by Guest » 06 Sep 2025, 13:08
Ich versuche, eindeutige Gematria (alphanumerische Chiffre) zu erstellen.
Code: Select all
ABCDEFGHIJKLMNOPQRSTUVWXYZ
alphabet1 = {
"A": 1,
"B": 2,
"C": 3,
"D": 4,
"E": 5,
"F": 6,
"G": 7,
"H": 8,
"I": 9,
"J": 10,
"K": 11,
"L": 12,
"M": 13,
"N": 14,
"O": 15,
"P": 16,
"Q": 17,
"R": 18,
"S": 19,
"T": 20,
"U": 21,
"V": 22,
"W": 23,
"X": 24,
"Y": 25,
"Z": 26
}
x=' ';
for i in range(1 + 1):
for product in itertools.product(alphabet.keys(), repeat=i):
value = 0
comb = ""
for char in product:
print([char],":",(x*1),alphabet[char],",",sep='')
< /code>
Ich suche alle Buchstaben: < /p>
KEY OF THREE (A,D,G) (A=1,D=2,G=3,J=4,M=5,P=6,S=7,V=8,Y=9,B=10,E=11,H=12,K=13,N=14, Q=15,T=16,W=17,Z=18,C=19,F=20,I=21,K=22,N=23,Q=24,T=25,W=26)
KEY OF FIVE (A,F,K) (A=1,F=2,K=3,P=4,U=5,Z=6,E=7,J=8,O=9,T=10,Y=11,D=12,I=13,N=14, S=15,X=16,C=17,H=18,M=19,R=20,W=21,B=22,G=23,L=24,Q=25,V=26)
KEY OF SEVEN (A,H,O) (A=1,H=2,O=3,V=4,C=5,J=6,Q=7,X=8,E=9,L=10,S=11,Z=12,G=13,N=14, U=15,B=16,I=17,P=18,W=19,D=20,K=21,R=22,Y=23,F=24,M=25,T=26)
KEY OF ELEVEN (A,J,S)(A=1,J=2,S=3,B=4,K=5,T=6,C=7,L=8,U=9,D=10,M=11,V=12,E=13,N=14, W=15,F=16,O=17,X=18,G=19,P=20,Y=21,H=22,Q=23,Z=24,I=25,R=26)
KEY OF ELEVEN (A,L,W) (A=1,L=2,W=3,H=4,S=5,D=6,O=7,Z=8,K=9,V=10,G=11,R=12,C=13,N=14, Y=15,J=16,U=17,F=18,Q=19,B=20,M=21,X=22,I=23,T=24,E=25,P=26)
< /code>
Diese Verschiebungen werden im Grunde genommen mit dem Buchstaben A /F /K beginnen und Anhänge aus den unteren Beispielen erstellen. /> Ausgabe ist: < /p>
alphabet1 = {
"A": 1,
"D": 2,
"G": 3,
"J": 4,
"M": 5,
"P": 6,
"S": 7,
"V": 8,
"Y": 9,
"B": 10,
"E": 11,
"H": 12,
"K": 13,
"N": 14,
"Q": 15,
"T": 16,
"W": 17,
"Z": 18,
"C": 19,
"F": 20,
"I": 21,
"K": 22,
"N": 23,
"Q": 24,
"T": 25,
"W": 26
}
Ich versuche, eindeutige Gematria (alphanumerische Chiffre) zu erstellen.[code]ABCDEFGHIJKLMNOPQRSTUVWXYZ
alphabet1 = {
"A": 1,
"B": 2,
"C": 3,
"D": 4,
"E": 5,
"F": 6,
"G": 7,
"H": 8,
"I": 9,
"J": 10,
"K": 11,
"L": 12,
"M": 13,
"N": 14,
"O": 15,
"P": 16,
"Q": 17,
"R": 18,
"S": 19,
"T": 20,
"U": 21,
"V": 22,
"W": 23,
"X": 24,
"Y": 25,
"Z": 26
}
x=' ';
for i in range(1 + 1):
for product in itertools.product(alphabet.keys(), repeat=i):
value = 0
comb = ""
for char in product:
print([char],":",(x*1),alphabet[char],",",sep='')
< /code>
Ich suche alle Buchstaben: < /p>
KEY OF THREE (A,D,G) (A=1,D=2,G=3,J=4,M=5,P=6,S=7,V=8,Y=9,B=10,E=11,H=12,K=13,N=14, Q=15,T=16,W=17,Z=18,C=19,F=20,I=21,K=22,N=23,Q=24,T=25,W=26)
KEY OF FIVE (A,F,K) (A=1,F=2,K=3,P=4,U=5,Z=6,E=7,J=8,O=9,T=10,Y=11,D=12,I=13,N=14, S=15,X=16,C=17,H=18,M=19,R=20,W=21,B=22,G=23,L=24,Q=25,V=26)
KEY OF SEVEN (A,H,O) (A=1,H=2,O=3,V=4,C=5,J=6,Q=7,X=8,E=9,L=10,S=11,Z=12,G=13,N=14, U=15,B=16,I=17,P=18,W=19,D=20,K=21,R=22,Y=23,F=24,M=25,T=26)
KEY OF ELEVEN (A,J,S)(A=1,J=2,S=3,B=4,K=5,T=6,C=7,L=8,U=9,D=10,M=11,V=12,E=13,N=14, W=15,F=16,O=17,X=18,G=19,P=20,Y=21,H=22,Q=23,Z=24,I=25,R=26)
KEY OF ELEVEN (A,L,W) (A=1,L=2,W=3,H=4,S=5,D=6,O=7,Z=8,K=9,V=10,G=11,R=12,C=13,N=14, Y=15,J=16,U=17,F=18,Q=19,B=20,M=21,X=22,I=23,T=24,E=25,P=26)
< /code>
Diese Verschiebungen werden im Grunde genommen mit dem Buchstaben A /F /K beginnen und Anhänge aus den unteren Beispielen erstellen. /> Ausgabe ist: < /p>
alphabet1 = {
"A": 1,
"D": 2,
"G": 3,
"J": 4,
"M": 5,
"P": 6,
"S": 7,
"V": 8,
"Y": 9,
"B": 10,
"E": 11,
"H": 12,
"K": 13,
"N": 14,
"Q": 15,
"T": 16,
"W": 17,
"Z": 18,
"C": 19,
"F": 20,
"I": 21,
"K": 22,
"N": 23,
"Q": 24,
"T": 25,
"W": 26
}
[/code]