Knight Probability in Chessboard

10 Prerequisite LeetCode Problems

“Knight Probability in Chessboard” is about finding the probability that a knight remains on the board after a specified number of moves. This problem involves using dynamic programming to keep track of the probability at each position on the board for each number of moves.

An isomorphic problem, i.e., a problem with the same underlying structure but with a different context, might be the “Robot Room Cleaner”.

Robot Room Cleaner involves a robot that is asked to clean a room represented as a grid (like a chessboard). The robot starts from a cell, and it can make a number of moves (similar to the knight) to clean cells, but it can’t go outside the room (like how the knight can’t leave the chessboard).

Both problems require tracking a position on a grid-like structure, making a specific number of moves, and handling boundary conditions. Instead of probability calculations as in the Knight Probability problem, the Robot Room Cleaner problem involves deciding the best path to clean all cells efficiently, thus employing a backtracking strategy.

From simpler to more complex:

1. “Flood Fill” (LeetCode) - It is a relatively straightforward problem involving depth-first search on a 2D grid.
2. “Number of Islands” (LeetCode) - Adds complexity by requiring identification of disconnected components on the grid.
3. “Robot Room Cleaner” (LeetCode) - Adds further complexity by including rules of movement and a requirement to cover all accessible cells.

The reasoning for these selections is in the similar mechanics of navigating a 2D grid. The problems increase in complexity due to the addition of movement rules and larger objectives like covering all spaces or counting components. Note that these are not exact mappings, but they follow similar patterns of problem-solving and use related techniques.

10 Prerequisite LeetCode Problems

This problem is about understanding how to apply dynamic programming (DP) and depth-first search (DFS) to a problem related to a probabilistic system.

Below are some simpler problems that cover similar themes and will help you prepare:

1. 70. Climbing Stairs: This problem introduces the basic concept of DP, where each step’s result depends on the previous ones.

2. 62. Unique Paths: Here, you will learn about counting the number of ways to reach a destination in a 2D grid, which is somewhat similar to the Knight Probability problem.

3. 64. Minimum Path Sum: This problem asks you to find the path with the minimum sum, which further reinforces the principles of DP on a 2D grid.

4. 63. Unique Paths II: This problem adds the obstacle in the grid, which will help you understand how to handle additional constraints.

5. 200. Number of Islands: This problem requires DFS to explore a 2D grid, similar to how the knight moves around the chessboard.

6. 279. Perfect Squares: This problem introduces BFS in the context of number theory, but the concept of exploring all possibilities until you find the best one is important for the Knight Probability problem.

7. 91. Decode Ways: This problem is another classic DP problem that could help you understand how to build solutions based on previous computations.

8. 322. Coin Change: This problem involves finding the minimum number of coins that you need to make up an amount. The DP approach used here is valuable.

9. 139. Word Break: This is a dynamic programming problem, where the answer to the current problem depends on the solution to smaller problems.

10. 416. Partition Equal Subset Sum: This problem is a classic 0/1 knapsack problem, which can help you understand DP with a binary decision.

The key to solving such problems is in understanding the principles of dynamic programming and DFS, as well as learning to deal with probability in a dynamic programming context.

Clarification Questions

Here are some clarification questions you might ask about this problem:

1. What should the output be if the initial position of the knight is at the corner of the chessboard and k is greater than 0? Is it assumed that the knight will move off the board after the first move, or is there a chance that it will not move?

2. Are all moves by the knight equally likely, or is there a bias towards certain directions?

3. If the knight is already off the board, should we consider further movements in the count of k moves?

4. How should the output probability be formatted? Is there a specific number of decimal places we need to consider?

5. What should we return if the knight starts off the board? Is this situation possible according to the problem constraints?

6. If k is 0, should we always return 1, indicating that the knight will definitely stay on the board as no moves are made, regardless of the starting position?

7. If n equals 1 (a 1x1 board) and k is greater than 0, should we assume that the knight will always move off the board after the first move?

8. What should we return in case of invalid input parameters, such as negative values for k, row, or column, or values for row or column that exceed n-1?

Problem Analysis and Key Insights

1. Problem is a Probabilistic Scenario: The knight’s moves are chosen uniformly at random, so the problem involves calculating probabilities based on possible outcomes.

2. Dynamic Programming Approach: As the knight can make ‘k’ moves, we need to remember the results of previous calculations. This hints towards a dynamic programming approach where we store the probability of the knight being on the board after ‘i’ moves from every possible position.

3. Grid Boundaries: The chessboard’s boundary conditions are important. If a move takes the knight off the chessboard, it stays off the board. We need to take this into account when calculating the probability.

4. Variable Chessboard Size: The chessboard is not fixed to the usual 8x8 size; it can be up to 25x25 according to the problem statement. This requires us to handle variable grid sizes.

5. Multiple Moves: The problem allows the knight to make ‘k’ moves, suggesting the solution will likely involve some form of iterative process or recursion to handle these multiple steps.

6. Knight’s Moves: A knight in chess has eight possible moves, but not all of them will be valid at all times. The set of valid moves changes depending on the knight’s position on the board, and especially when it’s near or at the edge of the board.

7. Starting Position: The knight’s starting position is an important factor. If it starts near or at the edge, it has a higher chance of moving off the board, especially when ‘k’ is large.

Problem Boundary

The scope of this problem is well-defined by the given problem statement and constraints. Specifically, the problem is to determine the probability that a chess knight remains on a given chessboard after a specified number of moves. The board is an n x n grid, and the knight has a specified starting cell. The size of the chessboard (n) can be between 1 and 25, inclusive, and the number of moves (k) can range from 0 to 100, inclusive.

Key areas within the scope include:

1. Implementing a way to represent the chessboard and the knight’s movements. Since a knight in chess can have up to eight possible moves from a given position, it’s essential to accurately implement these possible moves.

2. Designing a solution that efficiently handles the knight’s movements for up to 100 moves. Given that the number of moves can be relatively large, an efficient solution is necessary to avoid exceeding time limits. This suggests the use of techniques like dynamic programming to store intermediate results and avoid repeated calculations.

3. Taking into account the boundary conditions of the chessboard. The knight may move off the chessboard, which affects the probability calculations.

4. Calculating the probability that the knight remains on the board after k moves. This involves understanding and implementing probability calculations in the context of the problem.

It’s important to note that out-of-scope areas would include modifications to the rules of chess, changing the size of the chessboard beyond the specified limits, or considering other chess pieces or their movements.

Establishing the boundary of a problem helps to define the limits of what the solution should achieve, based on the problem statement and constraints. For this specific problem, we can establish boundaries as follows:

1. Chessboard Size: The size of the chessboard is defined to be a grid of size n x n, where n is between 1 and 25 inclusive. Solutions should be able to handle any grid size within this range.

2. Knight Moves: The knight can move in any of the eight standard ways a knight moves in chess, chosen uniformly at random. If a move takes the knight off the board, it cannot return. The solution must reflect this rule.

3. Number of Moves: The knight is expected to make exactly ‘k’ moves, where ‘k’ is an integer between 0 and 100 inclusive. The solution should be capable of handling the maximum number of moves without performance issues.

4. Starting Position: The knight starts at a specific cell on the board, denoted by the ‘row’ and ‘column’ inputs. The solution should be adaptable to any starting position.

5. Output: The output of the problem is a probability, i.e., a floating-point number between 0 and 1 inclusive. The solution must calculate this probability accurately, considering all possible move sequences for the knight.

6. Performance: The solution should be optimized to handle the upper constraints of the problem within a reasonable time limit. As the problem potentially involves a large number of calculations, the solution should employ techniques like dynamic programming to avoid redundant computations.

The boundaries ensure that the solution remains focused on the task, i.e., calculating the probability that the knight remains on the board after ‘k’ moves, given the defined rules and constraints. Other aspects, like the presence of other chess pieces or changes in the rules of chess knight movement, fall outside these boundaries.

Problem Classification

Domain of the Problem: This problem falls under the domain of Probability Theory and Combinatorial Game Theory in Mathematics. It is also related to Dynamic Programming in Computer Science, as it involves computing and storing the results of overlapping subproblems.

‘What’ Components:

1. Chessboard Configuration: An n x n chessboard with a knight chess piece. The top-left cell is (0,0) and the bottom-right cell is (n - 1, n - 1).

2. Knight Movement: The knight chooses one of its eight possible moves uniformly at random.

3. Knight’s Journey: The knight continues moving until it has made exactly k moves or has moved off the chessboard.

4. Probability Calculation: The problem is to calculate the probability that the knight remains on the chessboard after it has stopped moving.

