Search results
Results from the WOW.Com Content Network
For example, one variant of the block nested loop join reads an entire page of tuples into memory and loads them into a hash table. It then scans S {\displaystyle S} , and probes the hash table to find S {\displaystyle S} tuples that match any of the tuples in the current page of R {\displaystyle R} .
Keep only the last and current row of the DP table to save memory (((,)) instead of ()) The last and current row can be stored on the same 1D array by traversing the inner loop backwards; Store only non-zero values in the rows. This can be done using hash-tables instead of arrays.
Another option when there are multiple matches is to use some criterion to select one of the rows from the matching set, usually a spatial optimization criterion. [2] [8] For example, one could join the school building points (not the districts) to the student residents points by selecting the school that is nearest to each student. Not all ...
Consider the problem of assigning values, either zero or one, to the positions of an n × n matrix, with n even, so that each row and each column contains exactly n / 2 zeros and n / 2 ones. We ask how many different assignments there are for a given . For example, when n = 4, five possible solutions are
Motion planning, also path planning (also known as the navigation problem or the piano mover's problem) is a computational problem to find a sequence of valid configurations that moves the object from the source to destination.
In graph theory, Yen's algorithm computes single-source K-shortest loopless paths for a graph with non-negative edge cost. [1] The algorithm was published by Jin Y. Yen in 1971 and employs any shortest path algorithm to find the best path, then proceeds to find K − 1 deviations of the best path.
The hash join is an example of a join algorithm and is used in the implementation of a relational database management system.All variants of hash join algorithms involve building hash tables from the tuples of one or both of the joined relations, and subsequently probing those tables so that only tuples with the same hash code need to be compared for equality in equijoins.
Apply dynamic programming to this path decomposition to find a longest path in time (!), where is the number of vertices in the graph. Since the output path has length at least as large as d {\displaystyle d} , the running time is also bounded by O ( ℓ ! 2 ℓ n ) {\displaystyle O(\ell !2^{\ell }n)} , where ℓ {\displaystyle \ell } is the ...