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Thread: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

  1. #1
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    Post f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    In this cryptosystem we use the ordinary English alphabet A,...,Z with the letters numbered from 0 to 25. We number digraphs using enumaration to base 26. We encode digraphs using the affine transformation

    f(x) ≡ 213x + 111 (mod 676).

    How would you encode the message FINAL?
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    Note the following
    676 = 26^2

    na &\equiv nb \mod n^2 \iff a &\equiv b \mod n. (Show through definition to be convinced.)

    f(x) &\equiv 213x + 111 \mod 26 \iff f(x) &\equiv 5x + 7 \mod 26

    EDIT://
    Nevermind. Above isn't as useful as I thought.
    I'll go through one letter, F = 6, the straightforward way.

    f(6) = 213(6) + 111 = 1389 \mod 676 &\equiv 37 \mod 676
    Last edited by MacstersUndead; Jan 7th 2017 at 09:42 AM.
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    thankyou, but how is F = 6, shouldnt it be F=5?
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    If A=0, B=1 ... F=5
    If A=1, B=2 ... F=6
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    Yes, you're right. F=5
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    ok so,
    f(5) = 213 (5) + 111 = 1176 mod 676 ≡ 125 mod 676 ?

    what do i do after this step? am i able to convert F into another letter?
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    It wouldn't be another letter. Remember, you're using digraphs. 125 is the resulting digraph from the function.
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    right, so for the answer, i would have 5 numbers ( one for each letter)?
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    I think so, yes.
    I did some more research on affine transformations. Those that I could find use functions that involve mod 26. In that case, you could have a correspondence between letters to letters. ex. f(x) = x + 1 mod(26) is a simple Caesar cipher. A->B, etc.

    So for F,
    f(5) = 213 (5) + 111 = 1176 = 500 mod 676 (Not sure why you got 125 on second look)

    Hence FINAL can be mapped to 500-?-?-?-?

    One can imagine the number 500 being encoded into a single character under a mod(676) affine transformation.
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    Re: f(x) ≡ 213 x + 111 (mod 676). How would you encode the message FINAL

    Thankyou
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