q4.py

#!/usr/bin/env python3

# This problem is tricky.
# We were told that our output has to be both random and
# different each time we run.
#
# It will be difficult to make sure that our output is one
# that has never occurred before.

# If we use the "random" module, a determined attacker could
# set things up so that we get the same output as a previous run.
# We need cryptography-strength RNG.
import secrets

NUM_SEQUENCES = 5
SEQUENCE_LENGTH = 10
POSSIBLE_LETTERS = ["A", "C", "G", "T"]

# In order to truly ensure that we get a result different
# from all other times we've run, we need to keep a set of
# all prior results.
all_prior_results = set([
#AUTOGENERATED_CODE
    ('CCGAGTCCTT', 'AAGGGGACGG', 'CGATATTTTC', 'GCCACTGCAT', 'ATCGCTAGGG'),
    ('CTTTGTCCCC', 'CTGAATGGTT', 'CTATCAGAAG', 'CCCCGGCTAC', 'CCATGGTTAG'),
    ('ACTTCTATTT', 'TATCTGAGAA', 'GTGTGGTTTG', 'AGTCTTAGTT', 'GGATGTAAAG'),
    ('CGGGCGCGGT', 'ATCCCAGGGA', 'AAACTCAGTG', 'GGGCGAAGGC', 'AGGGCATGAA'),
#END_AUTOGENERATED_CODE
])

seqs = []
# Keep trying until we get a completely new sequence.
while True:
    # Create a new list of sequences
    for _ in range(NUM_SEQUENCES):
        seq = "".join(secrets.choice(POSSIBLE_LETTERS) for _ in range(SEQUENCE_LENGTH))
        seqs.append(seq)
    if not tuple(seqs) in all_prior_results:
        break

# Add our new entry to the results.
# Read this script
with open(__file__, "r") as f:
    lines = f.readlines()

# Write the script back, adding our new line.
with open(__file__, "w") as f:
    for l in lines:
        f.write(l)
        # Look for the magic string.
        if l == "#AUTOGENERATED_CODE\n":
            f.write("    %s,\n" % str(tuple(seqs)))

# Print
for (i, s) in enumerate(seqs):
    print("Random sequence %d: %s" % (i, s))