Compressing a DNA-like string

Question

Challenge:

Write a a program to take an array of 24 4-letter strings, and output a compressed (lossless) representation of the array. You should also provide a program to decompress your representations.

Input:

The input will be comprised of 4 letter words using each of [G,T,A,C] exactly once (e.g. GTAC, ATGC, CATG, etc). There are always exactly 24 words, so the full input is something like:

GTAC CATG TACG GACT GTAC CATG TACG GACT GTAC CATG TACG GACT GTAC CATG TACG GACT GTAC CATG TACG GACT GTAC CATG TACG GACT

So there are \$4!=24\$ possibilities for each word, and 24 words in total.

Output:

The output should be a printable ASCII string that can be decompressed back into the input string. Since this is lossless compression, each output should match to exactly one input.

Example:

>>> compress(["GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT"])
"3Ax/8TC+drkuB"

Winning Criterion

Your code should minimize the output length. The program with the shortest output wins. If it's not consistent, the longest possible output will be the score.

Edit: Based on helpful feedback comments, I would like to amend the criterion to add:

If two entries have the same results, the shortest code measured in characters wins. If an entry is length dependent, a length of 1 million words is the standard.

However, I fear the community angrily telling me I didn't say it right and downvoting me to the abyss, so I leave to community editors to see if it makes sense with code golf traditions

Answer 1

Python (output-length: 18 or 19, mean 18.80)

#list of the 24 possible words 
Lbase = ['GTAC','GTCA','GATC','GACT','GCTA','GCAT','TGAC',\
'TGCA','TAGC','TACG','TCGA','TCAG','AGTC','AGCT','ATGC',\
'ATCG','ACGT','ACTG','CGTA','CGAT','CTGA','CTAG','CAGT','CATG']

#base-64 alphabet (used only in the output string-representation) 
alph = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/'

#change a bijective list-representation base from b to newb 
def chg_bij_base(L, b, newb):
    n = sum([L[i]*b**i for i in range(len(L))])
    L = []
    while n > 0:
        L.append(1+(n-1)%newb)
        n = (n-1)//newb
    return L

#convert a bijective list-repr. to base-64 string-repr.
def list_to_str(Lin):
    L24 = [Lbase.index(w)+1 for w in Lin]
    L64 = chg_bij_base(L24,24,64) 
    return ''.join([alph[i-1] for i in L64])

#convert a base-64 string to the corresponding base-64 list-repr.
def str_to_list(s):
    L64 = [alph.index(c)+1 for c in s]
    L24 = chg_bij_base(L64,64,24)
    return [Lbase[i-1] for i in L24]

Test run:

Lin = ["GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT", "GTAC", "CATG", "TACG", "GACT"]

#convert input list to a base-64 string  
lts = list_to_str(Lin) 

#convert the base-64 str back to the list   
stl = str_to_list(lts)

print(len(lts))
print(lts)
print(stl == Lin)

Test output:

18
acFZIBuiiZJ+dwUn2V
True

This uses bijective numeration (which has no 0 digit) to get invertibility.

NB: When using base-64 with this method, all output strings for the above problem are either 18 or 19 characters. (A million random inputs gave an average of 18.80.) The same method works with other bases:

base         output-length
---------    -------------
56 - 58      19
59 - 69      18 or 19
70 - 73      18
74 - 88      17 or 18
89 - 96      17
...          ...
233 - 256    14