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Hm. I really appreciate the effort made to guide us towards solving homophone ciphers.

However, I feel that the warm up is actually harder than the original challenge:

- real English does have much more properties than the proposed xyzw-"plaintext";
not only groups of two letters do have typical frequency distributions, but also group of 3,4,5 and more letters.

- these frequency distributions are known for real languages

- real language does have a vocabulary, syntax and sematic, allowing to
decide whether a partial decoding is correct and even correcting it.
These are not present (or not known) in xyzw-speak.

Ok, I guess these ramblings just sum up my frustration that my method for solving homophone ciphers did not work here .

Hmm, I'm surprised because the technique you described in the 'solved' forum was spot on - I would have thought it would give good results for this too. It's basically what I did to check this was solvable, and the output was really clear-cut.

It might be a mistake to use a made-up language, but it does have the advantage that the pairwise relationships between the letters are very strong and consistent, which made the analysis easier for me.

Am I missing something here? Without knowing what those pairwise relationships are, how are we supposed to check our solution if only gibberish is coming out anyway?

I could try every possible way of "deciphering" this and still not have a clue which is the correct one.

The only thing I can think of so far is that some solutions can be excluded for having every possible pairing of letters, because it says in the challenge this doesn't occur.

Hmm, this xyz 'language' makes the decoding a little more abstract, but once you are on the right track...
The small amount of characters definitely made it easier to test my solver and even check answers with pen & paper

OK, I have read paper how to solve it efficiently, just to implement it. ...
But having automated solver could be helpful .

BTW: I am able to do warmup challenge with about 1/5 of ciphertext, but it is absolutely not related to decrypting English text ...

My code from the warmup helped me to find two different numbers, that represent the same letter in the final challange. It still took me hours to solve it mostly by hand.

Hippo wrote:OK, I have read paper how to solve it efficiently, just to implement it. ...
But having automated solver could be helpful .

BTW: I am able to do warmup challenge with about 1/5 of ciphertext, but it is absolutely not related to decrypting English text ...

Hmm, either the paper was crap or I used bad "language model" ... I have to use methods from another paper ...
and now I had big problems defining good language model ... now seems, I am finally on good path, but the solver is so slow ... . I bet it would work in "more risky" setting, but I don't want to interrupt it ...

Can someone help me clarify my confusion?

======================= 1 =======================

"The plaintext consists of four characters - "xyzw", in order of decreasing frequency"
What does this refer to?

1. When '0123456789ABCDEF' is replaced by 'xyzw', then f(x) > f(y) > f(z) > f(w)

2. When every given input string is replaced by 'xyzw', then f(x) > f(y) > f(z) > f(w)

Note: Here, f(n) is frequency is character 'n' in the string.


======================= 2 =======================

In '0123456789ABCDEF', is it a possible encryption that we replace:
4 characters with 'x'
6 characters with 'y'
4 characters with 'z'
2 characters with 'w'

Thanks in advance

2. is correct. The decrypted message contains more x than y, more y than z and more z than w.