| Back: | ⟨a, b | aababba=baaa⟩ |
|---|
Solved by morph:2/1. (See Morphocompletion.)
Checking up to 20 rules for overlaps.
Rewriting system is not complete: rulesLimit
| # | Rule |
|---|---|
| 1. | aababba ⇒ baaa |
| 2. | aababbbaaa ⇒ baabaaa |
| 3. | aababbbabaaa ⇒ baababaaa |
| 4. | aababbbaabaaa ⇒ baabaabaaa |
| 5. | aababbbababaaa ⇒ baabababaaa |
| 6. | aababbbaababaaa ⇒ baabaababaaa |
| 7. | aababbbabaabaaa ⇒ 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] | aba | 15 | lf:15,rf:14 |
| [3/2] | aaa | 12 | lf:6,ls:6,rf:7,rs:7,lsrs:6 |
| [3/3] | bab | 12 | lf:12,rf:5 |
| [3/4] | baa | 9 | lf:9,rf:16,rp:7 |
| [3/5] | bba | 8 | lf:7,ls:1 |
| [3/6] | abb | 7 | lf:7 |
| [4/0] | aaba | 17 | lf:10,lp:7,rf:9 |
| [4/1] | baaa | 12 | lf:6,ls:6,rf:7,rp:1,rs:7,re:1,lsrs:6 |
| [4/2] | abab | 9 | lf:9,rf:5 |
| [4/3] | babb | 7 | lf:7 |
| [4/4] | abaa | 6 | lf:6,rf:9,a:1 |
| [4/5] | bbba | 6 | lf:6,a:1 |
| [4/6] | abbb | 6 | lf:6,a:1 |
| [5/0] | aabab | 15 | lf:8,lp:7,rf:4 |
| [5/1] | abaaa | 10 | lf:5,ls:5,rf:6,rs:6,lsrs:5,a:1 |
| [5/2] | ababb | 7 | lf:7 |
| [5/3] | abbba | 6 | lf:6,a:1 |
| [5/4] | babbb | 6 | lf:6,a:1 |
| [5/5] | babaa | 4 | lf:4,rf:4,a:3 |
| [5/6] | baaba | 3 | lf:3,rf:9,rp:6,a:1 |
| [6/0] | aababb | 14 | lf:7,lp:7 |
| [6/1] | babaaa | 6 | lf:3,ls:3,rf:3,rs:3,lsrs:3,a:3 |
| [6/2] | babbba | 6 | lf:6,a:1 |
| [6/3] | ababbb | 6 | lf:6,a:1 |
| [6/4] | aabaaa | 4 | lf:2,ls:2,rf:3,rs:3,lsrs:2,a:1 |
| [6/5] | abbbaa | 3 | lf:3,a:1 |
| [6/6] | bbbaba | 3 | lf:3,a:3 |
| [7/0] | aababbb | 12 | lf:6,lp:6,a:1 |
| [7/1] | ababbba | 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] | babbbaa | 3 | lf:3,a:1 |
| [7/5] | abbbaba | 3 | lf:3,a:3 |
| [7/6] | babbbab | 3 | lf:3,a:3 |
Considering [length 2 / frequency 1] ba=c.
Rewriting system is complete. See ⟨a, b | aababba=baaa⟩.