Problem Classification: This problem can be classified in a few ways based on the above components:

1. Combinatorial Game: This problem involves calculating the probability of a specific outcome in a game with a large number of possible move sequences.

2. Probability Theory: The task requires the calculation of the probability of an event (the knight remaining on the board) after a series of random events (the moves).

3. Dynamic Programming: The problem involves solving overlapping subproblems (calculating the probability of staying on the board after each move) and combining these solutions to solve the main problem.

The classification of this problem helps guide the approach towards solving it. Recognizing it as a dynamic programming problem suggests that we can use techniques like memoization or tabulation to efficiently solve it, while the elements of probability theory inform the method of calculating the probabilities.

Distilling the Problem to Its Core Elements

Fundamental Concept: The fundamental concept of this problem is based upon Probability Theory. Probability Theory deals with uncertainty. Here, we need to calculate the probability of the knight staying on the board after a certain number of moves, each of which is chosen randomly from a set of possible moves.

Simplified Explanation: Imagine a game where a chess knight moves around on a chessboard making a certain number of moves, each chosen randomly from the possible moves a knight can make. If a move causes the knight to leave the board, it can’t return. Now, we want to know the chances, or probability, that the knight will still be on the board after it’s done making its moves.

Core Problem: The core problem we are trying to solve is to calculate the probability that a chess knight remains on an n x n chessboard after making k moves, chosen randomly from its possible moves.

Key Components:

1. Knight’s Moves: Understanding the possible moves a knight can make.

2. Chessboard: The environment in which the knight moves, including the understanding of boundaries.

3. Number of Moves (k): The exact number of moves the knight makes.

4. Probability Calculation: Determining the chance of the knight remaining on the board after making its moves.

Minimal Set of Operations:

1. Initialize a 3D array to store probabilities for each cell for every move from 0 to k.

2. For each of the k moves, update the probability of landing on each cell by averaging the probabilities of making a valid move from each of the eight possible positions.

3. At the end, the probability of the knight remaining on the board will be the sum of probabilities in all cells for the kth move.

Visual Model of the Problem

Visualizing this problem can be helpful for understanding its dynamics and the interactions between its components. Here’s how one might do it:

1. Chessboard: Start by drawing an n x n grid to represent the chessboard. For a simple visualization, you could consider a 3x3 or 4x4 grid. Mark the cells using row and column numbers, starting from (0,0) at the top-left cell to (n-1,n-1) at the bottom-right cell.

2. Knight’s Position: Mark the starting position of the knight on the chessboard with a different color or a symbol like “K”.

3. Knight’s Moves: Draw arrows from the knight’s starting cell to represent the eight possible moves of the knight.

4. Knight’s Journey: Now, consider the knight makes a move. Mark this new position and draw arrows from this new position to represent the next possible moves. If a move falls outside the grid, it represents the knight moving off the chessboard.

5. Probability: Assign equal probability (1/8) to each move since the moves are made randomly. For calculating the probability of staying on the board after k moves, sum up the probabilities of the paths that keep the knight on the board after k moves. Remember, once a knight moves off the board, it can’t return, so such paths are terminated.

This visualization helps in understanding the different states the problem could have and how the state changes as the knight makes its moves. You can see how the problem has many overlapping sub-problems, which indicates that dynamic programming might be an effective strategy for solving it.

Problem Restatement

Let’s distill the problem:

We have a knight chess piece on a square chessboard of size n x n. The knight begins at a specified cell on the board, and it will make exactly k moves. For each move, the knight picks one of its eight possible movements at random. If it steps off the board, it doesn’t return.

Our task is to compute the chance - or in other words, the probability - that after making its k moves, the knight is still on the chessboard.

The size of the chessboard, n, can be any integer between 1 and 25 inclusive. The number of moves, k, is an integer that lies between 0 and 100 inclusive. The initial position of the knight is specified by a ‘row’ and a ‘column’, both of which can be any integer between 0 and n - 1 inclusive.

Abstract Representation of the Problem

Let’s create an abstract representation of this problem:

Imagine a finite grid of size n x n. You have a unit located at a specified position within this grid. This unit is capable of moving in eight distinct directions from its current position, each direction chosen with equal probability.

The unit is scheduled to make k moves. With each move, it’s equally likely to proceed in any of the eight directions. If a move leads the unit outside the grid, it can’t return.

The task is to find out the probability that after k moves, the unit remains within the confines of the grid.

In this abstract representation, the grid corresponds to the chessboard, the unit corresponds to the knight, and the eight distinct directions correspond to the knight’s possible moves in chess. The movement of the unit captures the randomness of the knight’s movements and the condition of staying within the grid reflects the constraint of the knight remaining on the chessboard.

Terminology

There are several specialized terms and technical concepts crucial to understanding this problem:

1. Chessboard: In the context of this problem, a chessboard is a square grid of size n x n, where n can be any integer between 1 and 25. The chessboard is the environment where the knight moves.

2. Knight: In chess, the knight is a piece that moves in an L-shape: two squares in a straight line (vertically or horizontally) followed by one square perpendicular to that direction. It’s the only piece that can “jump over” other pieces. In this problem, the knight starts at a specific position and makes a number of moves.

3. Moves: A move refers to a change in position of the knight on the chessboard. Each move is chosen from one of the eight possible moves a knight can make, and is chosen uniformly at random. The knight continues to move until it has made exactly k moves, where k can be any integer from 0 to 100, or until it moves off the chessboard.

4. Probability: Probability is a measure of the likelihood that an event will occur. It’s calculated as the number of favorable outcomes divided by the total number of outcomes. In this problem, we want to calculate the probability that the knight remains on the chessboard after making exactly k moves.

5. 0-indexed: In the context of this problem, both rows and columns of the chessboard are 0-indexed. This means that numbering starts from 0 instead of 1. The top-left cell of the chessboard is (0, 0) and the bottom-right cell is (n - 1, n - 1).

6. Dynamic Programming: While not explicitly mentioned in the problem statement, dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It’s particularly useful when a problem has overlapping subproblems, as in this case. It involves storing the results of certain calculations, which are then reused rather than being recalculated. In the context of this problem, we use dynamic programming to store and reuse probabilities of reaching certain positions on the board after a given number of moves.

Problem Simplification and Explanation

Let’s simplify the problem and discuss it with an analogy:

Think of this problem as a game of blindfolded hopscotch played on a grid. Your game’s playground (the chessboard) is a square of n x n cells. Your starting position is a cell on this playground.

You’re blindfolded (to signify the random choice of direction), and you can hop in eight distinct ways from any cell (just like the knight’s L-shaped moves in chess). However, you can’t see where you’re hopping, and you can’t hop back onto the playground if you hop outside of it.

Now, you’re supposed to make exactly k hops, and these hops are in random directions. The question is: what is the likelihood (probability) that you’ll still be within the confines of the playground after you’ve made your k hops?

The key concepts here are:

1. Random Movement: This corresponds to the blindfolded hops you’re making. The direction is chosen at random from the eight possible ones.

2. Boundary Conditions: These are the edges of the playground. If you hop beyond these, you’re out of the game.

3. Probability: This is the chance of you remaining within the playground after k hops. Each hop you make affects this probability, and we’re interested in the cumulative effect after k hops.

In this analogy, each hop you make is independent and contributes to the overall probability. The goal is to calculate this final probability after k hops. This involves keeping track of all the valid movements you can make at each hop, which is a core part of the problem and often solved using dynamic programming.

Constraints

The problem statement and constraints offer some useful details that can help us form an efficient solution:

1. Finite Movements: The knight can make exactly k moves. The fact that the number of moves is finite and relatively small (maximum 100) means that we can iteratively calculate the probability for each move.

2. Limited Directions: The knight has only eight possible moves from any position on the board. This provides a finite number of cases we have to consider at each step.

3. Board Size: The size of the board is capped at 25 x 25, which limits the number of possible positions for the knight. This makes it feasible to use a memory-intensive method like dynamic programming.

4. 0-Indexed Board: The board is 0-indexed. While this doesn’t provide a computational advantage, it simplifies the process of computing the knight’s position after a move.

5. Independence of Moves: Each move is independent and chosen uniformly at random. This allows us to simply divide the probability by 8 for each subsequent move.

6. Overlapping Subproblems: The same positions will be visited multiple times with a different number of remaining steps. This overlap is a key characteristic that can be exploited by using dynamic programming to store results of subproblems to avoid redundant computations.

