| Back: | ⟨a, b | aaaabbba=baa⟩ |
|---|
Solved by morph:3/1,3/0. (See Morphocompletion.)
Checking up to 20 rules for overlaps.
Rewriting system is not complete: rulesLimit
| # | Rule |
|---|---|
| 1. | aaaabbba ⇒ baa |
| 2. | aaaabbbbaa ⇒ babaa |
| 3. | aaaabbbbbaa ⇒ babbaa |
| 4. | aaaabbbbabaa ⇒ bababaa |
| 5. | aaaabbbbbbaa ⇒ babbbaa |
| 6. | aaaabbbbabbaa ⇒ bababbaa |
| 7. | aaaabbbbbabaa ⇒ babbabaa |
| ... |
Collecting factors up to length 7, frequency 4:
| Length 2: | [2/0] aa, [2/1] bb, [2/2] ba, [2/3] ab |
|---|---|
| Length 3: | [3/0] aaa, [3/1] bbb, [3/2] baa, [3/3] bba |
| Length 4: | [4/0] aaaa, [4/1] bbbb, [4/2] bbba, [4/3] bbaa |
| Length 5: | [5/0] aaaab, [5/1] aabbb, [5/2] aaabb, [5/3] bbbba |
| Length 6: | [6/0] aaaabb, [6/1] aaabbb, [6/2] bbbbaa, [6/3] aabbbb |
Considering [length 3 / frequency 1] bbb=c.
Checking up to 20 rules for overlaps.
Rewriting system is not complete: roundsLimit
| # | Rule |
|---|---|
| 1. | baa ⇒ aaaaca |
| 2. | bcaa ⇒ caaaaca |
| 3. | bacaa ⇒ aaaaccaa |
| 4. | cb ⇒ bc |
| 5. | aaaacaaacaaaca ⇒ caa |
| 6. | bccaa ⇒ ccaaaaca |
| 7. | baccaa ⇒ aaaacccaa |
| 8. | bcacaa ⇒ caaaaccaa |
| 9. | bacacaa ⇒ aaaaccacaa |
| 10. | bbb ⇒ c |
| 11. | bcccaa ⇒ cccaaaaca |
| 12. | bacccaa ⇒ aaaaccccaa |
| 13. | bcaccaa ⇒ caaaacccaa |
| 14. | bacaccaa ⇒ aaaaccaccaa |
| 15. | bccacaa ⇒ ccaaaaccaa |
| 16. | baccacaa ⇒ aaaacccacaa |
| 17. | bcacacaa ⇒ caaaaccacaa |
| 18. | bccccaa ⇒ ccccaaaaca |
| 19. | baccccaa ⇒ aaaacccccaa |
| 20. | bcacccaa ⇒ caaaaccccaa |
| ... |
Collecting factors up to length 6, frequency 7:
| Length 2: | [2/0] aa, [2/1] ca, [2/2] ac, [2/3] bc, [2/4] cc, [2/5] ba, [2/6] bb |
|---|---|
| Length 3: | [3/0] caa, [3/1] bac, [3/2] aca, [3/3] cca, [3/4] bca, [3/5] cac, [3/6] bcc |
| Length 4: | [4/0] ccaa, [4/1] acaa, [4/2] bcac, [4/3] bacc, [4/4] baca, [4/5] caca, [4/6] ccca |
| Length 5: | [5/0] cccaa, [5/1] cacaa, [5/2] accaa, [5/3] bcacc, [5/4] baccc, [5/5] bcaca, [5/6] bacca |
Considering [length 3 / frequency 0] caa=d.
Rewriting system is complete. See ⟨a, b | aaaabbba=baa⟩.