We have a context free grammar with non-terminals A1..A57. The target sequence is a list of integers (terminals) separated by spaces: plus we need to produce exactly that sequence.

The initial symbol is A1. The current grammar includes many terminal productions and some expansions. Some terminal symbols have been removed from production rules (presumably originally there were more terminals). We need to insert terminal symbols into existing production rules (anywhere in RHS), but cannot delete, reorder, or create new production rules. The goal is to make it possible for the grammar to generate the target sequence.

We must minimize the total number of inserted terminal symbols across all production rules. So we need to find some minimal modifications that enable the language to include the given target string. We must output only the modified rules with insertions but not the unmodified ones.

Now we must read the current grammar and see if the target can be derived. If not, insert terminals.

First, let's parse the grammar.

But it's huge; let's try systematically. The target length is 36 (since we see 36 numbers; let's count: Actually sequence: 36,33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31,30,32,35,29,34. Count them: I'll count: 1:36,2:33,3:18,4:15,5:1,6:28,7:24,8:10,9:5,10:2,11:21,12:9,13:12,14:4,15:3,16:8,17:19,18:14,19:17,20:16,21:26,22:25,23:11,24:22,25:23,26:13,27:7,28:6,29:20,30:27,31:31,32:30,33:32,34:35,35:29,36:34. Yes 36 symbols.

Goal: Starting from A1 produce that exact sequence.

Given rule_1: A1 -> A2
rule_2: A2 -> A57

Thus, A1 => A57.

rule_109: A57 -> 36 A56 34

Thus A57 derives 36, then A56, then 34.

Thus start string: 36 ... 34.

Target starts with 36 and ends with 34. Good.

Now A56 -> A54 30 A54 35 (rule_108). So A56 yields A54, then terminal 30, then A54, then terminal 35.

Thus the full derived string is:

36 A54 30 A54 35 34

So target must be: 36 + something + 30 + something else + 35 + 34. Does target have 30 and 35 placed? Let's check target: [36,33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31,30,32,35,29,34]

We see 30 appears at position 32 (0-index??) it's the 32nd item (index 31). Check: after 31, the 30 is there near end. Indeed sequence contains "...31,30,32,35,...". So we have 30 and later 35. So target appears as:

36 [...something...] 30 ... 35 ... 34.

But the grammar forces that after 36, there will be two A54 blocks, each has a block of terminals separated by 30 and 35.

We need to map the target to that structure.

We have 36 at start, then the target after 36 is: 33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31,30,32,35,29,34.

So after 36 we have 33 ... then later there's a 30 and later a 35 as we saw.

In the grammar: 36 A56 34. A56 yields A54 30 A54 35. So structure:

36 (A54) 30 (A54) 35 34.

Thus A54 occurs twice: before the 30 and after the 30, before 35. The 35 appears after second A54. So the target must be decomposable as:

36 [block1] 30 [block2] 35 34.

Thus block1 is the sequence between the first 36 and the 30 (excluding 30). That is: 33,18,15,...,31 (from start after 36 up to before 30). Let's list up to just before the 30: target string scanning until 30 appears at position 32: count indices (starting at 1 after 36? Actually let's index the target positions 0..35). Let's rewrite with indices:

0:36
1:33
2:18
3:15
4:1
5:28
6:24
7:10
8:5
9:2
10:21
11:9
12:12
13:4
14:3
15:8
16:19
17:14
18:17
19:16
20:26
21:25
22:11
23:22
24:23
25:13
26:7
27:6
28:20
29:27
30:31
31:30
32:32
33:35
34:29
35:34

Thus 30 is at index 31. So block1 includes positions 1-30 inclusive: 33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31.

Thus block1 length = 30 symbols.

