Path with Maximum Probability
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Problem Classification
The problem falls under the category of Graph Theory in Computer Science, and it specifically deals with weighted graphs and the concept of finding paths between nodes. The problem is looking to find an optimal path, making it an Optimization problem as well.
‘What’ Components:
Graph: You’re given a weighted, undirected graph with ’n’ nodes represented as an edge list. The nodes are 0-indexed.
Edges: Each item in the ’edges’ list is an undirected edge connecting nodes ‘a’ and ‘b’.
Edge Weights (success probabilities): The graph is weighted with each edge having an associated probability of success. This is represented by the ‘succProb’ list where ‘succProb[i]’ corresponds to the probability of success of ’edges[i]’.
Start and End Nodes: You are given two specific nodes - ‘start’ and ’end’.
Success Probability: The objective is to find the path from ‘start’ to ’end’ that maximizes the overall probability of success. The success probability of a path is the product of the success probabilities of the traversed edges.
Output: The output is the maximum probability of successfully traversing from the ‘start’ node to the ’end’ node. If no path exists, the output should be 0.
Given the requirements and constraints, this is a Graph Traversal problem, which involves finding an optimal path between two nodes in a weighted graph. Since it involves finding the path with the maximum probability (product of weights), it can be classified as a variation of the Shortest Path problem (or rather Longest Path in terms of probability in this context), which is a well-known problem in graph theory. The specific algorithm used to solve it might fall under Greedy Algorithms or Dynamic Programming, depending on the approach. It also involves elements of Probability Theory, as the weights in this case represent probabilities.
Constraints
Here are some specific characteristics, patterns, or conditions that could be exploited to find an efficient solution:
Graph Structure: The graph is undirected, which means that traversing from node ‘a’ to node ‘b’ is the same as traversing from ‘b’ to ‘a’. This symmetry could be utilized during the graph traversal.
Edge Weights (Probabilities): The problem describes probabilities associated with each edge. Probabilities are always within the range [0, 1], and this range could potentially be used to simplify calculations. Also, multiplying probabilities corresponds to adding their logarithms, and adding is generally faster than multiplication in a computer. This might help avoid precision issues and improve performance.
Start and End Nodes: You are only interested in paths from the ‘start’ to the ’end’ node. This means that the graph traversal does not necessarily need to consider all nodes and paths, but can instead focus on the ones that could potentially lead to the ’end’ node. If a path does not lead to the ’end’ node, it can be ignored, potentially saving computational resources.
Probabilistic Interpretation: The problem requires finding a path with the maximum success probability. Because probabilities multiply along the path, this implies that we are looking for the path with maximum product of edge weights. This is a form of the “maximum product path” problem, which is a variant of the more common “shortest path” problem. Algorithms for shortest path problems, such as Dijkstra’s or Bellman-Ford, could potentially be adapted to this problem.
Constraints: The constraints provided can also be used to guide the choice of the solution. For example, since ’n’ can be as large as 10^4, an O(n^2) solution would be too slow, indicating that an efficient graph traversal algorithm will be needed. On the other hand, the fact that there is at most one edge between any two nodes can simplify the representation of the graph and potentially the traversal algorithm.
By taking these factors into account, one can develop an efficient algorithm to solve this problem.
Thought Process
From the problem statement, the main hint is that it’s about finding a path in a graph - that should immediately make you think about graph traversal algorithms. However, it’s not as simple as finding a shortest or longest path, because the “weight” of a path is the product of the probabilities on the edges, not their sum. That’s a key insight - we need to use a variant of a shortest path algorithm, but with multiplications instead of additions.
Moreover, we want the path with the maximum probability of success, which means we’re looking for the ’longest’ path in terms of the product of probabilities. Another important insight is that the probabilities are all between 0 and 1, so when we multiply them, we get a smaller number - that means that a path with more edges is not necessarily better.
Finally, there is a constraint that says there is at most one edge between every two nodes. This means our graph is simple (no parallel edges or self-loops), which can simplify our graph traversal.
Given all this, the Dijkstra’s algorithm comes to mind. It’s a classic algorithm for finding the shortest paths from a single source in a graph with non-negative weights. In our case, the ‘distance’ from the source to a node is the maximum product of probabilities on a path to that node, and the ‘distance’ between two nodes is the probability on the edge between them.
With that in mind, we can start implementing our solution in Python. We’ll use a priority queue to implement the Dijkstra’s algorithm, and we’ll represent our graph as a dictionary of lists for efficiency.
