cashAddressPolynomialModulo(v: readonly number[]): number
Perform the CashAddress polynomial modulo operation, which is based on the
Bech32 polynomial modulo operation, but the returned checksum is 40 bits,
rather than 30.
A.K.A. PolyMod
Remarks
Notes from C++ implementation:
This function will compute what 8 5-bit values to XOR into the last 8 input
values, in order to make the checksum 0. These 8 values are packed together
in a single 40-bit integer. The higher bits correspond to earlier values.
The input is interpreted as a list of coefficients of a polynomial over F
= GF(32), with an implicit 1 in front. If the input is [v0,v1,v2,v3,v4],
that polynomial is v(x) = 1x^5 + v0x^4 + v1x^3 + v2x^2 + v3*x + v4.
The implicit 1 guarantees that [v0,v1,v2,...] has a distinct checksum
from [0,v0,v1,v2,...].
The output is a 40-bit integer whose 5-bit groups are the coefficients of
the remainder of v(x) mod g(x), where g(x) is the cashaddr generator, x^8
[1]. g(x) is chosen in such a way that the resulting code is a BCH
code, guaranteeing detection of up to 4 errors within a window of 1025
characters. Among the various possible BCH codes, one was selected to in
fact guarantee detection of up to 5 errors within a window of 160
characters and 6 errors within a window of 126 characters. In addition,
the code guarantee the detection of a burst of up to 8 errors.
Note that the coefficients are elements of GF(32), here represented as
decimal numbers between []. In this finite field, addition is just XOR of
the corresponding numbers. For example, [27] + [13] = [27 ^ 13] = [22].
Multiplication is more complicated, and requires treating the bits of
values themselves as coefficients of a polynomial over a smaller field,
GF(2), and multiplying those polynomials mod a^5 + a^3 + 1. For example,
[5] * [26] = (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 = a^3 + 1 (mod a^5 + a^3 + 1) = [9].
During the course of the loop below, c contains the bit-packed
coefficients of the polynomial constructed from just the values of v that
were processed so far, mod g(x). In the above example, c initially
corresponds to 1 mod (x), and after processing 2 inputs of v, it
corresponds to x^2 + v0*x + v1 mod g(x). As 1 mod g(x) = 1, that is the
starting value for c.
Parameters
v: readonly number[]
Array of 5-bit integers over which the checksum is to be computed
Perform the CashAddress polynomial modulo operation, which is based on the Bech32 polynomial modulo operation, but the returned checksum is 40 bits, rather than 30.
A.K.A.
PolyModRemarks
Notes from C++ implementation: This function will compute what 8 5-bit values to XOR into the last 8 input values, in order to make the checksum 0. These 8 values are packed together in a single 40-bit integer. The higher bits correspond to earlier values.
The input is interpreted as a list of coefficients of a polynomial over F = GF(32), with an implicit 1 in front. If the input is [v0,v1,v2,v3,v4], that polynomial is v(x) = 1x^5 + v0x^4 + v1x^3 + v2x^2 + v3*x + v4. The implicit 1 guarantees that [v0,v1,v2,...] has a distinct checksum from [0,v0,v1,v2,...].
The output is a 40-bit integer whose 5-bit groups are the coefficients of the remainder of v(x) mod g(x), where g(x) is the cashaddr generator, x^8
Note that the coefficients are elements of GF(32), here represented as decimal numbers between []. In this finite field, addition is just XOR of the corresponding numbers. For example, [27] + [13] = [27 ^ 13] = [22]. Multiplication is more complicated, and requires treating the bits of values themselves as coefficients of a polynomial over a smaller field, GF(2), and multiplying those polynomials mod a^5 + a^3 + 1. For example, [5] * [26] = (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 = a^3 + 1 (mod a^5 + a^3 + 1) = [9].
During the course of the loop below,
ccontains the bit-packed coefficients of the polynomial constructed from just the values of v that were processed so far, mod g(x). In the above example,cinitially corresponds to 1 mod (x), and after processing 2 inputs of v, it corresponds to x^2 + v0*x + v1 mod g(x). As 1 mod g(x) = 1, that is the starting value forc.