Block2 is between the 30 and the 35: positions 32 (32) is after 30 before 35. So block2 is just [32] at index 32. Since 35 is at index 33. So block2 length = 1 symbol (32). After the 35 we have 29, and then 34. The grammar expects after second A54 (block2) comes terminal 35, then 34. So we have 35 (ok) then block2 must be just before 35? Wait according to grammar: A56 -> A54 30 A54 35. So after block2 via second A54 we have 35, then final 34 (coming from A57 outer). But our target is ...30 32 35 29 34. So after 35 we have 29 before 34, which is not allowed by grammar as currently defined, unless the second A54 (block2) continues beyond 30? Actually after second A54, we produce 35, but we have a 29 before 34. Grammar currently: A57 -> 36 A56 34; A56 -> A54 30 A54 35. So after A56 we produce the second A54's terminals then 35, then A57 adds 34 after that. There's no space for 29 after 35 before 34. 29 appears before final 34, but after 35. So we need to accommodate this 29: either need to insert a terminal 29 somewhere in the productions (like within second A54 after 35?). But we cannot reorder or delete symbols, only insert terminals anywhere in RHS. So we could insert terminal 29 after the 35 (or before 35) but must preserve order of existing symbols. The location for insertion can be anywhere in a RHS. That includes after 35 before reaching the final 34? But we cannot modify A57 rule (except by insertion). A57 currently: 36 A56 34. If we insert 29 before the 34, i.e., A57 -> 36 A56 29 34, then final string gets 29 before 34. That matches target. However we also see 35 should appear before 29 (as target: 30,32,35,29,34). Insert 29 after that would yield 30 block2 35 29 34. Good. So we could insert 29 into rule_109 A57: after A56 and before 34. That's one insertion.

Now we need block2 = [32] must be produced by second A54 production. A54 currently has rules: rule_105: A54 -> 32; rule_106: A54 -> A55 A52 A21 A51. So A54 can produce the terminal 32 directly via rule_105, which matches block2 exactly: 32. Good.

Thus second A54 can use rule_105 (or maybe multiple ways). Block2 = [32], matches.

Now block1 must be produced by the first A54. That is a longer sequence of 30 terminals: 33 18 15 1 28 24 10 5 2 21 9 12 4 3 8 19 14 17 16 26 25 11 22 23 13 7 6 20 27 31

We need to see if A54 can produce that. Possibilities:

- rule_105 gives single terminal 32. Not enough.
- rule_106: A54 -> A55 A52 A21 A51 (a concatenation of four nonterminals). Each presumably produce some subsequence.

Thus block1 can be segmented into sub-blocks generated by A55, A52, A21, A51 in order.

Now we have to examine each of those nonterminals:

- A55: rule_107: A55 -> 33 A26 (so produces terminal 33 then A26)
- A52: rule_103: A52 -> 28 A47 A53 (produces terminal 28 then A47 then A53)
- A21: rule_47: A21 -> A22 A23 (concatenation)
- A51: rule_102: A51 -> 26 A49 (produces terminal 26 then A49)

Thus the block1 is presumably: 33 (via A55) plus whatever A26 generates; then 28 (via A52) plus A47 and A53; then A22 A23 sub-block; then 26 (via A51) plus A49, and that's all.

Let's break down block1 accordingly:

Block1 = [33, 18, 15, 1, 28, 24, 10, 5, 2, 21, 9, 12, 4, 3, 8, 19, 14, 17, 16, 26, 25, 11, 22, 23, 13, 7, 6, 20, 27, 31].

But we have sub-block segmentation predicted:

- A55: 33 A26: So first element 33 matches. Then A26 must produce the next part.

- Then from A52: 28 A47 A53: after A26, next expected is 28; indeed block1's fifth element is 28 (position after first few). The A26 must produce a sequence that ends just before the 28. Then A52 yields 28, then A47, then A53.

- Then A21 yields A22 A23: after A47 A53, next part of block must match A22 A23 expansions.

- Then A51 yields 26 A49: after A21's expansions, the next must be 26 then A49.

Let's enumerate block1 and see if this segmentation works.

First element: index 0 (of block1) = 33 (position 1 of overall). So A55 matches 33 and then A26 must derive the rest: block1_rest = after 33: [18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31] (length 29). Then A52 will produce 28 as its first terminal after A26. So A26 must produce everything up to immediately before the 28. So we need to generate [18,15,1] maybe? Let's check: In A26, we need to find its rules.

