NACLO 2023 - Problem PCounting in Roon

In 1855, we see "nuru" = 2, "arzus safur nuru" = 32; the knee-jerk reaction is that "arzus safur" = 30. Similarly, "aresoyosier safur" (1955) and "ares kior beberin" (2012) are all 30. Furthermore "safur" (1955) = 10, so naturally "arzus" (1855) and "aresoyosier" (1955) are 3... or 20! This depends on whether the language is base 10 or base 20. I'm leaning towards the latter, because later we have "aresoyosier rim", in which if "aresoyosier" = 3, then "rim" would have to be some other base, which is unlikely. Anyway, we know that the system is at least somewhat base 10, because 2 and 32 have the same ending.

Another observation is that we have some compounds. For example, "onemerim" (1855) = 10, "onemenuru" (1855) = 7 are "onem-e-rim" and "onem-e-nuru" respectively, since we see "onem", "rim", and "nuru" in the data as well. So let's try to break down all the number phrases we see:

Number	1855	1955	2012
2	nuru	nuru	nuru
10	onem-e-rim	safur	safur
7	onem-e-nuru	rim-e-nuru	fik
32	arzus safur nuru	areso-yosier safur nuru	ares kior beberin nuru
a.	nuru × ŋokor	rim-e-yosier	yosier + rim
b.	onem × fak	iŋokor × rim-iŋokor	ares nuru beberin fiak
c.	safur onem + onem-e-fak	areso-yosier rim	(siu × nuru) + fik
d.	arzus di nuru yoser + safur lim	safur nuru + areso-nuru fak	ares rim beberin wonem
e.	fak	fak	fiak

The key observation is "rim-e-yosier" (1955) = "yosier + rim" (2012). This means that the "A-e-B" construction is A + B—and we can naturally extend this to the 1855 system. Therefore, from "onem-e-nuru" (1855) = "rim-e-nuru" (1955) = 7, we know that "onem-e-X" (1855) = "rim-e-X" (1955) = 5 + X. From "onem-e-rim" (1855) = 10, we know that "rim" (1855) = 5. We have a minor problem though: "onem" (1855) can also appear on its own, but "rim" (1855) and "onem" (1855) cannot both be 5, so "onem" (1855) must be something else, which only becomes "5+X" when it's in the "onem-e-X" construction.

If "rim-e-X" (1955) = 5 + X, then X should be between 1 and 4 (per our usual understanding of the base system). In a. "rim-e-yosier" (1955) = "2 × ŋokor" (1855), we only have two possibilities:

If "yosier" (1955) = 1, then "rim-e-yosier" (1955) = 6, and "ŋokor" (1855) = 3.
If "yosier" (1955) = 3, then "rim-e-yosier" (1955) = 8, and "ŋokor" (1855) = 4.

In b, we have "onem × fak" (1855) = "iŋokor × rim-iŋokor" (1955). Substituting in things we know, we get "onem × fak" = "iŋokor × (5 + iŋokor)". We know the following:

Because "iŋokor" (1955) looks very like "ŋokor" (1855), the former should also be either 3 or 4, depending on which alternative we choose above for "yosier" (1955).
"fak" (1855) must be no larger than 4, because in c we see "onem-e-fak" (1855).
We already know "nuru" (1855) = 2, so the factorization on the left-hand side cannot involve 2.
"onem" (1855) is less than 10, because in c we have "safur onem", so "onem" should be smaller than "safur" (1855) = 10. (This is not a 100% fool-proof logic, but it's very rare for a base word to be followed by a digit that's larger than the base, unless subtraction is performed.)

Let's test all possible values of "iŋokor" (1955):

If "iŋokor" (1955) = 3, then "onem × fak" (1855) = 24. This case is when "ŋokor" (1855) = 3, so "fak" (1855) cannot be 3, so the only factorization is "onem" (1855) = 6, "fak" (1855) = 4.
If "iŋokor" (1955) = 4, then "onem × fak" (1855) = 36. This case is when "ŋokor" (1855) = 4, so "fak" (1855) cannot be 4, so the only factorization is "onem" (1855) = 12, "fak" (1855) = 3. But "onem" (1855) should be less than 10, so this case is impossible.

Therefore, we arrive at the conclusion that "yosier" (1955) = 1, "ŋokor" (1855) = "iŋokor" (1955) = 3, "onem" (1855) = 6, and "fak" (1855) = 4.

In c, "safur onem + onem-e-fak" (1855) = "areso-yosier rim" (1955). Substituting in what we know, we get "16 + 9" = "areso-yosier + 5". This means that "areso-yosier" (1955) = 20. So this language is indeed base 20.

In d, "arzus di nuru yoser + safur lim" (1855) = "safur nuru + areso-nuru fak" (1955). Substituting in what we know, we get "arzus di nuru yoser" + 10 + "lim" (1855) = 12 + 44 (since "areso-yosier" (1955) = 20, presumably "areso-nuru" (1955) = 40, like "two-twenties"). We can hypothesize that "arzus di nuru" (1855) = 40 as well ("twenty of two"), and "yoser" (1855) = "yosier" (1955) = 1. So 41 + 10 + "lim" (1855) = 12 + 44, giving us "lim" (1855) = 5. So "rim" (1855) and "lim" (1855) are identical—this is okay, because they look so similar. The former is used in "onem-e-rim" while the latter is used standalone.

At this point, let's recap the number system for 1855 and 1955:

Number	1855	1955
1	yoser	yosier
2	nuru	nuru
3	ŋokor	iŋokor
4	fak	fak
5	lim	rim
6	onem	rim-e-yosier
7	onem-e-nuru	rim-e-nuru
8	onem-e-ŋokor	rim-iŋokor
9	onem-e-fak	rim-e-fak
10	onem-e-rim	safur
10 + X (1 ≤ X ≤ 9)	safur X	safur X
20	arzus	areso-yosier
20 + X (1 ≤ X ≤ 19)	arzus X	areso-yosier X
20Y + X (1 ≤ X ≤ 19, 1 ≤ Y ≤ 5)	arzus di Y X	areso-Y X

Now we can turn to 2012.

Number	2012
2	nuru
10	safur
7	fik
32	ares kior beberin nuru
6	yosier + rim
24	ares nuru beberin fiak
25	(siu × nuru) + fik
56	ares rim beberin wonem
4	fiak
3	kior
8	war
j.	safur fik

We have "(siu × nuru) + fik" = "(siu × 2) + 7" = 25, so "siu" (2012) = 9. The three large numbers tell us that "ares kior beberin" (2012) = 30, "ares nuru beberin" (2012) = 20, and "ares rim beberin" (2012) = 50. Knowing that "nuru" = 2, "kior" = 3, and "rim" = 5 (based on 1955), we deduce that "ares X beberin" (2012) = 10X, and the language has changed to base 10. Here are the numbers for 2012:

Number	2012
1	yosier
2	nuru
3	kior
4	fiak
5	rim
6	wonem
7	fik
8	war
9	siu
10	safur
10 + X (1 ≤ X ≤ 9)	safur X
10Y + X (1 ≤ X ≤ 9, 1 ≤ Y ≤ 10)	ares Y beberin X

To fill in the table, just apply the system above.