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Does the challenge title imply some algorithm to be used to decode the message? Or rather the author of the algorithm?

The substring "n Tropi" from the title sounds like "entropy".
So it's more likely a data compression rather than a crypto encoding.
One of my first guesses was a continued fraction, from which I got 24 integers below 128.
But these integers don't make sense at all, so I think it's the wrong way.
And what's about the number 25 from the challenge?

Seeing how teebee is back on the forum:
Could we please have a hint for this challenge? Only 4 out of 56 people have solved it, which means that it is pretty damn hard.

The hint is already given by TheBigBoss ...

teebee wrote:The hint is already given by TheBigBoss ...

Well, he never solved the challenge. Nor did anybody else in the past five years.

I know how the text is encoed. The big problem is to find the right alphabet and the frequencies. All my brute-force-solvers took too much time to find the right answer.

Oh, I see. I have reverted the changes from revision https://en.wikipedia.org/w/index.php?ti ... =659346887.

What did you do against rounding errors?
I can clearly see two familiar words at the beginning.

I blew the numbers up to calculate with sage (BigInteger library supporting fractions), but the end of the text looks quite random to me (ends with 'gaso').

By the way: the continued fraction is [0, 1, 6, 26, 2, 124, 1, 1, 4, 3, 3, 13, 1, 42, 1, 1, 8, 1, 55, 1, 16, 1, 3, 2, 2, 1, 5, 1, 45, 2, 1, 1, 1, 2].
I'm telling this, because this doesn't help at all :)

When I say, I know the encoding, I do not talk about continued fractions.

Me neither.
As I got the first two words, I should be on the right track.

Against the hot climate, you could use an air conditioner - maybe that's the reason for the challenge name.

I fed my algorithm with the familiar words too and came quite near to the number given in the challenge. But this approach is not very promising, because the Greedy-algorithm approach will not necessarily converge to the plain text in each and every case.
The NTL library does a good job for me. I have no trouble with rounding errors.

eulerscheZahl wrote:[...] (ends with 'gaso').

Use 2.360 instead of 2.361 ...

2.360 - that was really helpful.
Now both sage and pari/gp find the correct solution.

TheBigBoss wrote:I fed my algorithm with the familiar words too and came quite near to the number given in the challenge.

You know, that you can reverse the function, right?

teebee wrote:Use 2.360 instead of 2.361 ...

eulerscheZahl wrote:2.360 - that was really helpful.
Now both sage and pari/gp find the correct solution.

The whole challenge falls apart when one value is off by one thousandth? Now that seems a bit silly.

Especially when neither 2.360 nor 2.361 can be proven to be the absolute correct value. It really depends on which source you use; and there is no way of telling which source is the right one to use.

Compare the value for 'w' in wikipedia 2009 and wikipedia now on the on hand and wikipedia 2015 on the other hand.
The value was changed back to 2.360 just two days ago from a German IP (teebee, is that you ?).
Maybe that's the reason, that noone solved it for a long time.