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This function first builds the adjacency list representation of the graph. Then it initializes the success probability of the start node to 1 and all other nodes to 0, and starts Dijkstra’s algorithm with a priority queue. The algorithm iterates until there are no more nodes to visit (queue is empty), always visiting the unvisited node with the highest success probability first. If it finds a path to a node with a higher success probability than the current one, it updates the probability and adds the node to the queue. Finally, it returns the maximum success probability to reach the end node.
Maximizing Path Probability in a Graph: A Modified Dijkstra’s Algorithm Approach
The Dijkstra’s algorithm is traditionally used to find the shortest path in a graph. However, in the given problem, we’re tasked with finding the path with the maximum probability, not the shortest path. So, we need to modify the traditional Dijkstra’s algorithm to suit this problem. Here’s how we do it:
MaxHeap instead of MinHeap: In the traditional Dijkstra’s algorithm, we use a MinHeap to always select the node with the shortest distance from the source. However, for this problem, we want to select the node that has the highest probability. Therefore, we use a MaxHeap instead of a MinHeap.
Multiplication instead of Summation: In traditional Dijkstra’s, we add up the weights of the edges to calculate the total cost of a path. In this problem, however, the ‘cost’ of a path is determined by the product of the probabilities (weights) of the edges along the path.
The resulting solution, despite these modifications, is indeed considered as an application of Dijkstra’s algorithm. The core idea of Dijkstra’s, which is continually selecting the ‘best’ next node to visit, remains intact.
As for proving its correctness and time complexity, it has to do with the properties of the weights in the graph. In the traditional Dijkstra’s algorithm, the weights must be non-negative to guarantee that once we’ve found the shortest path to a node, we won’t find a shorter one later. Similarly, in this problem, since all probabilities are in the range of [0, 1], multiplying by another probability (taking another step in the path) will only decrease the total probability. Hence, once we’ve found the maximum probability path to a node, we won’t find a higher one later.
The time complexity remains O(E log V), where E is the number of edges and V is the number of vertices, because we’re still using a priority queue (MaxHeap) to select the next node to visit, and we potentially visit all edges and vertices.
In essence, this problem is equivalent to the classic shortest path problem, but viewed from a different perspective - instead of finding the shortest path (minimum sum), we’re finding the most probable path (maximum product).
General Why?
Dijkstra’s algorithm is a common algorithm for finding the shortest path in a graph. However, this problem asks us for the path with the maximum product of edge weights - a quite unusual request. Many people might get confused and just memorize the solution without understanding why it works. But the aim here is to understand the underlying principles.
The key principle behind Dijkstra’s algorithm is that once we determine the shortest path to a node, we never need to update it - this is because we always visit the nodes in increasing order of distance from the start. But when we try to find the longest path, this principle breaks down due to cycles. If a cycle has a positive weight, Dijkstra’s algorithm would get stuck in an infinite loop. Hence, the longest path problem is generally NP-hard.
Why Related To Problem?
Now, why does Dijkstra’s algorithm work for this problem? The trick here lies in the property of the edge weights. They are all probabilities and are hence less than or equal to 1.
In this problem, the probability of the start node is 1. For any node we visit from the start, the probability is going to be less than 1 (since we multiply the edge weights). So, once we visit a node, we can be sure that we’ve found the path with the maximum probability to that node, just like how we can be sure we’ve found the shortest path in the traditional Dijkstra’s algorithm.
So in essence, we can apply Dijkstra’s algorithm here because of the specific properties of the problem: edge weights are probabilities and hence less than or equal to 1.
If the edge weights were greater than 1, could we still use Dijkstra’s algorithm? Understanding why or why not is key to mastering the application of Dijkstra’s algorithm in a variety of problems.
Solution Approach and Analysis
In terms of the general approach to solving this problem, we can frame it as a graph traversal problem where we are trying to find the path with the maximum product of edge weights (probabilities). This is a bit like a typical shortest path problem, but instead of adding edge weights, we are multiplying them. Also, since we want to maximize the product, we’re looking for the ’longest’ path rather than the shortest one.
Here are the steps to solve the problem:
Graph Representation: First, represent the graph in a suitable data structure. Given the maximum constraint of 10^4 nodes, an adjacency list would be a suitable representation. Each node could be represented as a list of its adjacent nodes along with the probabilities (edge weights) associated with each edge.
