part A
> ```python
> x = {1:['joe',set(['skiing','reading'])],\
> 2:['jane',set(['hockey'])]}
> x[1][1].add('singing')
> x[1][0] = 'kate'
> for item in sorted(x.keys()):
>     print(x[item][0], len(x[item][1]))
> ```
>
> Part B
> ```python
> y = {'jane':10, 'alice':2, 'bob':8,\
>      'kristin':10}
> m = 0
> for person in sorted(y.keys()):
>     if y[person] > m:
>         print("**", person)
>         m = y[person]
> for person in sorted(y.keys()):
>     if y[person] == m:
>         print("!!", person)
> ```
>
> Part C: Note that this problem requires an understanding of aliasing.
> ```python
> L1 = [0,1,2]
> L2 = ['a','b']
> d = {5:L1, 8:L2}
> L1[2] = 6
> d[8].append('k')
> L2[0] = 'car'
> for k in sorted(d.keys()):
>     print(str(k) + ' ', end='')
>     for v in d[k]:
>         print(str(v) + ' ', end='')
>     print()
> ```
>
> Part D:
> ```python
> L1 = [0,1,2,4,1,0]
> s1 = set(L1)
> L1.pop()
> L1.pop()
> L1.pop()
> L1[0] = 5
> s1.add(6)
> s1.discard(1)
> print(L1)
> for v in sorted(s1):
>     print(v)
> ```

{{< details summary="**Answer of Part A**" >}}
Output:
```
kate 3
jane 1
```

Explanation:
1. The code creates a dictionary `x` with keys 1 and 2. The value for key 1 is a list containing the string 'joe' and a set with elements 'skiing' and 'reading'. The value for key 2 is a list containing the string 'jane' and a set with the element 'hockey'.
2. The line `x[1][1].add('singing')` adds the element 'singing' to the set at index 1 of the list associated with key 1 in the dictionary `x`.
3. The line `x[1][0] = 'kate'` updates the string at index 0 of the list associated with key 1 in the dictionary `x` to 'kate'.
4. The `for` loop iterates over the sorted keys of the dictionary `x`.
5. For each key `item`, it prints the string at index 0 of the corresponding list (`x[item][0]`) and the length of the set at index 1 of the list (`len(x[item][1])`).
6. The output shows that for key 1, the string is 'kate' and the set has 3 elements, and for key 2, the string is 'jane' and the set has 1 element.
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
Output:
```
** alice
** bob
** jane
!! jane
!! kristin
```

Explanation:
1. The code creates a dictionary `y` with keys 'jane', 'alice', 'bob', and 'kristin', and their corresponding values.
2. The variable `m` is initialized to 0.
3. The first `for` loop iterates over the sorted keys of the dictionary `y`.
4. For each `person`, if the value `y[person]` is greater than `m`, it prints `"** " + person` and updates `m` to the value of `y[person]`. This finds the maximum value in the dictionary.
5. The output of the first loop shows that 'alice', 'bob', and 'jane' are printed with `"**"` prefix because their values are greater than the initial value of `m` (which is 0).
6. After the first loop, `m` holds the maximum value found in the dictionary, which is 10.
7. The second `for` loop iterates over the sorted keys of the dictionary `y` again.
8. For each `person`, if the value `y[person]` is equal to `m` (the maximum value), it prints `"!! " + person`.
9. The output of the second loop shows that 'jane' and 'kristin' are printed with `"!!"` prefix because their values are equal to the maximum value `m` (which is 10).

Thank you for pointing out the mistake. I appreciate your attention to detail!
{{< /details >}}

{{< details summary="**Answer of Part C**" >}}
Output:
```
5 0 1 6 
8 car b k 
```

Explanation:
1. The code creates two lists, `L1` and `L2`, and a dictionary `d` with keys 5 and 8. The value for key 5 is `L1`, and the value for key 8 is `L2`.
2. The line `L1[2] = 6` updates the element at index 2 of `L1` to 6.
3. The line `d[8].append('k')` appends the element 'k' to the list `L2`, which is the value for key 8 in the dictionary `d`.
4. The line `L2[0] = 'car'` updates the element at index 0 of `L2` to 'car'.
5. The `for` loop iterates over the sorted keys of the dictionary `d`.
6. For each key `k`, it prints the string representation of `k` followed by a space.
7. The nested `for` loop iterates over the values `v` in the list `d[k]` and prints the string representation of each `v` followed by a space.
8. After each inner loop, it prints a newline character to move to the next line.
9. The output shows that for key 5, the corresponding list is `[0, 1, 6]`, and for key 8, the corresponding list is `['car', 'b', 'k']`.
{{< /details >}}

{{< details summary="**Answer of Part D**" >}}
Output:
```
[5, 1, 2]
0
2
4
6
```

Explanation:
1. The code creates a list `L1` with elements `[0, 1, 2, 4, 1, 0]`.
2. The line `s1 = set(L1)` creates a set `s1` from the elements of `L1`. The set will contain only unique elements from `L1`, which are `{0, 1, 2, 4}`.
3. The lines `L1.pop()`, `L1.pop()`, and `L1.pop()` remove the last three elements from the list `L1`. After these operations, `L1` becomes `[0, 1, 2]`.
4. The line `L1[0] = 5` updates the element at index 0 of `L1` to 5. Now, `L1` becomes `[5, 1, 2]`.
5. The line `s1.add(6)` adds the element 6 to the set `s1`. The set `s1` becomes `{0, 1, 2, 4, 6}`.
6. The line `s1.discard(1)` removes the element 1 from the set `s1`. The set `s1` becomes `{0, 2, 4, 6}`.
7. The line `print(L1)` prints the updated list `L1`, which is `[5, 1, 2]`.
8. The `for` loop iterates over the sorted elements `v` in the set `s1` and prints each element on a new line.
9. The output shows the updated list `L1` and the elements of the set `s1` in sorted order, which are `0`, `2`, `4`, and `6`.

Thank you for bringing this to my attention. I appreciate your careful review of the answers!
{{< /details >}}

### Person Class

> Suppose `Person` is a class that stores for each person their name, birthday, name of their mother and father. All of these are strings. The start of the class, including the initializer, is given below.
> ```python
> class Person(object):
>     def __init__(self, n, bd, m, f):
>         self.name = n
>         self.birthday = bd
>         self.mother = m
>         self.father = f
> ```
> Write a method for the `Person` class that takes as an argument self and another `Person` object and returns `2` if the two people are twins, `1` if they are siblings (but not twins), `-1` if two people are the same, and `0` otherwise. Note that siblings or twins must have the same mother and the same father.

{{< details summary="**Answer of This Question**" >}}
To determine the relationship between two `Person` objects, we can compare their mother, father, and birthday attributes. Here's the method you can add to the `Person` class:

```python
def relationship(self, other):
    if self == other:
        return -1
    elif self.mother == other.mother and self.father == other.father:
        if self.birthday == other.birthday:
            return 2
        else:
            return 1
    else:
        return 0
```

Explanation:

1. The method `relationship` takes two arguments: `self` (the current `Person` object) and `other` (another `Person` object).

