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I didn't analyze your solution completely, but it looks a bit too complicated in my opinion. I'd simply keep the last 120 request timestamps (as you do) and delay each request that would be less than 60 seconds after the 120th request before. Then you will be able to handle bursts but still ensure that there are never more than 120 requests within any 60-...


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You can think its better to have a gpu call per level of the tree, but I think its simpler if you just put it all in 1 call, do the steps in an iteration, and get your cores in operation back via amount of simultaneous whole accesses. Heres a setup to do a key store in a tree, on the gpu. use a 32bit uint texture, and use this format-> [NEXTFREEPOSITION]...


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Improving the pseudocode structure In line 7 it appears that you restart every time you find a new element to add to D. However, in pseudocode and mathematical language, you have no guarantee about the order in which foreach takes the elements of I. Therefore, you could as well continue the loop to the end of the foreach and add an outer loop: D ← {...


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The solution by Jonathan Eunice is pretty clean and easy to follow. When constructing the list of parents, I'd slice off the parent information from the value side of the parent dictionary. Instead of this: # contstruct list of parents parents = defaultdict(list) for p in people: parents[p[2]].append(p) show_val("parents", parents) I'd have this: # ...


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Late answer, but an explanation for how this works: The algorithm is based on Lucas sequence relations for Fibonacci numbers. https://en.wikipedia.org/wiki/Lucas_sequence#Other_relations Specifically these equations: F(m) = F(m-1) + F(m-2) F(m+n) = F(m+1) F(n) + F(m) F(n-1) F(2n) = F(n) L(n) = F(n) (F(n+1) + F(n-1)) = F(n)((F(n) + F(n-1)) + F(n-...


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Represent your (in)equations as a matrix: 1 0 0 0 0 .01 1 1 0 0 0 .02 1 1 1 0 0 .02 1 1 1 1 1 .05 Subtract an appropriate multiple (in this case 1) of the first row from each of the others in order to leave zeros in the rest of the column: 1 0 0 0 0 .01 0 1 0 0 0 .01 0 1 1 0 0 .01 0 1 1 1 1 .04 Subtract an appropriate multiple of the second row from each ...


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You can just do it like the lottery: shuffle and pick. It is possible that the same kind of item is picked two or more times, just as it is possible that sequential rolls with dice yield the same number, that's how random works :-) For the case where a number of prizes should be drawn until the 600th draw, put all of these prizes in a list together with ...


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It is a knapsack problem: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. Source: https://en.wikipedia.org/wiki/Knapsack_problem The value of each box is the amount of money ...


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Following Doc Brown's recommended steps of GUID generation/replacement, and this SO answer to recursively go through all properties of the object graph (https://stackoverflow.com/a/20554262/10340388) I came with the following solution. The function is recursive for collections of objects as well as classes that could hold GUID's. It can probably be cleaned ...


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Memberwise clone all the green and yellow objects Walk the list of green and yellow objects and update their Guids Add the old and new guid pairs to a Dictionary Walk the list of green and yellow objects again and swap the Refs using the Dictionary created in the previous step You can also serialize and then deserialize to a format that supports object ...


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The problem of copying the object graph and the problem of assigning new GUIDs to the object graph are very similar, but they can be handled separately and one after another, so let me start with copying first. In case the object graph is just a tree, there may be a simpler solution, but in a comment you mentioned several MainObjects sharing references to ...


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If I understand your question correctly, you want to filter a list based on the next item in the list instead of the item itself, i.e. in pseudocode: if(items[i] == stop && items[i + 1] == start) return items[i]; This question focuses on how to observe the previous item, but it's essentially doing the same thing and can be tailored to your ...


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There has been some research on detecting clock skew. Check out "Real Time Clock Skew Estimation over Network Delays" and detecting spoofed messages. Also, I don't have a good source for this, computers while running have very precise clocks. It's the battery powered RTC that runs when they are off that tends to skew.


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This answer is about a more advanced destination control system. In a building with multiple elevators where people indicate which floor they want to go to and the system assigns an elevator to them. The idea is fairly simple, theoretically you know where everyone and every elevator is going. So you can for each elevator, calculate When your expected ...


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The problem that you've posted is unclear in the following: Are you given one X and N ranges [L, R] or many X's for the same N ranges?   — This makes a big difference in efficiency considerations. Do you need to constantly modify the N ranges: adding/removing ranges, or adjusting L's and R's   — This also makes a huge difference in ...


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