This assignment is due on Wednesday, June 2, 2021 before 08:59AM.
Links to tutorials and other Python resources are posted on the schedule page in the Python Review parts.

# Homework 1: Python Skills & Uninformed Search [200 points]

## Instructions for Python Skills

In the first part of this assignment, you will answer some conceptual questions about Python and write a collection of basic algorithms and data structures.

A skeleton file homework1.py containing empty definitions for each question has been provided. Since portions of this assignment will be graded automatically, none of the names or function signatures in this file should be modified. However, you are free to introduce additional variables or functions if needed.

Unless explicitly stated otherwise, you may not import any of the standard Python modules (except for section 7) in this portion of the assignment, meaning your solutions should not include any lines of the form import x or from x import y. Accordingly, you may find it helpful to refresh yourself on Python’s built-in functions and data types.

You will find that in addition to a problem specification, each programming question also includes a pair of examples from the Python interpreter. These are meant to illustrate typical use cases, and should not be taken as comprehensive test suites.

You may submit as many times as you would like before the deadline, but only the last submission will be saved.

You are strongly encouraged to follow the Python style guidelines set forth in PEP 8, which was written in part by the creator of Python. However, your code will not be graded for style.

Once you have completed the assignment, you should submit your file on Gradescope.

You may submit as many times as you would like before the deadline, but only the last submission will be saved.

## 1. Python Concepts - Study Questions [5 points]

For each of the following questions, write your answers as triply-quoted strings using the indicated variables in the provided file.

1. [1 points] Explain what it means for Python to be both strongly and dynamically typed, and give a concrete example of each.

2. [2 points] You would like to create a dictionary that maps some 2-dimensional points to their associated names. As a first attempt, you write the following code:

points_to_names = {[0, 0]: "home", [1, 2]: "school", [-1, 1]: "market"}


However, this results in a type error. Describe what the problem is, and propose a solution.

3. [2 points] Consider the following two functions, each of which concatenates a list of strings.

def concatenate1(strings):
result = ""
for s in strings:
result += s
return result

def concatenate2(strings):
return "".join(strings)


One of these approaches is significantly faster than the other for large inputs. Which version is better, and what is the reason for the discrepancy?

## 2. Working with Lists [15 points]

1. [5 points] Consider the function extract_and_apply(l, p, f) shown below, which extracts the elements of a list l satisfying a boolean predicate p, applies a function f to each such element, and returns the result.

def extract_and_apply(l, p, f):
result = []
for x in l:
if p(x):
result.append(f(x))
return result


Rewrite extract_and_apply(l, p, f) in one line using a list comprehension.

2. [5 points] Write a function concatenate(seqs) that returns a list containing the concatenation of the elements of the input sequences. Your implementation should consist of a single list comprehension, and should not exceed one line.

>>> concatenate([[1, 2], [3, 4]])
[1, 2, 3, 4]

>>> concatenate(["abc", (0, [0])])
['a', 'b', 'c', 0, [0]]

3. [5 points] Write a function transpose(matrix) that returns the transpose of the input matrix, which is represented as a list of lists. Recall that the transpose of a matrix is obtained by swapping its rows with its columns. More concretely, the equality matrix[i][j] == transpose(matrix)[j][i] should hold for all valid indices i and j. You may assume that the input matrix is well-formed, i.e., that each row is of equal length. You may further assume that the input matrix is non-empty. Your function should not modify the input.

>>> transpose([[1, 2, 3]])
[[1], [2], [3]]

>>> transpose([[1, 2], [3, 4], [5, 6]])
[[1, 3, 5], [2, 4, 6]]


## 3. Sequence Slicing [9 points]

1. [3 points] The functions in this section should be implemented using sequence slices. Recall that the slice parameters take on sensible default values when omitted. In some cases, it may be necessary to use the optional third parameter to specify a step size.

Write a function copy(seq) that returns a new sequence containing the same elements as the input sequence.

