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Today, we're going to understand how to calculate the number of paths in a grid. Can anyone tell me how we might approach finding a path from point (0,0) to (i,j)?
We can move right or up?
Exactly! So if we denote the number of paths to a point (i,j) as paths(i,j), how might we express that mathematically?
Maybe it's the sum of paths from the left and below?
Right again! Thus, we get the formula: paths(i,j) = paths(i-1,j) + paths(i,j-1). Remember this as P = L + B!
What about the starting point? Is there only one path to (0,0)?
That's correct! There’s one trivial path that stays at (0,0). So, we always start with paths(0,0) = 1.
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Now, what happens if there's a hole in our path, say at point (2,4)?
Doesn't that mean no paths can go through there?
Exactly! So, we declare paths(2,4) as 0, and this will propagate to all neighbors. How does this help us?
It means that surrounding paths will not be counted from there!
Correct! This concept keeps our calculations accurate and dynamic.
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Let’s shift gears to our strategies. Can someone explain the difference between memoization and dynamic programming?
Memoization remembers previous computations, while dynamic programming is about solving all subproblems first.
Exactly! Memoization only stores results for future use, but dynamic programming solves the entire problem in a structured way. Why does this matter?
Because dynamic programming can be more efficient in terms of overall calculations?
Spot on! As we compute our grid paths, a systematic approach works to avoid redundancy.
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Now, let's illustrate our grid as a DAG. Why do we visualize it this way?
To clearly see the dependencies for each point?
That's correct! By recognizing the structures like this, what is our best approach to fill in the values efficiently?
We can fill values row by row or column by column!
Right again! This systematic filling accentuates the flexibility of topological sorting in path calculations.
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To wrap up our discussion, can someone summarize our findings on grid paths?
We learned about how to calculate paths, manage holes, and the differences between memoization and dynamic programming.
Excellent summary! And how can we visualize dependencies in path calculations?
Through a DAG to show how values depend on the points above and to the left.
Great team! This solid understanding will serve us well in our future computations.
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In this section, we delve into the intricacies of determining paths in a grid using dynamic programming. We discuss the inductive approach to count the number of paths, how to handle obstacles, and the pivotal difference between memoization and dynamic programming strategies.
Topological sorting is a computational technique primarily used in graph theory to order vertices in a directed acyclic graph (DAG) such that, for every directed edge UV from vertex U to vertex V, U comes before V in the ordering. In this section, we exemplify this concept within the context of grid paths from point (0,0) to any point (i,j).
We begin by understanding how to reach any point (i,j) in a grid. The paths can only come from two directions — from the left (i,j-1) or from below (i-1,j). Therefore, we derive the recursive relationship:
\[ ext{paths}(i,j) = ext{paths}(i-1,j) + ext{paths}(i,j-1) \]
This inductive formulation helps us count the total number of paths to each point while accounting for boundary conditions, especially in the leftmost column and the top row where paths are restricted.
Next, we discuss handling obstacles, referred to as holes in the grid. When a hole is present at a given point, the number of paths contributing to that point is set to zero. This not only applies to the point itself but also propagates to its neighboring points, ensuring dynamic programming integrity.
An essential aspect is understanding the distinction between memoization and dynamic programming. Memoization stores results of expensive function calls and reuses them when the same inputs occur again, while dynamic programming involves solving complex problems by breaking them down into simpler subproblems and solving each subproblem just once.
In dynamic programming, we visualize our grid as a Directed Acyclic Graph (DAG) where nodes represent grid points and edges signify dependencies (left and below). We can compute paths iteratively, filling from the base case (0,0) to larger values, regardless of holes, following a systematic approach incrementally.
Topological sorting in our grid paths emphasizes the relationships and dependencies inherent in choosing optimal paths. The choice of calculating through rows, columns, or even diagonally reflects the flexibility in dynamic programming approaches to solve path-counting problems.
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So, let us write paths (i,j) to denote the number of paths from (0,0) to the current point (i,j). So, what we have just seen from our inductive analysis on the problem is, that if you are at (i,j), then the paths come from left or from below. So, the paths to (i,j) are the sum of paths to (i minus1,j) and the paths to (i,j minus1).
To understand how many paths lead to any point (i,j) on a grid starting from (0,0), we define a function called paths(i,j). This function counts all the unique paths from the starting point (0,0) to (i,j). It shows that to get to (i,j), you can come from two possible routes: either directly from the left (i,j-1) or from below (i-1,j). Thus, the total number of paths to (i,j) is the sum of paths to (i-1,j) and paths to (i,j-1). This formulation uses the principle of recursion by looking at previous known values to determine current ones.
