It is known that the following ciphertext is a password of the key phrase, and the test is difficult to analyze:
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
analysis:
The probability of each letter is as follows:
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
Among them, the maximum is H, 47 times. So we guess DK (h) = e. The other letters K, R, X, J, Z, W, B, A are more than 20 times, so that these letters are desirable to correspond to T, A, O, I, N, S, H, R. You may wish to assume DK (k) = T. At the same time, it is noted that the letter X appears separately, guess DK (x) = a. And the 3 letter combination KWH appeared 6 times, we assume that DK (w) = h, the result is as follows:
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 the BAE BE HRJ Erijt Cibynedj RJ TB TIU TB TETEIDRAE RT THRJ RJ A TREERZMNT CIBYNED
Observing, two letters combined TB, guess DK (b) = O; three letters combined AIE, guess DK (i) = r, the replacement results are as follows:
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 the OAE OE HRJ er er TETERDRAE RT THRJ RJ A TREERZMNT CROYNED Assumment AOT Corresponds to NOT, so DK (a) = n; Observe To Ye, assume DK (Y) = B; EOR corresponds to For, so DK (e) = f; assume that RT corresponds to IT, so DK (R) = I, the results are as follows:
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 dam, one of hij firjt crobnedj j to tru to titterdine it thij ij a TiffizmNT CROBNED
Observing IJ, assume DK (j) = S; observe TIFFERENT, so DK (T) = D, the result is:
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 Qrists Is To Tru To Derdine IT this Is A DiffizMNT CROBNED
Observe Tru, assume DK (u) = y; deterDine, assuming DK (d) = m; Statistiz, so the DK (z) = C, the replacement results are as follows: 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 A DIFFICMNT CROBNEM
Observing the chacter, so it is assumed that DK (c) = p; reniabne, so suppose DK (n) = L; Messaoes, it is assumed that DK (o) = g, qHich, assuming DK (q) = W, the result is as follows:
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 Of His First Problems Is To try To Determine It this Is A DiffMLT Problem
Finally, DK (M) = U, DK (g) = Q, DK (L) = K, DK (P) = V, the result is as follows:
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 The One of His First Problems Is To try DETERMINE IT THIS A DIFFICULT ProBLEM Clemarant Table is as follows:
Bright
a
B
c
di
e
fly
G
hide
i
j
k
l
M
n
o
p
Qi
r
s
t
U
v
w
x
y
z
dense
X
Y
Z
T
Hide
E
O
W
R
L
N
Di
A
B
C
G
I
J
K
M
P
Qi
U
This replacement table is completed according to the law of the key quotation password.
Bright
a
B
c
di
e
fly
G
hide
i
j
k
l
M
n
o
p
Qi
r
s
t
U
v
w
x
y
z
dense
X
Y
Z
T
Hide
E
O
W
R
Fly
L
N
Di
A
B
C
G
I
J
K
M
P
Qi
S
U
V