Initialize: Create a probability array (let’s call it ‘prob’) where prob[i] represents the maximum success probability to reach node i from the ‘start’ node. Initialize prob[start] = 1 (since the probability of reaching the start node from itself is 1), and the rest of the prob array to 0.
Graph Traversal: Now, we will need to traverse the graph. A Depth-First Search (DFS) or Breadth-First Search (BFS) could work, but we are looking for a ’longest path’ and these algorithms are usually more suited to ‘shortest path’ problems. However, since we only have positive edge weights (probabilities), Dijkstra’s algorithm is a suitable choice.
- In this context, Dijkstra’s algorithm involves using a priority queue where the nodes with the highest success probability are given priority. We start from the ‘start’ node, and for each adjacent node, we calculate the success probability if we were to traverse the edge to this node. If this probability is higher than the current maximum probability for this node (stored in prob[i]), we update prob[i] and add the node to the priority queue. This continues until we have visited all nodes or the priority queue is empty.
Result: After the graph traversal, prob[end] will contain the maximum success probability to reach the ’end’ node. If the end node was not reachable, prob[end] will still be 0, which is consistent with the problem statement.
Let’s now discuss how specific operations or changes in the problem’s parameters would affect the solution.
Adding an edge: Adding an edge would potentially increase the success probability if the new edge forms part of a more successful path than the current maximum. This would involve updating the adjacency list and possibly re-running the Dijkstra’s algorithm.
Removing an edge: Similarly, removing an edge could potentially decrease the maximum success probability if the edge was part of the current most successful path. Again, this would involve updating the adjacency list and re-running the Dijkstra’s algorithm.
Changing the probability of an edge: This could either increase or decrease the maximum success probability, depending on whether the probability was increased or decreased and whether the edge forms part of the current most successful path.
As for an example, consider the first example from the problem statement:
n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2
Here, there are two paths from 0 to 2: directly from 0 to 2 with success probability 0.2, or from 0 to 1 to 2 with success probability 0.5 * 0.5 = 0.25. Using the approach above, we start from node 0 and update the success probabilities of nodes 1 and 2 to 0.5 and 0.2 respectively. Then, we take node 1 (since it has the highest success probability in the queue), and update the success probability of node 2 to 0.25. At this point, we have visited all nodes and the algorithm ends, giving a maximum success probability of 0.25, which is the correct answer.
Language Agnostic Coding Drills
Consider the following piece of complex software code.
Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.
Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.
Targeted Drills in Python
Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.
Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.
In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.
Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.
Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.
Q&A
A standard setup for Dijkstra’s algorithm is as follows:
We have a graph with weighted, non-negative edges (usually directed). The cost of a route is computed by summing up the weights along the route We are interested in finding the routes with minimal cost from a source node. In this question, however, we have quite a different setup:
The cost of the route is determined by multiplying the weights along the route, instead of a summation. We are interested in finding the routes that have maximal cost rather than minimal. The weights are between 0 and 1, as they represent probabilities.
How do we know Dijkstra’s algorithm is applicable to this problem?
Great question! Dijkstra’s algorithm, as you’ve stated, is designed to find the shortest path in a graph from a source node to all other nodes, where edges have non-negative weights. It works by iteratively selecting the node with the smallest path from the source node and relaxing all its edges.
However, Dijkstra’s algorithm can be adapted to handle this problem for several reasons:
Edge weights are non-negative: Dijkstra’s algorithm relies on this property to ensure that once a node is chosen, no shorter path to that node can be found. Even though the weights here represent probabilities between 0 and 1, they are still non-negative, thus preserving this important property.
Multiplicative costs can be converted to additive costs: This can be achieved through the use of logarithms. If we replace each probability p with -log(p), then the multiplication of probabilities (as required in this problem) turns into the addition of their logarithms. That’s because log(a*b) = log(a) + log(b).
Maximization of cost can be converted to minimization: We’re interested in the path with the maximum probability of success, but Dijkstra’s algorithm minimizes cost. However, using -log(p) as above flips this around - minimizing the sum of -log(p) values maximizes the product of the probabilities.
Given these considerations, Dijkstra’s algorithm is indeed applicable to this problem and provides an efficient solution when appropriately adapted. It’s important to note that this kind of flexibility is part of what makes Dijkstra’s algorithm and similar graph search algorithms so powerful.
I see that there are both the types of solutions; one with visited implementation and another without it. Both the solutions are accepted. I am not able to understand why we shouldn’t use visited in this case.