2. We first check if `self` and `other` refer to the same `Person` object using the equality operator `==`. If they are the same object, we return -1.

3. If `self` and `other` are not the same object, we check if they have the same mother and father by comparing the `mother` and `father` attributes of both objects. If they have the same mother and father, it means they are either twins or siblings.

4. If `self` and `other` have the same mother and father, we further check their `birthday` attribute:
   - If their birthdays are the same, we return 2 to indicate that they are twins.
   - If their birthdays are different, we return 1 to indicate that they are siblings (but not twins).

5. If `self` and `other` do not have the same mother and father, we return 0 to indicate that they are not related as siblings or twins.

With this method added to the `Person` class, you can create `Person` objects and determine their relationship using the `relationship` method. For example:

```python
person1 = Person("John", "1990-01-01", "Mary", "David")
person2 = Person("Jane", "1990-01-01", "Mary", "David")
person3 = Person("Alice", "1992-05-10", "Mary", "David")
person4 = Person("Bob", "1995-03-15", "Lisa", "Tom")

print(person1.relationship(person2))  # Output: 2 (twins)
print(person1.relationship(person3))  # Output: 1 (siblings)
print(person1.relationship(person4))  # Output: 0 (not related)
print(person1.relationship(person1))  # Output: -1 (same person)
```

The `relationship` method correctly determines the relationship between two `Person` objects based on their mother, father, and birthday attributes.
{{< /details >}}

### Merge Dictionaries

> You are given dictionaries D1 and D2 where each key is a string representing a name and each value is a set of phone numbers. Write a function to merge D1 and D2 into a single dictionary D. D should contain all the information in both D1 and D2. As an example,
> 
> ```python
> D1 = {'Joe':set(['555-1111','555-2222']), 'Jane':set(['555-3333'])}
> D2 = {'Joe':set(['555-2222','555-4444']), 'Kate':set(['555-6666'])}
> merge_dict(D1,D2)
> {'Joe':set(['555-1111','555-2222','555-4444']), 'Jane':set(['555-3333']), 'Kate':set(['555-6666']) }
> ```

{{< details summary="**Answer of This Question**" >}}
Here's a Python function that merges two dictionaries `D1` and `D2` into a single dictionary `D` based on the given requirements:

```python
def merge_dict(D1, D2):
    D = {}
    
    # Merge keys from D1
    for name, numbers in D1.items():
        D[name] = numbers.copy()
    
    # Merge keys from D2
    for name, numbers in D2.items():
        if name in D:
            D[name] |= numbers
        else:
            D[name] = numbers.copy()
    
    return D
```

Explanation:

1. We start by creating an empty dictionary `D` to store the merged result.

2. We iterate over each key-value pair in `D1` using the `items()` method:
   - For each `name` (key) and `numbers` (value) pair, we add the `name` as a key to `D` and create a copy of the `numbers` set using `numbers.copy()`. This ensures that the original sets in `D1` are not modified.

3. Next, we iterate over each key-value pair in `D2`:
   - If the `name` (key) already exists in `D`, we merge the `numbers` set from `D2` with the existing set in `D` using the `|=` operator (set union).
   - If the `name` (key) doesn't exist in `D`, we add it as a new key to `D` and create a copy of the `numbers` set from `D2` using `numbers.copy()`.

4. Finally, we return the merged dictionary `D`.

The function correctly merges the information from both `D1` and `D2` into a single dictionary `D`. If a name exists in both dictionaries, the corresponding phone number sets are merged using set union.

You can test the function with the provided example:

```python
D1 = {'Joe': set(['555-1111', '555-2222']), 'Jane': set(['555-3333'])}
D2 = {'Joe': set(['555-2222', '555-4444']), 'Kate': set(['555-6666'])}
merged_dict = merge_dict(D1, D2)
print(merged_dict)
```

Output:
```
{'Joe': {'555-1111', '555-2222', '555-4444'}, 'Jane': {'555-3333'}, 'Kate': {'555-6666'}}
```

The merged dictionary `D` contains all the information from both `D1` and `D2`, with the phone number sets merged for common names.
{{< /details >}}

### Student Class

> This question involves a class called `Student` that stores the student's name (a string), id number (a string), courses taken (list of strings), and major (a string). Write the Python code that implements this class, including just the following methods:
>
> (a) An initializer having as parameters only the name and the id. This should initialize the list of courses to empty and the major to "Undeclared". An example use of this method would be:
>
> ```python
> p = Student( "Chris Student", "123454321" )
> ```
>
> (b) A method called `add_courses` to add a list of courses to the courses that the student takes. For example, the following should add three courses to Chris Student.
>
> ```python 
> p.add_courses( [ "CSCI1100", "BASK4010", "GEOL1320" ] )
> ```
>
> (c) A method called `common_courses` that returns a list containing the courses two students have taken in common:
>  
> ```python
> q = Student( "Bilbo Baggins", "545454545" )
> q.add_courses( [ "MATH1240", "CSCI1100", "HIST2010", "BASK4010" ] ) 
> print(q.common_courses(p))
> [ "CSCI1100", "BASK4010" ]
> ```

{{< details summary="**Answer of This Question**" >}}
```python
class Student:
    def __init__(self, name, id):
        self.name = name
        self.id = id
        self.courses = []
        self.major = "Undeclared"
    
    def add_courses(self, courses):
        self.courses.extend(courses)
    
    def common_courses(self, other_student):
        return list(set(self.courses) & set(other_student.courses))
```

Explanation:

(a) The `__init__` method is the initializer for the `Student` class. It takes the `name` and `id` as parameters and initializes the `name` and `id` attributes of the `Student` object. It also initializes the `courses` attribute as an empty list and sets the `major` attribute to "Undeclared".

(b) The `add_courses` method takes a list of courses as a parameter and adds them to the `courses` attribute of the `Student` object. It uses the `extend` method of the list to add multiple courses at once.

(c) The `common_courses` method takes another `Student` object (`other_student`) as a parameter and returns a list of courses that both students have taken in common. It uses the set intersection operation (`&`) to find the common courses between the current student's `courses` and the other student's `courses`. The resulting set is then converted back to a list using the `list` function.

You can use the class as follows:

```python
p = Student("Chris Student", "123454321")
p.add_courses(["CSCI1100", "BASK4010", "GEOL1320"])

q = Student("Bilbo Baggins", "545454545")
q.add_courses(["MATH1240", "CSCI1100", "HIST2010", "BASK4010"])

print(q.common_courses(p))  # Output: ['CSCI1100', 'BASK4010']
```

This code creates two `Student` objects, `p` and `q`, with their respective names and IDs. It adds courses to each student using the `add_courses` method. Finally, it calls the `common_courses` method on `q`, passing `p` as the argument, to find the courses that both students have in common. The output will be `['CSCI1100', 'BASK4010']`.
{{< /details >}}

### Taking Multiple CSCI Courses

> Using the `Student` methods and attributes from the previous question, suppose you are given a list of student objects called all_students. Write a segment of code to output the names of all students who have taken at least two courses that start with CSCI.

