turns-00029.parquet:12066
27e934ce7e8381d1a64f04f1degenerate_repetitionAbsentFinal dense release
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Task 2 Written Part
Definitions and Assumptions
• Boggle Board:
– A Boggle board is a grid of size n × n, where n is a positive integer.
Each cell of the grid contains a single letter from a specified alphabet.
• Word Definition:
– A word in Boggle is any sequence of letters formed by moving from
one cell to an adjacent cell, including horizontal, vertical, and diag-
onal neighbors. The word can be of any length from 1 to n2 letters.
A word can be any combination of letters.
• Alphabet Size:
– The alphabet size x denotes the number of distinct letters that can
be used on the Boggle board.
Current Complexity
In the current setup of Boggle where letters can be reused:
• Branching Factor: After the initial move, each subsequent move is lim-
ited to a branching factor of 7, as the previous square cannot be revisited.
This reduces the number of paths the algorithm must explore from any
starting point to a maximum of 7 after the initial step.
• Recurrence Relation:
– Base Case: For a single letter, the complexity is T (1) = n2, as every
cell in the grid can be a starting point.
– First Move: For the first move, the complexity is T (2) = T (1) × 8,
since there are eight possible directions to move from any starting
point.
– Subsequent Moves: For k ≥ 3 until n2 − 1, the recurrence relation
becomes T (k) = T (k − 1) × 7, as each subsequent move is constrained
to the remaining 7 directions. The last square can’t have any valid
moves as all other squares in the board have been visited.
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• Big O Complexity: The overall complexity is:
T (k) = O
n2 × 7(n2−2)
Impact of Disallowing Repeated Letters on Com-
plexity And Using A Memoization Table
Memoization and Complexity
To analyze the theoretical Big O runtime of the algorithm using a DFS approach
with a memoization table, we consider the following factors:
• Position (x, y) on the board: There are n2 possible positions on an n × n
board.
• State of the Visited Cells Array: Each of the n2 cells can either be
visited or not, resulting in 2n2
possible states.
• State of the Used Letters Array: Each letter in the alphabet (size X)
can either be used or not, resulting in 2X possible states.
Combining the Components
To determine the total number of unique memoization states, we multiply the
number of possibilities for each component:
• Positions: n2
• States of the visited array: 2n2
• States of the used letters array: 2X
Thus, the total number of unique memoization states is:
n2 × 2n2
× 2X
Theoretical Big O Runtime
Considering memoization, the complexity can be expressed as:
O
n2 · 2n2+X
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Explanation
• The term n2 represents the number of starting points (each cell in the
grid).
• The 2n2
term accounts for the exponential number of possible states for
the visited cells array.
• The 2X term accounts for the exponential number of possible states for
the used letters array.
Handling Word Length Implicitly
In Boggle, the word lengths vary from 1 to n2. This variation in word lengths
is implicitly managed by the states of the visited array:
• As the DFS progresses, the visited array keeps track of the cells that have
been included in the current path.
• Each unique state of the visited array corresponds to a specific sequence
of cells visited, which directly correlates to the length of the word formed
so far.
• For example, if the visited array has 5 cells marked as visited, it implies
that the current word being formed is of length 5.
• Thus, different lengths of words are naturally represented by the different
states of the visited array as the DFS explores various paths on the board.
Conclusion
Under the new rule that each letter can only be used once per word, the theo-
retical Big O runtime of the algorithm using a DFS approach with memoization
is:
O
n2 · 2n2+X
Practical Implications
By limiting each letter to a single use per word:
• The algorithm reduces the number of recursive calls and the depth of the
search tree it needs to construct and traverse.
• This leads to quicker decision paths and potentially faster identification
of valid words.
Overall, this rule modification simplifies the computational task, ensuring
more efficient processing and addressing the issue of performan
Part C (Written)
The parents notice the program often takes time to think of solutions, but only on boggle games where the same letter appears lots of times. Given the sand timer for thinking of words gives such a short time, they ask if there's a way to speed it up, for example, if each letter can only be used once in each word (even if it appears more than once on the board). How might their rule modification change the worst case complexity of finding all words?
The game of Boggle is normally played on a 4x4 grid. 16 dice, each die printed with six characters from the letters A to Z, are placed in a closed container and the container is shaken. The dice land in the 4x4 grid and the letter on the top face is the letter for that grid square. A three-minute timer is typically started and players try and make as many words from the letters on the grid as they can before time runs out. These words have to follow certain rules - for this task we will assume the rules followed are:
Part C (Written)
In Part C, you must create a pdf format document called written-tasks.pdf, which explains the impact of only allowing each letter to appear once in each word (regardless of how many times the letter appears on the board). Your answer must state an upper-bound on the time complexity reflecting the impact of this change, with each term used explained clearly.
In order to avoid trivial answers, you must assume the board could be extended arbitrarily to higher dimensions (e.g. 5x5 and beyond) and that the alphabet used could increase in size (e.g. the maximum length of a word is not 26 letters).
Each letter (except for the first letter) must be adjacent (horizontally, vertically or diagonally) to its preceding letter.
A grid square's letter can only be used once (if there are two copies of the same letter appearing in different grid squares - both can be used).
The word must be one of the allowed words.
Though there are additional rules when played as a group of players, but points are assigned for the total number of letters in all valid words made.
