Tarou-tachi=wa gohan=o tabe-mashita
Taro-ASSOC=TOP rice=ACC eat-POLITE:PFV
‘Taro and his group ate rice.'
- Polysynthetic = High ratio .lg[(Central Yup’ik)] ← IOL loves this type.
tuntu-ssur-qatar-ni-ksaite-ngqiggte-uq
reindeer-hunt-FUT-say-NEG-again-3SG.IND
‘He had not yet said again that he was going to hunt reindeer.’
---
# Making words from morphemes
- Affixation
- Prefix: .ex[⟨me-⟩muni] .g[.sm[.sc[1pl.excl.subj]-drink ‘(We) drink.’]] .lg[(Lewo)]
- Suffix: .ex[lil⟨-its⟩] .g[.sm[cry-.sc[caus] ‘make (sb) cry.’]] .lg[(Chichewa)]
- Circumfix: .ex[ke-⟩besar⟨-an] .g[.sm[.sc[nmlz]-huge-.sc[nmlz] ‘hugeness’]] .lg[(Malay)]
- Infix: .ex[k⟨-amn-⟩aət] .g[.sm[born:.sc[nmlz] ‘birth’ ]] .lg[(Khmer)]
- Compounding: .ex[demir] .g[.sm[road]] + .ex[yol-u] .g[.sm[iron-.sc[poss]]] → .ex[demiryolu] .g[.sm[‘railway (lit. road of iron)’]] .lg[(Turkish)]
- Base modification: .ex[káʔba] .g[.sm[‘filth’]] → .ex[káʔbá] .g[.sm[filth:.sc[adjlz] ‘filthy’]] .lg[(Chalcatongo Mictec)]
- Reduplication
- Full reduplication: .ex[แดง] → .ex[แดงๆ] .g[.sm[‘kind of red’]] .lg[(Thai)]
- Partial reduplication: .ex[kuk] → .ex[k**uk**-**uk**] .g[.sm[bark-.sc[prog] ‘be barking’]] .lg[(Mangap-Mbula)]
- ‘‘Duplifix’’: .ex[jid] → .ex[ji**d**-a**d**] .g[.sm[street-.sc[pl] ‘streets’]] .lg[(Somali)]
???
Languages have many ways to build words out of morphemes.
---
# Discontinuous morphology
- Morphemes are usually combined in a (superficially) linear fashion.
- .ex[nat-] .g[(root)] → .ex[nat-ion] → .ex[nat-ion-al] → .ex[nat-ion-al-ity] → .ex[nat-ion-al-iti-es] .lg[(English)]
- ... But Semitic languages such as Arabic, Amharic, Hebrew and Maltese have .t[consonantal roots] (usually 3) and .t[transfixes].
Examples from Arabic:
- .ex[**k**a**t**a**b**a] ‘he wrote.’
- .ex[**k**a**t**a**b**at] ‘she wrote.’
- .ex[ya**kt**u**b**u] ‘he writes.’
- .ex[ta**kt**u**b**u] ‘she writes.’
