How to resolve the algorithm 9 billion names of God the integer step by step in the C programming language
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:
-
Header Inclusions:
- The code includes the necessary header files
<stdio.h>
for input and output and<gmp.h>
for the GMP library.
- The code includes the necessary header files
-
Constant Definition:
- A constant
N
is defined as 100,000, which represents the maximum index up to which the sequence will be calculated.
- A constant
-
Global Array:
- A global array
p
is declared to store the values of the sequence up to indexN
.
- A global array
-
calc
Function:- This function takes an integer
n
as input and calculates the value ofp[n]
based on the definition given above. - It uses GMP's
mpz_init_set_ui
to initializep[n]
to 0 and then iterates fromk = 1
tok <= n
. - Inside the loop, it calculates
d
and checks ifd
is non-negative. Ifd
is valid, it performs the operations defined in the sequence. - It uses GMP's
mpz_add
andmpz_sub
functions to add or subtract values fromp[n]
based on whetherk
is odd or even.
- This function takes an integer
-
main
Function:- The
main
function is the entry point of the program. - It initializes
idx
as an array of integers representing specific values ofn
for which the sequence will be printed. - It initializes
at
as 0, which keeps track of the index in theidx
array. - It initializes
p[0]
to 1 usingmpz_init_set_ui
.
- The
-
Calculation and Printing:
- The code enters a loop that runs from
i = 1
to the last value inidx
. - For each
i
, it calls thecalc
function to calculatep[i]
. - If
i
is not equal toidx[at]
, it moves to the next iteration. - Otherwise, it prints
i
and the value ofp[i]
usinggmp_printf
. - It then increments
at
to move to the next value in theidx
array.
- The code enters a loop that runs from
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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