The use of a “visited” set in graph algorithms typically helps to avoid redundant work or, in the worst case, infinite loops. However, whether to use a “visited” set or not depends on the specifics of the problem and the algorithm.
In the context of Dijkstra’s algorithm, a “visited” set is usually not necessary because of how the algorithm works. Specifically, Dijkstra’s algorithm uses a priority queue (or a heap) to always select the node with the smallest distance (or in this problem, highest probability) to process next. Once a node has been processed, there is no need to revisit it because we have already found the shortest path (or maximum probability path) to it.
In the case of this problem, where edge weights represent probabilities and are less than 1, once we have found the maximum probability path to a node, we know that any other path to this node would have a lower probability. This is because multiplying by a number less than 1 will only decrease the value. Thus, revisiting a node will not lead us to a path with a higher probability.
However, it’s also possible to maintain a “visited” set as a safety measure to ensure that no node gets processed more than once. This will not change the outcome of the algorithm, but it may add some unnecessary overhead, especially if the graph is large.
In short, while using a “visited” set might not hurt, it’s usually not necessary for Dijkstra’s algorithm and might add some unnecessary overhead.
Good problem to revise dijkstra algorithm. I started by reversing the logic , because this time we dont want minimum but maximum.So,i used a max heap and the probability of start node will always be 1 (maximum probabilty); All other nodes with probability of reaching =0.0000 in starting;
Yes, you’re absolutely correct! This problem is indeed an excellent one to revise and expand your understanding of Dijkstra’s algorithm. As you’ve noted, the logic here is reversed from the standard Dijkstra’s algorithm because we want the maximum probability rather than the minimum path cost.
Here, initializing the probability of reaching the start node to 1 is a key observation, as this reflects the fact that we’re already at the start node, so the probability is 100%. For all other nodes, we can start with a probability of 0 because we haven’t found any path to them yet.
Applying Dijkstra’s algorithm in this manner, with these initial probability values and using a max heap instead of a min heap, allows us to effectively find the path with the maximum probability from the start to the end node. This modified algorithm, as applied to this problem, provides a solid example of how classical algorithms can be adjusted to fit new problem constraints.
IS THERE A FLAW IN MY ALGORITHM?
My reasoning is that if Psx is the max probability of going from start to Node X and Pxe the probability of going directly from Node X to the end node then Psx * Pxe is the maximum probability of going from the start node to the end node via node X.
So all you need to do is calculate the maximum probabilities from the start Node to each node that has a direct path to the end node, multiply their probabilities and choose the maximum result. To do this, you need a visited array and a probability array. When a node is visited, it is marked visited and its probability initially set to zero (so that it won’t play a part in its own calculation). The start Node automatically is marked as visited and its probability set to 1.0 (it is certain you can get from start to start) before doing the dfs.
This algorithm worked for 9 out of the 18 testcases. Unfortunately, the testcase where it failed has 1000 nodes so code tracing is out.
Can anybody see something that I have missed?
Your approach is on the right track but there’s a small but significant flaw in it. The flaw lies in the assumption that the maximum probability path to the end node must pass through some intermediate node ‘X’ such that the path from the start node to ‘X’ has maximum probability. This is not necessarily true in this case.
Consider a situation where the path from the start node to an intermediate node ‘X’ has high probability but the path from ‘X’ to the end node has very low probability. There might be another path from the start node to ‘Y’ which has slightly lesser probability than the path to ‘X’, but the path from ‘Y’ to the end node might have high probability, making the overall path from the start node to the end node via ‘Y’ having higher probability than the path via ‘X’.
So, in this case, calculating the maximum probabilities from the start node to each node that has a direct path to the end node, multiplying their probabilities and choosing the maximum result won’t work.
This is why we need to use a modified Dijkstra’s algorithm to ensure that we consider all possible paths to find the path with maximum probability.
To avoid going into an infinite loop due to cycles, you can use a visited array or a distance array where you update the maximum probability of reaching each node. If during the traversal, you come across a state that you’ve visited before with a higher or equal probability, then you don’t need to visit it again. This ensures that each state is visited at most once, preventing infinite loops.
Can we use Shortest Path Faster Algorithm (SPFA) for this problem?
The Shortest Path Faster Algorithm (SPFA) is a solution to the single-source shortest path problem which features better performance than Dijkstra’s algorithm in the case of graphs with fewer edges and more nodes (sparse graphs). It is based on Bellman-Ford’s algorithm but uses a queue to determine the order in which nodes are evaluated to potentially cut down the number of relaxation operations.