- .ex[**k**i**t**aa**b**] ‘book’
- .ex[**k**aa**t**i**b**] ‘writer (masculine)’
- .ex[**k**aa**t**i**b**at] ‘writer (feminine)’
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
*tusʕēn* + *tusʕ* = *sudsēn*
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
*subʕēn* + *xumsēn* = `\(\frac{24}{35}\)`
]
.d8em[All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*___t___u___mn___* + *___t___u___mn___ēn* = *___t___a___l___a___t___t i___tm___ā___n___*
*___s___a___b___a___ʕ___t i___tl___ā___t___* + *___s___u___ds___* = *___ʕ___a___š___a___r___t i___rb___ā___ʕ___*
*___t___u___sʕ___ēn* + *___t___u___sʕ___* = *___s___u___ds___ēn*
*___x___a___m___a___s___t i___xm___ā___s___* + *___s___u___bʕ___* = *___t___a___m___a___n___t i___sb___ā___ʕ___*
*___s___u___bʕ___ēn* + *___x___u___ms___ēn* = `\(\frac{24}{35}\)`
]
.d8em[All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*___t___u___mn___* + *___t___u___mn___ēn* = *___t___a___l___a___t___t i___tm___ā___n___*
*___s___a___b___a___ʕ___t i___tl___ā___t___* + *___s___u___ds___* = *___ʕ___a___š___a___r___t i___rb___ā___ʕ___*
*___t___u___sʕ___ēn* + *___t___u___sʕ___* = *___s___u___ds___ēn*
*___x___a___m___a___s___t i___xm___ā___s___* + *___s___u___bʕ___* = *___t___a___m___a___n___t i___sb___ā___ʕ___*
*___s___u___bʕ___ēn* + *___x___u___ms___ēn* = `\(\frac{24}{35}\)`
]
.d8em[All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
.container[
.left-col[
#### Roots
.ipa[
1. t-m-n
2. t-l-t
3. s-b-ʕ
4. s-d-s
5. ʕ-š-r
6. r-b-ʕ
7. t-s-ʕ
8. x-m-s
]
]
.right-col[
#### Transfixes
.ipa[
1. Ø-u-Ø-Ø
2. Ø-u-Ø-ēn
3. Ø-a-a-t
4. i-Ø-ā-Ø
]
]
]
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
*tusʕēn* + *tusʕ* = *sudsēn*
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
___subʕēn + xumsēn = `\(\frac{24}{35}\)`___
]
.d8em[All words and sums are fractions. **Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.**
]
]
.right-col[
All denoms ≤ 10
∴ .ipa[subʕēn, xumsēn] denoms = 5 and 7.
∴ .ipa[{subʕēn, xumsēn}] = `{\(\frac{2}{5}\)`, `\(\frac{2}{7}\)}`
∴ .ipa[Ø-u-Ø-ēn] = `\(\frac{2}{\text{root}}\)`
.ipa[{s-b-ʕ, x-m-s} = {5, 7}]
∴ Transfixes = `\(\frac{x}{\text{root}}\)`
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
___tusʕēn + tusʕ = sudsēn___
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
*subʕēn + xumsēn = `\(\frac{24}{35}\)`*
]
.d8em[
All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
`\(\frac{\text{Ø-u-Ø-Ø} + 2}{\text{t-s-ʕ}} = \frac{2}{\text{s-d-s}}\)`
- Consider constraints
- Assume integer roots
- Assume unique words for integers
We get
- Ø-u-Ø-Ø = `\(\frac{1}{\text{root}}\)`
- t-s-ʕ = 6 or 9
- s-d-s = 4 or 6
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[___*tumn* + *tumnēn* = *talatt itmān*___
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
*tusʕēn* + *tusʕ* = *sudsēn*
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
*subʕēn + xumsēn = `\(\frac{24}{35}\)`*
]
.d8em[
All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
We know that:
- .ipa[Ø-u-Ø-Ø] = `\(\frac{1}{\text{root}}\)`
- .ipa[Ø-u-Ø-ēn] = `\(\frac{2}{\text{root}}\)`
`\(\frac{\text{Ø-u-Ø-Ø} + \text{Ø-u-Ø-ēn}}{\text{t-m-n}} = \frac{1+2}{\text{t-m-n}} = \frac{\text{t-l-t}}{\text{t-m-n}}\)`
∴ .ipa[t-l-t] = 3
Observe that .ipa[Ø-a-a-t] always occurs with .ipa[i-Ø-ā-n], and all the remaining fractions have them:
∴ .ipa[1a1a1t i22ā2n] = `\(\frac{\text{1-1-1}}{\text{2-2-2}}\)`
... when 1-1-1 > 2
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
*tusʕēn* + *tusʕ* = *sudsēn*
__*xamast ixmās* + *subʕ* = *tamant isbāʕ*__
*subʕēn + xumsēn = `\(\frac{24}{35}\)`*
]
.d8em[
All words and sums are fractions. Words and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
.ipa[
`\begin{align}
\frac{\text{x-m-s}}{\text{x-m-s}} + \frac{1}{\text{s-b-ʕ}} &= \frac{\text{t-m-n}}{\text{s-b-ʕ}} \\
\text{s-b-ʕ} + 1 &=\text{t-m-n}
\end{align}`]
We know that .ipa[s-b-ʕ] ∈ {5, 7}, so .ipa[t-m-n] ∈ {6, 8}
But we also know that 6 **must** be either .ipa[t-s-ʕ] or .ipa[s-d-s]. So .ipa[t-m-n] cannot be 6.
