memoization

问题 Any pointers on how to solve efficiently the following function in Haskell, for large numbers (n > 108) f(n) = max(n, f(n/2) + f(n/3) + f(n/4)) I've seen examples of memoization in Haskell to solve fibonacci numbers, which involved computing (lazily) all the fibonacci numbers up to the required n. But in this case, for a given n, we only need to compute very few intermediate results. Thanks 回答1: We can do this very efficiently by making a structure that we can index in sub-linear time. But

问题 I have built an app on ReactJS 16.8.5 and React-Redux 3.7.2. When the app loads the app mounts, initial store is set and database subscriptions are set up against a Firebase Realtime Database. The app contains a header, Sidebar and content section. I have implemented reselect along with React.memo to avoid rerendring when props change, but the Sidebar component is still re-rendering. Using React profiler API and a areEqual comparison function in React.memo I can see that the Sidebar is being

问题 I have built an app on ReactJS 16.8.5 and React-Redux 3.7.2. When the app loads the app mounts, initial store is set and database subscriptions are set up against a Firebase Realtime Database. The app contains a header, Sidebar and content section. I have implemented reselect along with React.memo to avoid rerendring when props change, but the Sidebar component is still re-rendering. Using React profiler API and a areEqual comparison function in React.memo I can see that the Sidebar is being

问题 I have this memoization technique to reduce the number of calls getting a Fibonacci sequence number: def fastFib(n, memo): global numCalls numCalls += 1 print 'fib1 called with', n if not n in memo: memo[n] = fastFib(n-1, memo) + fastFib(n-2, memo) return memo[n] def fib1(n): memo = {0:1, 1:1} return fastFib(n, memo) numCalls = 0 n = 6 res = fib1(n) print 'fib of', n,'=', res, 'numCalls = ', numCalls But i am stuck at here: memo[n] = fastFib(n-1, memo) + fastFib(n-2, memo) and here memo = {0

问题 I am a student studying javascript , and I encountered a problem while studying memoization pattern. This is the code : Function.prototype.memoization = function(key) { var arg = Array.prorotype.slice.call( arguments, 1 ); this.data = this.data || {} ; //THE code return this.data[key] !== undefined ? this.data[key] : this.data[key] = this.apply(this, arg); }; For me it is shown as just OR operation between an array and a blank array, and I cannot understand why such code is needed. 回答1: The |

问题 Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 22 days ago . There are many Javascript libraries out there for memoizing functions. What are some of the best ones (high quality, flexible, fast, etc.), for the various use-cases? 回答1: I've found that there are tons of different JavaScript memoization libraries out there, and they all operate in

问题 I will be happy to get some help. I have the following problem: I'm given a list of numbers seq and a target number and I need to write 2 things: A recursive solution that returns True if there is a sum of a subsequence that equals the target number and False otherwise. example: subset_sum([-1,1,5,4],0) # True subset_sum([-1,1,5,4],-3) # False Secondly, I need to write a solution using what I wrote in the previous solution but now with memoization that uses a dictionary in which the keys are

问题 I will be happy to get some help. I have the following problem: I'm given a list of numbers seq and a target number and I need to write 2 things: A recursive solution that returns True if there is a sum of a subsequence that equals the target number and False otherwise. example: subset_sum([-1,1,5,4],0) # True subset_sum([-1,1,5,4],-3) # False Secondly, I need to write a solution using what I wrote in the previous solution but now with memoization that uses a dictionary in which the keys are

问题 I came up with simple following recursive solution for Longest increasing sub-sequence. But, Can you help to include memoization into this recursive solution. public int findLIS(int a[], int maxSoFar, int item, int count) { if(item == a.length) { return count; } int length1 = findLIS(a,maxSoFar, item+1, count); int length2 = 0; if(a[item] > maxSoFar) { length2 = findLIS(a, a[item], item+1, count + 1); } return Math.max(length1, length2); } PS: This not a homework question, it is more of my

问题 Here is a generic memoization wrapper I wrote for functions. It makes use of tuplehash. template<typename R, typename... Args> class memofunc{ typedef R (*func)(Args...); func fun_; unordered_map<tuple<Args...>, R, tuplehash::hash<tuple<Args...> > > map_; public: memofunc(func fu):fun_(fu){} R operator()(Args&&... args){ auto key = make_tuple(std::forward<Args>(args)...); auto q = map_.find(key); if(q == map_.end()){ R res = fun_(std::forward<Args>(args)...); map_.insert({key,res}); return