This function will compute what 6 5-bit values to XOR into the last 6 input values, in order to * make the checksum 0. These 6 values are packed together in a single 30-bit integer. The higher * bits correspond to earlier values. */
| 35 | * make the checksum 0. These 6 values are packed together in a single 30-bit integer. The higher |
| 36 | * bits correspond to earlier values. */ |
| 37 | uint32_t PolyMod(const data& v) |
| 38 | { |
| 39 | // The input is interpreted as a list of coefficients of a polynomial over F = GF(32), with an |
| 40 | // implicit 1 in front. If the input is [v0,v1,v2,v3,v4], that polynomial is v(x) = |
| 41 | // 1*x^5 + v0*x^4 + v1*x^3 + v2*x^2 + v3*x + v4. The implicit 1 guarantees that |
| 42 | // [v0,v1,v2,...] has a distinct checksum from [0,v0,v1,v2,...]. |
| 43 | |
| 44 | // The output is a 30-bit integer whose 5-bit groups are the coefficients of the remainder of |
| 45 | // v(x) mod g(x), where g(x) is the Bech32 generator, |
| 46 | // x^6 + {29}x^5 + {22}x^4 + {20}x^3 + {21}x^2 + {29}x + {18}. g(x) is chosen in such a way |
| 47 | // that the resulting code is a BCH code, guaranteeing detection of up to 3 errors within a |
| 48 | // window of 1023 characters. Among the various possible BCH codes, one was selected to in |
| 49 | // fact guarantee detection of up to 4 errors within a window of 89 characters. |
| 50 | |
| 51 | // Note that the coefficients are elements of GF(32), here represented as decimal numbers |
| 52 | // between {}. In this finite field, addition is just XOR of the corresponding numbers. For |
| 53 | // example, {27} + {13} = {27 ^ 13} = {22}. Multiplication is more complicated, and requires |
| 54 | // treating the bits of values themselves as coefficients of a polynomial over a smaller field, |
| 55 | // GF(2), and multiplying those polynomials mod a^5 + a^3 + 1. For example, {5} * {26} = |
| 56 | // (a^2 + 1) * (a^4 + a^3 + a) = (a^4 + a^3 + a) * a^2 + (a^4 + a^3 + a) = a^6 + a^5 + a^4 + a |
| 57 | // = a^3 + 1 (mod a^5 + a^3 + 1) = {9}. |
| 58 | |
| 59 | // During the course of the loop below, `c` contains the bitpacked coefficients of the |
| 60 | // polynomial constructed from just the values of v that were processed so far, mod g(x). In |
| 61 | // the above example, `c` initially corresponds to 1 mod (x), and after processing 2 inputs of |
| 62 | // v, it corresponds to x^2 + v0*x + v1 mod g(x). As 1 mod g(x) = 1, that is the starting value |
| 63 | // for `c`. |
| 64 | uint32_t c = 1; |
| 65 | for (auto v_i : v) { |
| 66 | // We want to update `c` to correspond to a polynomial with one extra term. If the initial |
| 67 | // value of `c` consists of the coefficients of c(x) = f(x) mod g(x), we modify it to |
| 68 | // correspond to c'(x) = (f(x) * x + v_i) mod g(x), where v_i is the next input to |
| 69 | // process. Simplifying: |
| 70 | // c'(x) = (f(x) * x + v_i) mod g(x) |
| 71 | // ((f(x) mod g(x)) * x + v_i) mod g(x) |
| 72 | // (c(x) * x + v_i) mod g(x) |
| 73 | // If c(x) = c0*x^5 + c1*x^4 + c2*x^3 + c3*x^2 + c4*x + c5, we want to compute |
| 74 | // c'(x) = (c0*x^5 + c1*x^4 + c2*x^3 + c3*x^2 + c4*x + c5) * x + v_i mod g(x) |
| 75 | // = c0*x^6 + c1*x^5 + c2*x^4 + c3*x^3 + c4*x^2 + c5*x + v_i mod g(x) |
| 76 | // = c0*(x^6 mod g(x)) + c1*x^5 + c2*x^4 + c3*x^3 + c4*x^2 + c5*x + v_i |
| 77 | // If we call (x^6 mod g(x)) = k(x), this can be written as |
| 78 | // c'(x) = (c1*x^5 + c2*x^4 + c3*x^3 + c4*x^2 + c5*x + v_i) + c0*k(x) |
| 79 | |
| 80 | // First, determine the value of c0: |
| 81 | uint8_t c0 = c >> 25; |
| 82 | |
| 83 | // Then compute c1*x^5 + c2*x^4 + c3*x^3 + c4*x^2 + c5*x + v_i: |
| 84 | c = ((c & 0x1ffffff) << 5) ^ v_i; |
| 85 | |
| 86 | // Finally, for each set bit n in c0, conditionally add {2^n}k(x): |
| 87 | if (c0 & 1) c ^= 0x3b6a57b2; // k(x) = {29}x^5 + {22}x^4 + {20}x^3 + {21}x^2 + {29}x + {18} |
| 88 | if (c0 & 2) c ^= 0x26508e6d; // {2}k(x) = {19}x^5 + {5}x^4 + x^3 + {3}x^2 + {19}x + {13} |
| 89 | if (c0 & 4) c ^= 0x1ea119fa; // {4}k(x) = {15}x^5 + {10}x^4 + {2}x^3 + {6}x^2 + {15}x + {26} |
| 90 | if (c0 & 8) c ^= 0x3d4233dd; // {8}k(x) = {30}x^5 + {20}x^4 + {4}x^3 + {12}x^2 + {30}x + {29} |
| 91 | if (c0 & 16) c ^= 0x2a1462b3; // {16}k(x) = {21}x^5 + x^4 + {8}x^3 + {24}x^2 + {21}x + {19} |
| 92 | } |
| 93 | return c; |
| 94 | } |
no outgoing calls
no test coverage detected