7. Uniform Probability: Each of the knight’s moves is equally likely, simplifying the computation of the probabilities.

Recognizing these features will allow us to build an efficient solution based on dynamic programming, iteratively calculating probabilities for remaining on the board after each move, starting from the initial position and considering all possible subsequent positions.

Analyzing the constraints provides key insights that help in approaching the problem:

1. Finite and Limited Steps: The knight can make exactly k moves where k can go up to 100. This suggests that a brute force method of checking every possible sequence of moves would not be feasible as the number of sequences grows exponentially with k. However, since k is finite and relatively small, an iterative or recursive solution which calculates and combines results from each step is plausible.

2. Small Chessboard Size: The chessboard is at most 25 x 25, limiting the total number of positions. This hints that a method like dynamic programming, which may require a 3D array (considering position on the board and number of moves made) for storing intermediate results, is feasible in terms of memory usage.

3. Uniform Probability: The knight has eight possible moves from any position, each chosen uniformly at random. This simplifies the calculation of the probabilities, as each move contributes equally to the probability of staying on the board.

4. Start and End Points: The knight starts at a specific cell, and we’re interested in whether it ends on the board, not where it ends. This means we don’t need to calculate probabilities for every ending position, just whether those positions are on the board or not.

By understanding these constraints, we can leverage them to design a more efficient solution to the problem. Specifically, they suggest a dynamic programming approach where we iteratively calculate and store probabilities for each position and step count, which are then used to calculate probabilities for the next step.

Case Analysis

We can explore more examples, including edge and boundary conditions. Let’s divide them into three categories:

1. Base Cases:

   Case 1: n = 1, k = 0, row = 0, column = 0 The knight is at the only cell of a 1x1 chessboard and doesn’t need to make any move. Since the knight is already on the board and doesn’t need to move, the probability that it remains on the board is 1.00000.

2. Boundary Cases:

   Case 2: n = 3, k = 3, row = 0, column = 0 This case tests the condition where the knight is at the edge of the board and has to make multiple moves. As the knight is at the edge of the board and has to make multiple moves, it has a higher chance of moving off the board, so the resulting probability should be less than 1.

   Case 3: n = 25, k = 100, row = 12, column = 12 This case tests the upper limit of the constraints where the knight is at the center of the largest possible chessboard and has to make the maximum number of moves. Since the knight starts at the center of a large chessboard and has to make a lot of moves, it could end up anywhere on the board. Calculating the exact probability would require considering a large number of possibilities.

3. Edge Cases:

   Case 4: n = 2, k = 1, row = 0, column = 0 This case tests the situation where the knight is at the corner of a small 2x2 chessboard and has to make one move. Since all moves from a corner cell of a 2x2 chessboard take the knight off the board, the probability that the knight remains on the board is 0.

   Case 5: n = 3, k = 1, row = 1, column = 1 This case tests the scenario where the knight is at the center of a small 3x3 chessboard and has to make one move. The knight can move to any cell on the board from the center of a 3x3 chessboard, so the probability that it remains on the board after one move is 1.

These examples cover different aspects of the problem, such as the size of the chessboard, the position of the knight, and the number of moves the knight has to make, and help in understanding the problem better and ensuring that the solution handles all possible scenarios.

Let’s use a grid to represent the chessboard where each cell represents a possible position for the knight. We’ll mark the initial position of the knight with a ‘K’. For each case, we’ll also show the possible positions the knight can move to in the next step, marking them with an ‘X’. Let’s visualize each case:

1. Base Case:

   Case 1: n = 1, k = 0, row = 0, column = 0 Here, the knight does not move. The knight stays at the initial position.

   K
   

2. Boundary Cases:

   Case 2: n = 3, k = 3, row = 0, column = 0 The knight starts at the corner and can move to two positions.

   K 0 X
   0 X 0
   X 0 0
   

   Case 3: n = 25, k = 100, row = 12, column = 12 The knight starts at the center. For simplicity, let’s show a 5x5 slice of the 25x25 board around the knight. The knight can move to eight positions.

   0 X 0 X 0
   X 0 0 0 X
   0 0 K 0 0
   X 0 0 0 X
   0 X 0 X 0
   

3. Edge Cases:

   Case 4: n = 2, k = 1, row = 0, column = 0 The knight starts at the corner and all its moves take it off the board.

   K X
   X 0
   

   Case 5: n = 3, k = 1, row = 1, column = 1 The knight starts at the center and can move to any cell.

   X 0 X
   0 K 0
   X 0 X
   

The visualization helps to understand the possible moves the knight can make and how it affects the probability of the knight remaining on the board after it has stopped moving.

Analyzing the different cases brings us several key insights:

1. Initial Position Matters: The initial position of the knight on the chessboard significantly impacts the probability. If the knight starts from a corner or edge, it has fewer valid moves compared to when it starts from a cell not touching the board’s edge or corners.

2. Number of Moves Matters: The more moves the knight has to make, the higher the chances it will move off the board, especially if it starts near the edge or corners.

3. Chessboard Size Matters: On a larger board, the knight has a greater chance of staying on the board, especially if it starts from a cell not touching the edge or corners. Conversely, on a smaller board, the knight has a higher chance of moving off the board, regardless of the number of moves.

4. Corner Cases: For a board of size less than 3x3, the knight can’t make any move without moving off the board. It’s an important edge case to consider.

These insights give us a better understanding of the problem and can guide the development of an efficient solution strategy. They also highlight the need for the solution to account for varying board sizes, initial positions, and move counts.

Here’s how the output is computed from the given input:

1. Initialize: We start with a 3x3 chessboard (n=3), a knight at position (0,0), and we need to make exactly 2 moves (k=2). The knight has eight possible moves it can make, but since we are on the edge of the board, only two of these are valid: moving to (2,1) and moving to (1,2).

2. First Move: We make our first move. For each of the two possible positions, there are also two possible moves that the knight can make to remain on the board. This means that for each starting move, there are two more valid moves the knight can make.

3. Probability Calculation: Since the knight chooses one of eight possible moves uniformly at random, the probability for each move is 1/8. The total probability of the knight staying on the board after the first move is 2/8 = 0.25, since there are two valid moves.

4. Second Move: For the second move, from each of the two positions reached after the first move, there are also two moves that will keep the knight on the board. The total probability for this move is also 0.25.

5. Final Probability: The final probability is the product of the probabilities of the first and second move, i.e., 0.25 * 0.25 = 0.0625.

6. Output: Therefore, after 2 moves, the knight has a 0.0625 probability of remaining on the board.

This process can be visualized as a tree with branches representing the knight’s moves. Each level of the tree represents a move, and the valid moves form the branches to the next level. The probability of each path (from root to a leaf) is the product of the probabilities of the moves (branches) in the path. The final probability is the sum of the probabilities of all paths that keep the knight on the board.

Let’s imagine the visualization as a decision tree. For clarity, let’s refer to the positions the knight can move to as P1, P2, etc. and remember that the knight has only 2 valid moves from the starting position (0,0) on a 3x3 chessboard.

    Start
    /   \
  P1     P2 
 / \     / \
P3  P4  P5  P6

At the start, the knight is at (0,0).

• It has 2 options: move to Position 1 (P1 which is (2,1)) or Position 2 (P2 which is (1,2)).
• From P1, it can further move to two positions: P3 and P4.
• Similarly, from P2, it can move to two positions: P5 and P6.

Remember, each level in this tree represents a move by the knight, and each node represents a position on the chessboard.

The probability of each move is 1/8, and the total probability after each move is the sum of the probabilities of the valid moves. After 2 moves (i.e., reaching the second level in the tree), the total probability of the knight remaining on the board is (1/8 * 2) * (1/8 * 2) = 0.0625. This is represented by the total number of end-nodes (leaf nodes) in the tree that represent valid positions on the chessboard, divided by the total number of end-nodes the tree could have had if all moves were valid.

This tree structure is a simplified visualization. In practice, the tree would be much larger, considering all possible moves (valid and invalid) at each step.

Identification of Applicable Theoretical Concepts

A few mathematical and algorithmic concepts can be applied to this problem to make it more manageable:

1. Dynamic Programming (DP): This problem can be solved efficiently using dynamic programming, where we calculate the probabilities for each step recursively. Since the next state depends only on the current state and not on how we got there, we can store intermediate results to avoid repeated calculation. In our case, the state can be defined by the remaining moves and the current position.

2. Depth-First Search (DFS): DFS is a traversing or searching algorithm in tree/graph data structures. In this problem, each position the knight can move to can be considered as a node, and the path the knight takes forms a tree or a graph. The DFS approach can be used to simulate all possible moves the knight can make.