>>> copy("abc")
'abc'
>>> copy((1, 2, 3))
(1, 2, 3)

>>> x = [0, 0, 0]; y = copy(x)
>>> print(x, y); x[0] = 1; print(x, y)
[0, 0, 0] [0, 0, 0]
[1, 0, 0] [0, 0, 0]

2. [3 points] Write a function all_but_last(seq) that returns a new sequence containing all but the last element of the input sequence. If the input sequence is empty, a new empty sequence of the same type should be returned.

>>> all_but_last("abc")
'ab'
>>> all_but_last((1, 2, 3))
(1, 2)

>>> all_but_last("")
''
>>> all_but_last([])
[]

3. [3 points] Write a function every_other(seq) that returns a new sequence containing every other element of the input sequence, starting with the first. This function can be written in one line using the optional third parameter of the slice notation.

>>> every_other([1, 2, 3, 4, 5])
[1, 3, 5]
>>> every_other("abcde")
'ace'

>>> every_other([1, 2, 3, 4, 5, 6])
[1, 3, 5]
>>> every_other("abcdef")
'ace'


## 4. Combinatorial Algorithms [12 points]

The functions in this section should be implemented as generators. You may generate the output in any order you find convenient, as long as the correct elements are produced. However, in some cases, you may find that the order of the example output hints at a possible implementation.

Although generators in Python can be used in a variety of ways, you will not need to use any of their more sophisticated features here. Simply keep in mind that where you might normally return a list of elements, you should instead yield the individual elements.

Since the contents of a generator cannot be viewed without employing some form of iteration, we wrap all function calls in this section’s examples with the list function for convenience.

1. [6 points] The prefixes of a sequence include the empty sequence, the first element, the first two elements, etc., up to and including the full sequence itself. Similarly, the suffixes of a sequence include the empty sequence, the last element, the last two elements, etc., up to and including the full sequence itself. Write a pair of functions prefixes(seq) and suffixes(seq) that yield all prefixes and suffixes of the input sequence.

>>> list(prefixes([1, 2, 3]))
[[], [1], [1, 2], [1, 2, 3]]
>>> list(suffixes([1, 2, 3]))
[[1, 2, 3], [2, 3], [3], []]

>>> list(prefixes("abc"))
['', 'a', 'ab', 'abc']
>>> list(suffixes("abc"))
['abc', 'bc', 'c', '']

2. [6 points] Write a function slices(seq) that yields all non-empty slices of the input sequence.

>>> list(slices([1, 2, 3]))
[[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

>>> list(slices("abc"))
['a', 'ab', 'abc', 'b', 'bc', 'c']


## 5. Text Processing [20 points]

1. [5 points] A common preprocessing step in many natural language processing tasks is text normalization, wherein words are converted to lowercase, extraneous whitespace is removed, etc. Write a function normalize(text) that returns a normalized version of the input string, in which all words have been converted to lowercase and are separated by a single space. No leading or trailing whitespace should be present in the output.

>>> normalize("This is an example.")
'this is an example.'

>>> normalize("   EXTRA   SPACE   ")
'extra space'

2. [5 points] Write a function no_vowels(text) that removes all vowels from the input string and returns the result. For the purposes of this problem, the letter ‘y’ is not considered to be a vowel.

>>> no_vowels("This Is An Example.")
'Ths s n xmpl.'

>>> no_vowels("We love Python!")
'W lv Pythn!'

3. [5 points] Write a function digits_to_words(text) that extracts all digits from the input string, spells them out as lowercase English words, and returns a new string in which they are each separated by a single space. If the input string contains no digits, then an empty string should be returned.

