Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Good news everyone! The Boards.ie Subscription service is live. See here: https://subscriptions.boards.ie/

Another simple question: polynomial multiplication modulo 2

  • 10-07-2008 10:40PM
    #1
    Closed Accounts Posts: 12,382 ✭✭✭✭


    Right,

    Thanks everyone for the help so far.

    I'm implementing "binary polynomial multiplication" at the moment as part of some cryptography code I'm writing. Nearly finished...

    Anyway, this is my simple question:

    Multiply the following polynomials:

    (x^7 + x^5 + x^3 + x) * (x^27 + x^15 + x^13 + x^12 + x^11 + 1)

    The answer:

    x^34 + x^32 + x^30 + x^28 + x^22 + 2x^20 + x^19 + 3x^18 + x^17 + 3x^16 + x^15 + 2x^14 + x^13 + x^12 + x^7 + x^5 + x^3 + x

    Now, let's say I want to reduce this modulo 2. Should it become -

    A: x^34 + x^32 + x^30 + x^28 + x^22 + x^19 + x^18 + x^17 + x^16 + x^15 + x^13 + x^12 + x^7 + x^5 + x^3 + x

    or

    B: x^34 + x^32 + x^30 + x^28 + x^22 + x^19 + x^17 + x^15 + x^13 + x^12 + x^7 + x^5 + x^3 + x

    I think perhaps it should become A, as 3 modulo 2 is 1.

    In fact, I'm pretty sure of this, but I just want to double check before I move on to polynomial division and polynomial multiplicative inverse...

    Thanks :)


Comments

  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Assuming your multiplication is right then Answer A looks good.


Advertisement