How to resolve the algorithm 9 billion names of God the integer step by step in the C programming language

Published on 7 June 2024 03:52 AM
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How to resolve the algorithm 9 billion names of God the integer step by step in the C programming language

Table of Contents

Problem Statement

This task is a variation of the short story by Arthur C. Clarke. (Solvers should be aware of the consequences of completing this task.) In detail, to specify what is meant by a   “name”:

Display the first 25 rows of a number triangle which begins: Where row

n

{\displaystyle n}

corresponds to integer

n

{\displaystyle n}

,   and each column

C

{\displaystyle C}

in row

m

{\displaystyle m}

from left to right corresponds to the number of names beginning with

C

{\displaystyle C}

. A function

G ( n )

{\displaystyle G(n)}

should return the sum of the

n

{\displaystyle n}

-th   row. Demonstrate this function by displaying:

G ( 23 )

{\displaystyle G(23)}

,

G ( 123 )

{\displaystyle G(123)}

,

G ( 1234 )

{\displaystyle G(1234)}

,   and

G ( 12345 )

{\displaystyle G(12345)}

.
Optionally note that the sum of the

n

{\displaystyle n}

-th   row

P ( n )

{\displaystyle P(n)}

is the     integer partition function. Demonstrate this is equivalent to

G ( n )

{\displaystyle G(n)}

by displaying:

P ( 23 )

{\displaystyle P(23)}

,

P ( 123 )

{\displaystyle P(123)}

,

P ( 1234 )

{\displaystyle P(1234)}

,   and

P ( 12345 )

{\displaystyle P(12345)}

.

If your environment is able, plot

P ( n )

{\displaystyle P(n)}

against

n

{\displaystyle n}

for

n

1 … 999

{\displaystyle n=1\ldots 999}

.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm 9 billion names of God the integer step by step in the C programming language

This C code uses the GNU Multiple Precision Arithmetic Library (GMP) to calculate and print the values of a sequence of numbers defined as follows:

p[n] = 0
for k = 1 to n
 d = n - k * (3 * k - 1) / 2
 if d >= 0
   if k is odd
     p[n] += p[d]
   else
     p[n] -= p[d]

   d -= k
   if d >= 0
     if k is odd
       p[n] += p[d]
     else
       p[n] -= p[d]

Here's a detailed breakdown of the code:

  1. Header Inclusions:

    • The code includes the necessary header files <stdio.h> for input and output and <gmp.h> for the GMP library.
  2. Constant Definition:

    • A constant N is defined as 100,000, which represents the maximum index up to which the sequence will be calculated.
  3. Global Array:

    • A global array p is declared to store the values of the sequence up to index N.
  4. calc Function:

    • This function takes an integer n as input and calculates the value of p[n] based on the definition given above.
    • It uses GMP's mpz_init_set_ui to initialize p[n] to 0 and then iterates from k = 1 to k <= n.
    • Inside the loop, it calculates d and checks if d is non-negative. If d is valid, it performs the operations defined in the sequence.
    • It uses GMP's mpz_add and mpz_sub functions to add or subtract values from p[n] based on whether k is odd or even.
  5. main Function:

    • The main function is the entry point of the program.
    • It initializes idx as an array of integers representing specific values of n for which the sequence will be printed.
    • It initializes at as 0, which keeps track of the index in the idx array.
    • It initializes p[0] to 1 using mpz_init_set_ui.
  6. Calculation and Printing:

    • The code enters a loop that runs from i = 1 to the last value in idx.
    • For each i, it calls the calc function to calculate p[i].
    • If i is not equal to idx[at], it moves to the next iteration.
    • Otherwise, it prints i and the value of p[i] using gmp_printf.
    • It then increments at to move to the next value in the idx array.

This code essentially computes a sequence of values defined by a recursive formula and prints specific terms of the sequence based on the values provided in the idx array.

Source code in the c programming language

#include <stdio.h>
#include <gmp.h>

#define N 100000
mpz_t p[N + 1];

void calc(int n)
{
	mpz_init_set_ui(p[n], 0);

	for (int k = 1; k <= n; k++) {
		int d = n - k * (3 * k - 1) / 2;
		if (d < 0) break;

		if (k&1)mpz_add(p[n], p[n], p[d]);
		else	mpz_sub(p[n], p[n], p[d]);

		d -= k;
		if (d < 0) break;

		if (k&1)mpz_add(p[n], p[n], p[d]);
		else	mpz_sub(p[n], p[n], p[d]);
	}
}

int main(void)
{
	int idx[] = { 23, 123, 1234, 12345, 20000, 30000, 40000, 50000, N, 0 };
	int at = 0;

	mpz_init_set_ui(p[0], 1);

	for (int i = 1; idx[at]; i++) {
		calc(i);
		if (i != idx[at]) continue;

		gmp_printf("%2d:\t%Zd\n", i, p[i]);
		at++;
	}
}


  

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