| Back: | ⟨a, b | abbababba=ab⟩ |
|---|
Solved by morph:3/0. (See Morphocompletion.)
Checking up to 20 rules for overlaps.
Rewriting system is not complete: rulesLimit
| # | Rule |
|---|---|
| 1. | abbab ⇒ abbba |
| 2. | abbbab ⇒ abbbba |
| 3. | abbbbab ⇒ abbbbba |
| 4. | abbbbbaaa ⇒ ab |
| 5. | abbbbbbaaa ⇒ abb |
| ... |
Collecting factors up to length 8 / frequency 7:
| [2/0] | ab | 16 | lf:8,lp:5,ls:3,rf:5,rp:5,rs:1,re:1,lprp:5,lsrp:3,lprs:1,a:4 |
|---|---|---|---|
| [2/1] | bb | 15 | lf:15,rf:10,rs:1,a:4 |
| [2/2] | aa | 6 | lf:4,ls:2,a:18 |
| [2/3] | ba | 5 | lf:5,rf:3,rs:3,a:4 |
| [3/0] | abb | 10 | lf:5,lp:5,rf:4,rp:4,rs:1,re:1,lprp:4,lprs:1,a:4 |
| [3/1] | bbb | 10 | lf:10,rf:6,a:4 |
| [3/2] | bab | 6 | lf:3,ls:3,a:4 |
| [3/3] | bba | 5 | lf:5,rf:3,rs:3,a:4 |
| [3/4] | aaa | 4 | lf:2,ls:2,a:18 |
| [3/5] | baa | 2 | lf:2,a:18 |
| [4/0] | abbb | 8 | lf:4,lp:4,rf:3,rp:3,lprp:2,a:4 |
| [4/1] | bbab | 6 | lf:3,ls:3,a:4 |
| [4/2] | bbbb | 6 | lf:6,rf:3,a:13 |
| [4/3] | baaa | 4 | lf:2,ls:2,a:18 |
| [4/4] | bbba | 4 | lf:4,rf:3,rs:3,a:4 |
| [4/5] | abba | 2 | lf:1,lp:1,a:4 |
| [4/6] | bbaa | 2 | lf:2,a:18 |
| [5/0] | abbbb | 6 | lf:3,lp:3,rf:2,rp:2,lprp:1,a:13 |
| [5/1] | bbaaa | 4 | lf:2,ls:2,a:18 |
| [5/2] | bbbab | 4 | lf:2,ls:2,a:13 |
| [5/3] | bbbba | 3 | lf:3,rf:2,rs:2,a:13 |
| [5/4] | bbbbb | 3 | lf:3,rf:1,a:18 |
| [5/5] | abbba | 2 | lf:1,lp:1,rf:1,rp:1,rs:1,re:1,a:4 |
| [5/6] | bbbaa | 2 | lf:2,a:18 |
| [6/0] | bbbaaa | 4 | lf:2,ls:2,a:18 |
| [6/1] | abbbbb | 4 | lf:2,lp:2,rf:1,rp:1,a:18 |
| [6/2] | abbbba | 2 | lf:1,lp:1,rf:1,rp:1,rs:1,re:1,a:13 |
| [6/3] | bbbbaa | 2 | lf:2,a:18 |
| [6/4] | bbbbab | 2 | lf:1,ls:1,a:19 |
| [6/5] | bbbbba | 2 | lf:2,rf:1,rs:1,a:18 |
| [6/6] | bbbbbb | 1 | lf:1,a:23 |
| [7/0] | bbbbaaa | 4 | lf:2,ls:2,a:18 |
| [7/1] | abbbbba | 2 | lf:1,lp:1,rf:1,rp:1,rs:1,re:1,a:18 |
| [7/2] | bbbbbaa | 2 | lf:2,a:18 |
| [7/3] | abbbbbb | 2 | lf:1,lp:1,a:23 |
| [7/4] | bbbbbba | 1 | lf:1,a:23 |
Considering [length 3 / frequency 0] abb=c.
Rewriting system is complete. See ⟨a, b | abbababba=ab⟩.