∴ t-m-n = 8; s-b-ʕ = 7
Because {x-m-s, s-b-ʕ} = {5, 7}
∴ x-m-s = 5
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
___*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*___
*tusʕēn* + *tusʕ* = *sudsēn*
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
*subʕēn + xumsēn = `\(\frac{24}{35}\)`*
]
.d8em[
All words and sums are fractions. Words' and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
`\begin{align}\frac{\text{s-b-ʕ}}{\text{t-l-t}} + \frac{1}{\text{s-d-s}} &= \frac{\text{ʕ-š-r}}{\text{r-b-ʕ}} \\
\frac{7}{3} + \frac{1}{\text{s-d-s}} &= \frac{\text{ʕ-š-r}}{\text{r-b-ʕ}}
\end{align}`
Because s-d-s ∈ {4, 6}
`\begin{align}
\frac{7}{3} + \frac{1}{\text{s-d-s}} &= \frac{7}{3} + \frac{1}{4} = \frac{31}{12} \text{, or} \\
&= \frac{7}{3} + \frac{1}{6} = \frac{5}{2} = \frac{10}{4}
\end{align}`
Given the constraints and integer uniqueness, we get
- s-d-s = 6
- t-s-ʕ = 9
- ʕ-š-r = 10
- r-b-ʕ = 4
]
]
---
# IOL 2003, Problem 2: Arabic Arithmetic
.container[
.left-col[
.ipa[*tumn* + *tumnēn* = *talatt itmān*
*sabaʕt itlāt* + *suds* = *ʕašart irbāʕ*
*tusʕēn* + *tusʕ* = *sudsēn*
*xamast ixmās* + *subʕ* = *tamant isbāʕ*
*subʕēn + xumsēn = `\(\frac{24}{35}\)`*
]
.d8em[
All words and sums are fractions. Words' and sums' numerators and denominators ≤ 10 and there is no `\(\frac{x}{1}\)`.
]
]
.right-col[
And we're done!
.container[
.left-col[
#### Roots
.ipa[
1. t-m-n = 8
2. t-l-t = 3
3. s-b-ʕ = 7
4. s-d-s = 6
5. ʕ-š-r = 10
6. r-b-ʕ = 4
7. t-s-ʕ = 9
8. x-m-s = 5
]
]
.right-col[
#### Transfixes
.ipa[
1. Ø-u-Ø-Ø=`\(\frac{1}{\text{root}}\)`
2. i-Ø-ā-Ø=`\(\frac{2}{\text{root}}\)`
3. 1a1a1t 2u22ēn=`\(\frac{\text{1-1-1}}{\text{2-2-2}}\)`
]
]
]
]
]
---
# Derivation and inflection
- .t[Derivation] = morphological processes that change the *core* meaning of a word
- .ex[read] → .ex[re-read] (affixation)
- .ex[dɔ̀] .sm[.g[drink]] → .ex[tɔ] .sm[.g[drink:.sc[caus] ‘cause to drink’]] (base modification)
- .t[Inflection] = no change in the core meaning, change in grammatical features
- .ex[read] → .ex[(he) read-s] (affixation)
- .ex[paːrʃt̪] .sm[.g[bard:.sc[nom:sg:indf] ‘a bard’]] → .ex[vaːrʃt̪] .sm[.g[bard:.sc[gen:pl:indf] ‘of bards’]] (base modification)
.box[
### Universal!!! 🤩:
- Across spoken languages, derivational morphemes are (generally) closer to the root than inflectional ones.