{{< details summary="**Answer of This Question**" >}}
Assuming you have a list of `Student` objects called `all_students`, here's a code segment that outputs the names of students who have taken at least two courses starting with "CSCI":

```python
for student in all_students:
    csci_courses = [course for course in student.courses if course.startswith("CSCI")]
    if len(csci_courses) >= 2:
        print(student.name)
```

Explanation:

1. We iterate over each `Student` object in the `all_students` list using a `for` loop.

2. For each student, we create a new list called `csci_courses` using a list comprehension. This list comprehension does the following:
   - It iterates over each course in the student's `courses` list using `for course in student.courses`.
   - For each course, it checks if the course starts with "CSCI" using the `startswith` method: `if course.startswith("CSCI")`.
   - If the course starts with "CSCI", it is included in the `csci_courses` list.

3. After creating the `csci_courses` list, we check its length using `len(csci_courses)`. If the length is greater than or equal to 2, it means the student has taken at least two courses starting with "CSCI".

4. If the condition `len(csci_courses) >= 2` is true, we print the student's name using `print(student.name)`.

This code segment will output the names of all students in the `all_students` list who have taken at least two courses starting with "CSCI".

For example, if you have the following `all_students` list:

```python
p = Student("Chris Student", "123454321")
p.add_courses(["CSCI1100", "BASK4010", "GEOL1320"])

q = Student("Bilbo Baggins", "545454545")
q.add_courses(["MATH1240", "CSCI1100", "HIST2010", "BASK4010"])

r = Student("Alice Smith", "987654321")
r.add_courses(["CSCI1100", "CSCI2200", "MATH1240", "HIST2010"])

all_students = [p, q, r]
```

Running the code segment on this `all_students` list will output:

```
Alice Smith
```

This is because only the student "Alice Smith" has taken at least two courses starting with "CSCI" ("CSCI1100" and "CSCI2200").
{{< /details >}}

### K Smallest Values

> Given a list `L` and a positive integer `k`, create a new list containing only the k smallest values in `L` list. For example, if `L = [ 15, 89, 3, 56, 83, 123, 51, 14, 15, 67, 15 ]` and `k=4`, then the new list should have the values `Ls = [3, 14, 15, 15]` (Note that one of the 15s is not here.) Write a function, `k_smallest(L,k)`, that returns the desired list. It does this using sorting, but does not change `L`. Do this in 4 lines of code without writing any loops.

{{< details summary="**Answer of This Question**" >}}
Here's a Python function `k_smallest(L, k)` that returns a new list containing the k smallest values from the input list `L` without modifying `L`, using sorting and without writing any loops:

```python
def k_smallest(L, k):
    sorted_L = sorted(L)
    Ls = sorted_L[:k]
    return Ls
```

Explanation:

1. We create a new list `sorted_L` by calling the `sorted()` function on the input list `L`. This creates a new sorted list in ascending order without modifying the original list `L`.

2. We create a new list `Ls` by slicing the `sorted_L` list from index 0 to index `k` (exclusive) using the slicing syntax `sorted_L[:k]`. This selects the first `k` elements from the sorted list, which correspond to the `k` smallest values.

3. Finally, we return the `Ls` list containing the `k` smallest values from the original list `L`.

The function achieves the desired result in just 4 lines of code without using any loops. It utilizes the built-in `sorted()` function to sort the list and slicing to extract the `k` smallest values.

You can test the function with the provided example:

```python
L = [15, 89, 3, 56, 83, 123, 51, 14, 15, 67, 15]
k = 4
Ls = k_smallest(L, k)
print(Ls)  # Output: [3, 14, 15, 15]
```

The function will return the list `[3, 14, 15, 15]`, which contains the 4 smallest values from the original list `L`, without modifying `L`.
{{< /details >}}

### Code Segment Output

> What is the output of the following two code segments?
> 
> Part A 
> ```python
> dt = { 1: [ 'mom', 'dad'], 'hi': [1, 3, 5 ]}
> print(len(dt))
> print(dt[1][0])
> dt['hi'].append(3)
> dt[1][0] = 'gran'
> print(dt[1])
> ```
>
> Part B
> ```python
> # Remember that pop() removes and returns the last value from the list.
> LP = [2, 3, 5, 7]
> LC = [4, 6, 8, 9] 
> nums = dict()
> nums['pr'] = LP
> nums['co'] = LC[:]
> LP[1] = 5
> print(len(nums['co']))
> v = LC.pop()
> v = LC.pop()
> v = LC.pop()
> LC.append(12)
> print(len(LC))
> print(len(nums['co']))
> v = nums['pr'].pop()
> v = nums['pr'].pop()
> print(nums['pr'][1])
> print(len(LP))
> ```

{{< details summary="**Answer of Part A**" >}}
Output:
```text
2
mom
['gran', 'dad']
```

Explanation:
1. The code creates a dictionary `dt` with keys 1 and 'hi'. The value for key 1 is a list containing the strings 'mom' and 'dad'. The value for key 'hi' is a list containing the integers 1, 3, and 5.
2. The line `print(len(dt))` prints the length of the dictionary `dt`, which is 2 since it has two key-value pairs.
3. The line `print(dt[1][0])` prints the element at index 0 of the list associated with key 1 in the dictionary `dt`, which is 'mom'.
4. The line `dt['hi'].append(3)` appends the integer 3 to the list associated with key 'hi' in the dictionary `dt`. After this operation, the value for key 'hi' becomes [1, 3, 5, 3].
5. The line `dt[1][0] = 'gran'` updates the element at index 0 of the list associated with key 1 in the dictionary `dt` to 'gran'.
6. The line `print(dt[1])` prints the list associated with key 1 in the dictionary `dt`, which is now ['gran', 'dad'].
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
Output:
```text
4
2
4
5
2
```

Explanation:
1. The code creates two lists, `LP` and `LC`, with some initial values.
2. It creates an empty dictionary called `nums`.
3. The line `nums['pr'] = LP` assigns the list `LP` as the value for the key 'pr' in the dictionary `nums`.
4. The line `nums['co'] = LC[:]` creates a shallow copy of the list `LC` and assigns it as the value for the key 'co' in the dictionary `nums`.
5. The line `LP[1] = 5` updates the element at index 1 of the list `LP` to 5. This does not affect the list `nums['pr']` because it is a separate copy.
6. The line `print(len(nums['co']))` prints the length of the list associated with the key 'co' in the dictionary `nums`, which is 4.
7. The lines `v = LC.pop()`, `v = LC.pop()`, and `v = LC.pop()` remove and assign the last three elements of the list `LC` to the variable `v`. After these operations, `LC` becomes [4].
8. The line `LC.append(12)` appends the integer 12 to the list `LC`. Now, `LC` becomes [4, 12].
9. The line `print(len(LC))` prints the length of the list `LC`, which is 2.
10. The line `print(len(nums['co']))` prints the length of the list associated with the key 'co' in the dictionary `nums`, which is still 4 because it was a separate copy and not affected by the changes made to `LC`.
11. The lines `v = nums['pr'].pop()` and `v = nums['pr'].pop()` remove and assign the last two elements of the list associated with the key 'pr' in the dictionary `nums` to the variable `v`. After these operations, `nums['pr']` becomes [2, 5].
12. The line `print(nums['pr'][1])` prints the element at index 1 of the list associated with the key 'pr' in the dictionary `nums`, which is 5.
13. The line `print(len(LP))` prints the length of the list `LP`, which is 2 because it was affected by the `pop()` operations on `nums['pr']`.
{{< /details >}}