>>> digits_to_words("Zip Code: 19104")
'one nine one zero four'

>>> digits_to_words("Pi is 3.1415...")
'three one four one five'

4. [5 points] Although there exist many naming conventions in computer programming, two of them are particularly widespread. In the first, words in a variable name are separated using underscores. In the second, words in a variable name are written in mixed case, and are strung together without a delimiter. By mixed case, we mean that the first word is written in lowercase, and that subsequent words have a capital first letter. Write a function to_mixed_case(name) that converts a variable name from the former convention to the latter. Leading and trailing underscores should be ignored. If the variable name consists solely of underscores, then an empty string should be returned.

>>> to_mixed_case("to_mixed_case")
'toMixedCase'

>>> to_mixed_case("__EXAMPLE__NAME__")
'exampleName'


## 6. Polynomials [39 points]

In this section, you will implement a simple Polynomial class supporting basic arithmetic, simplification, evaluation, and pretty-printing. An example demonstrating these capabilities is shown below.

1. [3 points] In this problem, we will think of a polynomial as an immutable object, represented internally as a tuple of coefficient-power pairs. For instance, the polynomial 2x+1 would be represented internally by the tuple ((2, 1), (1, 0)). Write an initialization method __init__(self, polynomial) that converts the input sequence polynomial of coefficient-power pairs into a tuple and saves it for future use. Also write a corresponding method get_polynomial(self) that returns this internal representation.

>>> p = Polynomial([(2, 1), (1, 0)])
>>> p.get_polynomial()
((2, 1), (1, 0))

>>> p = Polynomial(((2, 1), (1, 0)))
>>> p.get_polynomial()
((2, 1), (1, 0))

2. [4 points] Write a __neg__(self) method that returns a new polynomial equal to the negation of self. This method will be used by Python for unary negation.

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = -p; q.get_polynomial()
((-2, 1), (-1, 0))


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = -(-p); q.get_polynomial()
((2, 1), (1, 0))


3. [3 points] Write an __add__(self, other) method that returns a new polynomial equal to the sum of self and other. This method will be used by Python for addition. No simplification should be performed on the result.

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = p + p; q.get_polynomial()
((2, 1), (1, 0), (2, 1), (1, 0))


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = Polynomial([(4, 3), (3, 2)])
>>> r = p + q; r.get_polynomial()
((2, 1), (1, 0), (4, 3), (3, 2))


4. [3 points] Write a __sub__(self, other) method that returns a new polynomial equal to the difference between self and other. This method will be used by Python for subtraction. No simplification should be performed on the result.

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = p - p; q.get_polynomial()
((2, 1), (1, 0), (-2, 1), (-1, 0))


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = Polynomial([(4, 3), (3, 2)])
>>> r = p - q; r.get_polynomial()
((2, 1), (1, 0), (-4, 3), (-3, 2))


5. [5 points] Write a __mul__(self, other) method that returns a new polynomial equal to the product of self and other. This method will be used by Python for multiplication. No simplification should be performed on the result. Your result does not need to match the examples below exactly, as long as the same terms are present in some order.

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = p * p; q.get_polynomial()
((4, 2), (2, 1), (2, 1), (1, 0))


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = Polynomial([(4, 3), (3, 2)])
>>> r = p * q; r.get_polynomial()
((8, 4), (6, 3), (4, 3), (3, 2))


6. [3 points] Write a __call__(self, x) method that returns the result of evaluating the current polynomial at the point x. This method will be used by Python when a polynomial is called as a function. Hint: This method can be written in one line using Python’s exponentiation operator, **, and the built-in sum function.

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> [p(x) for x in range(5)]
[1, 3, 5, 7, 9]


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = -(p * p) + p
>>> [q(x) for x in range(5)]
[0, -6, -20, -42, -72]


7. [8 points] Write a simplify(self) method that replaces the polynomial’s internal representation with an equivalent, simplified representation. Unlike the previous methods, simplify(self) does not return a new polynomial, but rather acts in place. However, because the fundamental character of the polynomial is not changing, we do not consider this to violate the notion that polynomials are immutable.

