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A non-recursive implementation of DFS with worst-case space complexity (| |), with the possibility of duplicate vertices on the stack: [6] procedure DFS_iterative(G, v) is let S be a stack S.push(v) while S is not empty do v = S.pop() if v is not labeled as discovered then label v as discovered for all edges from v to w in G.adjacentEdges(v) do ...
A depth-first search (DFS) is an algorithm for traversing a finite graph. DFS visits the child vertices before visiting the sibling vertices; that is, it traverses the depth of any particular path before exploring its breadth. A stack (often the program's call stack via recursion) is generally used when implementing the algorithm.
If G is a tree, replacing the queue of this breadth-first search algorithm with a stack will yield a depth-first search algorithm. For general graphs, replacing the stack of the iterative depth-first search implementation with a queue would also produce a breadth-first search algorithm, although a somewhat nonstandard one. [10]
While the stack is not empty Pop a cell from the stack and make it a current cell; If the current cell has any neighbours which have not been visited Push the current cell to the stack; Choose one of the unvisited neighbours; Remove the wall between the current cell and the chosen cell; Mark the chosen cell as visited and push it to the stack
The algorithm performs a depth-first search of the given graph G, maintaining as it does two stacks S and P (in addition to the normal call stack for a recursive function). Stack S contains all the vertices that have not yet been assigned to a strongly connected component, in the order in which the depth-first search reaches the vertices.
Heck, you could make the stack explicit in the recursive example by replacing "recursively call DFS(G,w)" with "stack.push(w); tailcall(DFS)" (or "goto top" if you prefer). What's there now is complicating things by changing the algorithm at the same time as it's changing where the stack is.
Since IDDFS, at any point, is engaged in a depth-first search, it need only store a stack of nodes which represents the branch of the tree it is expanding. Since it finds a solution of optimal length, the maximum depth of this stack is d {\displaystyle d} , and hence the maximum amount of space is O ( d ) {\displaystyle O(d)} .
The canonical application of topological sorting is in scheduling a sequence of jobs or tasks based on their dependencies.The jobs are represented by vertices, and there is an edge from x to y if job x must be completed before job y can be started (for example, when washing clothes, the washing machine must finish before we put the clothes in the dryer).