Cypher Chapter 5: MECHANISED CRYPTOGRAPHY

Chapter 5: MECHANISED CRYPTOGRAPHY

恩格玛机示意图：

谜题围绕恩格玛机展开。

PUZZLE1

Input/output:
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Scrambler I:
ABCDEFGHIJKLMNOPQRSTUVWXYZ UWYGADFPVZBECKMTHXSLRINQOJ

Reflector:
ABCDEFGHIJKLMNOPQRSTUVWXYZ YRUHQSLDPXNGOKMIEBFZCWVJAT

Ciphertext:ZYDNI

SOLVE1

根据恩格玛机反射器的对称性，用 ZYDNI 进行一次恩格玛机的加密过程即可。

• 恩格玛机的操作顺序是先转转子，再进行映射。

可以得到明文为 ULTRA。

PUZZLE2

This message was encrypted with the same prototype cipher machine, but the scrambler starting orientation is unknown. Messages from this command post always begin with a roman numeral 15.

Ciphertext:QHSGUWIG

SOLVE2

首先 \(15\) 的罗马数字表示为 XV，因此有 XQ，VH 的置换，由于只有一个转子，我们只关心第一个字母 X 能否得到即可。

经检验，转子以 F 开头，按顺序模拟可以得到明文为 XLPURPLE。

PUZZLE3

Spies have successfully captured a complete Enigma machine and codebook.The internal wirings for the remaining scramblers are as follows:

Scrambler II:
ABCDEFGHIJKLMNOPQRSTUVWXYZ AJPCZWRLFBDKOTYUQGENHXMIVS

Scrambler III:
ABCDEFGHIJKLMNOPQRSTUVWXYZ TAGBPCSDQEUFVNZHYIXJWLRKOM

The key this message was sent with is:
AB SZ UY GH LQ EN
II I III
AEB

Ciphertext:GYHRVFLRXY

SOLVE3

题目告诉我们 AB SZ UY GH LQ EN 接线，即相互映射，转子的顺序为 II->I->III，转子的初始状态分别为 A(II)E(I)B(III)，

可以列出初始恩格玛机状态：

Input/output:
ABCDEFGHIJKLMNOPQRSTUVWXYZ BACDNFHGIJKQMNOPLRZTYVWXUS

Scrambler II:
ABCDEFGHIJKLMNOPQRSTUVWXYZ AJPCZWRLFBDKOTYUQGENHXMIVS

Scrambler I:
EFGHIJKLMNOPQRSTUVWXYZABCD ADFPVZBECKMTHXSLRINQOJUWYG

Scrambler III:
BCDEFGHIJKLMNOPQRSTUVWXYZA AGBPCSDQEUFVNZHYIXJWLRKOMT

Reflector:
ABCDEFGHIJKLMNOPQRSTUVWXYZ YRUHQSLDPXNGOKMIEBFZCWVJAT

具体流程如下：

G(Input) -> (Scrambler II rotate) -> H(接线映射) -> I(Scrambler II) -> A(Scrambler I) -> B(Scrambler III) -> C(Reflector) -> L(Scrambler III) -> M(Scrambler I) -> B(Scrambler III) -> A(接线映射) -> B(Output)

Y(Input) -> (Scrambler II rotate) -> U(接线映射) -> W(Scrambler II) -> H(Scrambler I) -> N(Scrambler III) -> M(Reflector) -> H(Scrambler III) -> B(Scrambler I) -> S(Scrambler III) -> Q(接线映射) -> L(Output)

H(Input) -> (Scrambler II rotate) -> G(接线映射) -> J(Scrambler II) -> C(Scrambler I) -> J(Scrambler III) -> S(Reflector) -> S(Scrambler III) -> I(Scrambler I) -> L(Scrambler III) -> I(Output)

R(Input) -> (Scrambler II rotate) -> V(Scrambler II) -> Y(Scrambler I) -> Z(Scrambler III) -> N(Reflector) -> F(Scrambler III) -> V(Scrambler I) -> X(Scrambler III) -> T(Output)

V(Input) -> (Scrambler II spin) -> A(Scrambler II) -> Z(Scrambler I) -> G(Scrambler III) -> B(Reflector) -> X(Scrambler III) -> U(Scrambler I) -> X(Scrambler III) -> S(接线映射) -> Z(Output)

剩余略，容易猜出是单词 Blitzkrieg(闪电战)。

在操作流程中，转子II不会选装超过 \(26\) 格，因此不会带动其他转子旋转，模拟可得明文为：BLITZKRIEG。