After converting the given binary number into a base between L and R, calculate the number of prime numbers

The title "Count of prime numbers after converting a given binary number between L and R" refers to a mathematical problem involving converting a binary number to a base between L and R, Then count the number of prime numbers from between L and R. Convert. In mathematics, a prime number is an integer greater than 1 that is only divisible by 1 and itself.

To convert a binary number to a number in a different base, you need to write the number in a different number system. The base of a number system is the number of unique numbers, and conversion is accomplished by finding an equivalent representation of that number in the new base. Computing prime numbers after transformation is a difficult number theory problem that has uses in cryptography, computer science, and other fields. To solve this problem you need to know a lot about number theory, prime numbers and number systems.

What is a prime number?

A number is called a prime number only if it is divisible by 1 and the number itself. For example, the number 5 is prime because it is only divisible by the numbers 1 and 5, but 6 is not prime because it is also divisible by 2 and 3.

The number of primes is simply asking how many primes there are in a given set of numbers. For example, take a set of numbers {1,2,3,4,5,6,7,8,9}. In this set of numbers, the number of prime numbers is 4, and they are 2, 3, 5, and 7. Furthermore, 1 is not a prime number because its only positive factor is 1 itself.

method

There are two main ways to calculate the prime number problem, as follows −

• Violent method

• Prime Factorization

algorithm

Step 1 - Enter the binary number and the range for base L and R.

Step 2 - Iterate over each base between L and R (inclusive).

Step 3 - Convert the binary number to the current base.

Step 4 − Check whether the converted number is a prime number.

Step 5 - If the converted number is a prime number, increase the prime number count by 1.

Step 6 - Repeat steps 3-5 for all bases in the range L to R.

Step 7 − Return the total number of prime numbers obtained.

The pseudo code of the algorithm is given below -

input: binary number b, range of bases L and R
output: count of prime numbers in the given range

Number_of_prime = 0
for base = L to R
convert b to base
if number_is_prime(converted_number)
   Number_of_prime ++
return Number_of_prime

number_is_prime() is a method that accepts a number as input and returns a Boolean value showing whether the number is prime.

Method 1: Violent solution

Brute Force Approach involves converting binary numbers into each base from L to R and counting the number of prime numbers in each conversion. For larger numbers, all possible variations need to be checked, which can be time-consuming.

The following code contains three functions. The first function is "isPrime" which returns 1 if the input number is prime and 0 otherwise. The second function "binaryToDecimal" converts a binary number to a decimal number. The third function "countPrimes" counts the number of prime numbers obtained by converting binary numbers between the input ranges to decimal numbers. Finally, the main function takes in a binary number and a range of numbers, calls the "countPrimes" function and prints the count of primes.

The Chinese translation of

Example

is:

Example

This code provides predefined values ​​for binary numbers and ranges L and R. In this example I used the binary number 1010 and the range 5 to 20. You can change these values ​​in the main function as needed.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

// Function to check if a number is prime or not
int isPrime(int n) {
   int i;
   for(i = 2; i <= sqrt(n); i++) {
      if(n%i == 0) {
         return 0;
      }
   }
   return 1;
}

// Function to convert binary to decimal
int binaryToDecimal(int n) {
   int decimal = 0, i = 0, remainder;
   while(n != 0) {
      remainder = n % 10;
      n /= 10;
      decimal += remainder * pow(2, i);
      ++i;
   }
   return decimal;
}

// Function to count primes in a given range
int countPrimes(int L, int R) {
   int count = 0, i;
   for(i = L; i <= R; i++) {
      int decimal = binaryToDecimal(i);
      if(isPrime(decimal)) {
         count++;
      }
   }
   return count;
}

// Main function
int main() {
   int binary = 1010; // Example binary number
   int L = 5;         // Example range lower limit
   int R = 20;        // Example range upper limit

   // Count primes and print result
   int count = countPrimes(L, R);
   printf("Number of primes after converting %d to base between %d and %d is: %d\n", binary, L, R, count);

   return 0;
}

Output

Number of primes after converting 1010 to base between 5 and 20 is: 7

Method 2: Prime Factorization

Prime factorization involves finding the prime factors of the transformed number and checking whether they are in the prime range. It can be an efficient method for smaller numbers, but can be computationally expensive for larger numbers.

The following code defines two functions isPrime() and countPrimes(), which check whether a given number is a prime number or count the number of prime numbers before a given number. The main function accepts a binary number and a radix limit entered by the user, converts the binary number to decimal, and then converts it to a different radix within the given limits. For each conversion, the program looks for prime factors and, if they are within the current base limits, increments a counter. Finally, the program prints the number of primes found. The code imports the standard input/output and boolean libraries.

The Chinese translation of

Code

is:

code

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

bool isPrime(int n) {
   if (n <= 1) {
      return false;
   }
   int i;
   for (i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
         return false;
      }
   }
   return true;
}

int main() {
   int binaryNum = 110101; // Predefined binary number input
   int L = 3; // Predefined lower limit of base
   int R = 6; // Predefined upper limit of base

   int decimalNum = 0, base = 1;
   while (binaryNum > 0) {
      int digit = binaryNum % 10;
      decimalNum += digit * base;
      base *= 2;
      binaryNum <span>/</span>= 10;
   }

   int transformedNum, factor;
   int primeCount = 0;
   for (int baseNum = L; baseNum <= R; baseNum++) {
      transformedNum = decimalNum;
      while (transformedNum > 1) {
         for (int i = 2; i <= transformedNum; i++) {
            if (transformedNum % i == 0) {
               factor = i;
               break;
            }
         }
         transformedNum <span>/</span>= factor;
         if (isPrime(factor) && factor >= baseNum) {
            primeCount++;
         }
      }
   }
   printf("Count of primes after converting the given binary number in base between L to R is: %d", primeCount);
   return 0;
}

Output

Count of primes after converting the given binary number in base between L to R is: 4

in conclusion

In summary, we can determine the number of prime numbers by first converting a given binary number to a base between L and R, and then counting the number of prime numbers in that range.

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