# Annotated Solution to the Jotto Puzzle of the Day

## May 7, 2003

farer - 2
solar - 2
means - 2
muley - 1
sheth - 1
flews - 2
renew - 1

Hmm, 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

Jotto Puzzle of the Day

Jotto

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Last modified: 08 May 2003