However, in this specific problem, we’re not looking for the shortest path, but the path with the maximum probability of success, which is a bit of a twist. Moreover, we need to multiply the probabilities, not add them up, which changes the nature of the problem significantly.
Like Bellman-Ford, SPFA also allows for negative edge weights. But remember, this isn’t about negative edges; this is about finding a path where you multiply the edge weights (probabilities), not add them.
In this problem, it’s more like we’re looking for the longest path (as a higher probability is ‘better’), not the shortest. Finding the longest path in a graph is a NP-hard problem and doesn’t have an efficient solution like SPFA or Dijkstra’s.
So, to answer your question: No, the Shortest Path Faster Algorithm (SPFA) cannot be directly applied to this problem due to the differences in problem nature and operation on edge weights.
However, as discussed in previous explanations, a modified version of Dijkstra’s algorithm can be used effectively for this problem.
The hint is weird. Of course it is true that multiplying floating-point values will result in precision errors, but summing-up floating-point values also result in precision errors, and when one floating-point value is much smaller than the other value, multiplication is more reliable than addition.
You’re right, both multiplication and addition of floating-point values can result in precision errors. However, there are some differences that might make one operation more prone to errors than the other in certain contexts.
When you add two floating-point numbers, the relative error of the result is about the sum of the relative errors of the numbers being added. So if you’re adding a very small number to a very large number, the small number might not affect the large number at all due to the limitations of floating-point precision. This is known as catastrophic cancellation.
On the other hand, when you multiply two floating-point numbers, the relative error of the result is about the sum of the relative errors of the numbers being multiplied. So if one of your numbers has a large relative error, the result will also have a large relative error.
In the case of this problem, where we are multiplying probabilities together (which are always between 0 and 1), the results can get very small very quickly when dealing with a large number of edges. This can lead to a loss of precision. However, in this case, the precision of the floating-point representation is not the main concern.
The hint likely refers to the fact that using multiplication to calculate the cost of a path (as in this problem) fundamentally changes the problem from a standard shortest path problem, where costs are calculated by summing edge weights. This affects how the problem can be solved algorithmically. For example, algorithms like Dijkstra’s or the Bellman-Ford algorithm, which are designed for problems where path costs are calculated by summing edge weights, may not be directly applicable or may need modification.
I am very new to this type of problem.. I thought I could just do DFS/BFS for all possible paths, store all the probabilities for each path, then return the maximum probability.. Why is not a viable approach?
The approach you suggested, which involves using depth-first search (DFS) or breadth-first search (BFS) to find all possible paths from the start to the end node, then computing and comparing the probabilities for each path to find the maximum one, is theoretically correct. However, it is highly inefficient and may not be feasible for larger graphs due to the time and space complexity.
The reason is that in a graph with ’n’ nodes and ’m’ edges, there can be up to O(n!) different paths (considering a full graph scenario). Trying to explore all of them would require exponential time, which is not acceptable for larger inputs as per the problem constraints (2 <= n <= 10^4, succProb.length == edges.length <= 2*10^4).
This is why we employ algorithms like Dijkstra’s or variants of it. Dijkstra’s algorithm, for instance, works by iteratively picking the “most promising” node (the node with the maximum probability in this case) and updating the probabilities of its neighboring nodes. This way, it ensures that once it has processed a node, it has found the path with the maximum probability to that node. This strategy, known as a greedy strategy, allows Dijkstra’s algorithm to solve the problem in polynomial time (O(n^2) or O(m log m) with a binary heap), making it much more efficient for larger graphs.
In conclusion, while your approach is theoretically correct, it would not be efficient or practical for larger graphs due to its exponential time complexity. A more efficient approach would be to use a modified version of Dijkstra’s algorithm as discussed.
Would a DFS solution with memoization work in this case? I came up with one which isn’t working:
def _dfs(self, g: 'Graph', visited: set, cache: dict, curr: int, end: int) -> int:
if curr == end:
return 1
if curr in cache:
return cache[curr]
visited.add(curr)
max_prob = 0
for neighbor, weight in g.nodes[curr]:
if neighbor in visited:
continue
max_prob = max(max_prob, weight*self._dfs(g, visited, cache, neighbor, end))
cache[curr] = max_prob
return max_prob
Yes, a DFS solution with memoization can work in this case, but it needs to be carefully implemented. The idea is to store the results of previously solved subproblems (paths from a given node to the end) to avoid solving them again, which can significantly improve the efficiency.