- .ex[read-er-s] .sm[.g[read-.sc[nmlz]-.sc[pl]]]
]
---
# UR & SR
- Often, the same morphemes (same meaning) have different forms, changing some sounds:
- .ex[Buch] ~ .ex[Büch-er] .sm[.g[book-.sc[pl] ‘books’]] .lg[(German Umlaut)]
- .ex[**k**ami] .sm[.g[‘paper’]] ~ .ex[ori-**g**ami] .sm[.g[fold-paper ‘folding paper’]] .lg[(Japanese Rendaku)]
- Not random but predictable from sound and word structure, or from the specific words themselves.
- German Umlaut (simplified): .ex[ü] .g[[yː]] occurs in the morpheme when the word is plural, .ex[u] .g[[uː]] elsewhere
- Japanese Rendaku (simplified): .ex[g] .g[[g]] at the start of the second morpheme in a compound, when preceded by a vowel; .ex[k] .g[[k]] elsewhere
---
# UR & SR: German
- Sometimes it is useful to think of a word and its morphemes as having an ‘‘original’’ form (.t[underlying representation; UR]), and the derived form (.t[surface representation; SR])
Consider the German examples below:
.container[
.left-col[
#### Singular
1. [taːk] .sm[.g[‘day:.sc[sg]’]]
2. [los] .sm[.g[‘lot:.sc[sg]’]]
3. [ʔanʁuːf] .sm[.g[‘phone.call:.sc[sg]’]]
4. [ʔaʁm] .sm[.g[arm:.sc[sg]’]]
5. [hʊnt] .sm[.g[‘dog:.sc[sg]’]]
6. [boːt] .sm[.g[‘boat:.sc[sg]’]]
]
.right-col[
#### Plural
1. [taːg-ə] .sm[.g[‘day:.sc[pl]’]]
3. [loz-ə] .sm[.g[‘lot:.sc[pl]’]]
4. [ʔanʁuːf-ə] .sm[.g[‘phone.call:.sc[pl]’]]
5. [ʔaʁm-ə] .sm[.g[arm:.sc[pl]’]]
2. [hʊnd-ə] .sm[.g[‘dog:.sc[pl]’]]
6. [boːt-ə] .sm[.g[‘boat:.sc[pl]’]]
]]
What are the URs (original form)? What is the rule for deriving the SRs?
---
# UR & SR: German
- If the forms of the root are different, the UR is the ones with a final voiced obstruent .ipa[[g, d, z]], if any. (Why choose the voiced variant?)
- [taːg], [loz], [hund], [ʔaʁm]
- Otherwise, the root's UR is the same as its SR.
- [ʔanʁuːf], [boːt]
- The rule is:
.l2em[`\(\begin{bmatrix}
+\text{consonantal} \\
-\text{sonorant}
\end{bmatrix} \rightarrow{}
\begin{bmatrix}
-\text{voice}
\end{bmatrix}
/
\_\_\#
\)`]
where
- `\(\begin{bmatrix}
+\text{consonantal} \\
-\text{sonorant}
\end{bmatrix}\)` = obstruent consonants
- `\(\begin{bmatrix}
-\text{voice}
\end{bmatrix}\)` = voiceless
‘‘All obstruent consonants become voiceless at the end of the word (before pause, #).’’
---
# UR & SR: Indonesian
Isolate the prefix common to the words below. What is its UR? What are the rules to derive the SR? (No need for fancy notation.)
