## Using Abstractions

In this problem you will fill out three functions to complete the Group ADT and the Diner ADT. The goal is to organize how diners manage the groups that want to eat there and the tables where these groups sit.

It is important to take the time to read through the docstrings and the doctests. Additionally, make sure to not violate abstraction barriers for other ADTs, i.e. when implementing functions for the Diner ADT, do not violate abstraction barriers for the Group ADT, and vice versa.

``````# Diner ADT
def make_diner(name):
""" Diners are represented by their name and the number of free tables they have."""
return [name, 0]

def num_free_tables(diner):
return diner[1]

def name(diner):
return diner[0]
# You will implement add_table and serve which are part of the Diner ADT
def make_group(name):
""" Groups are represented by their name and their status."""
return [name, 'waiting']

def name(group):
return group[0]

def status(group):
return group[1]

def start_eating(group, diner):
group[1] = 'eating'

# You will implement finish_eating which is part of the Group ADT    ``````

### Question 1

Implement `add_table` which increases the diner's number of free tables by 1:

``````def add_table(diner):
"""
>>> num_free_tables(din)
0
>>> num_free_tables(din)
2
"""
diner[1] += 1``````

Use OK to test your code:

``python3 ok -q add_table``

### Question 2

Implement `serve` so that the diner uses one of its free tables to seat the group. If there are no free tables, return the string 'table not free'. If there are free tables, the group's status should be updated to 'eating' and the diner should have one less free table.

``````def serve(diner, group):
"""
>>> din = make_diner("Cafe 3")
>>> g1 = make_group("Vandana's Group")
>>> g2 = make_group("Shreya's Group")
>>> serve(din, g1)
>>> status(g1)
'eating'
>>> num_free_tables(din)
0
>>> serve(din, g2)
'table not free'
>>> status(g2)
'waiting'
"""
if diner[1] > 0:
diner[1] -= 1
start_eating(group, diner)
else:
return 'table not free'``````

Use OK to test your code:

``python3 ok -q serve``

### Question 3

Implement `finish_eating` which sets a group's status to 'finished' and frees the table they were using so that the diner has one more free table.

``````def finish_eating(group, diner):
"""
>>> din = make_diner("Foothill")
>>> g1 = make_group("Nick's Group")
>>> serve(din, g1)
>>> num_free_tables(din)
0
>>> finish_eating(g1, din)
>>> num_free_tables(din)
1
>>> status(g1)
'finished'
"""
group[1] = 'finished'

Use OK to test your code:

``python3 ok -q finish_eating``

## Dictionary Review

### Question 4: Replace All

Given a dictionary `d`, return a new dictionary where all occurences of `x` as a value (not a key) is replaced with `y`.

``````def replace_all(d, x, y):
"""
>>> d = {'foo': 2, 'bar': 3, 'garply': 3, 'xyzzy': 99}
>>> e = replace_all(d, 3, 'poof')
>>> e == {'foo': 2, 'bar': 'poof', 'garply': 'poof', 'xyzzy': 99}
True
"""

new = {}
for key in d:
if d[key] == x:
new[key] = y
else:
new[key] = d[key]
return new``````

Use OK to test your code:

``python3 ok -q replace_all``

### Question 5: Build the Full Rosters

Implement the function `common_players`. The `common_players` function identifies which keys from the `full_roster` share the same values. The function returns a new dictionary where each key is the value from the original dictionary, and the corresponding values of the new dictionary are list that contain keys that share the same value.

``````def common_players(roster):
"""Returns a dictionary containing values along with a corresponding
list of keys that had that value from the original dictionary.
>>> full_roster = {"bob": "Team A", "barnum": "Team B", "beatrice": "Team C", "bernice": "Team B", "ben": "Team D", "belle": "Team A", "bill": "Team B", "bernie": "Team B", "baxter": "Team A"}
>>> common_players(full_roster)
{'Team A': ['bob', 'belle', 'baxter'], 'Team B': ['barnum', 'bernice', 'bill', 'bernie'], 'Team C': ['beatrice'], 'Team D': ['ben']}
"""

result_dict = {}
for player in roster:
team = roster[player]
if team in result_dict:
result_dict[team] += [player]
else:
result_dict[team] = [player]
return result_dict``````

Use OK to test your code:

``python3 ok -q common_players``

## Submit

Make sure to submit this assignment by running:

``python3 ok --submit``