krypton3

We have 3 files, translated from english text that were encrypted from the same key. Since they are originally english, we can have a general idea of the key by counting the frequency of each character.

cat found* | grep -o . | sort | uniq -c

Website for more details

letter	** count**	** frequency**	** equals**
P	2	2.8409091e-3	Z
H	4	5.6818182e-3	Q
R	4	5.6818182e-3	X
O	12	0.017045455	J
I	19	0.026988636	K
F	28	0.039772727	V
A	55	0.078125	B
L	60	0.085227273	P
E	64	0.090909091	Y
K	67	0.095170455	G
X	71	0.10085227	F
T	75	0.10653409	W
Y	84	0.11931818	M
M	86	0.12215909	U
W	129	0.18323864	C
V	130	0.18465909	L
Z	132	0.1875	D
D	210	0.29829545	R
C	227	0.32244318	H
G	227	0.32244318	S
N	240	0.34090909	N
B	246	0.34943182	I
U	257	0.36505682	O
J	301	0.42755682	A
Q	340	0.48295455	T
S	456	0.64772727	E
TOTAL	704	1

Single Letter Counts and Frequencies:

ciphertext = "<<concatanted>>"
password = "KSVVW BGSJD SVSIS VXBMN YQUUK BNWCU ANMJS"
order = ['E','Q','T','S','O','R','I','N','H','C','L','D','U','P','M','F','W','G','Y','B','K','V','X','Q','J','Z']
 
def letter_order(string):
    counts = [0]*26
    order = ""
    for i in string:
        c = ord(i.upper()) - 65
        if -1 < c < 26:
            counts[c] += 1
    for i in range(26):
        m = (0,0)
        for j in range(26):
            if counts[j] > m[1]:
                m = (j,counts[j])
        counts[m[0]] = 0
        order += chr(65+m[0])
    return order
 
def genkeys(order):
    if len(order) == 0:
        return [""]
    keys = []
    for i in range(len(order[0])):
        if len(order[0]) > 1:
            neworder = [order[0][0:i] + order[0][i+1:]] + order[1:]
        else:
            neworder = order[1:]
        for j in genkeys(neworder):
            keys.append(order[0][i]+j)
    return keys
 
def solve(s,orig,ceasar):
    solved = ""
    for i in s:
        if 64 < ord(i.upper()) < 91:
            solved += orig[ceasar.find(i.upper())]
    return solved
 
cipherorder = letter_order(ciphertext)
for i in genkeys(order):
    print(i)
    print(cipherorder)
    print(solve(password, i, cipherorder))

EQTSORINHCLDUPMFWGYBKVXQJZ
SQJUBNCGDZVWMYTXKELAFIOHRP
WELLDONETHELEVELFOURPQSSWORDISBRUTE

The password is BRUTE