Finding a Path through Difficult Times

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The publication is still readable, it is, in fact, quite nice. One of the reasons that it is so nice was that I designed it without pencil and paper. I learned later that one of the advantages of designing without pencil and paper is that you are almost forced to avoid all avoidable complexities. Eventually that algorithm became, to my great amazement, one of the cornerstones of my fame. Dijkstra thought about the shortest path problem when working at the Mathematical Center in Amsterdam in as a programmer to demonstrate the capabilities of a new computer called ARMAC. He designed the shortest path algorithm and later implemented it for ARMAC for a slightly simplified transportation map of 64 cities in the Netherlands 64, so that 6 bits would be sufficient to encode the city number.

Let the node at which we are starting be called the initial node.

Let the distance of node Y be the distance from the initial node to Y. Dijkstra's algorithm will assign some initial distance values and will try to improve them step by step. When planning a route, it is actually not necessary to wait until the destination node is "visited" as above: the algorithm can stop once the destination node has the smallest tentative distance among all "unvisited" nodes and thus could be selected as the next "current".

Suppose you would like to find the shortest path between two intersections on a city map: a starting point and a destination. Dijkstra's algorithm initially marks the distance from the starting point to every other intersection on the map with infinity. This is done not to imply that there is an infinite distance, but to note that those intersections have not been visited yet. Some variants of this method leave the intersections' distances unlabeled.

Now select the current intersection at each iteration. For the first iteration, the current intersection will be the starting point, and the distance to it the intersection's label will be zero. For subsequent iterations after the first , the current intersection will be a closest unvisited intersection to the starting point this will be easy to find. From the current intersection, update the distance to every unvisited intersection that is directly connected to it. This is done by determining the sum of the distance between an unvisited intersection and the value of the current intersection and then relabeling the unvisited intersection with this value the sum if it is less than the unvisited intersection's current value.

In effect, the intersection is relabeled if the path to it through the current intersection is shorter than the previously known paths. After you have updated the distances to each neighboring intersection , mark the current intersection as visited and select an unvisited intersection with minimal distance from the starting point — or the lowest label—as the current intersection. Intersections marked as visited are labeled with the shortest path from the starting point to it and will not be revisited or returned to.

Continue this process of updating the neighboring intersections with the shortest distances, marking the current intersection as visited, and moving onto a closest unvisited intersection until you have marked the destination as visited. Once you have marked the destination as visited as is the case with any visited intersection , you have determined the shortest path to it from the starting point and can trace your way back following the arrows in reverse. In the algorithm's implementations, this is usually done after the algorithm has reached the destination node by following the nodes' parents from the destination node up to the starting node; that's why we also keep track of each node's parent.


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This algorithm makes no attempt of direct "exploration" towards the destination as one might expect. Rather, the sole consideration in determining the next "current" intersection is its distance from the starting point. This algorithm therefore expands outward from the starting point, interactively considering every node that is closer in terms of shortest path distance until it reaches the destination. When understood in this way, it is clear how the algorithm necessarily finds the shortest path.

However, it may also reveal one of the algorithm's weaknesses: its relative slowness in some topologies. The variable alt on line 18 is the length of the path from the root node to the neighbor node v if it were to go through u. If this path is shorter than the current shortest path recorded for v , that current path is replaced with this alt path. The prev array is populated with a pointer to the "next-hop" node on the source graph to get the shortest route to the source.

Now we can read the shortest path from source to target by reverse iteration:. Now sequence S is the list of vertices constituting one of the shortest paths from source to target , or the empty sequence if no path exists. A more general problem would be to find all the shortest paths between source and target there might be several different ones of the same length. Then instead of storing only a single node in each entry of prev[] we would store all nodes satisfying the relaxation condition.

For example, if both r and source connect to target and both of them lie on different shortest paths through target because the edge cost is the same in both cases , then we would add both r and source to prev[ target ]. When the algorithm completes, prev[] data structure will actually describe a graph that is a subset of the original graph with some edges removed. Its key property will be that if the algorithm was run with some starting node, then every path from that node to any other node in the new graph will be the shortest path between those nodes in the original graph, and all paths of that length from the original graph will be present in the new graph.

Then to actually find all these shortest paths between two given nodes we would use a path finding algorithm on the new graph, such as depth-first search. As mentioned earlier, using such a data structure can lead to faster computing times than using a basic queue. Other data structures can be used to achieve even faster computing times in practice.

Invariant hypothesis : For each visited node v , dist[v] is considered the shortest distance from source to v ; and for each unvisited node u , dist[u] is assumed the shortest distance when traveling via visited nodes only, from source to u. This assumption is only considered if a path exists, otherwise the distance is set to infinity. The base case is when there is just one visited node, namely the initial node source , in which case the hypothesis is trivial.