The simplification process should begin by combining terms with a common power. Then, terms with a coefficient of zero should be removed, and the remaining terms should be sorted in decreasing order based on their power. In the event that all terms have a coefficient of zero after the first step, the polynomial should be simplified to the single term $$0⋅x^0$$, i.e. (0, 0).

python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = -p + (p * p); q.get_polynomial()
((-2, 1), (-1, 0), (4, 2), (2, 1),
(2, 1), (1, 0))
>>> q.simplify(); q.get_polynomial()
((4, 2), (2, 1))


python
>>> p = Polynomial([(2, 1), (1, 0)])
>>> q = p - p; q.get_polynomial()
((2, 1), (1, 0), (-2, 1), (-1, 0))
>>> q.simplify(); q.get_polynomial()
((0, 0),)


8. [10 points] Write a __str__(self) method that returns a human-readable string representing the polynomial. This method will be used by Python when the str function is called on a polynomial, or when a polynomial is printed.

In general, your function should render polynomials as a sequence of signs and terms each separated by a single space, i.e. "sign1 term1 sign2 term2 ... signN termN", where signs can be "+" or "-", and terms have the form "ax^b" for coefficient a and power b. However, in adherence with conventional mathematical notation, there are a few exceptional cases that require special treatment:

* The first sign should not be separated from the first term by a space, and should be left blank if the first term has a positive coefficient.
* The variable and power portions of a term should be omitted if the power is 0 , leaving only the coefficient.
* The power portion of a term should be omitted if the power is 1.
* Coefficients with magnitude 0 should always have a positive sign.
* Coefficients with magnitude 1 should be omitted, unless the power is 0.

<br/>
You may assume that all polynomials have integer coefficients and non-negative integer powers.

python
>>> p = Polynomial([(1, 1), (1, 0)])
>>> qs = (p, p + p, -p, -p - p, p * p)
>>> for q in qs: q.simplify(); str(q)
...
'x + 1'
'2x + 2'
'-x - 1'
'-2x - 2'
'x^2 + 2x + 1'


python
>>> p = Polynomial([(0, 1), (2, 3)])
>>> str(p); str(p * p); str(-p * p)
'0x + 2x^3'
'0x^2 + 0x^4 + 0x^4 + 4x^6'
'0x^2 + 0x^4 + 0x^4 - 4x^6'
>>> q = Polynomial([(1, 1), (2, 3)])
>>> str(q); str(q * q); str(-q * q)
'x + 2x^3'
'x^2 + 2x^4 + 2x^4 + 4x^6'
'-x^2 - 2x^4 - 2x^4 - 4x^6'



## 7. Python Packages [4 points]

In this section, you are allowed to use external python packages which are widely used in Artificial Intelligence.

1. [2 point] Numpy

Install the numpy first:

pip3 install numpy


Use numpy package to implement the sort_array(list_of_matrics) which takes in a list of matrices with various dimensions and return a sorted 1D array with decreasing order contains all the values in these matrices. The data type of the returned array is int.

>>> import numpy as np
>>> matrix1 = np.array([[1, 2], [3, 4]])
>>> matrix2 = np.array([[5, 6, 7], [7, 8, 9], [0, -1, -2]])
>>> sort_array([matrix1, matrix2])
array([ 9,  8,  7,  7,  6,  5,  4,  3,  2,  1,  0, -1, -2])


Check this tutorial for some hints.

2. [2 points] NLTK Install the nltk first:

pip3 install nltk


>>> import nltk


Implement the POS_tag(sentence) function which takes in a string and return the Part-of-Speech(POS) tag of each word in the sentence. To complete this task, you need to fulfill the requirements shown below:

1. Convert all the characters in the sentence to lower case.

2. Tokenize the sentence.

3. Remove the stop words and punctuation.

4. Conduct the pos tagging and return a list of tuples.

Here is a test case:

>>> import nltk
>>> sentence = 'The Force will be with you. Always.'
>>> POS_tag(sentence)
[('force', 'NN'), ('always', 'RB')]


In the following section, you will explore three classic puzzles from the perspective of uninformed search.

