farer - 2 solar - 2 means - 2 muley - 1 sheth - 1 flews - 2 renew - 1Hmm, not an obvious place to start. renew has a repeated letter and a score of 1, so that breaks down into only 4 possibilities. Let's start there. e is the most common letter, and it's also shared by a lot of the other words, so we'll start by assuming there's an e. This eliminates r, n, and w, so let's reduce from there:
e(fa=1)(sola=1)(mas=1)(muly=0)(sht=0)(fls=1)
Reducing with the new zeros gives us:
e(fa=1)(oa=1)af
af gives us a contradiction (fa=1), so we know there is no e
. Let's try to r next. Reducing gives us:
r(far=1)(sola=1)(mas=2)(muly=1)(sht=1)(fls=2)
We know there can't be both an a and an s (sola=1), so there must be an m (mas=2). Reducing gives us:
rm(far=1)(sola=1)(as=1)(uly=0)(sht=1)(fls=2)
Since (as=1), (sola=1) tells us that there can't be an o or an l (since the 1 in (sola=1) has to come from either the a or the s). With this information and the new zero, we reduce again to get:
rm(far=1)(as=1)(sht=1)(fls=2)
Let's see what we get when we assume an a:
rma(ht=1)fl
That's 6 letters, so a is out. That means it has to be the s (as=1). Let's reduce:
rms(fr=1)(fl=1)
We now have two choices. We can exclude rmsrl because it has no vowels. That leaves us with rmsf. A quick look at our original words tells us that we've seen every vowel except i. Adding that in gives us rmsfi. A little anagramming gives us the solution: firms.
kevin@aq.org