.container[
.left-col[
#### Verb stems
.ipa[
1. bat͡ʃa
2. ɲaɲi
3. antuk
4. ŋat͡ʃo
5. t͡ʃut͡ʃi
6. lempar
7. pukul
8. masak
9. gambar
10. tulis
11. d͡ʒawab
12. isi
]]
.right-col[
#### Prefixed forms
.ipa[
1. məmbat͡ʃa
2. məɲaɲi
3. məŋantuk
4. məŋat͡ʃo
5. meɲt͡ʃut͡ʃi
6. məlempar
7. məmukul
8. məmasak
9. məŋgambar
10. mənulis
11. məɲd͡ʒawab
12. məŋisi
]
]
]
---
# UR & SR: Indonesian
The UR is **.ipa[məŋ]**. The rules are:
- Change the place of articulation of ŋ to be the same as the following consonant
`\(\text{ŋ} \rightarrow{} \begin{bmatrix} α\text{place} \end{bmatrix} / \_\_ \begin{bmatrix}+\text{consonantal} \\ α\text{place}\end{bmatrix} \)`
- Delete the nasal before a sonorant (.ipa[n, ŋ, m, l])
`\(\begin{bmatrix}+\text{nasal} \end{bmatrix} \rightarrow{} ∅ / \_\_ \begin{bmatrix}+\text{consonantal} \\ +\text{sonorant}\end{bmatrix} \)`
- Delete .ipa[p, t, k] after the nasal
`\(\begin{bmatrix}+\text{consonantal} \\ -\text{voice} \\ -\text{delayed release} \end{bmatrix} \rightarrow{} ∅ / \begin{bmatrix}+\text{nasal}\end{bmatrix} \_\_ \)`
... where .ipa[+] = morpheme boundaries.
Do these rules have to be ordered? Why?
---
# UR & SR: Indonesian
Why not choose other forms as URs, e.g. .ipa[**mən, məm, mə** or **məɲ**]?
---
# UR & SR: Indonesian
Why not choose other forms as URs, e.g. .ipa[**mən, məm, mə** or **məɲ**]?
Because of cases like these:
- antuk → məŋantuk
- isi → məŋisi
You may have another set of rules.
These are fine as long as your rules account for 100% of the data, but there are probably more rules/they are more complicated.
.box[
### All unpredictable things in the UR:
Posit a UR that has all unpredictable information. In this case, we cannot predict what nasals will occur between vowels, so we put [ŋ] in the UR.
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. aman
2. aptra
4. cĕt
5. çit
6. čühen
7. hupăn
9. kaç
10. kăvakar
11. kuç
12. puçtarăn
13. shăn
14. taptan
15. tupăn
16. uçăn
17. ük
18. vĕre
18. vĕren
19. vitĕn
20.
21.
]
]
.right-col[
.ipa[
1. amant
2.
4. çĕter
5.
6. čühe
7.
9.
10. kăvakart
11. kuçar
12. puçtar
13. shănt
14. tapta
15. tup
16. uç
17. üker
18. vĕret
18. vĕrent
19. vit
20. kĕrt
21. pytar
]
]
]
]
.right-col[
Instructions:
- Fill in the blanks
- Indicate which blanks cannot be filled with certainty (i.e. more than one possible forms).
]
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. __aman__
2. aptra
4. __cĕt__
5. çit
6. čühen
7. hupăn
9. kaç
10. __kăvakar__
11. __kuç__
12. puçtarăn
13. __shăn__
14. taptan
15. tupăn
16. uçăn
17. __ük__
18. __vĕre__
18. __vĕren__
19. vitĕn
20.
21.
]
]
.right-col[
.ipa[
1. amant
2.
4. çĕter
5.
6. __čühe__
7.
9.
10. kăvakart
11. kuçar
12. __puçtar__
13. shănt
14. __tapta__
15. __tup__
16. __uç__
17. üker
18. vĕret
18. vĕrent
19. __vit__
20. kĕrt
21. pytar
]
]
]
]
.right-col[
The key is just finding the right URs for the stems!
Sometimes the UR is on the left, sometimes on the right. (Hax: The UR is just the shorter one.)
]
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. aman
2. aptra
4. cĕt
5. çit
6. čühen
7. hupăn
9. kaç
10. kăvakar
11. __.r[kuç]__
12. __.r[puçtarăn]__
13. shăn
14. taptan
15. tupăn
16. uçăn
17. ük
18. vĕre
18. vĕren
19. vitĕn
20.
21.
]
]
.right-col[
.ipa[
1. amant
2.
4. çĕter
5.
6. čühe
7.
9.
10. kăvakart
11. __kuçar__
12. __puçtar__
13. shănt
14. tapta
15. tup
16. uç
17. üker
18. vĕret
18. vĕrent
19. vit
20. kĕrt
21. pytar
]
]
]
]
.right-col[
The key is just finding the right URs for the stems!