Otherwise, assume the hypothesis for n-1 visited nodes. After processing u it will still be true that for each unvisited nodes w , dist[w] will be the shortest distance from source to w using visited nodes only, because if there were a shorter path that doesn't go by u we would have found it previously, and if there were a shorter path using u we would have updated it when processing u. How tight a bound is possible depends on the way the vertex set Q is implemented.

For any implementation of the vertex set Q , the running time is in. The simplest implementation of Dijkstra's algorithm stores the vertex set Q as an ordinary linked list or array, and extract-minimum is simply a linear search through all vertices in Q. It must be noted that if the implementation stores the graph as an adjacency list, the running time for a dense graph i.

2. You’ve Overcome Challenges Before

To perform decrease-key steps in a binary heap efficiently, it is necessary to use an auxiliary data structure that maps each vertex to its position in the heap, and to keep this structure up to date as the priority queue Q changes. With a self-balancing binary search tree or binary heap, the algorithm requires. The Fibonacci heap improves this to. In common presentations of Dijkstra's algorithm, initially all nodes are entered into the priority queue.

This is, however, not necessary: the algorithm can start with a priority queue that contains only one item, and insert new items as they are discovered instead of doing a decrease-key, check whether the key is in the queue; if it is, decrease its key, otherwise insert it.

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Moreover, not inserting all nodes in a graph makes it possible to extend the algorithm to find the shortest path from a single source to the closest of a set of target nodes on infinite graphs or those too large to represent in memory. The resulting algorithm is called uniform-cost search UCS in the artificial intelligence literature [7] [15] [16] and can be expressed in pseudocode as.

When arc weights are small integers bounded by a parameter C , a monotone priority queue can be used to speed up Dijkstra's algorithm. Finally, the best algorithms in this special case are as follows.

Travelling salesman problem

The functionality of Dijkstra's original algorithm can be extended with a variety of modifications. For example, sometimes it is desirable to present solutions which are less than mathematically optimal. To obtain a ranked list of less-than-optimal solutions, the optimal solution is first calculated. A single edge appearing in the optimal solution is removed from the graph, and the optimum solution to this new graph is calculated.

Each edge of the original solution is suppressed in turn and a new shortest-path calculated. The secondary solutions are then ranked and presented after the first optimal solution. Unlike Dijkstra's algorithm, the Bellman—Ford algorithm can be used on graphs with negative edge weights, as long as the graph contains no negative cycle reachable from the source vertex s.

The presence of such cycles means there is no shortest path, since the total weight becomes lower each time the cycle is traversed. It is possible to adapt Dijkstra's algorithm to handle negative weight edges by combining it with the Bellman-Ford algorithm to remove negative edges and detect negative cycles , such an algorithm is called Johnson's algorithm. This approach can be viewed from the perspective of linear programming : there is a natural linear program for computing shortest paths , and solutions to its dual linear program are feasible if and only if they form a consistent heuristic speaking roughly, since the sign conventions differ from place to place in the literature.

Ask them how they knew this path was right for them. Ask them what they would do if they were you. I learned this exercise from successful writer Jeff Goins. A worldview statement helps you shape your perception on the world and can provide guidance on the direction you should head towards. My world view statement is "All people should build a life around their natural talents and strengths. You think all of the time, but it's distracted, unfocused thinking. Blocking out time to do nothing else but think can lead to some serious discoveries.

How to Get Through Rough Patches in Life - Jocko Willink

Bill Gates goes on a two week retreat every year just to think. Maybe you can start with 30 minutes -- to be alone with your thoughts and work through the choices you want to make for your future. You can use regret to determine if you should heard in a certain direction. Look to the end of your life. If you don't pursue what you're thinking about, will you ultimately regret it when it's all said and done?


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There's a second part to this test. Again, look to the end of your life. If you tried and things didn't work out, would you feel at peace knowing you gave it a shot? If you can answer yes both questions, you're probably on the right track. I learned about this technique while listening to a podcast interview with Pat Flynn. Here's how the airport test works. Imagine it's five years from now and you run into an old friend.

Your friend asks you how you're doing and what you've been up to.

Dijkstra's algorithm - Wikipedia

Given that you've spent the past 5 years building your dream life, what would your response be? The airport test takes the boring "5 year plan" and makes it sexier. Fear and self-doubt get a bad rap. If you're pursuing something that doesn't scare you it's not worth your time. The more you feel like an impostor, the more you feel terrified, and the more you doubt yourself, the more likely you're on the right track.

You don't need to conquer fear.