Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Show that for one-dimensional cellular automata, CA-Predecessor is in P. Hint:
think about using recursion on a one-bit-shorter prefix of the input configuration.
Suppose that the CA follows the rule that the configuration shrinks by 2 cells on each update, since
the leftmost cell has no left neighbor and the rightmost cell has no right neighbor. Therefore, on
input a configuration of length n, we are asking whether there is a predecessor configuration of
length n + 2.
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- The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ ?? if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destroying ????return extracted (1) Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its…arrow_forwardThe off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ Kj if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destoying Kjreturn extracted Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its insertion.…arrow_forwardLet x[n] = {1,2,3}. Let y[n] = {1, 2, 3, 4, 5). Circularly convolve x and y where N = 6. Develop the flow-graph of N = 2 x 2 point Decimation-in-time FFT algorithm.arrow_forward
- Perform Dijkstra’s algorithm (while (Q is not empty){...}, where Q isa min priority queue) on the following graph for source node “s”, writing outevery execution of the RELAX function, the update to the priority queue and theupdates of distance from the source node and parent for the currently visited node.arrow_forwardShow that for one-dimensional cellular automata, CA-Predecessor is in P. Hint:think about using recursion on a one-bit-shorter prefix of the input configuration.Suppose that the CA follows the rule that the configuration shrinks by 2 cells on each update, sincethe leftmost cell has no left neighbor and the rightmost cell has no right neighbor. Therefore, oninput a configuration of length n, we are asking whether there is a predecessor configuration oflength n + 2. By the way 2^n is not polynomial at all. Also,show how your algorithm works for input =010011 which follows rule 111 110 101 100 011 010 001 000 0 0 0 1 1 1 1 0arrow_forwardSuppose that in applying the minimization algorithm in Section 2.6, weestablish some fixed order in which to process the pairs, and we follow thesame order on each pass a. What is the maximum number of passes that might be required?Describe an FA, and an ordering of the pairs, that would require thisnumber.b. Is there always a fixed order (depending on M) that would guaranteethat no pairs are marked after the first pass, so that the algorithmterminates after two passes?arrow_forward
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