3. Probabilities: The problem deals with probabilities. Understanding how to calculate probabilities can simplify the problem. In this case, since each move is equally likely, the probability of moving to each valid next position is 1 divided by the number of valid moves.

4. Memoization: Given the nature of the problem, a lot of states will be visited multiple times. Therefore, we can use a memoization table to avoid recalculation, which will significantly improve the efficiency of the solution.

Applying these concepts, we can break the problem into smaller sub-problems, solve them, and then combine their solutions to solve the original problem. This bottom-up approach will make the problem more manageable and computationally efficient.

Simple Explanation

Let’s imagine we’re playing a game on a square grid or a checkerboard. In this game, you’re controlling a knight piece from a game of chess. Now, the knight in chess is a unique piece because of the way it moves. It can leap to any square that is two squares away horizontally and one square vertically, or two squares vertically and one square horizontally, forming an L shape.

For this game, your knight starts on a certain square of the board, and it can make exactly ‘k’ number of moves. The catch here is that the knight might sometimes leap out of the board, and if that happens, it’s out of the game.

Now, after making exactly ‘k’ leaps or moves, what you have to figure out is what are the chances that your knight will still be on the board and not leap out. You have to consider all the possible leaps your knight could make for each move, even if it means the knight might jump out of the board.

So in essence, this game is about predicting the chances of your knight remaining on the board after making ‘k’ leaps, starting from a certain square on the board.

To simplify it further with a metaphor, think of it like a game of hopscotch, where some squares lead to you falling into water if you step on them. Your goal is to figure out your chances of staying dry after a certain number of hops.

Problem Breakdown and Solution Methodology

Here’s a step-by-step approach to the problem:

1. Understanding the Problem: First, we need to comprehend the problem. We’re given a chessboard of size n x n, a knight, and a number of moves, k. The knight can move in 8 possible directions, and the objective is to find the probability that the knight will still be on the board after k moves, given it starts at a particular location.

2. Identifying the Key Elements: The key elements in the problem are the size of the chessboard, the starting position, the number of moves, and the 8 possible directions a knight can move.

3. Setting Up the Problem: We can set up the problem by creating a 3D array dp of size [k+1][n][n] to hold our dynamic programming states. The third dimension k is for the number of moves. We initialize all cells of dp to -1 to signify that they’re not computed yet.

4. Applying Dynamic Programming: Now we apply a top-down dynamic programming approach. We start from the initial position, and for each move, we try all possible 8 directions. If the new position is inside the board, we add its probability to the current position’s probability. If it’s not, we don’t add anything.

5. Calculating Probability: Each move from a position is equally likely. So, the probability of moving to a particular position is the probability at the current position divided by the number of possible moves.

6. Memoization: As we calculate the probabilities, we store the result in our dp array. If we come across a state that we’ve already computed, we can simply return the stored result.

7. Handling Base Cases: If the knight has made all k moves, we return 1, indicating a successful sequence of moves. If the knight moves outside the board before making all k moves, we return 0, indicating a failed sequence.

8. Obtaining the Final Result: After considering all k moves from the initial position, the result in dp[k][row][column] will be the probability that the knight remains on the board.

As an example, consider a 3x3 board where the knight starts at position (0, 0) and is allowed to make 2 moves. Initially, the dp array is filled with -1. After calculating probabilities for all possible moves, we find that there are only 2 valid positions the knight can move to for the first move. For each of these positions, there are also 2 valid moves for the second move. Thus, the final probability is 2/8 * 2/8 = 0.0625, which is the value stored in dp[2][0][0].

Changing the problem’s parameters, such as the size of the board, the starting position, or the number of moves, would result in different probabilities. Larger boards or fewer moves would generally increase the probability of the knight remaining on the board, while smaller boards, more moves, or starting positions near the edges could decrease this probability.

This problem is essentially about traversing a graph (the chessboard) using depth-first search, while using dynamic programming to avoid recalculating the same states. It’s a beautiful combination of these two fundamental computer science concepts.

Inference of Problem-Solving Approach from the Problem Statement

Let’s identify key terms and concepts in this problem:

1. Knight’s Move: In chess, a knight moves two squares in one direction (horizontally or vertically) and then one square perpendicular to that direction. Understanding this move helps in defining the possible steps the knight can take from any position.

2. Probability: The problem asks for the probability of the knight staying on the board after a certain number of moves. Probability is calculated as the ratio of the number of favorable outcomes to the total number of outcomes. In this case, a favorable outcome is a valid move that keeps the knight on the board, and total outcomes are all possible knight moves, valid or invalid.

3. Recursive Problem: The problem has a recursive nature since the knight can make a series of moves, and the probability of remaining on the board depends on the outcome of each previous move. Recognizing this property guides us towards using a recursive approach or dynamic programming.

4. Grid or Board: The knight moves on a chessboard, which is essentially a 2D grid. Understanding this guides us to use a 2D array for representing the board and calculating probabilities.

To visualize these concepts, consider creating a table or 2D grid representing the chessboard. Mark the starting point of the knight and plot the possible moves it can make. Each cell of this table can be filled with the probability of the knight reaching that cell. This table can be updated iteratively for each move, thus helping you understand and solve the problem.

The problem statement presents a scenario where the final outcome (the knight staying on the board) depends on a series of decisions (the knight’s moves), and each decision affects the next one. This is a hint that we could use a recursive approach to solve the problem, as we can break it down into smaller, overlapping sub-problems.

Dynamic Programming (DP) is a strategy to solve recursive problems with overlapping sub-problems in an efficient way. The key indication that DP may be applicable to this problem is the fact that the probability of the knight being on the board after ‘k’ moves depends on the probabilities of being on the board after ‘k-1’ moves.

This overlapping of sub-problems (where the same sub-problems may be calculated multiple times) is a characteristic of many dynamic programming problems. By storing the results of these sub-problems, we can avoid redundant calculations, thereby optimizing the solution.

In summary, the presence of a decision-making process, the need to remember past decisions, and the overlapping sub-problems suggested by the problem statement are strong indicators that dynamic programming could be a viable solution strategy.

Simple Explanation of the Proof

Let’s break down the Knight Probability in Chessboard problem and the way a dynamic programming approach provides a solution.

Our goal is to find the probability that the knight remains on the board after ‘k’ moves starting from a given position.

The crux of the proof for the dynamic programming approach lies in understanding two fundamental points:

1. Transition probability: At any given point, the knight has 8 potential moves, and it chooses any move with equal probability (1/8). Hence, the transition from one state (i.e., position on the board) to another has a fixed probability of 1/8, as long as the move is valid (i.e., stays within the board).

2. Memoryless property: The probability of ending up in a certain state after ‘k’ moves only depends on the states reachable after ‘k-1’ moves and does not depend on the path the knight took to reach there.

Now, let’s consider how the algorithm works:

1. We start with a 3D array dp[i][j][step] that represents the probability of the knight being on the cell (i, j) after ‘step’ moves.

2. Initialize dp[r][c][0] = 1 as the knight starts from cell (r, c) and zero moves have been made. The rest of the dp table is initialized to 0.

3. Now we start filling our dp table iteratively for each move from 0 to ‘k’. For each cell (i, j), we consider all possible cells from where the knight could have arrived at (i, j) in the previous step. We add up the transition probabilities from all those cells to (i, j) which is the probability of the knight being on a certain cell in the previous step divided by 8 (since each move is equally likely). This gives us the total probability of being on cell (i, j) after ‘step’ moves.

4. Finally, we sum up the probabilities in the dp table after ‘k’ moves for all cells within the board, which gives us the final answer.

This way, we’re dividing the problem into sub-problems (calculating probabilities for smaller numbers of moves), which we solve first. The solutions to these sub-problems then combine to give us the solution to our original problem.

The proof of correctness of this approach lies in the validity of the transition probabilities and the memoryless property, which guarantees that the final probability is the sum of the probabilities of all valid paths the knight could take.

Stepwise Refinement

Here’s how we can distill our approach into more granular, actionable steps:

Step 1: Define a 3-dimensional array ‘dp’ with size n x n x k+1 to store the intermediate results, where dp[i][j][step] represents the probability of the knight being on cell (i, j) after ‘step’ moves.

Step 2: Initialize the ‘dp’ array, where dp[row][column][0] = 1 as the knight starts at cell (row, column) and has made zero moves. All other entries in the dp array are initially 0.

