recursion

问题 I am new to Prolog and came across this practice excercise. The question asks to define a predicate zipper([[List1,List2]], Zippered). //this is two lists within one list. This predicate should interleave elements of List1 with elements of List2. For example, zipper([[1,3,5,7], [2,4,6,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8]. zipper([[1,3,5], [2,4,6,7,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8]. So far I have a solution for two different list: zipper ([],[],Z). zipper([X],[],[X]).

问题 I am new to Prolog and came across this practice excercise. The question asks to define a predicate zipper([[List1,List2]], Zippered). //this is two lists within one list. This predicate should interleave elements of List1 with elements of List2. For example, zipper([[1,3,5,7], [2,4,6,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8]. zipper([[1,3,5], [2,4,6,7,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8]. So far I have a solution for two different list: zipper ([],[],Z). zipper([X],[],[X]).

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 TL;DR: Is there a way to specify a conditional so that an opening element MUST match its paired closing element? Example is located on regex101.com. ===== Balancing elements in regex is typically handled through recursion. This means that nested {...{...{...}...}...} can be located. Also, PCRE allows the (?(DEFINE)...) construct, which lets you define various patterns without actually starting the match. In the regular expression # Define the opening and closing elements before the recursion

问题 TL;DR: Is there a way to specify a conditional so that an opening element MUST match its paired closing element? Example is located on regex101.com. ===== Balancing elements in regex is typically handled through recursion. This means that nested {...{...{...}...}...} can be located. Also, PCRE allows the (?(DEFINE)...) construct, which lets you define various patterns without actually starting the match. In the regular expression # Define the opening and closing elements before the recursion

问题 I 've seen lots of answers that turn a Json response to a Ul Tree structure. The thing is that all these subjects where around a nested response pointing out the relationship between parent object and child object. I have a non nested Json response indicating the relation by reference to the objects property. The following is part of the response: {"id":"1","parent_id":null,"name":"Item-0"}, {"id":"2","parent_id":"1","name":"Item-1"}, {"id":"3","parent_id":"2","name":"Item-2"}, {"id":"4",

问题 I am having trouble with recursive searching of list and creation of list of lists from the result.. The knowledge base contains team name, number of wins and zone they are in, all associated withe their team number. I am passing list of team numbers in Teams and I am searching for a matching pair with findMinMax/3 . The result I need is... List of lists of paired teams (ex. X = [[gonzaga, washington], [iowa, oklahoma], …] ) And 1 unmatched team (resulted from odd number of teams) or 0 (in

问题 I am having trouble with recursive searching of list and creation of list of lists from the result.. The knowledge base contains team name, number of wins and zone they are in, all associated withe their team number. I am passing list of team numbers in Teams and I am searching for a matching pair with findMinMax/3 . The result I need is... List of lists of paired teams (ex. X = [[gonzaga, washington], [iowa, oklahoma], …] ) And 1 unmatched team (resulted from odd number of teams) or 0 (in

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 I'm trying to split the following recursive modules into separate compilation units. Specifically, I'd like B to be in its own b.ml, to be able to reuse it with other A's. module type AT = sig type b type t = Foo of b | Bar val f : t -> b list end module type BT = sig type a type t = { aaa: a list; bo: t option } val g : t -> t list end module rec A : (AT with type b = B.t) = struct type b = B.t type t = Foo of b | Bar let f = function Foo b -> [ b ] | Bar -> [] end and B : (BT with type a =