However, from your provided code, it seems that there’s a mistake in how you are handling the visited set. You add a node to visited when you start processing it, but you never remove it once you’re done. This can lead to scenarios where a potentially optimal path is missed because a node is marked as visited even though we have backtracked from it.
To correct this, you should remove a node from visited once you have explored all of its neighbors:
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By removing the current node from visited after visiting all its neighbors, we allow the algorithm to visit the node again in a different context (part of a different path), which ensures we don’t miss any valid paths.
Remember that DFS with memoization, while it can be more efficient than plain DFS, is still not as efficient as Dijkstra’s algorithm for this particular problem. This is because the Dijkstra algorithm avoids unnecessary explorations by focusing on the most promising paths (those with the highest probabilities) first.
Even i am not able to understand why dfs + memo doesn’t work. Gives wrong answer when n=5000
class Solution { public: map<int, map<int,double»mp; double dfs(int curr, int end, vector<vector<pair<int,double»>&adj, vector&vis){ if(curr==end)return 1; vis[curr] = 1; double ans = 0; for(auto child: adj[curr]){ if(vis[child.first]!=1){ if(mp[curr].find(child.first)!=mp[curr].end()){ ans = max(ans, mp[curr][child.first]); continue; } mp[curr][child.first] = child.second*dfs(child.first, end, adj, vis); ans = max(mp[curr][child.first], ans); } } vis[curr] = 0; return ans; }
double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start, int end) {
vector<vector<pair<int, double>>>adj(n);
int m = edges.size();
for(int i=0; i<m; i++){
adj[edges[i][0]].push_back({edges[i][1], succProb[i]});
adj[edges[i][1]].push_back({edges[i][0], succProb[i]});
}
vector<int>vis(n, 0);
return dfs(start, end, adj, vis);
}
};
Your DFS + memoization approach is not working due to a slight issue in the way the recursion is handled. Specifically, the DFS and memoization do not correctly account for cases where a node might be revisited along a higher probability path after it has been first visited.
DFS + memoization tends to work well in problems where the state depends only on the current node and not on the path taken to reach the current node. However, in this problem, the state (probability) at each node depends on the path taken to reach that node. This path-dependence is a feature often found in optimization problems like shortest or longest path problems.
This is why Dijkstra’s algorithm works well in this case, because it gradually builds up the maximum probability path from the starting node to each of the other nodes by considering the nodes in order of their distances (probabilities) from the start node. The priority queue ensures that at any given moment, the algorithm is considering the node that it can reach with the highest probability.
Your code needs to be modified to use a priority queue and follow a Dijkstra-like process to find the path with the highest product of probabilities. Here is the modified code in Python:
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This code is more efficient than the DFS + memoization approach and will correctly handle the path-dependence in the problem. It has a time complexity of O(ElogV), where E is the number of edges and V is the number of vertices (nodes), which makes it suitable for handling large inputs.
This path requires one to understand that shortest path is a minimization problem but given problem is a maximization problem. So how to convert from maximization problem to minimization problem with following constraints:
a) Resulting graph can’t have a negative weight cycle.
As this is a undirected graph hence there would be cycles so straightforward conversion from positive probability to negative probabiity doesnt work. So two steps:
a) Take log of the probability. This will be negative as probability max-value is 1.0 b) Negate the value obtained from above step. Resulting step is positive.
Post above algorithm is running shortest-parth.
Indeed, the original problem is a maximization problem which typically would not be a direct candidate for Dijkstra’s algorithm as it is fundamentally designed to solve minimization problems (finding the shortest path). But we can transform the maximization problem into a minimization one with a few mathematical operations and make it suitable for Dijkstra’s algorithm.
The steps are:
Take the logarithm: By taking the natural logarithm of each probability, we convert the multiplication of probabilities along a path into a sum of logarithms, since
log(ab) = log(a) + log(b). This makes it fit the nature of Dijkstra’s algorithm better, which is designed to work with the summation of weights. Note that the logarithm of a probability will be negative since probabilities range between 0 and 1.Negate the value: As you mentioned, we can’t have negative cycles in the graph when applying Dijkstra’s algorithm, so we negate the value obtained from the previous step. This converts the problem into a minimization problem because minimizing the negative of a value is the same as maximizing the original value.
