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If you take the numbers one by one from 1 to n, there may be many combinations.
For example, if we take only one number at a time, the number of combinations will be nC1.
If we take two numbers at a time, the number of combinations will be nC2. Hence, the total number of combinations will be nC1 nC2 … nCn.
To find the sum of all combinations, we must use an efficient method. Otherwise, the time and space complexity will become very high.
Find the sum of the products of all combinations of numbers taken each time from 1 to N.
N is a given number.
enter
N = 4
Output
f(1) = 10 f(2) = 35 f(3) = 50 f(4) = 24
Explanation
f(x) is the sum of the product of all combinations taken x at a time. f(1) = 1 + 2+ 3+ 4 = 10 f(2) = (1*2) + (1*3) + (1*4) + (2*3) + (2*4) + (3*4) = 35 f(3) = (1*2*3) + (1*2*4) +(1*3*4) + (2*3*4) = 50 f(4) = (1*2*3*4) = 24
enter
N = 5
Output
f(1) = 15 f(2) = 85 f(3) = 225 f(4) = 274 f(5) = 120
The brute force method is to generate all combinations recursively, find their products, and then find the corresponding sum.
The following is a recursive C program that is used to find the sum of the products taken each time in all combinations (from 1 to N)
#include <bits/stdc++.h> using namespace std; //sum of each combination int sum = 0; void create_combination(vector<int>vec, vector<int>combinations, int n, int r, int depth, int index) { // if we have reached sufficient depth if (index == r) { //find the product of the combination int prod = 1; for (int i = 0; i < r; i++) prod = prod * combinations[i]; // add the product to sum sum += prod; return; } // recursion to produce a different combination for (int i = depth; i < n; i++) { combinations[index] = vec[i]; create_combination(vec, combinations, n, r, i + 1, index + 1); } } //Function to print the sum of products of //all combinations taken 1-N at a time void get_combinations(vector<int>vec, int n) { for (int i = 1; i <= n; i++) { // vector for storing combination //int *combi = new int[i]; vector<int>combinations(i); // call combination with r = i // combination by taking i at a time create_combination(vec, combinations, n, i, 0, 0); // displaying sum of the product of combinations cout << "f(" << i << ") = " << sum << endl; sum = 0; } } int main() { int n = 5; //creating vector of size n vector<int>vec(n); // storing numbers from 1-N in the vector for (int i = 0; i < n; i++) vec[i] = i + 1; //Function call get_combinations(vec, n); return 0; }
f(1) = 15 f(2) = 85 f(3) = 225 f(4) = 274 f(5) = 120
By creating the recursion tree of this approach, it is visible that the time complexity is exponential. Also, many steps get repeated which makes the program redundant. Hence, it is highly inefficient.
An effective solution would be to use dynamic programming and remove the redundancies.
Dynamic programming is a technique in which a problem is divided into subproblems. The subproblems are solved, and their results are saved to avoid repetitions.
Below is a C Program using Dynamic Programming to find the sum of all combinations taken (1 to N) at a time .
#include <bits/stdc++.h> using namespace std; //Function to find the postfix sum array void postfix(int a[], int n) { for (int i = n - 1; i > 0; i--) a[i - 1] = a[i - 1] + a[i]; } //Function to store the previous results, so that the computations don't get repeated void modify(int a[], int n) { for (int i = 1; i < n; i++) a[i - 1] = i * a[i]; } //Function to find the sum of all combinations taken 1 to N at a time void get_combinations(int a[], int n) { int sum = 0; // sum of combinations taken 1 at a time is simply the sum of the numbers // from 1 - N for (int i = 1; i <= n; i++) sum += i; cout << "f(1) = " << sum <<endl; // Finding the sum of products for all combination for (int i = 1; i < n; i++) { //Function call to find the postfix array postfix(a, n - i + 1); // sum of products taken i+1 at a time sum = 0; for (int j = 1; j <= n - i; j++) { sum += (j * a[j]); } cout << "f(" << i + 1 << ") = " << sum <<endl; //Function call to modify the array for overlapping problem modify(a, n); } } int main() { int n = 5; int *a = new int[n]; // storing numbers from 1 to N for (int i = 0; i < n; i++) a[i] = i + 1; //Function call get_combinations(a, n); return 0; }
f(1) = 15 f(2) = 85 f(3) = 225 f(4) = 274 f(5) = 120
In this article, we discussed the problem of finding the sum of the products of all combinations from 1 to N.
We started with a brute force approach with exponential time complexity and then modified it using dynamic programming. C programs for both methods are also provided.
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