[AdmSmith]

They need 26 letters + 10 numbers + “several punctuation marks”. The latter could be 15 - 20 or so, so we get about 46 characters. 4**2 = 16 and 4**3 = 64, so go for a triplet code. Could they have used a simple substitution ciphers? What will a frequency analysis give?

OK Freepers give it a try!

[AdmSmith]

A 3- letter frequency

http://www.richkni.co.uk/php/crypta/freq.php

ata => 48
act => 31
taa => 30
agc => 29
tga => 28
aat => 27
cta => 26
ctg => 25
tag => 25
tat => 25
cgg => 23
gct => 22
tca => 22
cat => 21
aca => 21
gat => 19
aac => 19
cac => 18
ttt => 18
tac => 18
ccg => 17
gcc => 17
cgt => 17
gtc => 17
att => 17
caa => 16
tgt => 16
aaa => 16
tgc => 16
gac => 15
ttg => 15
ggc => 14
atc => 14
gta => 14
gca => 14
atg => 14
tcg => 13
aag => 12
gag => 12
gga => 12
gtt => 12
gtg => 12
gaa => 12
acg => 12
ggt => 12
aga => 11
ttc => 11
tta => 11
agt => 9
gcg => 9
cag => 9
agg => 9
tct => 9
cca => 8
ctt => 8
ctc => 8
cga => 7
acc => 7
tcc => 6
tgg => 6
ggg => 6
cct => 5
ccc => 5
cgc => 4