# ECIP 1026: Modexp Precompiled Contract Source

Author Wei Tang Draft Standards Track Core 2017-06-28

## Rationale

This allows for efficient RSA verification inside of the EVM, as well as other forms of number theory-based cryptography. Note that adding precompiles for addition and subtraction is not required, as the in-EVM algorithm is efficient enough, and multiplication can be done through this precompile via a * b = ((a + b)**2 - (a - b)**2) / 4.

The bit-based exponent calculation is done specifically to fairly charge for the often-used exponents of 2 (for multiplication) and 3 and 65537 (for RSA verification).

## Specification

### Algorithm

At address 0x00……05, add a precompile that expects input in the following format:

<length_of_BASE> <length_of_EXPONENT> <length_of_MODULUS> <BASE> <EXPONENT> <MODULUS>

Where every length is a 32-byte left-padded integer representing the number of bytes to be taken up by the next value. Call data is assumed to be infinitely right-padded with zero bytes, and excess data is ignored. Consumes floor(max(length_of_MODULUS, length_of_BASE) ** 2 * max(ADJUSTED_EXPONENT_LENGTH, 1) / GQUADDIVISOR) gas, and if there is enough gas, returns an output (BASE**EXPONENT) % MODULUS as a byte array with the same length as the modulus.

• If length_of_EXPONENT <= 32, and all bits in EXPONENT are 0, return 0
• If length_of_EXPONENT <= 32, then return the index of the highest bit in EXPONENT (eg. 1 -> 0, 2 -> 1, 3 -> 1, 255 -> 7, 256 -> 8).
• If length_of_EXPONENT > 32, then return 8 * (length_of_EXPONENT - 32) plus the index of the highest bit in the first 32 bytes of EXPONENT (eg. if EXPONENT = \x00\x00\x01\x00.....\x00, with one hundred bytes, then the result is 8 * (100 - 32) + 253 = 797). If all of the first 32 bytes of EXPONENT are zero, return exactly 8 * (length_of_EXPONENT - 32).

For example, the input data:

0000000000000000000000000000000000000000000000000000000000000001
0000000000000000000000000000000000000000000000000000000000000020
0000000000000000000000000000000000000000000000000000000000000020
03
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2e
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f

Represents the exponent 3**(2**256 - 2**32 - 978) % (2**256 - 2**32 - 977). By Fermat’s little theorem, this equals 1, so the result is:

0000000000000000000000000000000000000000000000000000000000000001

Returned as 32 bytes because the modulus length was 32 bytes. The ADJUSTED_EXPONENT_LENGTH would be 255, and the gas cost would be 32**2 * 255 / 100 = 2611 gas (note that this is ~five thirds of the cost of using the EXP opcode to compute a 32-byte exponent). A 4096-bit RSA exponentiation would cost 512**2 * 4095 / 100 = 10734796 gas in the worst case, though RSA verification in practice usually uses an exponent of 3 or 65537, which would reduce the gas consumption to 2621 or 41943, respectively.

This input data:

0000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000020
0000000000000000000000000000000000000000000000000000000000000020
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2e
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f

Would be parsed as a base of 0, exponent of 2**256 - 2**32 - 978 and modulus of 2**256 - 2**32 - 978, and so would return 0. Notice how if the length_of_BASE is 0, then it does not interpret any data as the base, instead immediately interpreting the next 32 bytes as length_of_EXPONENT.

This input data:

0000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000020
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffe
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffd

Would parse a base length of 0, a modulus length of 32, and an exponent length of 2**256 - 1, where the base is empty, the modulus is 2**256 - 2 and the exponent is (2**256 - 3) * 256**(2**256 - 33) (yes, that’s a really big number). It would then immediately fail, as it’s not possible to provide enough gas to make that computation.

This input data:

0000000000000000000000000000000000000000000000000000000000000001
0000000000000000000000000000000000000000000000000000000000000002
0000000000000000000000000000000000000000000000000000000000000020
03
ffff
8000000000000000000000000000000000000000000000000000000000000000
07

Would parse as a base of 3, an exponent of 65535, and a modulus of 2**255, and it would ignore the remaining 0x07 byte.

This input data:

0000000000000000000000000000000000000000000000000000000000000001
0000000000000000000000000000000000000000000000000000000000000002
0000000000000000000000000000000000000000000000000000000000000020
03
ffff
80

Would also parse as a base of 3, an exponent of 65535 and a modulus of 2**255, as it attempts to grab 32 bytes for the modulus starting from 0x80, but then there is no further data so it right pads it with 31 zeroes.