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Transposition (CT) makes use of a key inclusive of K1. A (columnar) width N, and K2. A transposition t ¼ (t0, t1, . . . , tN21), zero, 1, . . . , N 2 1. machine protection and Cryptography. via Alan G. Konheim Copyright # 2007 John Wiley & Sons, Inc. 18 a permutation of the integers 2.2 the foundations OF COLUMNAR TRANSPOSITION ENCIPHERMENT 19 The encipherment of the plaintext x ¼ (x0, x1, . . . , xn21) of size n ¼ (r 2 1)N þ l ! N (0 , l N ) proceeds in steps: CT1. The plaintext x ¼ (x0, x1, . . .

As follows: P ‘ N(i, ‘) , nÀ1 P N(‘, i) , p^ 2 (i) ¼ ‘ nÀ1 N(i, j) , P( j=i) ; P ‘ N(i, ‘) p^ 1 (i) ; zero i,m (2:13) zero i,m (2:14) zero i, j , m: (2:15) We imagine the pattern measurement n is huge sufficient in order that p^ 1 (i) ¼ p^ (i) ¼ p (i) for zero that p satisﬁes p ( j) ¼ mÀ1 X p (i)P( j=i), zero j , m: i , m and (2:16) i¼0 To end up Equation (2.16), we begin with Equations (2.13) to (2.15), writing mÀ1 X i¼0 P( j=i)^ p1 (i) ¼ ( mÀ1 X i¼0 N(i, j) PmÀ1 ‘¼0 N(i, ‘) ) mÀ1 N(i, ‘) 1 X ¼ N(i, j).

P (h)P( =h) Â p (u)P(t=u) Â Á Á Á Â p (i)P(y=i) p (d)p (o) Â p (h)p ( ) Â p (u)p (t) Â Á Á Á Â p (i)p (y) five ¼ 1 P(o=d) Â P( =h) Â P(t=u) Â Á Á Á Â P(y=i) : p (o) Â p ( ) Â p (t) Â Á Á Á Â p (y) five The computation of the chances rating calls for numerous extra modiﬁcations: 1. Multiplying various chances or ratios of percentages is probably going to reason underﬂow, resulting in error within the scoring. to prevent underﬂow, the Markov 38 bankruptcy 2 COLUMNAR TRANSPOSITION odds rating may be.

2.36– 2.41 comprises the pairs ðdði; jÞ; IMPði; jÞÞ in terms of the adjacency ADJði; jÞ; a rating dði; jÞ and the variety of very unlikely letter-pairs IMPði; jÞ. merely the optimistic column entries are underlined. 44 bankruptcy 2 COLUMNAR TRANSPOSITION desk 2.36 Width N five three Markov Log-Odds ratings for cipherEx2.5 zero zero 1 2 21.0571 (0) 0.8745 (0) 1 2 21.2851 (1) 21.0275 (1) 21.4839 (6) 21.0863 (4) desk 2.37 Width N five four Markov Log-Odds rankings for cipherEx2.5 zero zero 1 2 three 20.8352 (3) 20.6623 (2).

Letter t with zero t , 26. we start with the subsequent remark: u(t) ¼ s if and provided that u(t 2 i þ i) 2 i ¼ s 2 i, from which it follows that desk 6.2 desk of Rotor Conjugates A B C D E F G H I J ok L M N O P Q R S T U V W X Y Z zero 1 2 three four five 6 7 eight nine 10 eleven 12 thirteen 14 15 sixteen 17 18 19 20 21 22 23 24 25 f p r d t v h p u a e x j m b s i ok l y q z w o g n Q s e u w i q v b f y okay n c t j l m z r a x p h o g t f v x j r w c g z l o d u okay m n a s b y q i p h r g w y okay s x d h a m p e v l n o b t c z r.