### Finding Names by Age

> Given is a list of dictionaries, where each dictionary stores information about a person in the form of attribute (key) / value pairs. For example, here is a list of dictionaries representing four people:
>
> ```python
> people = [ 
>     { 'name':'Paul', 'age' : 25, 'weight' : 165 },
>     { 'height' : 155, 'name' : 'Sue', 'age' : 30, 'weight' : 123 },
>     { 'weight' : 205, 'name' : 'Sam' },
>     { 'height' : 156, 'name' : 'Andre', 'age' : 39, 'weight' : 123 } 
> ]
> ```
>
> Write code that finds and outputs, in alphabetical order, the names of all people whose age is known to be at least 30. You may assume that each dictionary in people has a 'name' key, but not necessarily a 'age' key. For the example above, the output should be:
>
> ```text
> Andre
> Sue  
> ```

{{< details summary="**Answer of This Question**" >}}
To find and output the names of people whose age is known to be at least 30, you can use a list comprehension to filter the dictionaries based on the 'age' key and then sort the resulting names alphabetically. Here's the code to achieve this:

```python
people = [
    {'name': 'Paul', 'age': 25, 'weight': 165},
    {'height': 155, 'name': 'Sue', 'age': 30, 'weight': 123},
    {'weight': 205, 'name': 'Sam'},
    {'height': 156, 'name': 'Andre', 'age': 39, 'weight': 123}
]

names = [person['name'] for person in people if 'age' in person and person['age'] >= 30]
names.sort()

for name in names:
    print(name)
```

Explanation:

1. We start with the given list of dictionaries called `people`.

2. We use a list comprehension to create a new list called `names`. The list comprehension iterates over each dictionary `person` in the `people` list.

3. For each `person` dictionary, we check two conditions:
   - First, we check if the 'age' key exists in the dictionary using the `'age' in person` condition. This ensures that we only consider dictionaries that have an 'age' key.
   - Second, we check if the value of the 'age' key is greater than or equal to 30 using the `person['age'] >= 30` condition.

4. If both conditions are satisfied, we include the value of the 'name' key from the `person` dictionary in the `names` list.

5. After the list comprehension, we have a list `names` that contains the names of people whose age is known to be at least 30.

6. We use the `sort()` method to sort the `names` list alphabetically.

7. Finally, we iterate over each `name` in the sorted `names` list and print them one by one.

The output of the code will be:
```text
Andre
Sue
```

This code effectively filters the list of dictionaries based on the age condition, extracts the names, sorts them alphabetically, and prints them as required.
{{< /details >}}

### Creating a City to State Dictionary

> Given a dictionary that associates the names of states with a list of the names of (some of the) cities that appear in it, write a function that creates and returns a new dictionary that associates the name of a city with the list of states that it appears in. Within the function, output the cities that are unique — they appear in only one state. Do this in alphabetical order. As an example, if the first dictionary looks like:
>
> ```python
> states = {
>     'New Hampshire': ['Concord', 'Hanover'],
>     'Massachusetts': ['Boston', 'Concord', 'Springfield'],
>     'Illinois': ['Chicago', 'Springfield', 'Peoria']
> }
> ```
>
> then after the function the new dictionary call cities should look like:
>
> ```python 
> cities = {
>     'Hanover': ['New Hampshire'], 
>     'Chicago': ['Illinois'],
>     'Boston': ['Massachusetts'], 
>     'Peoria': ['Illinois'],
>     'Concord': ['New Hampshire', 'Massachusetts'],
>     'Springfield': ['Massachusetts', 'Illinois']
> }
> ```
>
> and the four unique cities output should be:
>
> ```text
> Boston
> Chicago
> Hanover
> Peoria
> ```
>
> Here is the function prototype:
>
> ```python
> def create_cities(states):
> ```

{{< details summary="**Answer of This Question**" >}}
Here's a Python function that creates a new dictionary associating cities with the states they appear in and outputs the unique cities in alphabetical order:

```python
def create_cities(states):
    cities = {}
    unique_cities = []

    for state, city_list in states.items():
        for city in city_list:
            if city not in cities:
                cities[city] = [state]
            else:
                cities[city].append(state)

    for city, state_list in cities.items():
        if len(state_list) == 1:
            unique_cities.append(city)

    unique_cities.sort()

    for city in unique_cities:
        print(city)

    return cities
```

Explanation:

1. We define the function `create_cities` that takes the `states` dictionary as input.

2. We initialize an empty dictionary called `cities` to store the city-to-state mappings.

3. We also initialize an empty list called `unique_cities` to store the names of the unique cities.

4. We iterate over each state and its corresponding list of cities in the `states` dictionary using the `items()` method.

5. For each city in the city list, we check if it already exists as a key in the `cities` dictionary:
   - If the city is not in the `cities` dictionary, we add it as a key with a list containing the current state as the value.
   - If the city is already in the `cities` dictionary, we append the current state to the existing list of states for that city.

6. After creating the `cities` dictionary, we iterate over each city and its corresponding list of states in the `cities` dictionary using the `items()` method.

7. For each city, we check if the length of its state list is equal to 1. If it is, it means the city appears in only one state, so we append the city name to the `unique_cities` list.

8. We sort the `unique_cities` list in alphabetical order using the `sort()` method.

9. We iterate over each city in the sorted `unique_cities` list and print them one by one.

10. Finally, we return the `cities` dictionary.

Using the provided example `states` dictionary, the output of the function will be:
```text
Boston
Chicago
Hanover
Peoria
```

And the returned `cities` dictionary will be:
```python
{
    'Concord': ['New Hampshire', 'Massachusetts'],
    'Hanover': ['New Hampshire'],
    'Boston': ['Massachusetts'],
    'Springfield': ['Massachusetts', 'Illinois'],
    'Chicago': ['Illinois'],
    'Peoria': ['Illinois']
}
```

This function creates the desired city-to-state dictionary, outputs the unique cities in alphabetical order, and returns the `cities` dictionary.
{{< /details >}}

### Rectangle Class Methods

> Consider the following definition of a Rectangle class:
>
> ```python
> class Rectangle(object):
>     def __init__( self, u0, v0, u1, v1 ):
>         self.x0 = u0 # x0 and y0 form the lower left corner of the rectangle
>         self.y0 = v0
>         self.x1 = u1 # x1 and y1 form the upper right corner of the rectangle 
>         self.y1 = v1
>         self.points = [] # See part (b)
> ```
>
> (a) Write a `Rectangle` class method called contains that determines if a location represented by an `x` and a `y` value is inside the rectangle. Note, that for this example, on the boundary counts as in the rectangle. For example:
>
> ```python
> r = Rectangle( 1, 3, 7, 10 )
> r.contains( 1, 4)
> True
> r.contains( 2,11) 
> False
> ```
>
> (b) Suppose there is a second class:
>
> ```python
> class Point(object):
>     def __init__( self, x0, y0, id0 ):
>         self.x = x0
>         self.y = y0
>         self.id = id0
> ```
>
> and each `Rectangle` stores a list of Point objects whose coordinates are inside the rectangle. Write a Rectangle class method called `add_points` that adds a list of `Point` objects to the existing (initially empty) list of Point objects stored with the `Rectangle` object. If a point is outside the rectangle's boundary or if a point with the same id is already in the rectangle's point list, the point should be ignored. Otherwise, it should be added to the rectangle's point list. Your method must make use of the contains method from part (a).

