Printing a number in sequence using C - Algorithm

Question

Sample pattern is given,

input :    16
output:    16 11 6 1 -4 1 6 11 16

If the input is 10, then program should print output as

10 5 0 5 10

Note: The above sequences decrement/increment by 5.

The challenge is to not declare any variables or loops. use only recursion.

I have tried with the following code.

void sequence(int input, int base){

    input = input - (input > 0?5:-5); //main execution
    printf("input:%d\n",input);
    if(input == base)return;
    sequence(input,base);
}

//for eg. input and base(initial Value) is 16. The above method recurse itself until input = base.

I can print upto this sequence (in Bold)

16 11 6 1 -4 1 6 11 16

How to complete the sequence. In the above method, in main execution line, i need to check condition as input = input - (input < 0?5:-5); to print the remaining sequence. But i am not sure how can i do this without any variables or loops. Is there any algorithm available or any other better solution.

Answer 1

Some example code to my comment, which would match if it doesn't have to be strictly left- or right-recursive:

void sequence(int n)
{
    printf("%d ", n);
    if (n > 0)
    {
        sequence(n-5);
        printf("%d ", n);
    }
}

Further notes:

1.) This seems to be about functional programming , where a key concept is that you can never assign a variable ... see how it is avoided here. (Strictly speaking, it's not functional because of the printf side effects)

2.) It's not strictly left- or right-recursive (meaning the recursion happens in the middle of evaluation), so it can't be easily transformed to something iterative.

Answer 2

It seems that the recursion should stop when it reaches 0 or below. So try this condition in you recursive function (reduced to only one argument):

void sequence(input) {
    // print input
    if (input > 0) {
        // recursive call
        sequence(input);
    }
    // print input again (sometimes...)
}

Update: Let's talks about Symmetry

For the sake of symmetry, also the reverse version of Felix's solution can be used

void sequence(int n)
{
    if (n > 0) {
        printf("%d ", n);
        sequence(n-5);
    }
    printf("%d ", n);
}

Both feature a kind of asymmetry that seems not to fit to the palindromic structure of the problem, and there is another problem: the blemish of a trailing space. So let's introduce to you a obvious solution that deals with these tiny(?) flaws:

void sequence(int n)
{
    if (n > 0) {
        printf("%d ", n); sequence(n-5); printf(" %d", n);
    } else {
        printf("%d", n);
    }
}

Answer 3

Use two different functions to cout up and down:

void down(int input, int base) {
    printf("input:%d\n",input);
    if(input > 0) down(input - 5, base);
    else up(input + 5, base);
}

void up(int input, int base) {
    printf("input:%d\n",input);
    if(input == base) return;
    up(input + 5, base);
}

Start the calculation by calling down .

Live Demo

Answer 4

#include <stdio.h>

using namespace std;

void printSequence(int input) {
    printf("%d ", input);
    if (input <= 0)
        return;
    printSequence(input - 5);
    printf("%d ", input);
}

int main ()
{
    int input = 0;
    scanf("%d", &input);

    printSequence(input);

    return 0;
}

Answer 5

The base case of your recursive code seems wrong. I think your code goes into infinite recursion the way it is. You can use a flag to achieve your sequence:

void sequence(int input, int base){
    static int flag = 0;
    //input = input - (input > 0?5:-5); //main execution, wrong
    printf("input:%d\n",input);
    input -= (flag == 0)?5:-5;          //use flag for base case of sequence
    if(input <= 0)
        flag = 1;
    if(input == base){
        printf("input:%d\n",input);
        return;
    }
    sequence(input,base);
}