math-buffer

big integer math operations implemented on top of node buffers

npm install math-buffer
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math-buffer

use node buffers as big integers.

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byte order & base

Buffers and interpreted as base 256 numbers, where each place is 256 times larger than the previous one, instead of 10 times larger. value = Math.pow(256, i).

Also note that this means bytes in a buffer are interpreted in little endian order. This means that a number such as 0x12345678 is represented in a buffer as new Buffer([0x78, 0x56, 0x34, 0x12]). This greatly simplifies the implementation because the least significant byte (digit) also has the lowest index.

api

in general, all methods follow this pattern:

result = op(a, b, m?) where m is an optional buffer that the result will be stored into. If m is not provided, then it will be allocated.

bool = isZero(x)

return true if this buffer represents zero.

fromInt: result = fromInt(int_n, length?)

convert int_n into a buffer, that is at least 4 bytes, but if you supply a longer length value, then it will be zero filled to that length.

add: result = add(a, b, m?)

add a to b and store the result in m (if m is not provided a new buffer will be allocated, and returned) in some cases, a may === m, in other cases, it must be a different buffer.

m may be a, b

subtract: result = subtract(a, b, m?)

subtract b from a and store the result in m (if m is not provided a new buffer will be allocated, and returned)

only positive integers are supported (currently) so a must be larger than b.

m may be a, b

multiply: result = multiply (a, b, m?)

multiply a by b.

m must not be a, b

modInt: int_result = modInt(a, int_n)

get the modulus of dividing a big number with a 32 bit int.

compare: order = compare(a, b)

Compare whether a is smaller than (-1), equal (0), or greater than b (1) this the same signature as is expected by Array.sort(comparator)

shift: result = shift(a, bits, m?)

move a big number across by bits. bits can be both positive or negative. a positive number of bits is the same as Math.pow(a, bits) This is an essential part of divide

m may be a

mostSignificantBit: bits = mostSignificantBit (a)

Find the position of the largest valued bit in the number. (since the buffer may have trailing zeros, it's not necessaryily in the last byte)

divide: {quotient, remainder} = divide (a, b, q?, r?)

Divide a by b, updating the q (quotient) and r (remainder) buffers.

a,b,q, and r must all be different buffers.

square: result = square (a, m?)

multiply a by itself.

m must not be a.

power: result = power (e, x, mod?)

Raise e to the power of x, If mod is provided, the result will be e^x % mod.

gcd: result = gcd(u, v, mutate=false)

Calculate the greatest common divisor using the binary gcd algorithm

If mutate is true, the inputs will be mutated, by default, new buffers will be allocated.

inverse: x = inverse(a, m)

Calculate the modular multiplicative inverse This is the inverse of a*x % m = 1.

My implementation is based on sjcl's implementation Unfortunately I was unable to find any reference explaining this particular algorithm, (the benefit this algorithm is that it doesn't require negative numbers, which I havn't implemented)

TODO

  • isProbablyPrime (needed for RSA)

License

MIT

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