A26 rule_58: A26 -> A27 A28. So A26 yields A27 followed by A28.

- A27: rule_59: A27 -> 18

Thus A27 yields terminal 18.

- A28: rule_60: A28 -> 15 A29 ; rule_61: A28 -> 17 A30.

Thus A28 can produce either "15" then A29, or "17" then A30.

A29: rules 62-64: A29 -> 1 | 2 | 3. So A29 produces one of 1,2,3.

A30: rules 65-67: A30 -> 1 | 2 | 3.

Thus A28 can produce "15 x" where x=1,2,3, or "17 x" where x=1,2,3.

Thus A26 can produce: "18 15 x" or "18 17 x". Also maybe other combos? No other rules.

Thus, from A26, we can produce 3 symbols total: 18 then 15/17 then a digit 1-3.

Thus block after 33 will be: [some triple], then we need 28.

Our block1 after 33 is: [18,15,1,28,...]. That matches A26 producing 18 (A27), then 15 (if pick rule 60 for A28 with 15), then 1 (if A29->1). That's exactly first three symbols after 33: 18,15,1. Good. Then next is 28 which matches A52's 28.

Thus A26 can produce 18,15,1 (or 18,15,2, etc). We need 18,15,1 exactly. So we must choose rule_60 for A28 (15 A29) and rule_62 (A29->1). That matches. Good.

Thus after A26, the next symbol is 28. So far we have matched 33, (A26->[18,15,1]), then A52->28.

Now A52: rule_103: A52 -> 28 A47 A53.

Thus after 28, we need to generate A47 and then A53.

Thus after 28, the rest of block1 continues: block after 28: [24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31] (let's verify length). Let's compute block1 after 28: original block1 after 33 and 18,15,1 and 28: we have originally block1 = [33, 18, 15, 1, 28, 24, 10, 5, 2, 21, 9, 12, 4, 3, 8, 19, 14, 17, 16, 26, 25, 11, 22, 23, 13, 7, 6, 20, 27, 31].

Thus after 28, the remaining sequence is: [24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31] length 25.

Now we need to produce A47 A53. Let's consider A53 rule_104: A53 -> 19. So A53 yields terminal 19.

Thus the last symbol of the A52 region should be 19. In the block after 28, we see somewhere a 19 at position... after 8 (24,10,5,2,21,9,12,4,3,8,19,...). Indeed 19 appears after 8. So A47 must generate all symbols up to (including?) before the 19? Actually A52 yields 28 A47 A53: 28 then whatever A47 yields, then 19 (by A53). After A53 returns, we move on to A21.

Thus block after 28 should be: some symbols generated by A47, then 19, then after that next comes A21's expansion.

Thus the sequence after 28 is: [24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31]; but after 8 we have 19, then 14,17,... So A47 must generate [24,10,5,2,21,9,12,4,3,8] (10 symbols). Then A53 yields 19. Good. Then after that is [14,17,16,26,25,11,22,23,13,7,6,20,27,31] which should be matched by A21 and A51.

Now analyze A47: rule_98: A47 -> A48 A44 A13 A31.

Thus A47 expands to concatenation of A48, A44, A13, A31. So we need to produce the sequence [24,10,5,2,21,9,12,4,3,8] via these four nonterminals.

Let's break down:

- A48 (rule_99): A48 -> 24 A7
- A44 (rules 93-94): A44 -> 21 A45 OR 22 A46. Actually rule_93: A44 -> 21 A45, rule_94: A44 -> 22 A46.
- A13 (rules 30-31): A13 -> 11 OR A14 A15 (both are alternatives). Actually rule_30: A13 -> 11, rule_31: A13 -> A14 A15.
- A31 (rules 68-79): A31 -> A32 | A33 | A34 | A35 | A36 | A37 | A38 | A39 | A40 | A41 | A42 | A43 (actually many rules). Let's see: rules have multiple expansions: rule_68: A31 -> A32; rule_69: A31 -> A33; etc up to rule_79: A31 -> A43. So essentially A31 yields one of many nonterminals A32..A43. Each of those yields terminal 8 (or maybe also other options). For each of A32,.. A43 there's a rule that yields terminal 8, plus for A36 also a rule to 20. So A31 can yield 8 (via any of them), or 20 via A36.