The skeleton file homework1.py contains empty definitions for each question has been provided. Since portions of this assignment will be graded automatically, none of the names or function signatures in this file should be modified. However, you are free to introduce additional variables or functions if needed.

For the remaining questions, you may import definitions from any standard Python library, and are encouraged to do so in cases where you find yourself reinventing the wheel. If you are unsure where to start, consider taking a look at the data structures and functions defined in the collections, itertools, math, and random modules.

## 8. N-Queens [25 points]

In this section, you will develop a solver for the $n$-queens problem, wherein $n$ queens are to be placed on an $n \times n$ chessboard so that no pair of queens can attack each other. Recall that in chess, a queen can attack any piece that lies in the same row, column, or diagonal as itself.

A brief treatment of this problem for the case where $n = 8$ is given on this Wikipedia article, which you may wish to consult for additional information.

1. [5 points] Rather than performing a search over all possible placements of queens on the board, it is sufficient to consider only those configurations for which each row contains exactly one queen.

Therefore, without taking any of the chess-specific constraints between queens into account, we want to first consider the number of possible placements of $n$ queens on an $n \times n$ board without or with the restriction that each row contains exactly one queen.

Implement the function num_placements_all(n), which returns the number of all possible placements of $n$ queens on an $n \times n$ board, and the function num_placements_one_per_row(n) that calculates the number of possible placements of $n$ queens on an $n \times n$ board such that each row contains exactly one queen.

Think carefully about why this restriction is valid, and note the extent to which it reduces the size of the search space. You should assume that all queens are indistinguishable for the purposes of your calculations.

2. [5 points] With the answer to the previous question in mind, a sensible representation for a board configuration is a list of numbers between $0$ and $n - 1$, where the $i$th number designates the column of the queen in row $i$ for $0 \le i \lt n$. A complete configuration is then specified by a list containing $n$ numbers, and a partial configuration is specified by a list containing fewer than $n$ numbers. Write a function n_queens_valid(board) that accepts such a list and returns True if no queen can attack another, or False otherwise. Note that the board size need not be included as an additional argument to decide whether a particular list is valid.

>>> n_queens_valid([0, 0])
False
>>> n_queens_valid([0, 2])
True

>>> n_queens_valid([0, 1])
False
>>> n_queens_valid([0, 3, 1])
True

3. [15 points] Write a function n_queens_solutions(n) that returns a list of all valid placements of $n$ queens on an $n \times n$ board, using the representation discussed above. The output may be in any order you see fit. Your solution should be implemented as a depth-first search, where queens are successively placed in empty rows until all rows have been filled. You may find it helpful to define a helper function n_queens_helper(n, board) that yields all valid placements which extend the partial solution denoted by board.

>>> n_queens_solutions(6)
[[1, 3, 5, 0, 2, 4], [2, 5, 1, 4, 0, 3],
[3, 0, 4, 1, 5, 2], [4, 2, 0, 5, 3, 1]]
>>> len(n_queens_solutions(8))
92


## 9. Lights Out [40 points]

The Lights Out puzzle consists of an $m \times n$ grid of lights, each of which has two states: on and off. The goal of the puzzle is to turn all the lights off, with the caveat that whenever a light is toggled, its neighbors above, below, to the left, and to the right will be toggled as well. If a light along the edge of the board is toggled, then fewer than four other lights will be affected, as the missing neighbors will be ignored.

In this section, you will investigate the behavior of Lights Out puzzles of various sizes by implementing a LightsOutPuzzle class. Once you have completed the problems in this section, you can test your code in an interactive setting using the provided GUI. See the last section for more details.

