Lovelace's Note G Program in C · GitHub
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sinclairtarget/bernoulli.c
Created<br>August 17, 2018 20:22
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Lovelace's Note G Program in C
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bernoulli.c
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#include
/*
* Calculates what Ada Lovelace labeled "B7", which today we would call the 8th
* Bernoulli number.
*/
int main(int argc, char* argv[])
// ------------------------------------------------------------------------
// Data
// ------------------------------------------------------------------------
float v1 = 1; // 1
float v2 = 2; // 2
float v3 = 4; // n
// ------------------------------------------------------------------------
// Working Variables
// ------------------------------------------------------------------------
float v4 = 0;
float v5 = 0;
float v6 = 0; // Factors in the numerator
float v7 = 0; // Factors in the denominator
float v8 = 0;
float v10 = 0; // Terms remaining count, basically
float v11 = 0; // Accumulates v6 / v7
float v12 = 0; // Stores most recent calculated term
float v13 = 0; // Accumulates the whole result
// ------------------------------------------------------------------------
// Result Variables
// ------------------------------------------------------------------------
float v21 = 1.0f / 6.0f; // B1
float v22 = -1.0f / 30.0f; // B3
float v23 = 1.0f / 42.0f; // B5
float v24 = 0; // B7, not yet calculated
// ------------------------------------------------------------------------
// Calculation
// ------------------------------------------------------------------------
// ------- A0 -------
/* 01 */ v4 = v5 = v6 = v2 * v3; // 2n
/* 02 */ v4 = v4 - v1; // 2n - 1
/* 03 */ v5 = v5 + v1; // 2n + 1
// In Lovelace's diagram, the below appears as v5 / v4, which is incorrect.
/* 04 */ v11 = v4 / v5; // (2n - 1) / (2n + 1)
/* 05 */ v11 = v11 / v2; // (1 / 2) * ((2n - 1) / (2n + 1))
/* 06 */ v13 = v13 - v11; // -(1 / 2) * ((2n - 1) / (2n + 1))
/* 07 */ v10 = v3 - v1; // (n - 1), set counter?
// A0 = -(1 / 2) * ((2n - 1) / (2n + 1))
// ------- B1A1 -------
/* 08 */ v7 = v2 + v7; // 2 + 0, basically a MOV instruction
/* 09 */ v11 = v6 / v7; // 2n / 2
/* 10 */ v12 = v21 * v11; // B1 * (2n / 2)
// A1 = (2n / 2)
// B1A1 = B1 * (2n / 2)
// ------- A0 + B1A1 -------
/* 11 */ v13 = v12 + v13; // A0 + B1A1
/* 12 */ v10 = v10 - v1; // (n - 2)
// On the first loop this calculates B3A3 and adds it on to v13.
// On the second loop this calculates B5A5 and adds it on.
while (v10 > 0)
// ------- B3A3, B5A5 -------
while (v6 > 2 * v3 - (2 * (v3 - v10) - 2))
{ // First Loop:
/* 13 */ v6 = v6 - v1; // 2n - 1
/* 14 */ v7 = v1 + v7; // 2 + 1
/* 15...