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11.4. THE KNAPSACK CIPHER SYSTEM 225
did know the W ’s, we would be stuck solving a hard knapsack problem.
This hints that this cipher is a good one.
However, if we do know e = 39 and P = 101 we may use the Euclidean
Algorithm to find that d = 57 is solution to 39 × d ≡ 1 (mod 101). Then,
as in Decimation Ciphers, since multiplying by e enciphered, multiplying
by d must (mostly) decipher.
Starting with 136, first multiply by d: (d × 136 = 57 × 136)%101 = 76.
Then write 76 as a sum of the original W ’s:
76 = 3 + 24 + 49
= 1 · W 1 + 0 · W 2 + 0 · W 3 + 1 · W 4 + 1 · W 5
→ 10011 = s.
So 136 → s. For the other letters we do the same. First, 190 → d · 190 =
(57 · 190)%101 = 88, and then
88 = 3 + 12 + 24 + 49
= 1 · W 1 + 0 · W 2 + 1 · W 3 + 1 · W 4 + 1 · W 5
→ 10111 = w.
So 88 → w. Similarly, 43 → i and 45 → m.
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This cipher system (also known as the Merkle–Hellman knapsack to differentiate
it from other knapsack systems is a public key cipher. We haven’t seen
any public key ciphers since Section 3.6, so let us briefly remind ourselves of the
advantage of such a cipher. Each person who wishes to be part of a conversation
may compute and make public a set of U’s, as in the Knapsack Cipher Setup.
As we will see, this allows anyone to send them a secret message, even though
they have never before communicated. There is no need to somehow pass a
secret private key: just look up the person’s public key and start enciphering!
Examples: Using the values W={5, 7, 15, 30, 59}, P = 179, e = 33:
(1) Encipher dark cave.
(2) Decipher 302, 260, 294, 157, 417, 459, 260, 294, 252, 52, 294,
417, 302. 6
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Now with five weights the hard knapsack problem only has 2 5 = 32 possible
answers to check, and so it is not really very hard. (In the previous example,
we could have determined that 136 consisted of U 1 , U 4 and U 5 with only a little
6 (1) 137, 157, 260, 304, 252, 157, 397, 294. (2) treasure chest.