Can All Numbers Of An Array Be Made Equal In C Program

Making elements of an array equal is a common problem in programming, often requiring specific operations or checks for inherent properties. In this article, you will learn how to determine if all numbers in a C array can be made equal by reducing them through division by specific prime factors.

Problem Statement Given an array of integers, the challenge is to ascertain if all its elements can be made equal. This often implies a process of reduction, such as repeatedly dividing elements by a predefined set of prime numbers (e.g., 2 and 3) until they reach a common base value. If all numbers, after such normalization, converge to the same value, then it's considered possible to make them equal. This problem is relevant in scenarios involving resource synchronization or checking for underlying commonalities in data sets.

Example Consider the array [12, 18, 72].

• 12 reduced by 2s and 3s: 12 -> 6 -> 3. Base value is 3.
• 18 reduced by 2s and 3s: 18 -> 9 -> 3. Base value is 3.
• 72 reduced by 2s and 3s: 72 -> 36 -> 18 -> 9 -> 3. Base value is 3.

Since all elements reduce to 3, it is possible to make them equal.

For the array [12, 15, 72]:

• 12 reduces to 3.
• 15 reduces to 5 (15 is not divisible by 2, 15/3 = 5, 5 is not divisible by 2 or 3).
• 72 reduces to 3.

Since 15 reduces to 5, which is different from 3, it is not possible to make all elements equal using this method.

Background & Knowledge Prerequisites To understand and implement the solutions, readers should be familiar with:

• C Programming Basics: Variables, data types, loops (for, while), conditional statements (if-else).
• Arrays: Declaring, initializing, and iterating through arrays.
• Functions: Defining and calling user-defined functions.
• Modulo Operator (%): For checking divisibility.

Use Cases or Case Studies The ability to check if numbers can be made equal (or share a common base form) is useful in several areas:

• Resource Normalization: In systems where resources are allocated in varying quantities but need to be normalized to a common 'unit' by dividing out common factors.
• Data Validation: Ensuring that a set of data points, after specific transformations, conform to a consistent base pattern.
• Algorithm Design: As a sub-problem in more complex algorithms that involve simplifying numbers or finding equivalences.
• Game Development: For balancing game mechanics where different item values or character stats need to be compared after applying common reduction rules.
• Number Theory Problems: When exploring properties of numbers related to their prime factorization.

Solution Approaches We will explore two approaches: first, a basic check for direct equality, and second, a more robust method involving reduction to a common base value by dividing out factors of 2 and 3.

Approach 1: Direct Equality Check

This approach simply verifies if all elements in the array are already identical. It's the simplest form of "making equal," where no operations are needed.

• One-line summary: Checks if all array elements have the exact same value.

// Check for Direct Equality
#include <stdio.h>
#include <stdbool.h> // For boolean type

// Function to check if all elements in an array are equal
bool areAllEqual(int arr[], int n) {
    // Step 1: Handle empty or single-element arrays
    if (n <= 1) {
        return true;
    }

    // Step 2: Compare each element with the first element
    int firstElement = arr[0];
    for (int i = 1; i < n; i++) {
        if (arr[i] != firstElement) {
            return false; // Found a different element
        }
    }

    // Step 3: All elements are equal
    return true;
}

int main() {
    // Example 1: All elements are equal
    int arr1[] = {5, 5, 5, 5};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);

    printf("Array: [");
    for (int i = 0; i < n1; i++) {
        printf("%d%s", arr1[i], (i == n1 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (areAllEqual(arr1, n1)) {
        printf("Result: All numbers can be made equal (they already are).\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal.\\n\\n");
    }

    // Example 2: Elements are not equal
    int arr2[] = {1, 2, 3, 4};
    int n2 = sizeof(arr2) / sizeof(arr2[0]);

    printf("Array: [");
    for (int i = 0; i < n2; i++) {
        printf("%d%s", arr2[i], (i == n2 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (areAllEqual(arr2, n2)) {
        printf("Result: All numbers can be made equal (they already are).\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal.\\n\\n");
    }

    return 0;
}

Run

• Sample output:

Array: [5, 5, 5, 5]
Result: All numbers can be made equal (they already are).

Array: [1, 2, 3, 4]
Result: Not all numbers can be made equal.

• Stepwise explanation for clarity:

1. The areAllEqual function takes an integer array and its size.
2. It first handles edge cases where the array is empty or has only one element, returning true as they are trivially "equal."
3. It stores the value of the first element.
4. It then iterates from the second element to the end of the array.
5. In each iteration, it compares the current element with the stored firstElement. If a mismatch is found, it immediately returns false.
6. If the loop completes without finding any mismatches, it means all elements are identical, and the function returns true.

Approach 2: Reducing to a Common Base Value by Dividing by 2 and 3

This approach determines if numbers can be made equal by repeatedly dividing them by 2 or 3 until they are no longer divisible by these primes. If all numbers normalize to the same "core" value, they are considered capable of being made equal.

• One-line summary: Checks if all numbers reduce to the same value after repeatedly dividing by 2 and 3.