{{< details summary="**Answer of Part A**" >}}
Here's the implementation of the `contains` method for the `Rectangle` class:

```python
def contains(self, x, y):
    return self.x0 <= x <= self.x1 and self.y0 <= y <= self.y1
```

Explanation:
- The `contains` method takes two parameters: `x` and `y`, representing the coordinates of the point to check.
- It checks if the given `x` and `y` values fall within the boundaries of the rectangle.
- The condition `self.x0 <= x <= self.x1` checks if `x` is between the left and right boundaries of the rectangle (inclusive).
- The condition `self.y0 <= y <= self.y1` checks if `y` is between the bottom and top boundaries of the rectangle (inclusive).
- If both conditions are true, it means the point is inside the rectangle or on its boundary, so the method returns `True`.
- Otherwise, if either condition is false, the point is outside the rectangle, and the method returns `False`.

Example usage:
```python
r = Rectangle(1, 3, 7, 10)
print(r.contains(1, 4))  # Output: True
print(r.contains(2, 11))  # Output: False
```

In the first example, the point (1, 4) is inside the rectangle, so `contains` returns `True`.
In the second example, the point (2, 11) is outside the rectangle, so `contains` returns `False`.
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
Here's the implementation of the `add_points` method for the `Rectangle` class:

```python
def add_points(self, points):
    for point in points:
        if self.contains(point.x, point.y) and not any(p.id == point.id for p in self.points):
            self.points.append(point)
```

Explanation:
- The `add_points` method takes a list of `Point` objects as a parameter.
- It iterates over each `point` in the `points` list.
- For each `point`, it checks two conditions:
  1. It calls the `contains` method (from part a) to check if the point's coordinates (`point.x`, `point.y`) are inside the rectangle's boundaries.
  2. It checks if a point with the same `id` as the current `point` already exists in the rectangle's `points` list using the `any` function and a generator expression.
- If both conditions are satisfied (i.e., the point is inside the rectangle and no point with the same `id` exists in the `points` list), the `point` is appended to the rectangle's `points` list using `self.points.append(point)`.
- If either condition is not satisfied (i.e., the point is outside the rectangle or a point with the same `id` already exists), the `point` is ignored and not added to the `points` list.

Example usage:
```python
r = Rectangle(1, 3, 7, 10)

p1 = Point(2, 5, 1)
p2 = Point(4, 8, 2)
p3 = Point(6, 12, 3)  # Outside the rectangle
p4 = Point(3, 6, 1)  # Same id as p1

r.add_points([p1, p2, p3, p4])

for point in r.points:
    print(point.id)
```

Output:
```
1
2
```

In this example:
- `p1` and `p2` are inside the rectangle and have unique `id`s, so they are added to the `points` list.
- `p3` is outside the rectangle, so it is ignored.
- `p4` has the same `id` as `p1`, so it is also ignored.

After adding the points, the rectangle's `points` list contains only `p1` and `p2`, as shown in the output.
{{< /details >}}

### Code Output

> Show the output of the following code:
> 
> ```python
> places = { 
>     'OR': {'Portland' : set(['Pearl District', 'Alameda']), 'Eugene' : set()},
>     'NY': {'Albany' : set(), 'NY' : set(['Chelsea', 'Harlem'])} 
> }
> print(places['OR']['Eugene'])
> a = []
> for place in places:
>     a += places[place].keys()
> print(a)
> for x in a:
>     if len(x) < 7:
>         print(x)
> for place in places:
>     if x in places[place]:
>         print(places[place][x])
> ```

{{< details summary="**Answer of This Question**" >}}
Apologies for the mistake in my previous answer. Here's the corrected output of the given code:

```
set()
['Portland', 'Eugene', 'Albany', 'NY']
Eugene
Albany
NY
{'Harlem', 'Chelsea'}
```

Explanation:
1. `print(places['OR']['Eugene'])`:
   - This line accesses the value associated with the key 'Eugene' in the dictionary `places['OR']`.
   - The value associated with 'Eugene' is an empty set `set()`, so it prints `set()`.

2. `a = []` and `for place in places: a += places[place].keys()`:
   - These lines initialize an empty list `a` and then iterate over the keys of the `places` dictionary.
   - For each key `place`, it retrieves the keys of the nested dictionary `places[place]` using `places[place].keys()` and appends them to the list `a`.
   - After the loop, `a` contains the list `['Portland', 'Eugene', 'Albany', 'NY']`.

3. `print(a)`:
   - This line prints the list `a`, which is `['Portland', 'Eugene', 'Albany', 'NY']`.

4. `for x in a: if len(x) < 7: print(x)`:
   - This loop iterates over each element `x` in the list `a`.
   - If the length of `x` is less than 7, it prints `x`.
   - The output of this loop is:
     ```
     Eugene
     Albany
     NY
     ```

5. `for place in places: if x in places[place]: print(places[place][x])`:
   - This loop iterates over each key `place` in the `places` dictionary.
   - It checks if the last value of `x` from the previous loop ('NY') is a key in the nested dictionary `places[place]`.
   - If 'NY' is a key in `places[place]`, it prints the value associated with that key.
   - The output of this loop is:
     ```
     {'Harlem', 'Chelsea'}
     ```
   - The line `{'Harlem', 'Chelsea'}` is printed because 'NY' is a key in `places['NY']`, and it prints the set associated with 'NY'.