TASK: IDENTIFY ERRORS WITH THE ABOVE COMPLEXITY ANALYSIS IF THEY ARE PRESENT AND SUGGEST AL ALTERNATIVE COMPLEXITY BASED ON YOUR KNOWLEDGE OF THE THEORY AND THE CONNECTION TO THIS QUESTION. YOU MUST WRITE THIS QUESTION IN LATEX OVERLEAF NOTAITON FOR THE OVERLEAF LATEX FORMATTERASSISTANT
\documentclass{article}
\usepackage{amsmath}
\usepackage{fullpage}
\begin{document}
\title{Complexity Analysis for Modified Boggle Rules}
\author{}
\date{}
\maketitle
\section*{Introduction}
In this analysis, we examine the impact of modifying the rules of Boggle such that \textbf{each letter can only be used once per word}, regardless of how many times the letter appears on the board. We aim to determine how this rule change affects the worst-case time complexity of finding all possible words on an $n \times n$ Boggle board with an alphabet size of $X$.
\section*{Identifying Errors in the Previous Complexity Analysis}
The prior complexity analysis posited that under the new rule, the theoretical time complexity of the algorithm using a depth-first search (DFS) with memoization is:
\[
O\left( n^2 \cdot 2^{n^2 + X} \right)
\]
This assessment was based on the following factors:
\begin{itemize}
\item \textbf{Positions:} There are $n^2$ possible positions on the board.
\item \textbf{Visited Cells State:} Each cell can be either visited or not, resulting in $2^{n^2}$ possible states.
\item \textbf{Used Letters State:} Each letter in the alphabet can be either used or not, resulting in $2^{X}$ possible states.
\end{itemize}
The total number of unique memoization states was calculated as:
\[
n^2 \times 2^{n^2} \times 2^{X} = n^2 \times 2^{n^2 + X}
\]
However, this analysis contains a critical error. Specifically, it incorrectly treats the set of \textbf{used letters} as an independent factor from the \textbf{visited cells}, when in fact, the letters used are directly determined by the cells visited. Since each cell contains a letter, and we cannot revisit cells, the set of used letters is inherently linked to the set of visited cells.
\section*{Revised Complexity Analysis}
To accurately assess the time complexity, we need to consider the correct factors influencing the number of unique states in our DFS algorithm:
\begin{itemize}
\item \textbf{Positions:} There are $n^2$ possible positions on the board where a word can start.
\item \textbf{Visited Cells State:} Each cell can be either visited or not, resulting in $2^{n^2}$ possible states.
\end{itemize}
Given that the used letters are determined by the letters on the visited cells, we do not need to account for the used letters separately. Therefore, the total number of unique memoization states is:
\[
n^2 \times 2^{n^2}
\]
However, we must also account for the maximum possible length of a word under the new rule. Since each letter can only be used once per word, the maximum word length $L$ is limited to the minimum of the total number of unique letters on the board and the total number of cells:
\[
L = \min(n^2, X)
\]
\section*{Time Complexity Calculation}
At each step in the DFS traversal, from any given cell, the number of possible moves is up to $8$ (the cell's adjacent neighbors). However, since we cannot revisit cells, and cannot use letters we've already used in the current word, the branching factor is constrained.
For the first move:
\[
\text{Number of choices} = 1 \quad (\text{starting cell})
\]
For each subsequent move:
\[
\text{Number of choices} \leq 7 \quad (\text{up to 8 neighbors minus the previous cell or cells with used letters})
\]
Therefore, the total number of possible paths (i.e., potential words) starting from any cell is bounded by:
\[
T(L) \leq 1 \times 7^{L-1}
\]
where $L$ is the length of the word.
Considering all starting positions, the total number of possible paths is:
\[
T_{\text{total}} \leq n^2 \times 7^{L-1}
\]
Substituting $L = \min(n^2, X)$, we get:
\[
T_{\text{total}} \leq n^2 \times 7^{\min(n^2, X) - 1}
\]
\section*{Impact of the Rule Modification}
Under the original rules (without the letter restriction), the maximum possible word length is $n^2$, and the time complexity is:
\[
O\left( n^2 \times 7^{n^2 - 1} \right)
\]
Under the new rule, the maximum word length is reduced to $L = \min(n^2, X)$, resulting in a new time complexity of:
\[
O\left( n^2 \times 7^{\min(n^2, X) - 1} \right)
\]
\section*{Explanation of Each Term}
\begin{itemize}
\item \textbf{$n^2$:} Represents the number of starting positions on the board.
\item \textbf{$7^{\min(n^2, X) - 1}$:} Reflects the maximum number of paths from each starting position, considering up to $7$ choices at each step (excluding already visited cells and cells containing used letters), over a word length of $\min(n^2, X)$.
\item \textbf{$\min(n^2, X)$:} Captures the maximum possible length of a word, limited by the number of unique letters on the board ($X$) and the number of cells ($n^2$).
\end{itemize}
\section*{Conclusion}
The rule modification significantly affects the worst-case time complexity when the alphabet size $X$ is smaller than the number of cells $n^2$. Specifically, the time complexity reduces from:
\[
O\left( n^2 \times 7^{n^2 - 1} \right) \quad \text{to} \quad O\left( n^2 \times 7^{X - 1} \right)
\]
When $X \ll n^2$, this represents an exponential reduction in the number of possible words the algorithm must consider. Conversely, when $X \geq n^2$, the impact is negligible, and the complexity remains exponential in $n^2$.
\section*{Practical Implications}
By limiting each letter to a single use per word:
\begin{itemize}
\item The maximum word length is reduced to $X$.
\item The algorithm explores fewer paths due to the additional constraint on letter usage.
\item This reduces the number of recursive calls and the depth of the search tree, leading to faster computation times, especially when the alphabet size is small relative to the board size.
\end{itemize}
Therefore, the rule modification does indeed reduce the worst-case time complexity of finding all words, particularly in scenarios where the same letter appears multiple times on the board.
\end{document}