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fix: indenting
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README.md
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README.md
@ -43,17 +43,23 @@ the input sequence.
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## Encryption Process
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Let $P$ represent the plaintext composed of a sequence of characters:
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$$P = p_{1},p_{2},\ldots,p_{n}$$
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The ciphertext $C$ is produced by applying the reversal transformation:
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$$C = \text{ reverse}(P) = p_{n},p_{n - 1},\ldots,p_{1}$$ For example,
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if $P = \text{ sula}$, then: $$C = \text{ alus }$$
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$$C = \text{ reverse}(P) = p_{n},p_{n - 1},\ldots,p_{1}$$
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For example, if $P = \text{ sula}$, then:
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$$C = \text{ alus }$$
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## Decryption Process
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Given that the reversal operation is an involution (its own inverse),
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the decryption process involves applying the same transformation. Let
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$C$ be the ciphertext; then the plaintext $P$ is recovered as:
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$$P = \text{ reverse}(C)$$
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# Implementation Considerations
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5
main.typ
5
main.typ
@ -69,11 +69,15 @@ the decryption operation -- they both consist of reversing the input sequence.
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== Encryption Process
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Let $P$ represent the plaintext composed of a sequence of characters:
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$ P = p_1, p_2, ..., p_n $
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The ciphertext $C$ is produced by applying the reversal transformation:
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$ C = "reverse"(P) = p_n, p_(n-1), ..., p_1 $
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For example, if $P = "sula"$, then:
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$ C = "alus" $
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== Decryption Process
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@ -81,6 +85,7 @@ $ C = "alus" $
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Given that the reversal operation is an involution (its own inverse), the
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decryption process involves applying the same transformation.
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Let $C$ be the ciphertext; then the plaintext $P$ is recovered as:
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$ P = "reverse"(C) $
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= Implementation Considerations
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