Instantly share code, notes, and snippets.

# sinclairtarget/bernoulli.c

Created August 17, 2018 20:22
Show Gist options
• Save sinclairtarget/ad18ac65d277e453da5f479d6ccfc20e to your computer and use it in GitHub Desktop.
Lovelace's Note G Program in C
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters
 #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 */ v8 = v6 / v7; // (2n - 1) / 3 /* 16 */ v11 = v8 * v11; // (2n / 2) * ((2n - 1) / 3) // Second Loop: // 17 v6 = v6 - v1; 2n - 2 // 18 v7 = v1 + v7; 3 + 1 // 19 v8 = v6 / v7; (2n - 2) / 4 // 20 v11 = v8 * v11; (2n / 2) * ((2n - 1) / 3) * ((2n - 2) / 4) } // A better way to do this might be to use an array for all of the // "Working Variables" and then index into it based on some calculated // index. Lovelace might have intended v14-v20 to be used on the // second iteration of this loop. // // Lovelace's program only has the version of the below line using v22 // in the multiplication. if (v10 == 2) { /* 21 */ v12 = v22 * v11; // B3 * A3 } else { /* 21 */ v12 = v23 * v11; // B5 * A5 } // B3A3 = B3 * (2n / 2) * ((2n - 1) / 3) * ((2n - 2) / 4) // ------- A0 + B1A1 + B3A3, A0 + B1A1 + B3A3 + B5A5 ------- /* 22 */ v13 = v12 + v13; // A0 + B1A1 + B3A3 (+ B5A5) /* 23 */ v10 = v10 - v1; // (n - 3), (n - 4) } /* 24 */ v24 = v13 + v24; // Store the final result in v24 /* 25 */ v3 = v1 + v3; // Move on to the next Bernoulli number! // This outputs a positive number, but really the answer should be // negative. There is some hand-waving in Lovelace's notes about the // Analytical Engine sorting out the proper sign. printf("A0 + B1A1 + B3A3 + B5A5: %.2f\n", v24); }

### switham commented Jul 11, 2020

I wonder whether there were people at the time who got through Ada's gigantic screed finally to that page and got it. "OMG those are the exact steps and what would be in each register and you would punch it into cards and it would just do it!!" ... or maybe nobody had that reaction.

### ziroby commented Jul 24, 2020

... or maybe nobody had that reaction.

I bet Alan Turing had that reaction.