Some Classical Recursive Functions

Some Classical Recursive Functions

Table of Contents

• 1. Find the maximum element of an array recursively.
• 2. Recursively output the digits of a number num under base base. e.g., 7 under base 2 is "111".
• 3. Compute "n choose k".

In this post, I will list the commonly met recursive functions that can light the road for further investigation into the magic world of computer science.
I am not an excellent programmer myself, but through delight practice, the little but essential part of the RECURSIVE PROBLEMS can be understood. So I will present some examples to illustrate what I mean.

1 Find the maximum element of an array recursively.

One possible solution to this problem is as follows, the basic idea is to use the function f_max to find the rest of the array (excluding the first element of the array), and a max macro is used to return the first element and the rest of the "maximum" element.

#define max(a,b)     (a) > (b) ? (a) : (b)

int f_max(int *a,int start,int end)
{
    if (start == end) {
        return a[start];
    }
    return  max(a[start],f_max(a,start + 1,end));
}

void test_max(void)
{
    int a[] = {
        12,1,22,56,4
    };
    printf("max=%d
",f_max(a,0,4));
}

Another way to find the maximum element is to use the binary method, which means to first split the array into two sub-arrays, and find the maximum element in each sub-array and return the bigger one of these two sub-max-maximum number.

int f_max_2(int *a,int start,int end)
{
    if (start == end) {
        return a[start];
    } else {
        return max(f_max_2(a,start,(start + end) / 2),f_max_2(a,(start + end) / 2 + 1,end));
    }
}

2 Recursively output the digits of a number num under base base. e.g., 7 under base 2 is "111".

void get_digits(int num,int base)
{
    if (num / base == 0) {
        printf("%d",num % base);
    } else {
        get_digits(num/base,base);
        printf("%d",num % base);
    }
}

3 Compute "n choose k".

It is important to get the recursive definition of c_n^k, one way to think about is "to be or not to be", the kth element can or can't be selected in a certain selecting process.

  int F(int n,int k)
{
    if (n == k) {
        return 1;
    }
    if (k == 0) {
        return 1;
    }
    return F(n - 1, k) + F(n - 1,k - 1);
}

The dynamic programming method is as follows:

void print_c_n_k_table(int table[][20],int num_of_rows)
{
    printf("table elements:
");
    for (int i = 0;i < num_of_rows;i++) {
        for (int j = 0;j <= i;j++) {
            printf("%d ",table[i][j]);
        }
        printf("
");
    }
}

int dp_c_n_k(int n,int k)
{
    int table[20][20];
    for (int i = 0;i < 20;i++) {
        for (int j = 0;j < 20;j++) {
            table[i][j] = 0;
            if (i == j) {
                table[i][j] = 1;
            }
            if (j == 0) table[i][j] = 1;
        }
    }
    print_c_n_k_table(table,20);
    for (int i = 0;i < 20;i++) {
        for (int j = 0;j < 20;j++) {
            if (table[i][j] == 0) {
                table[i][j] = table[i - 1][j] + table[i - 1][j - 1];                
            }
            print_c_n_k_table(table,i);
            if ((i == n) && (j == k)) {
                return table[i][j];
            }
        }
        printf("
");
    }
    return 0;
}

Author: wujing

Created: 2014-08-26 二 10:48

Emacs 24.3.1 (Org mode 8.2.6)

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