Step 3: Now we need to fill our dp array in a step-by-step manner. For each step from 1 to ‘k’, and for each cell on the board, we calculate dp[i][j][step] by considering all the previous positions from which the knight could have moved to cell (i, j). For each of these valid previous positions, we add the corresponding dp value divided by 8 to dp[i][j][step].

Step 4: Finally, once the dp array is fully filled, we sum up the values in dp[i][j][k] for all cells (i, j) within the board. This sum will be our answer, the total probability of the knight remaining on the board after ‘k’ moves.

Independent Parts & Repeatable Patterns

In terms of parts that can be solved independently, each cell’s probability for each step is calculated independently, based on previous results. That is, dp[i][j][step] does not depend on any other cell’s probability for the current step, but only on the previous step.

The repeatable pattern here is the calculation of transition probabilities for each cell and each step, which involves the same operation (adding up probabilities from previous positions divided by 8). This operation is repeated for all cells and all steps.

Refinement

This solution could be refined by noting that the dp array for step ‘k’ only depends on the dp array for step ‘k-1’. Therefore, we could reduce the space complexity from O(n^2 * k) to O(n^2) by keeping only the 2D dp arrays for the current and previous steps. This refinement takes advantage of the fact that the problem has optimal substructure and overlapping subproblems, which is the hallmark of dynamic programming.

Solution Approach and Analysis

Our goal in this problem is to find out the probability of a knight remaining on an n x n chessboard after making exactly k moves, starting from a given cell. We’re dealing with a knight’s movements on a chessboard, and we need to deal with probabilities as well. Here is a step-by-step breakdown of how we can approach solving it:

Step 1: Understand the Knight’s Movements

First, we need to understand the knight’s movements on a chessboard. A knight can move in a “L” shape - that is, it can move two squares in any cardinal direction (north, south, east, or west), and then one square in an orthogonal direction (perpendicular to the initial direction). In total, a knight has 8 possible moves it can make.

Step 2: Establish a Probabilistic Model

Next, we need to think about how the knight moves probabilistically. At each step, the knight has 8 potential moves it can make, and it chooses among them uniformly at random. This means that each potential move is equally likely, so each has a probability of 1/8.

Step 3: Dynamic Programming Table

Now we set up a dynamic programming table to track the probabilities of landing on each cell of the chessboard after each move. This table is a 3-dimensional array ‘dp’ with size n x n x (k+1), where dp[i][j][step] represents the probability of the knight being on cell (i, j) after ‘step’ moves.

Step 4: Base Case

We initialize our dynamic programming table by setting dp[row][column][0] = 1, because the knight starts at cell (row, column) and has made zero moves, so the probability is 1. All other entries in the dp array are initially 0.

Step 5: Transition Function

Now we need to fill our dp table. For each step from 1 to ‘k’, and for each cell on the board, we calculate dp[i][j][step] by considering all the previous positions from which the knight could have moved to cell (i, j). For each of these valid previous positions, we add the corresponding dp value divided by 8 to dp[i][j][step].

Step 6: Final Calculation

Once the dp table is fully filled, we sum up the probabilities in dp[i][j][k] for all cells (i, j) within the board. This sum will be our answer - the total probability of the knight remaining on the board after ‘k’ moves.

Example and Explanation

Let’s consider an example: n = 3, k = 2, row = 0, column = 0

• After 0 moves, our knight starts at cell (0,0), so the probability is 1 for cell (0,0) and 0 for all other cells.
• After 1 move, the knight can move to either cell (1,2) or cell (2,1) from cell (0,0). So, dp[1][2][1] and dp[2][1][1] are each 1/8, and all other cells have probability 0.
• After 2 moves, the knight can move to four possible cells from each of the cells (1,2) and (2,1). But only two of them are on the board from each cell. So, each of these cells will add 1/8 * 2/8 = 1/32 to the total probability.
• Summing up all the probabilities in dp[i][j][2] for all cells (i, j) within the board, we get 0.0625.

This dynamic programming approach gives us a way to compute the probabilities incrementally, reusing previous results to calculate the current one, which makes it efficient. And as the size of the chessboard (n) or the number of moves (k) increases, the time and space complexity of our solution will scale accordingly.

Identify Invariant

Invariants are conditions or properties in a problem or algorithm that remain unchanged across iterations or throughout the execution. They are useful for reasoning about the correctness and behavior of a problem or an algorithm.

For this particular problem, we have a probability invariant. It is:

The sum of probabilities of all possible positions of the knight after ‘k’ moves equals 1.

This is because the knight must be in one of the possible positions (including off-board positions) after ‘k’ moves. This holds true for every ‘k’, from 0 to the given ‘k’.

This invariant is important in this problem as it helps us understand the nature of probabilities involved and ensure that our computations are correct as the sum of all possibilities should always be 1. It also assists in building our dynamic programming transition as the probabilities at step ‘k’ are derived from those at step ‘k - 1’.

Identify Loop Invariant

A loop invariant is a condition or set of conditions that remain unchanged throughout each iteration of a loop.

In the context of the “Knight Probability in Chessboard” problem, the loop invariant could be stated as:

At each step ‘k’ for all positions on the chessboard, the probability stored for each position is the correct probability that the knight will be at that position after ‘k’ moves.

This holds true for each iteration ‘k’, provided the probabilities are correctly initialized before the loop begins (at k = 0, probability is 1 at the starting position and 0 everywhere else).

This invariant is central to the correctness of the dynamic programming solution. It allows us to correctly update the probabilities for each position at each step, building on the previously computed values, with the confidence that they represent the correct probabilities up to that point. This ensures that at the end of our loop (when k equals the number of moves the knight is to make), the probabilities we have stored indeed represent the correct probabilities that the knight ends at each position after those many moves.

Thought Process

In order to solve the “Knight Probability in Chessboard” problem, we need to carefully consider the problem statement. The essential point is that the knight is making exactly ‘k’ moves, and each move has equal probability, regardless of the board’s boundaries.

The first step is to clearly understand how a knight moves on the chessboard, which is unique compared to other pieces. For each possible position of the knight, we have a maximum of 8 possible moves.

To start solving this problem, it’s crucial to identify its underlying structure, which is a Markov Decision Process. This recognition leads us to use a dynamic programming (DP) approach. We’re dealing with probabilities and multiple states (positions on the board), and the outcome at each state is independent of how we got there, depending only on the current state.

The DP solution can be outlined as follows:

1. Initialize a 3D DP table. Dimensions are ’n’ x ’n’ x ‘k’, where ’n’ is the size of the chessboard and ‘k’ is the number of moves the knight makes. Initially, fill this table with -1, representing that these states have not yet been computed.

2. Create a recursive function, which takes the current row, column, and number of remaining moves as inputs. This function returns the probability of the knight remaining on the board.

3. In the recursive function, first check if the current state (row, column, moves left) has already been computed. If it has, return the computed value. If not, and if there are no remaining moves, return 1 if the current position is on the board, else return 0.

4. If there are remaining moves, and the current position is on the board, calculate the average of the probabilities of moving to each possible next position. Save this average probability in the DP table for the current state, and return it.

5. Finally, the answer is the return value when calling the recursive function with the initial row, column, and ‘k’ moves.

Below is the Python code for this approach:

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class Solution:
    def knightProbability(self, n: int, k: int, row: int, column: int) -> float:
        # Initialize a 3D DP table
        dp = [[[-1 for _ in range(k+1)] for _ in range(n)] for _ in range(n)]

        # Call the recursive function
        return self.findKnightProbability(n, k, row, column, dp)

    def findKnightProbability(self, n, k, row, column, dp):
        # Define the possible moves for a knight
        moves = [[-2, -1], [-1, -2], [1, -2], [2, -1], [-2, 1], [-1, 2], [1, 2], [2, 1]]

        # Check if the knight is out of the chessboard
        if row < 0 or row >= n or column < 0 or column >= n:
            return 0.0

        # Check if there are no more moves left
        if k == 0:
            return 1.0

        # Check if the current state has already been computed
        if dp[row][column][k] != -1:
            return dp[row][column][k]

        # Initialize the probability for the current cell
        prob = 0.0

        # Try all possible moves
        for move in moves:
            prob += 0.125 * self.findKnightProbability(n, k-1, row+move[0], column+move[1], dp)

        # Store the result in the dp table
        dp[row][column][k] = prob

        return prob

By adopting a DP approach, we’re able to effectively reduce repeated computations, providing a viable solution to the problem.

Establishing Preconditions and Postconditions

1. Parameters:

   • The method knightProbability takes four inputs: n, k, row, column.
   • The types of these parameters are all integers.
   • In the context of the problem, n is the dimension of the chessboard, k is the number of moves the knight makes, row and column are the starting position of the knight on the chessboard.

