closest pair of points algo #943
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| Edge case: closest points lie on different sides of partition | ||
| This case handled by forming a strip of points | ||
| which are at a distance (< closest_pair_dis) from mid-point. | ||
| (It is a proven that strip contains at most 6 points) |
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No, it's wrong. Please review the theoretical part of this part more.
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| def euclidean_distance(point1, point2): | ||
| return math.sqrt(pow(point1[0] - point2[0], 2) + pow(point1[1] - point2[1], 2)) |
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It's OK but actually sqrt is unnecessary. For the optimization, you can use sqrt only once after all calculations have been done.
| cross_strip.append(point) | ||
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| cross_strip = column_based_sort(cross_strip, 1) | ||
| closest_in_strip = dis_btw_closest_pair(cross_strip, |
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Don't use brute force for the points inside the strip. You have to calculate this part in O(n).
Dharni0607
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The codes are now fine. I'm going to merge it but consider that we should reduce the closest pair of points time complexity to O(n log n), in future (now it's O(n * log n * log n)). |
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Ok. |
| return min_dis | ||
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| points = [[2, 3], [12, 30], [40, 50], [5, 1], [12, 10]] | ||
| points = ((2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (0, 2), (5, 6), (1, 2)) |
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Don't use a tuple of tuples. Use a list of tuples: [(1, 2), (3, 4), (5, 6)]
| """ divide and conquer approach | ||
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| Parameters : | ||
| points, points_count (tuple(tuple(int, int)), int) |
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Don't use a tuple of tuples. Use a list of tuples.
| """ closest pair of points in strip | ||
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| Parameters : | ||
| points, points_count, min_dis (tuple(tuple(int, int)), int, int) |
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Don't use a tuple of tuples. Use a list of tuples.
| if current_dis < min_dis: | ||
| min_dis = current_dis | ||
| return min_dis | ||
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| #divide and conquer approach | ||
| def closest_pair_of_points(points, no_of_points): | ||
| def dis_btw_closest_in_strip(points, points_counts, min_dis = float("inf")): |
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It's not important but if you have a better name for this function, please update it. Long function names are also fine. Using "between" is better than "btw". Also you can use "brute force" word in its name.
These were only some suggestions. It's up to you.
| Time complexity: O(n * (logn) ^ 2) | ||
| min(closest_pair_dis, closest_in_strip) would be the final answer. | ||
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| Time complexity: O(n * (logn)^2) |
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If you want you can put a reference to it. Are there any references in CLRS book?
added closest pair of points algorithm under divide and conquer.
Please review.
;)