recursion

How to use tail recursion correctly?

大城市里の小女人 提交于 2020-01-06 18:57:43
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?

Finding the total sum of numbers in an array in Java, recursively

倖福魔咒の 提交于 2020-01-06 16:24:12
问题 I understand that finding the sum of numbers in an array is much more easily done through iteration instead of recursion, but if I were to use recursion to write such a function, what would be wrong with this code? public static double sum (double[] a) { if (a.length == 0) return 0.0; else{ return sumHelper (a, 1, a[0]); } } private static double sumHelper (double[] a, int i, double result) { if (i < a.length) { result = result + sumHelper (a, i + 1, result); } return result; } Everything

Number of ways such that sum of k elements equal to p

[亡魂溺海] 提交于 2020-01-06 15:59:08
问题 Given series of integers having relation where a number is equal to sum of previous 2 numbers and starting integer is 1 Series ->1,2,3,5,8,13,21,34,55 find the number of ways such that sum of k elements equal to p.We can use an element any number of times. p=8 k=4. So,number of ways would be 4.Those are, 1,1,1,5 1,1,3,3 1,2,2,3 2,2,2,2 I am able to sove this question through recursion.I sense dynamic programming here but i am not getting how to do it.Can it be done in much lesser time??? EDIT

Printing an integer each digit at a time in english form recursively

时间秒杀一切 提交于 2020-01-06 15:42:09
问题 I've been given an assignment where we must print out an integer digit by digit in english form using a recursive method. E.G. 534 prints out "five three four". This is what I've got: int englishInt(int num) { if(num < 10) { switch(num) { case 0: cout << "zero "; case 1: cout << "one "; case 2: cout << "two "; case 3: cout << "three "; case 4: cout << "four "; case 5: cout << "five "; case 6: cout << "six "; case 7: cout << "seven "; case 8: cout << "eight "; case 9: cout << "nine "; } } else

Graph or Relational DB specifically recursion

大城市里の小女人 提交于 2020-01-06 15:21:33
问题 I am about to develop a solution for a customer where the basic entity is a member and members can have different multiple social relationships with other members. For instance Lets say we have four types of members Doctors, Specialist, Nurses and Patients. So one or more Doctors can consult one or more Specialists, One or more Doctors can treat one or more Patients. One or more Doctor is in Charge of one or more Nurses. So if I were to use a Relational DB a high degree of recursion would be

print binary numbers in ascending order

巧了我就是萌 提交于 2020-01-06 14:57:11
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

Erlang exercise, creating lists

你离开我真会死。 提交于 2020-01-06 14:11:08
问题 I`m learning Erlang and doing exercises from the book: "Write a function which returns a list of the format [1,2,..,N-1,N]. create(3) -> [1,2,3]." I have a solution, which is: create(N) -> create(1, N). create (M,M) -> [M]; create(M,N) -> [M | create(M+1, N)]. I went through it dozen of times, but I simply can`t understand what is happening on line 2 of the solution. Can please somebody explain? Thank you. EDIT. Ok, so i think I`m on the right track for understanding it. In line 2, the new

Meteor throws exception: Exception from Tracker recompute function: too much recursion

﹥>﹥吖頭↗ 提交于 2020-01-06 14:00:42
问题 I have a recursive template in a Meteor application that builds tree structure with unlimited levels. The template works ok with small trees but when I test application on real data with many tree levels then I get the following two exceptions in console and no part of the tree gets rendered: > Exception from Tracker recompute function: Node was not found > meteor....3d493c5 (line 883) > > Exception from Tracker recompute function: too much recursion > _.forEach@http://localhost:3000/packages

Meteor throws exception: Exception from Tracker recompute function: too much recursion

守給你的承諾、 提交于 2020-01-06 14:00:42
问题 I have a recursive template in a Meteor application that builds tree structure with unlimited levels. The template works ok with small trees but when I test application on real data with many tree levels then I get the following two exceptions in console and no part of the tree gets rendered: > Exception from Tracker recompute function: Node was not found > meteor....3d493c5 (line 883) > > Exception from Tracker recompute function: too much recursion > _.forEach@http://localhost:3000/packages

Recursive graph traversal: how to generate and return all paths?

陌路散爱 提交于 2020-01-06 13:59:47
问题 Here is my code right now: hasht= {"A":["B", "D", "E"], "B":["C"], "C":["D", "E"], "D":["C", "E"], "E":["B"]} paths=[] def recusive(start, finish, val): if start==finish and val!=1: return start else: for i in hasht[start]: path= start+ recusive(i,finish,2) paths.append(path) print (recusive("C","C",1)) print paths Desired output: [CDC, CDEBC, CEBC] I am trying to generate a table like the one above, but I am running into a problem where the string and the array are not being concatenated.