Sometimes the UR is on the left, sometimes on the right. (Hax: The UR is just the shorter one.)
Choosing the wrong UR will result in unpredictability.
]
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. __ama.b[n]__
2. aptra
4. __cĕ.r[t]__
5. çit
6. čühen
7. hupăn
9. kaç
10. __kăvaka.b[r]__
11. __ku.r[ç]__
12. puçtarăn
13. __shă.b[n]__
14. taptan
15. tupăn
16. uçăn
17. __ü.r[k]__
18. __vĕr.b[e]__
18. __vĕre.b[n]__
20. vitĕn
21.
21.
]
]
.right-col[
.ipa[
1. __ama.b[n]__***t***
2.
4. __çĕ.r[t]__***er***
5.
6. čühe
7.
9.
10. __kăvaka.b[r]__***t***
11. __ku.r[ç]__***ar***
12. puçtar
13. __shă.b[n]__***t***
14. tapta
15. tup
16. uç
17. __ü.r[k]__***er***
18. __vĕr.b[e]__***t***
18. __vĕre.b[n]__***t***
19. vit
20. kĕrt
21. pytar
]
]
]
]
.right-col[
For URs on the left side, there are two suffixes:
- .ipa[-Vr] (.ipa[e] or .ipa[a] followed by .ipa[r]), used when the stem ends with an .r[obstruent].
- The choice of the vowel depends on vowel harmony.
.mw80[]
- .ipa[-t], used when the stem ends with a .b[sonorant] (vowels, .ipa[n], .ipa[r]).
]
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. aman
2. aptra
4. cĕt
5. çit
6. __čühe***n***__
7. hupăn
9. kaç
10. kăvakar
11. kuç
12. __puçta.r[r]***.b[ă]n***__
13. shăn
14. __tapta***n***__
15. __tu.r[p]***.b[ă]n***__
16. __u***.r[ç].b[ă]n***__
17. ük
18. vĕre
18. vĕren
19. __vi.r[t]***.b[ĕ]n***__
20.
21.
]
]
.right-col[
.ipa[
1. amant
2.
4. çĕter
5.
6. __čühe__
7.
9.
10. kăvakart
11. kuçar
12. __puçta.r[r]__
13. shănt
14. __tapta__
15. __tu.r[p]__
16. __u.r[ç]__
17. üker
18. vĕret
18. vĕrent
19. __vi.r[t]__
20. kĕrt
21. pytar
]
]
]
]
.right-col[
For URs on the right side, there is one suffix:
- .ipa[-(V̆)n] (.ipa[n], might be preceded by .ipa[ĕ] or .ipa[ă]): The .b[vowel] is harmonic and is inserted when the stem ends with a .r[consonant].
]
]
---
# IOL 2004, Problem 5 Chuvash verbs
.container[
.left-col[
.container[
.left-col[
.ipa[
1. aman
2. aptra
4. cĕt
5. çit
6. čühen
7. hupăn
9. kaç
10. kăvakar
11. kuç
12. puçtarăn
13. shăn
14. taptan
15. tupăn
16. uçăn
17. ük
18. vĕre
18. vĕren
19. vitĕn
20. __ker, kertĕn__
21. __pyt, pytarăn__
]
]
.right-col[
.ipa[
1. amant
2. __aptrat__
4. çĕter
5. __çiter__
6. čühe
7. __hup, hupănt__
9. __kaçar__
10. kăvakart
11. kuçar
12. puçtar
13. shănt
14. tapta
15. tup
16. uç
17. üker
18. vĕret
18. vĕrent
19. vit
20. kĕrt
21. pytar
]
]
]
]
.right-col[
We can now fill in the blanks. Some blanks are ambiguous because the data are compatible with the UR on either side.
]
]
---
# Notes on how to answer
When you write your answers, you can use formulae, made-up notations (a → b), technical terms (URs, obstruents, sonorants, etc.) and abbreviations (C for consonants, V for vowels, etc.). But you must **define** them in the answers.
Next module is **grammatical features**. Have fun!