1. [2 points] A natural representation for this puzzle is a two-dimensional list of Boolean values, where True corresponds to the on state and False corresponds to the off state. In the LightsOutPuzzle class, write an initialization method __init__(self, board) that stores an input board of this form for future use. Also write a method get_board(self) that returns this internal representation. You additionally may wish to store the dimensions of the board as separate internal variables, though this is not required.

>>> b = [[True, False], [False, True]]
>>> p = LightsOutPuzzle(b)
>>> p.get_board()
[[True, False], [False, True]]

>>> b = [[True, True], [True, True]]
>>> p = LightsOutPuzzle(b)
>>> p.get_board()
[[True, True], [True, True]]

2. [3 points] Write a top-level function create_puzzle(rows, cols) that returns a new LightsOutPuzzle of the specified dimensions with all lights initialized to the off state.

>>> p = create_puzzle(2, 2)
>>> p.get_board()
[[False, False], [False, False]]

>>> p = create_puzzle(2, 3)
>>> p.get_board()
[[False, False, False],
[False, False, False]]

3. [5 points] In the LightsOutPuzzle class, write a method perform_move(self, row, col) that toggles the light located at the given row and column, as well as the appropriate neighbors.

>>> p = create_puzzle(3, 3)
>>> p.perform_move(1, 1)
>>> p.get_board()
[[False, True, False],
[True,  True, True ],
[False, True, False]]

>>> p = create_puzzle(3, 3)
>>> p.perform_move(0, 0)
>>> p.get_board()
[[True,  True,  False],
[True,  False, False],
[False, False, False]]

4. [5 points] In the LightsOutPuzzle class, write a method scramble(self) which scrambles the puzzle by calling perform_move(self, row, col) with probability $1/2$ on each location on the board. This guarantees that the resulting configuration will be solvable, which may not be true if lights are flipped individually. After importing the random module with the statement import random, the expression random.random() < 0.5 generates the values True and False with equal probability.
5. [2 points] In the LightsOutPuzzle class, write a method is_solved(self) that returns whether all lights on the board are off.

>>> b = [[True, False], [False, True]]
>>> p = LightsOutPuzzle(b)
>>> p.is_solved()
False

>>> b = [[False, False], [False, False]]
>>> p = LightsOutPuzzle(b)
>>> p.is_solved()
True

6. [3 points] In the LightsOutPuzzle class, write a method copy(self) that returns a new LightsOutPuzzle object initialized with a deep copy of the current board. Changes made to the original puzzle should not be reflected in the copy, and vice versa.

>>> p = create_puzzle(3, 3)
>>> p2 = p.copy()
>>> p.get_board() == p2.get_board()
True

>>> p = create_puzzle(3, 3)
>>> p2 = p.copy()
>>> p.perform_move(1, 1)
>>> p.get_board() == p2.get_board()
False

7. [5 points] In the LightsOutPuzzle class, write a method successors(self) that yields all successors of the puzzle as (move, new-puzzle) tuples, where moves themselves are (row, column) tuples. The second element of each successor should be a new LightsOutPuzzle object whose board is the result of applying the corresponding move to the current board. The successors may be generated in whichever order is most convenient.

>>> p = create_puzzle(2, 2)
>>> for move, new_p in p.successors():
...     print(move, new_p.get_board())
...
(0, 0) [[True, True], [True, False]]
(0, 1) [[True, True], [False, True]]
(1, 0) [[True, False], [True, True]]
(1, 1) [[False, True], [True, True]]

>>> for i in range(2, 6):
...     p = create_puzzle(i, i + 1)
...     print(len(list(p.successors())))
...
6
12
20
30

8. [15 points] In the LightsOutPuzzle class, write a method find_solution(self) that returns an optimal solution to the current board as a list of moves, represented as (row, column) tuples. If more than one optimal solution exists, any of them may be returned. Your solver should be implemented using a breadth-first graph search, which means that puzzle states should not be added to the frontier if they have already been visited, or are currently in the frontier. If the current board is not solvable, the value None should be returned instead. You are highly encouraged to reuse the methods defined in the previous exercises while developing your solution.

