I was reading this gist and am not sure I am understanding/reading the hex of the multisig's redeem script correctly (line 14). It is:


So I see the three public key pairs, beginning in bold text. I understand that the last byte (0xae) is OP_CHECKMULTISIG. But I am confused about the bytes (0x53, before 0xae) and the 5241 at the beginning. I am under the impression that before the public keys there should be two variables pushed onto the stack; one because of the checkmultisig bug and another to identify the m of n required signatures to unlock the bitcoin. And then n should come right before the last opcode (3 in this case).

What am I not understanding correctly? What's up with those fives and the 41 at the beginning? Thanks for the help!

Link to gist if broken: https://gist.github.com/gavinandresen/3966071

1 Answer 1


You are on the right track. 0x52 and 0x53 are the multisig elements ("2 of 3"), and 0xae terminates the structure. The "off by one error" comes before the signatures, which is in lines 58-60 of the gist. A good explanation of the multisig is (of course) in Andreas' book "Mastering Bitcoin". I have decoded the structure below:

    52: OP_2:                the number 2 is pushed onto stack 
    41: OP_DATA_0x41:        uncompressed pub key (65 Bytes)
        MultiSig's uncompressed Public Key (X9.63 form)
        corresponding bitcoin address is:    139FpKh63Vn4Y73ijtyqq8A6XESH8brxqs
    41: OP_DATA_0x41:        uncompressed pub key (65 Bytes)
        MultiSig's uncompressed Public Key (X9.63 form)
        corresponding bitcoin address is:    1PNvbXZFysxvx3252w9JHMa7zbG95snqnm
    41: OP_DATA_0x41:        uncompressed pub key (65 Bytes)
        MultiSig's uncompressed Public Key (X9.63 form)
        corresponding bitcoin address is:    1jqo3ptYSnUhCJq75MMyRuwC2zNyQqRy3
    53: OP_3:                the number 3 is pushed onto stack
        ##### --> 2-of-3 Multisig 
    AE: OP_CHECKMULTISIG:    terminating multisig

        corresponding bitcoin address is:    3QJmV3qfvL9SuYo34YihAf3sRCW3qSinyC
  • Ok, that makes sense. I was unclear on how OP_2-16 worked but your answer I believe clears that up. Hypothetically then if this was a m=10 and n=13 scheme then we would have 0x5a and 0x5d? Our highest possible value for field n being 0x60, no? Thanks for the help! Commented Feb 19, 2018 at 22:59
  • the codes are listed here: en.bitcoin.it/wiki/Script#Constants, and yes, 0x5a and 0x5d. Highest value = 0x60 (OP_16). Hint: multisigs have their limitations, you cannot easily create a 10-of-13 multisig, cause it exceeds verification limits / size. Pieter explained it here: bitcoin.stackexchange.com/questions/23893/… - I think with segwit the limits are shifted higher. Commented Feb 19, 2018 at 23:07

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