| Back: | ⟨a, b | aabbaba=baaa⟩ |
|---|
Solved by morph:2/1. (See Morphocompletion.)
Checking up to 20 rules for overlaps.
Rewriting system is not complete: rulesLimit
| # | Rule |
|---|---|
| 1. | aabbaba ⇒ baaa |
| 2. | aabbabbaaa ⇒ baabaaa |
| 3. | aabbabbabaaa ⇒ baababaaa |
| 4. | aabbabbaabaaa ⇒ baabaabaaa |
| 5. | aabbabbababaaa ⇒ baabababaaa |
| 6. | aabbabbaababaaa ⇒ baabaababaaa |
| 7. | aabbabbabaabaaa ⇒ baababaabaaa |
| ... |
Collecting factors up to length 8 / frequency 7:
| [2/0] | aa | 35 | lf:22,lp:7,ls:6,rf:23,rs:7,lprs:7,lsrs:6 |
|---|---|---|---|
| [2/1] | ba | 23 | lf:22,ls:1,rf:21,rp:7,lsrp:1 |
| [2/2] | ab | 22 | lf:22,rf:14 |
| [2/3] | bb | 13 | lf:13 |
| [3/0] | aab | 17 | lf:10,lp:7,rf:9 |
| [3/1] | bba | 13 | lf:13 |
| [3/2] | abb | 13 | lf:13 |
| [3/3] | aaa | 12 | lf:6,ls:6,rf:7,rs:7,lsrs:6 |
| [3/4] | bab | 12 | lf:12,rf:5 |
| [3/5] | aba | 10 | lf:9,ls:1,rf:14 |
| [3/6] | baa | 9 | lf:9,rf:16,rp:7 |
| [4/0] | aabb | 14 | lf:7,lp:7 |
| [4/1] | abba | 13 | lf:13 |
| [4/2] | baaa | 12 | lf:6,ls:6,rf:7,rp:1,rs:7,re:1,lsrs:6 |
| [4/3] | bbab | 10 | lf:10 |
| [4/4] | baba | 7 | lf:6,ls:1,rf:5 |
| [4/5] | abaa | 6 | lf:6,rf:9,a:1 |
| [4/6] | babb | 6 | lf:6,a:1 |
| [5/0] | aabba | 14 | lf:7,lp:7 |
| [5/1] | abaaa | 10 | lf:5,ls:5,rf:6,rs:6,lsrs:5,a:1 |
| [5/2] | abbab | 10 | lf:10 |
| [5/3] | babba | 6 | lf:6,a:1 |
| [5/4] | bbabb | 6 | lf:6,a:1 |
| [5/5] | bbaba | 5 | lf:4,ls:1 |
| [5/6] | babaa | 4 | lf:4,rf:4,a:3 |
| [6/0] | aabbab | 14 | lf:7,lp:7 |
| [6/1] | babaaa | 6 | lf:3,ls:3,rf:3,rs:3,lsrs:3,a:3 |
| [6/2] | bbabba | 6 | lf:6,a:1 |
| [6/3] | abbabb | 6 | lf:6,a:1 |
| [6/4] | abbaba | 5 | lf:4,ls:1 |
| [6/5] | aabaaa | 4 | lf:2,ls:2,rf:3,rs:3,lsrs:2,a:1 |
| [6/6] | babbaa | 3 | lf:3,a:1 |
| [7/0] | aabbabb | 12 | lf:6,lp:6,a:1 |
| [7/1] | abbabba | 6 | lf:6,a:1 |
| [7/2] | baabaaa | 4 | lf:2,ls:2,rf:3,rp:1,rs:3,re:1,lsrs:2,a:1 |
| [7/3] | ababaaa | 4 | lf:2,ls:2,rf:3,rs:3,lsrs:2,a:3 |
| [7/4] | bbabbaa | 3 | lf:3,a:1 |
| [7/5] | babbaba | 3 | lf:3,a:3 |
| [7/6] | bbabbab | 3 | lf:3,a:3 |
Considering [length 2 / frequency 1] ba=c.
Rewriting system is complete. See ⟨a, b | aabbaba=baaa⟩.