The other day we had a very nice oriental-style BBQ at my place. Susi made a Couscous salad, while I made some hummus, a bell pepper salad, a salsa with Habanero chilis, and some marinated olives. On the meaty side, we had lamb chops and chicken wings. Very nice food indeed!<\/p>\n
The evening ended with us all sitting still in the semi-dark of my terrace watching out for shooting stars. And we even saw some. Even one, which we will remember forever \ud83d\ude09<\/p>\n
In the other life (work), I’ve written an algorithm for finding the bordering elements of contiguous ranges in a given array of ordered values. I’ve used rapid prototyping in Python before transferring the solution to C++. The Python code goes like this:<\/p>\n
n = [1, 2, 3, 4, 5, 6, 7,\u00a0\u00a0\u00a0 9, 10, 11, 12,\u00a0\u00a0\u00a0\u00a0\u00a0 14, 15, 16]\r\nend = len(n) - 1\r\n\r\nranges = []\r\n\r\ndef get_right(left, step):\r\n\u00a0\u00a0\u00a0 if left == end:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 return left\r\n\u00a0\u00a0\u00a0 leftmost_r = None\r\n\u00a0\u00a0\u00a0 i = last_i = end\r\n\u00a0\u00a0\u00a0 while True:\r\n match = n[i] - n[left] == i - left\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if (i == end and match) or i == leftmost_r:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 return last_i\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 last_i = i \r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if match:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 i += step\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 else:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 leftmost_r = i\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 i -= step\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 if step != 1:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 step \/= 2\r\n\r\nleft = 0\r\nwhile True:\r\n\u00a0\u00a0\u00a0 right = get_right(left, (len(n) - left) \/ 2)\r\n\u00a0\u00a0\u00a0 ranges.append((n[left], n[right]))\r\n\u00a0\u00a0\u00a0 if right == end:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 break\r\n\u00a0\u00a0\u00a0 else:\r\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 left = right + 1<\/pre>\nThe basic assumption for the algorithm to work is that
n<\/code> contains sorted values. This allows us to do a kind of binary search, if the currently looked at range (fromleft<\/code> toi<\/code>) does not fit the criterion for a match (namely thatn[i] - n[left] != i - left<\/code>). Maybe this is useful for someone!<\/p>\n<\/p>\n
NP: Agnostic Front – Gotta go<\/p>\n","protected":false},"excerpt":{"rendered":"
The other day we had a very nice oriental-style BBQ at my place. Susi made a Couscous salad, while I made some hummus, a bell pepper salad, a salsa with Habanero chilis, and some marinated olives. On the meaty side, … Continue reading