2. Preconditions:

   • Before this method is called, no specific state of the program is required.
   • Constraints on the input parameters are: 1 <= n <= 25, 0 <= k <= 100, 0 <= row, column <= n - 1.
   • There is no requirement for the program to be in a specific state before this method is called.

3. Method Functionality:

   • This method is expected to return the probability that the knight remains on the chessboard after it has made exactly k moves.
   • It interacts with the inputs and calculates the probability using a dynamic programming approach.

4. Postconditions:

   • After the method has been called and has returned, the calculated probability is returned. The state of the program remains unchanged as no global variables or states are altered.
   • The return value represents the probability that the knight remains on the board after it has stopped moving.
   • There are no side effects as all computations are done in local variables and returned.

5. Error Handling:

   • The method assumes that the preconditions are met and does not include specific error handling.
   • If the preconditions are not met, it will either return incorrect results or raise an exception, depending on how it’s used. For example, if n, k, row, or column are outside of their respective ranges, the method may not work correctly. But within this context, we expect these conditions to be met.

Problem Decomposition

1. Problem Understanding:

   • The problem is about finding the probability that a knight remains on an n x n chessboard after making exactly k moves. The knight starts at the position specified by row and column. The key components are the chessboard, the knight’s starting position, the number of moves, and the eight possible moves of the knight.

2. Initial Breakdown:

   • Major parts or stages of the problem:
     1. Understanding the knight’s movements.
     2. Simulating the knight’s movements for k turns.
     3. Calculating the probability of the knight being on the board.

3. Subproblem Refinement:

   • The subproblems can be further broken down as follows:
     1. Understanding the knight’s movements:
        • List all the eight possible movements of the knight.
     2. Simulating the knight’s movements for k turns:
        • For each move, calculate the new position of the knight.
     3. Calculating the probability of the knight being on the board:
        • Determine if the knight is still on the board after making a move.
        • If yes, add the probability of making that move to the total probability.

4. Task Identification:

   • The tasks that are repeated or very similar are the simulation of the knight’s movement and the determination of whether the knight is still on the board after each move.

5. Task Abstraction:

   • For the task of simulating the knight’s movement, it is abstracted enough as it involves iterating over all the possible moves and updating the knight’s position accordingly.
   • The task of determining if the knight is still on the board is also abstracted enough as it involves checking the coordinates of the new position against the dimensions of the board.

6. Method Naming:

   • For each task, potential method names could be:
     • simulateKnightMove for simulating the knight’s movements.
     • isKnightOnBoard for determining if the knight is still on the board.

7. Subproblem Interactions:

   • The subproblems interact with each other in a sequence. First, we simulate the knight’s movement, then we check if the knight is still on the board. This is done for each of the k moves.
   • The tasks need to be performed in the order mentioned above, and the task of checking if the knight is still on the board is dependent on the results of the simulation of the knight’s movement.

From Brute Force to Optimal Solution

Let’s start with a brute force approach.

Brute Force Approach:

1. From the current position, a knight can make 8 different moves. For each move, we recursively calculate the probability of the knight remaining on the board.

2. If at any point, the knight moves off the board, we stop further recursion and return a probability of 0.

3. If the knight has made k moves and is still on the board, we stop further recursion and return a probability of 1.

4. The final probability is the average of the probabilities returned by the 8 moves from the current position.

The pseudo code for the brute force solution can be written as follows:

def knightProbability(n: int, k: int, row: int, column: int) -> float:
    if row < 0 or row >= n or column < 0 or column >= n:
        return 0.0
    elif k == 0:
        return 1.0
    else:
        result = 0.0
        for each possible move:
            result += 0.125 * knightProbability(n, k-1, new_row, new_column)
        return result

This solution would work, but it’s highly inefficient. The main issue is that we’re recalculating the same probabilities many times.

Optimized Approach (Dynamic Programming):

To avoid redundant calculations, we can utilize a technique called dynamic programming (DP). The idea is to store the results of subproblems so that when the same subproblem arises, we can reuse the stored result instead of recalculating it. This technique will significantly speed up our solution.

In this problem, the subproblems are the probabilities for the knight to remain on the board after k moves from every possible position on the board. Thus, we can use a 3D DP array, dp[k][i][j], to store the probability for the knight to remain on the board after k moves from position (i, j).

The pseudo code for the optimized solution can be written as follows:

def knightProbability(n: int, k: int, row: int, column: int) -> float:
    dp = [[[0.0 for _ in range(n)] for _ in range(n)] for _ in range(k+1)]
    dp[0][row][column] = 1.0
    
    for steps in range(1, k+1):
        for i in range(n):
            for j in range(n):
                for each possible move:
                    new_row = i + move[0]
                    new_column = j + move[1]
                    if new_row >= 0 and new_row < n and new_column >= 0 and new_column < n:
                        dp[steps][i][j] += dp[steps-1][new_row][new_column] * 0.125
    return sum(dp[k][i][j] for i in range(n) for j in range(n))

Impact on Time and Space Complexity:

The brute force solution has a time complexity of O(8^k), which is exponential and hence infeasible for large inputs.

In contrast, the optimized solution with dynamic programming has a time complexity of O(n^2 * k), as it iterates over all cells of the n x n board for k steps. This is significantly faster than the brute force approach for large inputs.

As for the space complexity, the dynamic programming solution uses a 3D array of size nnk, so the space complexity is O(n^2 * k). Despite this, it is a necessary trade-off to improve time efficiency. The space complexity remains manageable as long as n and k are not extremely large.

Code Explanation and Design Decisions

1. Initial Parameters:

   The initial parameters are n, k, row, and column. n represents the dimension of the chessboard. k represents the number of moves the knight will make. row and column specify the starting position of the knight. These parameters are essential to the problem statement as they define the environment (the chessboard and the number of moves) and the initial state (the starting position).

2. Primary Loop:

   The outermost loop (for steps in range(1, k+1)) represents each move the knight makes up to k moves. In each iteration, the loop updates the probability of the knight staying on the board after steps moves from every cell of the chessboard. This iterative process is necessary to accurately calculate the final probability after k moves.

3. Conditions within the Loop:

   The conditions within the loop (if new_row >= 0 and new_row < n and new_column >= 0 and new_column < n) check if the knight’s new position after a move is still within the board boundaries. If the new position is off the board, the knight’s move isn’t valid, and we don’t update the probability for that move. These conditions ensure that we accurately represent the knight’s possible movements.

4. Updates within the Loop:

   Within the loop, the statement dp[steps][i][j] += dp[steps-1][new_row][new_column] * 0.125 updates the probability of the knight staying on the board after steps moves from cell (i, j). The update is based on the probability of the previous step at the new position and the probability of making the current move (which is 0.125 or 1/8 since a knight has 8 possible moves). These updates reflect changes in the state of the solution, accounting for the cumulative probabilities at each step.

5. Invariant:

   An invariant in this problem is the set of possible moves the knight can make from a given position. This set remains constant throughout the computation and guides the update of probabilities at each step.

6. Final Output:

   The final output is the sum of the probabilities in dp[k], which represents the probability of the knight remaining on the board after k moves from every cell. This value satisfies the problem’s requirement as it gives the desired solution: the probability that the knight will still be on the board after k moves, starting from the specified position.

Coding Constructs

1. High-Level Strategies:

   This code is using a dynamic programming strategy to solve the problem. It is exploiting the overlapping subproblems (calculating the probability of the knight being on the board after k moves) by storing intermediate results in a data structure (in this case, a 3-dimensional list). This allows the algorithm to avoid redundant computation, improving the time efficiency.

2. Explaining to a Non-programmer:

   Suppose you had a chessboard, and you had a knight that could make a certain number of moves, but the catch is, it can’t step off the board. This program is like a game simulator that calculates the probability that the knight stays on the board after making those moves, starting from a given position.

3. Logical Constructs:

   • Looping: This code uses multiple loops to iterate over the moves and the chessboard’s cells.
   • Conditional branching: It checks whether the knight’s new position is valid (within the chessboard).
   • Updating a data structure: It updates the probabilities in the dynamic programming table based on previous results.

4. Algorithmic Approach in Plain English:

   This code basically repeats the same process for the specified number of moves. In each move, it looks at every square on the board and calculates the chance of ending up there, given the 8 possible knight moves. If the knight would move off the board, the code ignores that move. It saves these probabilities and uses them to calculate probabilities for the next move, until it finishes all the moves.

