tail-recursion

问题 I am trying to rewrite this piece of code from https://github.com/lspector/gp/blob/master/src/gp/evolvefn_zip.clj to use recur: (defn random-code [depth] (if (or (zero? depth) (zero? (rand-int 2))) (random-terminal) (let [f (random-function)] (cons f (repeatedly (get function-table f) #(random-code (dec depth))))))) The problem is, I have absolutely no idea how to do that. The only thing I can think of is something like this: (defn random-code [depth] (loop [d depth t 0 c []] (if (or (zero?

问题 I was trying to print binary numbers in ascending order of 0's (00 then 01, 10, 11). Such that the zeros are before. I tried using the below code from here but this does not give the right op (running sample) void test2() { final int grayCodeLength = 4; // generate matrix final int grayCodeCount = 1 << grayCodeLength; // = 2 ^ grayCodeLength int grayCodeMatrix[][] = new int[grayCodeCount][grayCodeLength]; for (int i = 0; i < grayCodeCount; i++) { int grayCode = (i >> 1) ^ i; for (int j =0; j

问题 How can I pass a list as a parameter to a function adding elements to it recursively,and have it unmodified when it comes out of recursion? I want to use the list at each level of recursion with the list having the values added by deeper recursion levels. To be more specific I want to do a DFS search on a graph and I want to store in the list the nodes I visited. 回答1: One method of doing this is just to return the list so you have access to it at higher levels of recursion. Another method is

问题 How can I pass a list as a parameter to a function adding elements to it recursively,and have it unmodified when it comes out of recursion? I want to use the list at each level of recursion with the list having the values added by deeper recursion levels. To be more specific I want to do a DFS search on a graph and I want to store in the list the nodes I visited. 回答1: One method of doing this is just to return the list so you have access to it at higher levels of recursion. Another method is

问题 I have a naive implementation of a gameloop let gameLoop gamestate = let rec innerLoop prev gamestate = let now = getTicks() let delta = now - prev gamestate |> readInput delta |> update delta |> render delta |> innerLoop delta innerLoop 0L gamestate This implementation throws a stackoverflowexception. In my mind this should be tail recursive. I could make a work around like this let gameLoop gamestate = let rec innerLoop prev gamestate = let now = getTicks() let delta = now - prev let

问题 Recursion makes backtracking easy as it guarantees that you won't go through the same path again. So all ramifications of your path are visited just once. I am trying to convert a backtracking tail-recursive (with accumulators) algorithm to iteration. I heard it is supposed to be easy to convert a perfectly tail -recursive algorithm to iteration. But I am stuck in the backtracking part. Can anyone provide a example through code so myself and others can visualize how backtracking is done? I

问题 I need to find a path of the number 1 in a matrix[R][C] starting from matrix[0][0] until it gets to matrix[R-1][C-1] , using recursion. I can only go down or right. In most cases, I don't have a problem. My problem is only when there is nowhere to go, and I need to take a step backwards. Per example, this is the matrix I'm getting from a file: 1 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 1 The problem is when it gets to matrix[4][0] . I don't understand why it won't

问题 I have written this recursive function in Kotlin: fun recursive(target: String, population: Population, debugFun: (String) -> Unit) : Population { if (population.solution(target).toString() == target) { return population } debugFun.invoke(population.solution(target).toString()) return recursive(target, population.evolve(target), debugFun) } It will run an indeterminate amount of times (because I'm using randomness to converge on solution in evolutionary algorithm). I am frequently getting

问题 Suppose I have a mechanism for long-running computations that can suspend themselves to be resumed later: sealed trait LongRunning[+R]; case class Result[+R](result: R) extends LongRunning[R]; case class Suspend[+R](cont: () => LongRunning[R]) extends LongRunning[R]; The simplest way how to run them is @annotation.tailrec def repeat[R](body: LongRunning[R]): R = body match { case Result(r) => r case Suspend(c) => { // perhaps do some other processing here println("Continuing suspended

问题 I was trying out the following example provide in the talk to understand the tail recursion in java8. @FunctionalInterface public interface TailCall<T> { TailCall<T> apply(); default boolean isComplete() { return false; } default T result() { throw new Error("not implemented"); } default T get() { return Stream.iterate(this, TailCall::apply).filter(TailCall::isComplete) .findFirst().get().result(); } } Utility class to use the TailCall public class TailCalls { public static <T> TailCall<T>