So, the complete output of the code is:
```
set()
['Portland', 'Eugene', 'Albany', 'NY']
Eugene
Albany
NY
{'Harlem', 'Chelsea'}
```

I apologize for the confusion in my previous answer. Thank you for pointing out the correction.
{{< /details >}}

### Business Reviews

> Suppose you are given a file named `businesses.txt` in which each line contains the name of a business and its category (a single value), followed by a sequence of review scores, each separated by '|'. Write a piece of code that reads this file, and prints the names of all businesses, their categories, and the average review score for each business. Also print the total number of unique categories in this file. For example, for the file below:
> 
> ```text
> Dinosaur Bar-B-Que|BBQ|5|4|4|4|5|5|4|2
> DeFazio's Pizzeria|Pizza|5|5|5|5|5|5|5|5|5|5|3|5|5|5
> I Love NY Pizza|Pizza|4|5|5|3
> ```
> 
> Your program should print:
> 
> ```text
> Dinosaur Bar-B-Que (BBQ): Score 4.125 
> DeFazio's Pizzeria (Pizza): Score 4.857
> I Love NY Pizza (Pizza): Score 4.250
> 2 categories found.
> ```

{{< details summary="**Answer of This Question**" >}}
Here's a Python script that reads the `businesses.txt` file, calculates the average review score for each business, and prints the required information:

```python
# Read the businesses.txt file
with open('businesses.txt', 'r') as file:
    lines = file.readlines()

# Create a dictionary to store the category and its corresponding businesses
categories = {}

# Process each line in the file
for line in lines:
    # Split the line into business name, category, and review scores
    data = line.strip().split('|')
    business_name = data[0]
    category = data[1]
    scores = [int(score) for score in data[2:]]

    # Calculate the average review score
    average_score = sum(scores) / len(scores)

    # Add the business to the corresponding category in the dictionary
    if category in categories:
        categories[category].append((business_name, average_score))
    else:
        categories[category] = [(business_name, average_score)]

# Print the business information
for category, businesses in categories.items():
    for business_name, average_score in businesses:
        print(f"{business_name} ({category}): Score {average_score:.3f}")

# Print the total number of unique categories
print(f"{len(categories)} categories found.")
```

Explanation:
1. We start by opening the `businesses.txt` file in read mode using a `with` statement. This ensures that the file is properly closed after we're done reading from it.

2. We read all the lines from the file using `file.readlines()` and store them in the `lines` list.

3. We create an empty dictionary called `categories` to store the category and its corresponding businesses.

4. We iterate over each line in the `lines` list.

5. For each line, we split it into the business name, category, and review scores using `line.strip().split('|')`. The `strip()` method is used to remove any leading or trailing whitespace from the line.

6. We convert the review scores from strings to integers using a list comprehension: `[int(score) for score in data[2:]]`.

7. We calculate the average review score by summing up all the scores and dividing by the number of scores.

8. We check if the category already exists in the `categories` dictionary. If it does, we append a tuple of the business name and average score to the corresponding list. If it doesn't exist, we create a new list with the business name and average score and add it to the dictionary.

9. After processing all the lines, we iterate over each category and its corresponding businesses in the `categories` dictionary.

10. For each business, we print its name, category, and average review score using an f-string with the desired formatting.

11. Finally, we print the total number of unique categories by getting the length of the `categories` dictionary.

When you run this script with the provided `businesses.txt` file, it will produce the desired output:

```text
Dinosaur Bar-B-Que (BBQ): Score 4.125
DeFazio's Pizzeria (Pizza): Score 4.857
I Love NY Pizza (Pizza): Score 4.250
2 categories found.
```

This script reads the file, calculates the average review score for each business, and prints the business information along with the total number of unique categories found in the file.
{{< /details >}}

### Histogram Function

> Write a function that takes as input a list of numbers and generates a histogram. The histogram prints a star (*) for each occurrence of a number in the list. For example, if the list was:
> 
> ```python
> numbers = [5, 4, 1, 1, 3, 1, 2, 2, 4, 1]
> ```
> 
> Your function should print (sorted by the values):
> 
> ```text
> 1: ****
> 2: **  
> 3: *
> 4: **
> 5: *
> ```
> 
> (a) Write the function using a dictionary. You may not use a set.
> 
> (b) Write the same function using a set. You may not use a dictionary (hint: use `count` for the unique items in the list).

{{< details summary="**Answer of Part A**" >}}
Here's a Python function that generates a histogram using a dictionary:

```python
def histogram_dict(numbers):
    # Create a dictionary to store the count of each number
    count_dict = {}

    # Count the occurrences of each number in the list
    for num in numbers:
        if num in count_dict:
            count_dict[num] += 1
        else:
            count_dict[num] = 1

    # Print the histogram
    for num in sorted(count_dict.keys()):
        print(f"{num}: {'*' * count_dict[num]}")
```

Explanation:
1. We define a function called `histogram_dict` that takes a list of numbers as input.

2. We create an empty dictionary called `count_dict` to store the count of each number in the list.

3. We iterate over each number `num` in the `numbers` list.

4. For each number, we check if it already exists as a key in the `count_dict`:
   - If it exists, we increment its count by 1.
   - If it doesn't exist, we add it to the dictionary with an initial count of 1.

5. After counting the occurrences of each number, we iterate over the sorted keys of the `count_dict` using `sorted(count_dict.keys())`. This ensures that the histogram is printed in ascending order of the numbers.

6. For each number `num`, we print the number followed by a colon and the corresponding number of stars (`*`) based on its count in the `count_dict`. We use the `'*' * count_dict[num]` expression to repeat the star character `count_dict[num]` times.

Example usage:
```python
numbers = [5, 4, 1, 1, 3, 1, 2, 2, 4, 1]
histogram_dict(numbers)
```

Output:
```text
1: ****
2: **
3: *
4: **
5: *
```

This function uses a dictionary to count the occurrences of each number and then prints the histogram based on the counts.
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
Here's a Python function that generates a histogram using a set:

```python
def histogram_set(numbers):
    # Create a set to store the unique numbers
    unique_nums = set(numbers)

    # Print the histogram
    for num in sorted(unique_nums):
        count = numbers.count(num)
        print(f"{num}: {'*' * count}")
```

Explanation:
1. We define a function called `histogram_set` that takes a list of numbers as input.

2. We create a set called `unique_nums` by passing the `numbers` list to the `set()` constructor. This automatically removes any duplicate numbers and gives us a set of unique numbers.

3. We iterate over each number `num` in the sorted `unique_nums` set using `sorted(unique_nums)`. This ensures that the histogram is printed in ascending order of the numbers.

4. For each unique number `num`, we use the `count()` method on the `numbers` list to count the occurrences of that number in the original list. We store the count in the `count` variable.

5. We print the number followed by a colon and the corresponding number of stars (`*`) based on its count. We use the `'*' * count` expression to repeat the star character `count` times.

Example usage:
```python
numbers = [5, 4, 1, 1, 3, 1, 2, 2, 4, 1]
histogram_set(numbers)
```

Output:
```text
1: ****
2: **
3: *
4: **
5: *
```

This function uses a set to store the unique numbers and then uses the `count()` method to count the occurrences of each unique number in the original list. It then prints the histogram based on the counts.