5. Key Operations:

   • Iterating over each move.
   • For each move, iterating over each cell on the board.
   • For each cell, calculating the new positions based on the knight’s 8 possible moves.
   • Checking if the new positions are within the board.
   • Updating the cell’s probability for the current move based on the probabilities of the new positions from the previous move.

   These operations allow the code to simulate all possible move sequences of the knight and calculate the final probability of the knight remaining on the board.

6. Algorithmic Patterns:

   The key algorithmic pattern used in this code is dynamic programming. This pattern is commonly used when a problem can be divided into overlapping subproblems, and the solution to a subproblem can be used to solve larger problems. In this case, the dynamic programming table stores the probabilities after each move from every cell, which are used to calculate probabilities for the subsequent moves.

Language Agnostic Coding Drills

1. Identifying Distinct Concepts:

Here are the distinct coding concepts contained in this problem:

• Basic Input/Output operations: Understanding how to accept inputs and provide outputs. This is a fundamental concept in almost every programming language.
• Variables and Data Types: Knowing how to define and manipulate variables is crucial.
• Control Structures (loops and conditionals): Using loops for iteration and conditionals for decision making is fundamental to problem-solving in coding.
• Arrays/List manipulations: In this case, it’s a 3-dimensional list, which requires understanding how to declare, initialize, and manipulate data in multi-dimensional arrays or lists.
• Basic Math operations: Addition, division are used for calculating probabilities.
• Defining and calling functions: The solution might involve creating helper functions for code reusability and maintainability.
• Dynamic Programming: This is a more advanced concept where we solve the problem by breaking it down into smaller subproblems and storing the results of these subproblems to avoid repeated computation.

2. Order of Difficulty:

   • Basic Input/Output operations: This is the most basic level of coding. It’s about understanding how to accept inputs in the correct format and print or return outputs.
   • Variables and Data Types: Understanding variables and the types of data they can hold is a bit more complex than just input/output but still quite fundamental.
   • Control Structures (loops and conditionals): Using loops and conditionals require a bit more logical thinking, hence it’s next in the line of complexity.
   • Basic Math operations: Arithmetic operations are not difficult but can be tricky if not careful with precedence and associativity rules.
   • Arrays/List manipulations: Operating on multi-dimensional lists/arrays adds another level of complexity due to multiple indices and the potential for nested loops.
   • Defining and calling functions: While not difficult, writing and using functions require a good understanding of program flow and scope.
   • Dynamic Programming: This is the most advanced concept in this list. Dynamic Programming involves a good deal of logical reasoning, understanding of recursion, and problem decomposition.

3. Problem-solving Approach:

   • Understand the problem: The first step is to fully understand the problem, the inputs, and the desired outputs. In this case, we’re working with a chessboard, a knight, and a certain number of moves.
   • Break down the problem: Realize that the knight’s moves are limited and are the same for every square. There are only 8 possible moves, and some of them will lead off the board.
   • Identify subproblems: Notice that the probability of the knight being in a certain square after k moves depends on the probabilities of it being in the neighboring squares after k-1 moves.
   • Choose a data structure to store intermediate results: Here, a 3-dimensional list or array is used, where the first dimension is the move number, and the other two dimensions are the board’s coordinates.
   • Initialize the data structure: Initially, the probability of the knight being in the starting position is 1, and it’s 0 for all other squares.
   • Build up the solution iteratively: For each move from 1 to k, calculate the new probabilities based on the old ones. For each square, add up 1/8 of the probabilities of the neighboring squares from the previous move.
   • Get the final answer: Sum up the probabilities of all squares after k moves. That’s the total probability of the knight staying on the board.
   • Note about Dynamic Programming: At its core, the Dynamic Programming concept here is about building up the solution iteratively and storing intermediate results to avoid redundant computation. This is why Dynamic Programming is an essential part of the problem-solving process.

Targeted Drills in Python

Let’s break down these concepts into Python-based coding drills:

1. Basic Input/Output operations:

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def basic_io():
    print("Enter a number: ")
    x = int(input())
    print("You entered: ", x)

This function prints a prompt for the user, reads the user’s input, converts it to an integer, and then prints it.

2. Variables and Data Types:

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def variable_data_types():
    n = 5  # Integer
    s = "Hello"  # String
    b = True  # Boolean
    print(n, s, b)

This function creates and prints variables of different data types.

3. Control Structures (loops and conditionals):

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def control_structures():
    for i in range(5):
        if i % 2 == 0:
            print(i, "is even")
        else:
            print(i, "is odd")

This function demonstrates a basic for loop and conditional statements by printing whether each number from 0 to 4 is even or odd.

4. Basic Math operations:

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def basic_math():
    x = 10
    y = 3
    print("x + y =", x + y)
    print("x / y =", x / y)

This function demonstrates basic arithmetic operations.

5. Arrays/List manipulations:

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def array_manipulation():
    a = [[0 for _ in range(5)] for _ in range(5)]  # Initialize a 5x5 2D array with zeroes
    a[0][0] = 1  # Set the value at position (0,0) to 1
    print(a)

This function shows how to declare, initialize, and modify a 2-dimensional list.

6. Defining and calling functions:

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def say_hello(name):
    return "Hello, " + name + "!"

def call_functions():
    print(say_hello("World"))

This demonstrates defining a function and calling it.

7. Dynamic Programming:

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def dynamic_programming():
    # Initialize a 1D array dp with zeroes
    dp = [0 for _ in range(5)]
    
    # Base case
    dp[0] = 1
    
    # Iterative calculation
    for i in range(1, 5):
        dp[i] = dp[i-1] * 2
        
    print(dp)

This function shows a simple use case of dynamic programming where each value depends on the previous value.

Assembling the Drills

After understanding and practicing each of these drills, you’ll have a grasp of the concepts required to assemble the final solution for the initial problem. Starting from understanding the problem and breaking it down, to using control structures to perform iterations, conditional checks, defining and calling functions for specific tasks, making calculations using math operations, manipulating lists, and finally using the concept of dynamic programming to solve the problem by breaking it down and storing intermediate results.

Assembling these drills into a final solution would involve creating a 3-dimensional list to store the probabilities, using nested loops to iterate over the moves and the board, and using conditional checks to ensure we’re not going off the board. These tasks will be encapsulated in a function, which would be a combination of all the drills we’ve practiced. By following the problem-solving approach discussed earlier and putting together these drills, we can solve the problem.

Q&A

There are 8 possible directions that the knight can take each time. At each move, Probability of landing on each of the 8 possible squares = Probability of the current square / 8.

Imagine the knight as a person standing in the center of a room with 8 doors, each door leading to a different room. This person wants to walk into another room, but he can’t decide which one. So, he decides to roll an 8-sided die to choose the door. The chance of him choosing any specific door is the same - 1 in 8, or 1/8.

Now, consider each “room” as a square on the chessboard, and each “door” as a possible move the knight can make. From the starting square (room), the knight (person) has 8 possible moves (doors). The probability of the knight landing on any specific square (entering a particular room) after one move is equally likely - 1 in 8, or 1/8.

Then, as the knight continues to make moves, we consider each new square as a new starting point, with its own 8 doors. This is why we calculate the probability of each next square as the probability of the current square divided by 8 - because from each square, there are 8 equally likely moves the knight could make next.

In essence, this explanation means that from any given square, the knight has an equal chance of moving to any of its accessible squares. We divide by 8 to represent this equal distribution of probability.

I think the probability calculation of this question is wrong:

Probability = Number of desired outcomes/ total number of outcomes

Here that would be:

Probability of Knight on board after stopped moving = Number of moves where knight is on board at the last move/ Total number of moves (until either K moves or until knight is off board)

In the given example [Input: 3, 2, 0, 0] :

Number of moves where knight is on board at the last move = 4 Total number of moves = 6 moves that end in first round as they go off board

6 moves from (1,2) that go off board + 2 moves where the knight is on board and K moves are complete 6 moves from (2,1) that go off board + 2 moves where the knight is on board and K moved are complete = 6 + 6 +2 + 6 + 2 = 22 So probability = 4/22 = 0.181818 (not 0.0625)

Am I missing something here?

Your understanding of the general concept of probability is correct. However, the context of this problem changes things a bit.

In the problem, we’re asked to calculate the probability that the knight remains on the board after making K moves, assuming that the knight always makes a move if it can. So, the denominator isn’t just the total number of moves after K steps. Rather, it’s the total number of all possible sequences of K moves.