Here’s how you can modify the Python code accordingly:
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This code has the same complexity as before but now works by converting the maximization problem into a minimization problem through the application of logarithms and negation. We also convert back from log-probability to probability at the end by using math.exp(-max_prob[end]).
…or just compare (1 - probability) and now it’s a minimization problem without any logarithms.
Yes, you’re right! It’s another valid approach to transform this problem to a minimization one. Specifically, since the probabilities are all between 0 and 1, subtracting each probability from 1 will give you a new weight for each edge. The edge with the highest probability becomes the edge with the smallest weight in the new graph.
Hence, the problem can now be viewed as a shortest path problem, and Dijkstra’s algorithm can be applied.
However, please note that there’s a small difference between these two methods: when we use logarithms, we are transforming multiplications into additions, which fits directly into Dijkstra’s algorithm structure (summing up edge weights). If you just use (1 - probability), you’ll still have to deal with the product of probabilities when calculating the cost for a path, which might require an additional step to correctly maintain and compare these products.
Here’s how you might modify the Python code accordingly:
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This code also has the same complexity as before, and it works by converting probabilities into 1 - probability, which is more intuitive and doesn’t involve logarithms. However, as noted earlier, this approach might require some additional steps to handle the product of probabilities correctly.
Can someone help, I cannot understand why my code is giving Wrong Answer. class Solution { unordered_map<int,bool> um; public: double maxProbability(int n, vector<vector>& edges, vector& succProb, int start, int end) { vector<vector> adj; vector<vector> weights; vector visited; for(int i=0;i<n;i++){ vector v1; vector v2; adj.push_back(v1); weights.push_back(v2); if(i==end){ visited.push_back(1); }else{ visited.push_back(-1); } } for(int i=0;i<edges.size();i++){ adj[edges[i][0]].push_back(edges[i][1]); adj[edges[i][1]].push_back(edges[i][0]); weights[edges[i][0]].push_back(succProb[i]); weights[edges[i][1]].push_back(succProb[i]); } return DFS(start,adj,weights,visited,end); } double DFS(int start,vector<vector>& adj, vector<vector>& weights, vector& visited, int end){ if(visited[start]!=-1){ return visited[start]; } um[start]=true; for(int i=0;i<adj[start].size();i++){ if(um.count(adj[start][i])==0){ visited[start]=max(visited[start],weights[start][i]*DFS(adj[start][i],adj,weights,visited,end)); } } um.erase(start); if(visited[start]==-1){ visited[start]=0; } return visited[start]; } };
This problem might be difficult to solve by standard Depth-First Search (DFS) for several reasons.
DFS is not an optimal algorithm for path finding problems. It might not always find the shortest (or longest in this case) path between nodes.
The way you’ve structured the DFS doesn’t account for the nature of the problem. While DFS explores all possible paths, it doesn’t have a built-in way to prioritize paths with higher probabilities.
It’s not quite right to use a visited array for tracking the maximum probabilities, because the same node can be visited multiple times through different paths, and each of those paths may have a different probability.
Instead, the algorithm you’re looking for is a modification of Dijkstra’s algorithm. This algorithm is designed to find the shortest paths between nodes in a graph, which is similar to what you’re trying to do here (find the highest-probability paths).
Also, note that your visited vector seems to be misused. In your code, you are pushing a 1 for the end node and -1 for the rest, then updating these values with probabilities as you traverse the graph. However, ‘visited’ typically means a node has already been visited or not, which is a boolean value. For this specific problem, a more suitable data structure might be a ‘probabilities’ vector to track the maximum probability to reach each node.
Overall, using a priority queue with Dijkstra’s algorithm is the appropriate way to solve this problem. DFS might work for smaller test cases, but will not perform well on larger and more complex graphs due to its lack of an efficient way to prioritize paths based on their probabilities.
Your code seems to correctly implement a variation of Dijkstra’s algorithm, transforming the probabilities into a minimization problem by taking the logarithm and changing the sign. However, there may be an issue with your priority queue comparator.
You’re currently using a less-than comparison in the custom comparator for your priority queue. This is the default behavior for a priority queue in C++, which always gives the maximum element. However, Dijkstra’s algorithm needs a min-heap, not a max-heap.
Therefore, your comparator function should return p1.second < p2.second instead of p1.second > p2.second.
Here’s the corrected code:
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This change should make your priority queue behave as a min-heap, which is necessary for Dijkstra’s algorithm to work correctly.