1. Explain using algebra why any digraph ending in A using the standard alphabet ordering
will necessarily end in the same letter no matter what multiplier is used for affine
encoding, but this will not happen for any other letters. Find how many digraphs map
to themselves under affine encoding with a general multiplier a and shift b and under
what conditions a matrix encoding can have a digraph which maps to itself. Give a
non-trivial matrix M which has this property and find all digraphs such that M maps
the digraph to itself. [6]
2. You have received the segment of text selected in class. Given that it is a palindrome
(a sequence of letters that reads the same backwards as forwards) encoded using the
affine digraph method:
(a) Determine the affine multiplier using the fact that it is a palindrome of odd length
and the given second letter of the plain text. [3]
(b) Using what “JA” encodes to, find the affine shift and thus the decoding parameters.
Manually apply them to find the first 6 letters of your text. If part (a)
defeats you, let me know and I can supply what “ME” encodes to and then you
can proceed, having lost some marks. [6]
3. Let the last 3 digits of your registration number be r, s and t. Find the solution of
x ≡ r (mod 14), x ≡ s (mod 15) and x ≡ t (mod 17) using the Chinese Remainder
Theorem. Give examples which show all the essentially different ways that the CRT
can have different types of solutions modulo mn if m and n are not relatively prime. [5]
L “cjsrxsnr,rcsfr,joz” JA -> ” df”
O “qapmxaqbqaimbaqd” JA -> “al”
O “erxvk.ovcr. ” JA -> “tv”
N “wvridwgnqweitvhp” JA -> “m.”
E “yzpxxsdspxxzvk” JA -> “re”
A “qscageiars i” JA -> “hs”
I “nucv.wqwvvzubb” JA -> “lw”
A “utdtdsbeyevsutdtix” JA -> “yt”
A “xzipfzwzop,zbh” JA -> “dz”
O “,cbnmpmpjnqcwm” JA -> “ba

 

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