已知下述密文为密钥词组密码，试破译分析之：
XNKWBMOW KWH JKXRJKRZJ RA KWRJ ZWXCKHI XIH IHNRXYNH EBI THZRCWHIRAO DHJJXOHJ JHAK RA HAONRJW KWH IHXTHI JWBMNT ABK EBIOHK KWXK KWH JKXKRJKRZJ EBI XABKWHI NXAOMXOH XIH GMRKH NRLHNU KB YH TREEHIHAK WBQHPHI HGMRPXNHAK JKXKRJXRZJ XIH XPXRNXYNH EBI BKWHI NXAOMXOHJ RE KWH ZIUCKXAXNUJK TBHJ ABK LABQ KWH NXAOMXOH RA QWRZW KWH DHJJXOH QXJ QIRKKHA KWHA BAH BE WRJ ERIJK CIBYNHDJ RJ KB KIU KB THKHIDRAH RK KWRJ RJ X TREERZMNK CIBYNHD
分析：
各个字母出现的概率如下：
Z : 8 D : 5 O : 12 K : 41 R : 31 W : 22 H : 47 C : 5 P : 3
B : 20 Q : 5 L : 2 G : 2 I : 23 M : 8 A : 20 U : 4 Y : 5
J : 28 T : 7 N : 17 E : 11 X : 29
其中，出现最多的是H，为47次。所以我们猜测dk(H) = e。其他的字母K、R、X、J、Z、W、B、A分别出现20次以上，所以希望这些字母对应的是t、a、o、i、n、s、h、r。不妨假设dk(K) = t。同时，注意到字母X单独出现，猜测dk(X) = a。而且3字母组合KWH出现6次，我们假设为the，则dk(W) = h，结果如下：
aNthBMOh the JtaRJtRZJ RA thRJ ZhaCteI aIe IeNRaYNe EBI TeZRCheIRAO DeJJaOeJ JeAt RA eAONRJh the IeaTeI JhBMNT ABt EBIOet that the JtatRJtRZJ EBI aABtheI NaAOMaOe aIe GMRte NRLeNU tB Ye TREEeIeAt hBQePeI eGMRPaNeAt JtatRJaRZJ aIe aPaRNaYNe EBI BtheI NaAOMaOeJ RE the ZIUCtaAaNUJt TBeJ ABt LABQ the NaAOMaOe RA QhRZh the DeJJaOe QaJ QIRtteA theA BAe BE hRJ ERIJt CIBYNeDJ RJ tB tIU tB TeteIDRAe Rt thRJ RJ a TREERZMNt CIBYNeD
观察发现，两字母组合tB，猜测dk(B) = o；三字母组合aIe，猜测dk(I) = r，替换后结果如下：
aNthoMOh the JtaRJtRZJ RA thRJ ZhaCter are reNRaYNe Eor TeZRCherRAO DeJJaOeJ JeAt RA eAONRJh the reaTer JhoMNT Aot EorOet that the JtatRJtRZJ Eor aAother NaAOMaOe are GMRte NRLeNU to Ye TREEereAt hoQePer eGMRPaNeAt JtatRJaRZJ are aPaRNaYNe Eor other NaAOMaOeJ RE the ZrUCtaAaNUJt ToeJ Aot LAoQ the NaAOMaOe RA QhRZh the DeJJaOe QaJ QrRtteA theA oAe oE hRJ ERrJt CroYNeDJ RJ to trU to TeterDRAe Rt thRJ RJ a TREERZMNt CroYNeD
假设Aot对应not，所以dk(A) = n；观察to Ye，假设dk(Y) = b;Eor对应for，所以dk(E) = f；假设Rt对应it，所以dk(R) = i，结果如下：
aNthoMOh the JtaiJtiZJ in thiJ ZhaCter are reNiabNe for TeZiCherinO DeJJaOeJ Jent in enONiJh the reaTer JhoMNT not forOet that the JtatiJtiZJ for another NanOMaOe are GMite NiLeNU to be Tifferent hoQePer eGMiPaNent JtatiJaiZJ are aPaiNabNe for other NanOMaOeJ if the ZrUCtanaNUJt ToeJ not LnoQ the NanOMaOe in QhiZh the DeJJaOe QaJ Qritten then one of hiJ firJt CrobNeDJ iJ to trU to TeterDine it thiJ iJ a TiffiZMNt CrobNeD
观测iJ，假设dk(J) = s；观察Tifferent，所以dk(T) = d，结果为：
aNthoMOh the staistiZs in this ZhaCter are reNiabNe for deZiCherinO DessaOes sent in enONish the reader shoMNd not forOet that the statistiZs for another NanOMaOe are GMite NiLeNU to be different hoQePer eGMiPaNent statisaiZs are aPaiNabNe for other NanOMaOes if the ZrUCtanaNUst does not LnoQ the NanOMaOe in QhiZh the DessaOe Qas Qritten then one of his first CrobNeDs is to trU to deterDine it this is a diffiZMNt CrobNeD
观察trU，假设dk(U) = y；deterDine，假设dk(D) = m；statistiZs，所以假设dk(Z) = c，替换后结果如下：
aNthoMOh the staistics in this chaCter are reNiabNe for deciCherinO messaOes sent in enONish the reader shoMNd not forOet that the statistics for another NanOMaOe are GMite NiLeNy to be different hoQePer eGMiPaNent statisaics are aPaiNabNe for other NanOMaOes if the cryCtanaNyst does not LnoQ the NanOMaOe in Qhich the messaOe Qas Qritten then one of his first CrobNems is to try to determine it this is a difficMNt CrobNem
观察chaCter，故假设dk(C) = p；reNiabNe，所以假设dk(N) = l；messaOes，所以假设dk(O) = g，Qhich，假设dk(Q) = w，结果如下：
althoMgh the staistics in this chapter are reliable for deciphering messages sent in english the reader shoMld not forget that the statistics for another langMage are GMite liLely to be different howePer eGMiPalent statisaics are aPailable for other langMages if the cryptanalyst does not Lnow the langMage in which the message was written then one of his first problems is to try to determine it this is a difficMlt problem
最后，dk(M) = u，dk(G) = q，dk(L) = k，dk(P) = v，结果如下：
although the staistics in this chapter are reliable for deciphering messages sent in english the reader should not forget that the statistics for another language are quite likely to be different however equivalent statisaics are available for other languages if the cryptanalyst does not know the language in which the message was written then one of his first problems is to try to determine it this is a difficult problem
明密代替表如下：
明
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z
密
X
Y
Z
T
H
E
O
W
R
L
N
D
A
B
C
G
I
J
K
M
P
Q
U
根据密钥词组密码的规律，补全该代替表：
明
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z
密
X
Y
Z
T
H
E
O
W
R
F
L
N
D
A
B
C
G
I
J
K
M
P
Q
S
U
V
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