A key point here is that the knight has 8 possible moves at each step, not just the ones that keep it on the board. For each starting point, there are 8 possible moves, and for each of those, another 8, and so on for K steps. That’s why the total number of sequences of K moves is 8^K, which is the denominator in our probability calculation.

As for the numerator, we’re only counting sequences of K moves that keep the knight on the board for all K steps. So, we’re summing up the probabilities of landing on each valid (i.e., on-board) square after K steps. That’s what we’re calculating with our dynamic programming algorithm.

In the given example, the algorithm calculates the total probability as 0.0625, meaning that there’s a 6.25% chance that the knight will remain on the board after making 2 moves from position (0,0), given that it always makes a move if possible. This is consistent with the problem statement.

This problem can also be reframed as “What is the probability the knight can make K moves without stepping off the board?”,

Yes, that’s a great alternative way to rephrase the problem: “What is the probability that the knight can make K moves without stepping off the board?” This framing still captures the essence of the problem and emphasizes the part that the knight should stay within the limits of the board.

It’s essential to recognize that for each move, the knight has 8 different possibilities (directions it could move to). Each time we make a move, we are equally likely to choose any one of these directions. In this sense, it is not a standard probability problem where we have a fixed number of total outcomes. The total number of outcomes (i.e., the paths the knight could take) grows exponentially with each move, and it’s dependent on the knight’s position on the board.

With this in mind, we approach the problem with dynamic programming. We track the probabilities of reaching each cell on the board after a certain number of moves, building up to our final solution. The desired probability is then computed as the sum of probabilities of being on all valid positions after making K moves.

“Return the probability that the knight remains on the board after it has stopped moving.”

Incorrect. It should be:

“Return the probability that the knight has never stepped off the board after K moves.”

The problem statement provided by Leetcode allows for stepping off the board and then back on, which is not intended.

You’re correct. The problem statement from LeetCode does seem to be ambiguous, and it could be misinterpreted as you’ve mentioned. However, the intended meaning is indeed “Return the probability that the knight has never stepped off the board after K moves.”

In the context of this problem, once the knight moves off the board, it can’t move back onto the board, so we’re interested in calculating the probability that the knight remains within the boundaries of the chessboard throughout all of its K moves. The phrase “after it has stopped moving” in the original statement can be misleading. Your alternative phrasing makes the objective clearer.

In such problems, clarifying such ambiguities in the problem statement is a crucial first step towards the correct solution. It’s important to thoroughly understand the requirements and constraints of the problem before diving into solving it.

Input: 3, 2, 0, 0 Output: 0.0625 Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board. From each of those positions, there are also two moves that will keep the knight on the board. The total probability the knight stays on the board is 0.0625.

So here is how I compute the prob: As in the explanation, there are in total 4 chances that the knight stays on the board after stop moving. In addition, there are in total 22 (18 + 4) paths after 2 moves. Then the prob is 4 / 22.

I do not understand how 0.0625 comes…. What does it mean “stops moving”? In the question, it says the knight stops moving when exhausting K moves or move out of the board. So the prob that I understand is 1 - ( # of positions that are out of the board / # of ways that the knight stops). Could someone please help point out which part I am wrong?

I understand where your confusion is coming from, and it’s rooted in the nature of how probabilities are calculated. In this problem, the probability isn’t calculated based on the final number of positions, but rather on the transitions from one step to the next.

When the problem states “the knight stops moving when exhausting K moves”, it means that we consider the positions of the knight after it has made exactly K moves, and it has stayed on the board for all of these moves.

To clarify the computation, let’s break down the steps:

1. Step 0: The knight starts at position (0,0).

2. Step 1: From (0,0), the knight has 2 valid moves (to positions (1,2) and (2,1)), out of the total 8 potential moves a knight can make. So, the probability of the knight staying on the board after 1 move is 2/8 = 0.25.

3. Step 2: From either (1,2) or (2,1), the knight again has 2 valid moves, out of the total 8 potential moves. So, the probability of the knight staying on the board after 2 moves is (2/8) * (2/8) = 0.0625. We multiply probabilities here because we’re dealing with independent events - the outcome of the second move doesn’t depend on the outcome of the first move.

The key point here is that we’re not counting the number of final positions, but rather calculating the probability of making K moves in a row, all of which keep the knight on the board. The total number of final positions doesn’t factor into this probability calculation. Instead, we consider the probabilities at each step.

(r,c) = (0,0)

N=2, K=1 P(2,1) = 0 valid out of 8 total = 0/8 = 0

N=3, K=1 P(3,1) = 2 valid out of 8 total = 2/8 = 0.25

N=3, K=2 Case K=1: 2 valid out of 8 total = 2/8 = 0.25 Case K=2: 2 valid out of 8 total = 2/8 = 0.25 P(3,2) = 0.25*0.25 = 0.0625

N=3, K=3 P(3,3) = 0.0625*0.25 = 0.01563

N=3, K=4 P(3,4) = 0.01563*0.25 = 0.00391

N=4, K=1 P(4,1) = 2 valid out of 8 tot = 2/8 = 0.25

N=4, K=2 Case K=1: 2 valid out of 8 tot = 2/8 = 0.25 Case K=2: 4 valid out of 8 tot = 4/8 = 0.5 P(4,2) = 0.25*0.5 = 0.125

and so on.

Yes, you’ve correctly interpreted the problem and how to compute the probabilities.

For each turn, you calculate the probability of the knight remaining on the board by dividing the number of valid moves by the total number of moves (which is 8). And to get the overall probability after K turns, you multiply the probabilities together, because the events are independent of each other (i.e., the outcome of one move doesn’t affect the outcomes of the other moves).

This process can be summarized in these steps:

1. Identify the number of valid moves the knight can make from the current position without going off the board.

2. Divide this number by 8 to get the probability for the current move.

3. Multiply all such probabilities for each move until you’ve done K moves.

Doing so will give you the overall probability of the knight staying on the board after K moves.

As you have shown, with an increase in K (the number of moves), the probability decreases. This is because with each additional move, there’s a higher chance that the knight will step off the board.

Why is the total number of possible outcomes == 8 ** k?

Shouldn’t the 8 possible moves be denied when the Knight has moved off the chessboard?

Yes, you’re correct. In reality, the total number of possible outcomes isn’t always 8^k, as the knight can’t continue moving once it has stepped off the board.

The expression 8^k is a simplification used for this problem. It assumes that the knight always has 8 possible moves, which would be true if the board were infinitely large or if the knight could move off the board and then back on.

In this particular problem, we’re interested in the probability of the knight staying on the board, which involves only the paths where the knight doesn’t step off. Therefore, the individual probabilities we calculate at each step are based on the remaining valid moves (those that keep the knight on the board), not all 8 possible moves.

So while the expression 8^k doesn’t accurately represent the total number of outcomes, it does provide a useful conceptual model for how we approach the problem. The calculation of the actual probability accounts for the restriction of the knight staying on the board.

Similar Problems

Here are 10 problems that use similar underlying concepts:

1. Out of Boundary Paths: This problem also involves computing probabilities using dynamic programming, similar to our original problem. Here, instead of a knight, it’s a ball that can move in 4 directions.

2. Unique Paths II: Similar to the knight problem, this involves finding paths on a grid. However, in this case, it’s a 2D grid and some of the cells are blocked.

3. Minimum Path Sum: In this problem, we’re also traversing a grid and looking for the minimum path sum, which can be solved by dynamic programming in a similar manner to our original problem.

4. Cherry Pickup: This problem also involves traversing a grid in a certain manner to maximize the picked cherries. It requires the use of dynamic programming and also deals with grid-based movements.

5. Coin Change: This problem involves finding the minimum number of coins that make a certain amount, which can be solved by dynamic programming by breaking down the problem into subproblems.

6. Maximum Subarray: This problem involves finding a contiguous subarray that has the greatest sum, which can be solved by dynamic programming by breaking the problem into smaller subproblems.

7. Longest Increasing Subsequence: The dynamic programming approach to this problem involves breaking down the problem into subproblems in a similar manner to our original problem.

8. 0/1 Knapsack Problem: The underlying problem here is the well-known Knapsack problem. Dynamic programming is used to find the maximum total of values given a certain weight limit.

9. Climbing Stairs: This problem involves finding the number of ways to climb a staircase, which can be solved using dynamic programming by breaking down the problem into subproblems, like how many ways to reach a certain step.

10. Target Sum: This problem involves finding the number of ways to assign symbols to make the sum of numbers equal to target. It also can be solved using dynamic programming by breaking the problem into smaller subproblems, similar to the original problem.