Hint: For efficient testing of duplicate states, consider using tuples representing the boards of the LightsOutPuzzle objects being explored rather than their internal list-based representations. You will then be able to use the built-in set data type to check for the presence or absence of a particular state in near-constant time.

>>> p = create_puzzle(2, 3)
>>> for row in range(2):
...     for col in range(3):
...         p.perform_move(row, col)
...
>>> p.find_solution()
[(0, 0), (0, 2)]

>>> b = [[False, False, False],
...      [False, False, False]]
>>> b[0][0] = True
>>> p = LightsOutPuzzle(b)
>>> p.find_solution() is None
True


Once you have implemented the functions and methods described in this section, you can play with an interactive version of the Lights Out puzzle using the GUI provided.

## 10. Linear Disk Movement [30 points]

In this section, you will investigate the movement of disks on a linear grid.

The starting configuration of this puzzle is a row of $\ell$ cells, with disks located on cells $0$ through $n - 1$. The goal is to move the disks to the end of the row using a constrained set of actions. At each step, a disk can only be moved to an adjacent empty cell, or to an empty cell two spaces away if another disk is located on the intervening square. Given these restrictions, it can be seen that in many cases, no movements will be possible for the majority of the disks. For example, from the starting position, the only two options are to move the last disk from cell $n - 1$ to cell $n$, or to move the second-to-last disk from cell $n - 2$ to cell $n$.

1. [15 points] Write a function solve_identical_disks(length, n) that returns an optimal solution to the above problem as a list of moves, where length is the number of cells in the row and n is the number of disks. Each move in the solution should be a two-element tuple of the form (from, to) indicating a disk movement from the cell from to the cell to.

As suggested by its name, this function should treat all disks as being identical.

Your solver for this problem should be implemented using a breadth-first graph search. The exact solution produced is not important, as long as it is of minimal length.

Unlike in the previous two sections, no requirement is made with regards to the manner in which puzzle configurations are represented. Before you begin, think carefully about which data structures might be best suited for the problem, as this choice may affect the efficiency of your search.

>>> solve_identical_disks(4, 2)
[(0, 2), (1, 3)]
>>> solve_identical_disks(5, 2)
[(0, 2), (1, 3), (2, 4)]

>>> solve_identical_disks(4, 3)
[(1, 3), (0, 1)]
>>> solve_identical_disks(5, 3)
[(1, 3), (0, 1), (2, 4), (1, 2)]

2. [15 points] Write a function solve_distinct_disks(length, n) that returns an optimal solution to the same problem with a small modification: in addition to moving the disks to the end of the row, their final order must be the reverse of their initial order. More concretely, if we abbreviate length as $\ell$, then a desired solution moves the first disk from cell $0$ to cell $\ell - 1$, the second disk from cell $1$ to cell $\ell - 2$, $\cdots$, and the last disk from cell $n - 1$ to cell $\ell - n$.

Your solver for this problem should again be implemented using a breadth-first graph search. As before, the exact solution produced is not important, as long as it is of minimal length.

>>> solve_distinct_disks(4, 2)
[(0, 2), (2, 3), (1, 2)]
>>> solve_distinct_disks(5, 2)
[(0, 2), (1, 3), (2, 4)]

>>> solve_distinct_disks(4, 3)
[(1, 3), (0, 1), (2, 0), (3, 2), (1, 3), (0, 1)]
>>> solve_distinct_disks(5, 3)
[(1, 3), (2, 1), (0, 2), (2, 4), (1, 2)]


## 11. Feedback [3 points]

1. [1 points] Approximately how long did you spend on this assignment?

2. [1 points] Which aspects of this assignment did you find most challenging? Were there any significant stumbling blocks?

3. [1 points] Which aspects of this assignment did you like? Is there anything you would have changed?