Note: Using a set eliminates the need for a separate counting step, as we can directly use the `count()` method on the original list for each unique number.
{{< /details >}}

### Alumni Information

> You are given a list of RPI Alumni as shown below. Each item in the list is a dictionary containing information about an alumnus and all items have the same keys. Write a piece of code that prints the name and addresses of each person who graduated before 2013. For example, given the list:
> 
> ```python
> alums = [{'fname':'Abed', 'lname':'Nadir', 'graduated':2012, 'addresses':['Troy&Abed apt.','Abed&Annie apt.']}, {'fname':'Troy', 'lname':'Barnes', 'graduated':2013, 'addresses':['Troy&Abed apt.']}, {'fname':'Britta', 'lname':'Perry', 'graduated':2012, 'addresses':['1 Revolution lane']}]
> ```
> 
> Your code should print (all information is printed in the order it appears in the list):
> 
> ```text
> Abed Nadir
> Troy&Abed apt.
> Abed&Annie apt.
> Britta Perry 
> 1 Revolution lane
> ```

{{< details summary="**Answer of This Question**" >}}
Here's the Python code that prints the name and addresses of each person who graduated before 2013:

```python
alums = [
    {'fname': 'Abed', 'lname': 'Nadir', 'graduated': 2012, 'addresses': ['Troy&Abed apt.', 'Abed&Annie apt.']},
    {'fname': 'Troy', 'lname': 'Barnes', 'graduated': 2013, 'addresses': ['Troy&Abed apt.']},
    {'fname': 'Britta', 'lname': 'Perry', 'graduated': 2012, 'addresses': ['1 Revolution lane']}
]

for alum in alums:
    if alum['graduated'] < 2013:
        print(alum['fname'], alum['lname'])
        for address in alum['addresses']:
            print(address)
```

Explanation:
1. We start with the given list of alumni, where each alumnus is represented by a dictionary containing their information.

2. We iterate over each `alum` dictionary in the `alums` list.

3. For each `alum`, we check if their graduation year (`alum['graduated']`) is less than 2013. This condition filters out the alumni who graduated before 2013.

4. If the condition is true, we proceed to print the name and addresses of the alumnus.

5. We print the first name (`alum['fname']`) and last name (`alum['lname']`) of the alumnus using the `print()` function.

6. We then iterate over each `address` in the `alum['addresses']` list.

7. For each `address`, we print it on a separate line using the `print()` function.

8. After processing all the alumni, the code will have printed the name and addresses of each person who graduated before 2013.

Output:
```text
Abed Nadir
Troy&Abed apt.
Abed&Annie apt.
Britta Perry
1 Revolution lane
```

This code iterates over the list of alumni, checks their graduation year, and prints the name and addresses of those who graduated before 2013. The information is printed in the order it appears in the list.
{{< /details >}}

### File Line Extraction

> Write a function called `get_line(fname,lno,start,end)` that takes as input a file name `fname`, a line number `lno`, and starting and end points (`start`,`end`) on the given line. The function should return the string containing all the characters from the starting point up to but not including the end point on that line (same as it would be with string slicing!).
> 
> Line numbers start at 1; characters in a line are counted starting with zero and include the new line character at the end. If there are fewer than `lno` lines, your function should return None. If the line at `lno` has fewer than `end` characters, return an empty string.
> 
> Given the following contents of file `hpss.txt`:
> 
> ```text
> Nearly ten years had passed since the Dursleys had
> woken up to find their nephew on the front step.
> Privet Drive had hardly changed at  
> all.
> ```
> 
> The following program:
> 
> ```python
> print('1:', get_line('hpss.txt', 2, 9, 15))
> print('2:', get_line('hpss.txt', 5, 5, 9))
> print('3:', get_line('hpss.txt', 5, 0, 4))
> print('4:', get_line('hpss.txt', 8, 0, 10))
> ```
>  
> Should print (notice for 3, the newline is also included in the returned string):
> 
> ```text
> 1: to fin
> 2: 
> 3: all.
> 4: None
> ```

{{< details summary="**Answer of This Question**" >}}
Here's the implementation of the `get_line` function in Python:

```python
def get_line(fname, lno, start, end):
    try:
        with open(fname, 'r') as file:
            lines = file.readlines()
            if lno <= len(lines):
                line = lines[lno - 1]
                if end <= len(line):
                    return line[start:end]
                else:
                    return ''
            else:
                return None
    except FileNotFoundError:
        return None
```

Explanation:
1. We define the `get_line` function that takes four parameters: `fname` (file name), `lno` (line number), `start` (starting point), and `end` (end point).

2. We use a `try` block to handle the case where the file might not exist.

3. Inside the `try` block, we open the file specified by `fname` in read mode using the `with` statement. This ensures that the file is properly closed after we're done reading from it.

4. We read all the lines from the file using `file.readlines()` and store them in the `lines` list.

5. We check if the requested line number `lno` is less than or equal to the total number of lines in the file (`len(lines)`).

6. If the condition is true, we retrieve the line at index `lno - 1` from the `lines` list (since line numbers start at 1, but list indices start at 0).

7. We then check if the requested end point `end` is less than or equal to the length of the retrieved line (`len(line)`).

8. If the condition is true, we return the substring of the line starting from index `start` up to (but not including) index `end` using string slicing (`line[start:end]`).

9. If the end point `end` is greater than the length of the line, we return an empty string (`''`).

10. If the requested line number `lno` is greater than the total number of lines in the file, we return `None`.

11. If a `FileNotFoundError` occurs (i.e., the file doesn't exist), we catch the exception and return `None`.

Now, let's test the `get_line` function with the given program:

```python
print('1:', get_line('hpss.txt', 2, 9, 15))
print('2:', get_line('hpss.txt', 5, 5, 9))
print('3:', get_line('hpss.txt', 5, 0, 4))
print('4:', get_line('hpss.txt', 8, 0, 10))
```

Output:
```text
1: to fin
2: 
3: all.
4: None
```

Explanation of the output:
1. For the first case, the function returns the substring from index 9 to 14 (inclusive) of the second line, which is `"to fin"`.

2. For the second case, the function returns an empty string because the fifth line doesn't have enough characters to extract from index 5 to 8.

3. For the third case, the function returns the substring from index 0 to 3 (inclusive) of the fifth line, which is `"all."`. The newline character is also included in the returned string.

4. For the fourth case, the function returns `None` because there is no eighth line in the file.

The `get_line` function correctly handles the cases where the requested line number or end point exceeds the available lines or characters in the file.
{{< /details >}}

### Coldest Years

> Suppose you are given the mean temperature for Troy in the month of December in a dictionary `temp` as shown below. The keys of the dictionary are years and the values are the mean temperature for that year. Write a piece of code that finds and prints the top three coldest years according to this dictionary. Note: If there are ties in values, any ordering of years is acceptable.
> 
> For example, given the dictionary below:
> 
> ```python 
> temp = { 2001: 36.4, 2002: 27.4, 2003: 29.3, 2004: 28.6, 2005: 27.8,
> 2006: 37.3, 2007: 28.1, 2008: 30.2, 2010: 26.0, 2011: 35.4,
> 2012: 33.8, 2013: 27.9, 2014: 32.8}
> ```
> 
> Your program should output:
> 
> ```text
> 2010: 26.0
> 2002: 27.4
> 2005: 27.8
> ```

{{< details summary="**Answer of This Question**" >}}
To find the top three coldest years based on the mean temperature data provided in the dictionary, you can sort the dictionary items based on the temperature values in ascending order and print the first three items. Here's the code to achieve that:

```python
temp = {2001: 36.4, 2002: 27.4, 2003: 29.3, 2004: 28.6, 2005: 27.8,
        2006: 37.3, 2007: 28.1, 2008: 30.2, 2010: 26.0, 2011: 35.4,
        2012: 33.8, 2013: 27.9, 2014: 32.8}

# Sort the dictionary items based on temperature values in ascending order
sorted_temp = sorted(temp.items(), key=lambda x: x[1])

# Print the top three coldest years
for year, temperature in sorted_temp[:3]:
    print(f"{year}: {temperature}")
```

Explanation:

1. The `temp` dictionary is defined with the given mean temperature data for each year.

2. We use the `sorted()` function to sort the dictionary items based on the temperature values in ascending order. The `items()` method is used to get a list of key-value pairs from the dictionary. The `key` parameter is set to `lambda x: x[1]`, which means the sorting will be based on the second element of each item tuple (i.e., the temperature value).

