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In order to determine the full histogram for all matchings of a given size, we need to generate every single possible matching in a unique way. This is where the inductive description from the introduction becomes useful, as it provides a way to do so recursively: We can generate all arc diagrams with n arcs from all arc diagrams with (n - 1) arcs by adding one arc to each of them in precisely 2n 1 ways. To this end, we take an arc diagram with (n - 1) arcs, insert one new point at the left end and one more point somewhere to the right of it (2n − 1 options), and then match the newly inserted points to obtain the additional arc. The only "problem" is that we need to relabel some points in doing so. - - Inserting a point to the left implies that the indices of the other points all have to be increased by one. Moreover, if we insert another point at some position m, then all the indices with values m and larger again have to be increased by one. . = 1, 2, 3. This implies that all indices with values m and larger need 3 we find {(1, 2), (0,3)}. For example, if we start with an arc diagram with precisely one arc, {(0, 1)}, then inserting a point to the left gives this one index zero and increases the other indices, leading to {(1, 2), (0, ·)} with still to be inserted. We can now insert a second point at positions m = to be increased. For m = 1 this leads to {(2, 3), (0, 1)}, for m = 2 we get {(1, 3), (0, 2)}, and finally for m Using this approach, write a function get_all_matchings that takes as argument a non-negative integer n and returns a list of all (2n − 1)!! matchings on n arcs. You may do this recursively or iteratively. Test your function by verifying that len(get_all_matchings(n)) is equal to (2n − 1)!! for n = 0, 1, . . ., 7.

In order to determine the full histogram for all matchings of a given size, we need to generate every single possible matching in a unique way. This is where the inductive description from the introduction becomes useful, as it provides a way to do so recursively: We can generate all arc diagrams with n arcs from all arc diagrams with (n - 1) arcs by adding one arc to each of them in precisely 2n 1 ways. To this end, we take an arc diagram with (n - 1) arcs, insert one new point at the left end and one more point somewhere to the right of it (2n − 1 options), and then match the newly inserted points to obtain the additional arc. The only "problem" is that we need to relabel some points in doing so. - - Inserting a point to the left implies that the indices of the other points all have to be increased by one. Moreover, if we insert another point at some position m, then all the indices with values m and larger again have to be increased by one. . = 1, 2, 3. This implies that all indices with values m and larger need 3 we find {(1, 2), (0,3)}. For example, if we start with an arc diagram with precisely one arc, {(0, 1)}, then inserting a point to the left gives this one index zero and increases the other indices, leading to {(1, 2), (0, ·)} with still to be inserted. We can now insert a second point at positions m = to be increased. For m = 1 this leads to {(2, 3), (0, 1)}, for m = 2 we get {(1, 3), (0, 2)}, and finally for m Using this approach, write a function get_all_matchings that takes as argument a non-negative integer n and returns a list of all (2n − 1)!! matchings on n arcs. You may do this recursively or iteratively. Test your function by verifying that len(get_all_matchings(n)) is equal to (2n − 1)!! for n = 0, 1, . . ., 7.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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In order to determine the full histogram for all matchings of a given size, we need to generate every single possible matching in a unique way. This is where the inductive description from the introduction becomes useful, as it provides a way to do so recursively: We can generate all arc diagrams with n arcs from all arc diagrams with (n - 1) arcs by adding one arc to each of them in precisely 2n 1 ways. To this end, we take an arc diagram with (n - 1) arcs, insert one new point at the left end and one more point somewhere to the right of it (2n − 1 options), and then match the newly inserted points to obtain the additional arc. The only "problem" is that we need to relabel some points in doing so. - - Inserting a point to the left implies that the indices of the other points all have to be increased by one. Moreover, if we insert another point at some position m, then all the indices with values m and larger again have to be increased by one. . = 1, 2, 3. This implies that all indices with values m and larger need 3 we find {(1, 2), (0,3)}. For example, if we start with an arc diagram with precisely one arc, {(0, 1)}, then inserting a point to the left gives this one index zero and increases the other indices, leading to {(1, 2), (0, ·)} with still to be inserted. We can now insert a second point at positions m = to be increased. For m = 1 this leads to {(2, 3), (0, 1)}, for m = 2 we get {(1, 3), (0, 2)}, and finally for m Using this approach, write a function get_all_matchings that takes as argument a non-negative integer n and returns a list of all (2n − 1)!! matchings on n arcs. You may do this recursively or iteratively. Test your function by verifying that len(get_all_matchings(n)) is equal to (2n − 1)!! for n = 0, 1, . . ., 7.
Transcribed Image Text:In order to determine the full histogram for all matchings of a given size, we need to generate every single possible matching in a unique way. This is where the inductive description from the introduction becomes useful, as it provides a way to do so recursively: We can generate all arc diagrams with n arcs from all arc diagrams with (n - 1) arcs by adding one arc to each of them in precisely 2n 1 ways. To this end, we take an arc diagram with (n - 1) arcs, insert one new point at the left end and one more point somewhere to the right of it (2n − 1 options), and then match the newly inserted points to obtain the additional arc. The only "problem" is that we need to relabel some points in doing so. - - Inserting a point to the left implies that the indices of the other points all have to be increased by one. Moreover, if we insert another point at some position m, then all the indices with values m and larger again have to be increased by one. . = 1, 2, 3. This implies that all indices with values m and larger need 3 we find {(1, 2), (0,3)}. For example, if we start with an arc diagram with precisely one arc, {(0, 1)}, then inserting a point to the left gives this one index zero and increases the other indices, leading to {(1, 2), (0, ·)} with still to be inserted. We can now insert a second point at positions m = to be increased. For m = 1 this leads to {(2, 3), (0, 1)}, for m = 2 we get {(1, 3), (0, 2)}, and finally for m Using this approach, write a function get_all_matchings that takes as argument a non-negative integer n and returns a list of all (2n − 1)!! matchings on n arcs. You may do this recursively or iteratively. Test your function by verifying that len(get_all_matchings(n)) is equal to (2n − 1)!! for n = 0, 1, . . ., 7.
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