Imagine you're trying to reach a particular room in a building. You can only come from either the room to your left or the one directly below you. If you know how many ways there are to get to those rooms, you can add those ways together to find out how many ways there are to reach your desired room.
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So, if you look at our grid, in general, then if we look at the leftmost column, let us start the bottom row. So, we start the bottom row, right, then we know, that this is of the form (0,0), then (1,0) and so on to (5,0), right. So, anything of the form (i,0) derives its value only from the left because there is no corresponding row from the left.
In order to properly compute the number of paths to any point on a grid, we must establish boundary conditions. For instance, if we're at the first column (i,0), we can only get there from the left (i-1,0) because there are no path options above. Similarly, for the first row (0,j), we can only come from the point to the left (0,j-1). These scenarios define our starting points, which are crucial for the accurate calculation of paths to subsequent points in the grid.
Think of navigating a city where the leftmost road represents the first column and the topmost road represents the first row. If you're on the edge of the city (the grid), your movement options are very limited: you can only proceed from the road that runs parallel to the edge, just like how paths can only come from one direction in our grid.
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So, how do we deal with holes in this setup? That is easy, we just say, that whenever there is a hole at a given point we just declare paths at (i,j) be 0.
In our path calculations, if a hole exists in the grid at a certain point (i,j), it renders that point unreachable. Therefore, we set the number of paths at (i,j) to zero. This rule applies no matter what values come from the neighboring points, as that hole effectively blocks any potential path, propagating the zero value to its adjacent points accordingly.
Imagine walking on a path where there are construction zones (holes) blocking access. If there's a construction site in front of you, that means you can't advance to that part of the path, so you're left with fewer options – just like setting the number of paths to that hole as zero.
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So, we have seen before, we have two technologies to deal with this. So, one is memoization where we just make sure that we have never paths (i,j) the same value of i and j more than once. The other way is to, is to anticipate the sub problems, figure out how they depend on each other, then solve them directly iteratively in a suitable order and this is what we call dynamic programming.
Two strategies are commonly used to optimize the calculation of paths: memoization and dynamic programming. Memoization involves storing the results of previously calculated paths (i,j) to avoid redundant calculations. In contrast, dynamic programming seeks to understand the overlapping sub-problems and computes all values in a systematic manner, ensuring that all paths are calculated once in a specific order, making it more efficient overall.
Consider doing your math homework. If you repeatedly solve the same problem each time (memoization), you're wasting time. Instead, if you plan your approach and solve similar problems in a systematic way (dynamic programming), you'll find it more efficient to complete your work.
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So, how would we use dynamic programming on the grid? So, this is our grid with the two holes at the (2,4) and (4,4). The first thing is to identify DAG structure of the dependencies, right. So, we know, that every (i,j) depends on, it is left and bottom neighbor.
When applying dynamic programming, it's crucial to recognize the direction of dependencies within the grid, represented as a Directed Acyclic Graph (DAG). Each point (i,j) in the grid calculates its paths based on its left and bottom neighbors. This structured approach allows us to compute all necessary values in a valid topological order without re-evaluating previously calculated paths.
Think of organizing a team project. Each task (grid point) relies on the completion of different preceding tasks (dependencies). By following the correct order of task completion, you ensure that all necessary work is done efficiently, just like calculating paths across a grid.
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Key Concepts
Paths in a Grid: The total number of movements required to get from the start to an end point in a grid.
DAG Structure: Visual representation of dependencies that allows comprehension of how calculations depend on one another.
Dynamic Programming: Efficient problem-solving technique achieved by storing intermediate results.
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To calculate the number of paths from (0,0) to (2,2) in a 3x3 grid without any obstacles, we find the number of movements required is 2 (right and down). Thus, paths(2,2) = paths(1,2) + paths(2,1).
In a grid where (1,1) is a hole, any calculations involving that point should consider it as 0, affecting surrounding calculations.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
In the grid of two by two, paths can be few, right and down, that's the way through!
Once upon a grid, with obstacles galore, paths ran amok, but with a rule, they were sure to ignore holes just like chores!
P = L + B, remember paths always, left and below happily!
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Review the Definitions for terms.
Term: Grid Path
Definition:
The combination of movements one can take to reach from one coordinate to another on a grid.
Term: Inductive Formulation
Definition:
A method to define functions in terms of previously defined values or simpler cases.
Term: Dynamic Programming
Definition:
An optimization technique that solves problems by breaking them down into simpler subproblems, storing the results for future use.
Term: Memoization
Definition:
A technique where previously computed results are stored to avoid redundant calculations.
Term: Directed Acyclic Graph (DAG)
Definition:
A graph that consists of vertices and edges, having no cycles and directional flows, used to represent dependencies.