// Reduce to Common Base Value
#include <stdio.h>
#include <stdbool.h> // For boolean type

// Function to normalize a number by repeatedly dividing by 2 and 3
int normalize(int num) {
    // Step 1: Handle non-positive numbers if necessary, here 0 remains 0.
    if (num == 0) return 0;

    // Step 2: Repeatedly divide by 2 until it's no longer divisible
    while (num > 0 && num % 2 == 0) {
        num /= 2;
    }
    // Step 3: Repeatedly divide by 3 until it's no longer divisible
    while (num > 0 && num % 3 == 0) {
        num /= 3;
    }
    return num; // Return the normalized base value
}

// Function to check if all array elements can be made equal
// by normalizing them to a common base value
bool canBeMadeEqualByNormalization(int arr[], int n) {
    // Step 1: Handle empty or single-element arrays
    if (n <= 1) {
        return true;
    }

    // Step 2: Normalize the first element to get the target base value
    int targetBaseValue = normalize(arr[0]);

    // Step 3: Iterate through the rest of the array and normalize each element
    for (int i = 1; i < n; i++) {
        if (normalize(arr[i]) != targetBaseValue) {
            return false; // Found an element that normalizes to a different base
        }
    }

    // Step 4: All elements normalize to the same base value
    return true;
}

int main() {
    // Example 1: Can be made equal
    int arr1[] = {12, 18, 72};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);

    printf("Array: [");
    for (int i = 0; i < n1; i++) {
        printf("%d%s", arr1[i], (i == n1 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (canBeMadeEqualByNormalization(arr1, n1)) {
        printf("Result: All numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    }

    // Example 2: Cannot be made equal
    int arr2[] = {12, 15, 72};
    int n2 = sizeof(arr2) / sizeof(arr2[0]);

    printf("Array: [");
    for (int i = 0; i < n2; i++) {
        printf("%d%s", arr2[i], (i == n2 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (canBeMadeEqualByNormalization(arr2, n2)) {
        printf("Result: All numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    }

    // Example 3: Already equal
    int arr3[] = {6, 6, 6};
    int n3 = sizeof(arr3) / sizeof(arr3[0]);

    printf("Array: [");
    for (int i = 0; i < n3; i++) {
        printf("%d%s", arr3[i], (i == n3 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (canBeMadeEqualByNormalization(arr3, n3)) {
        printf("Result: All numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    }

    // Example 4: Contains zero
    int arr4[] = {0, 12, 18};
    int n4 = sizeof(arr4) / sizeof(arr4[0]);

    printf("Array: [");
    for (int i = 0; i < n4; i++) {
        printf("%d%s", arr4[i], (i == n4 - 1) ? "" : ", ");
    }
    printf("]\\n");
    if (canBeMadeEqualByNormalization(arr4, n4)) {
        printf("Result: All numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    } else {
        printf("Result: Not all numbers can be made equal by reducing factors of 2 and 3.\\n\\n");
    }

    return 0;
}

Run

• Sample output:

Array: [12, 18, 72]
Result: All numbers can be made equal by reducing factors of 2 and 3.

Array: [12, 15, 72]
Result: Not all numbers can be made equal by reducing factors of 2 and 3.

Array: [6, 6, 6]
Result: All numbers can be made equal by reducing factors of 2 and 3.

Array: [0, 12, 18]
Result: Not all numbers can be made equal by reducing factors of 2 and 3.

• Stepwise explanation for clarity:

1. normalize(int num) function:
   • Takes an integer num.

It handles the case of 0 separately, as 0 divided by any non-zero number is 0.

It enters a while loop that continues as long as num is positive and divisible by 2. Inside the loop, num is divided by 2. This removes all factors of 2.

Similarly, it enters another while loop that continues as long as num is positive and divisible by 3. Inside, num is divided by 3, removing all factors of 3.

The function returns the final num, which is its normalized base value.

1. canBeMadeEqualByNormalization(int arr[], int n) function:
   • Handles empty or single-element arrays by returning true.

It normalizes the first element of the array using the normalize function. This result becomes the targetBaseValue.

It then iterates through the remaining elements of the array (from the second element onwards).

For each element, it calls normalize to get its base value.

If any element's normalized base value does not match targetBaseValue, the function immediately returns false because they cannot all be made equal.

If the loop completes without finding any mismatches, it means all elements, when reduced, share the same base value, and the function returns true.

Conclusion Determining if all numbers in an array can be made equal depends heavily on the definition of "made equal" and the allowed operations. While a simple check for direct equality addresses trivial cases, more complex scenarios often involve reducing numbers to a common base form. The method of normalizing numbers by repeatedly dividing by prime factors like 2 and 3 provides a robust way to ascertain if an underlying commonality exists, allowing all elements to eventually converge to the same value.

Summary

• Problem: Check if array elements can be made equal under specific reduction rules.
• Direct Equality: Simplest case; all elements are already identical.
• Normalization Method: A common technique involves reducing numbers by dividing out factors of 2 and 3.
• Normalization Steps:

1. Create a function to repeatedly divide a number by 2 then 3 until it's no longer divisible by either. Handle 0 separately.
2. Normalize the first array element to establish a target base value.
3. Normalize all subsequent elements and compare their base values to the target.
4. If all match, the numbers can be made equal by this method.
   • Key Insight: This method identifies if numbers share the same prime factors other than 2 and 3.

Usefulness: Applicable in data validation, algorithm design, and resource normalization.