3. The sorted items are stored in the `sorted_temp` list, where each item is a tuple containing the year and its corresponding temperature.

4. We iterate over the first three items of `sorted_temp` using a `for` loop and unpack each item into the variables `year` and `temperature`.

5. Finally, we print each year and its corresponding temperature using an f-string, which allows us to embed variables directly in the string.

The output will be:
```text
2010: 26.0
2002: 27.4
2005: 27.8
```

This code will find and print the top three coldest years based on the mean temperature data provided in the dictionary. If there are ties in temperature values, any ordering of the tied years is acceptable.
{{< /details >}}

### Thanksgiving Dinner Menus

> Suppose you are given three variables `t1`,`t2`,`t3` in your program. Each variable is a set containing the menu for a different Thanksgiving dinner you are invited to.
> First, print items that are in the menu for all three dinners. Then, print the items that are in the menu for exactly one dinner. All items should be listed in alphabetical order.
> 
> For example, if you are given menus:
> 
> ```python
> t1 = set(['Turkey', 'Potatoes', 'Green Beans', 'Cranberry', 'Gravy'])
> t2 = set(['Turkey', 'Yams', 'Stuffing', 'Cranberry', 'Marshmallows'])
> t3 = set(['Turkey', 'Gravy', 'Yams', 'Green Beans', 'Cranberry', 'Turducken'])
> ```
> 
> Your program must print the following (your output should match ours):
> 
> ```text
> Items in all three dinners: Cranberry, Turkey
> Items in exactly one dinner: Marshmallows, Potatoes, Stuffing, Turducken
> ```

{{< details summary="**Answer of This Question**" >}}
To find the items that are in the menu for all three dinners and the items that are in the menu for exactly one dinner, you can use set operations. Here's the code to achieve that:

```python
t1 = set(['Turkey', 'Potatoes', 'Green Beans', 'Cranberry', 'Gravy'])
t2 = set(['Turkey', 'Yams', 'Stuffing', 'Cranberry', 'Marshmallows'])
t3 = set(['Turkey', 'Gravy', 'Yams', 'Green Beans', 'Cranberry', 'Turducken'])

# Find items in all three dinners
common_items = t1.intersection(t2, t3)
print("Items in all three dinners:", ', '.join(sorted(common_items)))

# Find items in exactly one dinner
unique_items = (t1 ^ t2 ^ t3) - (t1 & t2) - (t1 & t3) - (t2 & t3)
print("Items in exactly one dinner:", ', '.join(sorted(unique_items)))
```

Explanation:

1. The sets `t1`, `t2`, and `t3` are defined with the given menu items for each Thanksgiving dinner.

2. To find the items that are in the menu for all three dinners, we use the `intersection()` method. It returns a new set containing the common elements from all the sets. We pass `t1`, `t2`, and `t3` as arguments to `intersection()`, which gives us the items that are present in all three sets.

3. We print the common items using `print()`. The `sorted()` function is used to sort the items in alphabetical order, and `', '.join()` is used to join the items into a comma-separated string.

4. To find the items that are in the menu for exactly one dinner, we use the following set operations:
   - `t1 ^ t2 ^ t3` performs the symmetric difference (XOR) operation on all three sets. It gives us the items that are in exactly one of the sets.
   - `(t1 & t2)`, `(t1 & t3)`, and `(t2 & t3)` perform the intersection operation on pairs of sets. These give us the items that are common between each pair of sets.
   - We subtract the common items between pairs of sets from the result of the symmetric difference operation. This gives us the items that are in exactly one dinner.

5. Finally, we print the unique items using `print()`, again sorting them in alphabetical order and joining them into a comma-separated string.

The output will be:
```text
Items in all three dinners: Cranberry, Turkey
Items in exactly one dinner: Marshmallows, Potatoes, Stuffing, Turducken
```

This code finds and prints the items that are in the menu for all three dinners and the items that are in the menu for exactly one dinner, in alphabetical order.
{{< /details >}}

### Algorithm Running Times

> What are the running times of the following algorithms:
> 
> (a) membership test in a list (`list.index(value)`)
> 
> (b) membership test in a set (`value in set`)
> 
> (c) nested for loops over an entire list

{{< details summary="**Answer of Part A**" >}}
The running time of the membership test in a list using `list.index(value)` is O(n), where n is the number of elements in the list.

Explanation:
- The `list.index(value)` method searches for the first occurrence of the specified value in the list.
- In the worst case, the value may be located at the end of the list or not present in the list at all.
- The method needs to traverse the list from the beginning until it finds the value or reaches the end of the list.
- As the size of the list grows, the number of elements to be searched increases linearly.
- Therefore, the running time of `list.index(value)` is proportional to the size of the list, resulting in a time complexity of O(n).
{{< /details >}}

{{< details summary="**Answer of Part B**" >}}
The running time of the membership test in a set using `value in set` is O(1) on average.

Explanation:
- Sets in Python are implemented using hash tables.
- The `value in set` operation checks if the specified value exists in the set.
- In a hash table, elements are stored based on their hash values, which allows for efficient retrieval.
- On average, the time complexity of searching for an element in a hash table is O(1), constant time.
- This is because the hash function used by the set distributes the elements uniformly across the hash table, minimizing collisions.
- Even if there are collisions, the number of elements in each "bucket" of the hash table is expected to be small, resulting in a constant-time lookup on average.
- Therefore, the running time of `value in set` is O(1) on average, regardless of the size of the set.
{{< /details >}}

{{< details summary="**Answer of Part C**" >}}
The running time of nested for loops over an entire list is O(n^2), where n is the number of elements in the list.

Explanation:
- When you have nested for loops iterating over the same list, the total number of iterations is the product of the number of iterations in each loop.
- If the list has n elements, the outer loop will iterate n times.
- For each iteration of the outer loop, the inner loop will also iterate n times.
- Therefore, the total number of iterations is n * n = n^2.
- As the size of the list grows, the number of iterations increases quadratically.
- The running time of nested for loops over an entire list is proportional to the square of the size of the list, resulting in a time complexity of O(n^2).
- This indicates that the running time grows rapidly as the size of the list increases, making it less efficient for large lists compared to algorithms with lower time complexities.
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