Thus we need to map the segment [24,10,5,2,21,9,12,4,3,8] to concatenated expansions: A48 (24 A7), A44 (21 ... or 22 ...), A13 (something), A31 (something). Let's see details.

A48 yields "24" followed by A7. So A48 will produce "24" then ... what about A7? A7 rule_16: A7 -> A8 A9.

So A7 -> A8 A9. Evaluate A8: rule_17: A8 -> 10. So A8 yields 10. A9: rule_18-20: A9 -> 4 A10 OR 5 A11 OR 7 A12. So A9 can start with 4,5,7. A10 (3 rules) produce 1/2/3; A11 same; A12 same. So A7 -> A8 A9 yields "10" followed by (4/5/7) then some digit (1/2/3). That matches the next three symbols in the segment: after "24", we need "10", "5", "2". Indeed we have 10,5,2. So we can pick A9->5 A11, and A11->2. That matches 10,5,2. Good.

Thus A48 A7 yields exactly "24 10 5 2". So far we have matched the first four? Actually sequence "24 10 5 2" corresponds to first four elements: 24,10,5,2. Good.

Now A44 should generate the next few terminals: after "24 10 5 2" we need "21 9". Actually segment after "24,10,5,2" is "21,9". Indeed segment [24,10,5,2,21,9...]. So we need A44 to produce "21 9". Let's see.

A44 rule_93: A44 -> 21 A45. A45 -> 9 (rule_95). So A44 -> 21 9 yields exactly "21 9". That's good. So choose the alternative A44 -> 21 A45.

Thus now we've matched up to "21 9".

Now A13: we must generate the next few terminals: after "21,9" the segment continues "12,4,3,8". Actually after "21 9", the segment has "12,4,3,8". So we need A13 to produce "12 4 3"? Wait A13 alternatives: either 11, or A14 A15. Let's analyze A14 and A15.

- A14 rule_32: A14 -> 12.

- A15 rule_33-35: A15 -> 4 A16 OR 5 A17 OR 7 A18.

Thus A14 A15 yields "12" then (4 or 5 or 7) then A16/A17/A18.

A16, A17, A18 produce digits 1,2,3. So we can produce "12 4 1" or "12 4 2" etc.

Our segment after A44 is "12,4,3,8". We have 12 from A14, 4 from A15 (choice 4 A16), and then we need 3 as the terminal from A16 (which can produce 1,2,3). Choose A16->3. So A13 -> A14 A15 -> (12) (4 A16) where A16 -> 3 => yields "12 4 3". Good. That matches the three terminals "12 4 3" after 21,9.

Thus A13 yields 12,4,3.

Now A31 should produce the final terminal(s) of the original segment: after "12 4 3", we see "8". So A31 must be able to emit "8". Indeed A31 can choose A32 which yields terminal 8 (rule_80). So we can have A31 -> A32 -> 8. That works.

Thus A47 -> A48 A44 A13 A31 yields exactly "24 10 5 2 21 9 12 4 3 8". Good.

Thus A47 matches that prefix.

Now after A47 is A53 (19). So the segment after the "8" is indeed "19". So A53->19 matches.

Thus up to now the sequence matched is: 33 (A55) [then A26 -> 18 15 1] 28 (A52's 28) 24 10 5 2 21 9 12 4 3 8 19 (A53). So far we have matched:

33 (index after 36).
18,15,1 (A26)
28 (A52)
24,10,5,2,21,9,12,4,3,8 (A47)
19 (A53)

Up to which point? Let's list them all together:

Sequence so far:

33,
18,15,1,
28,
24,10,5,2,21,9,12,4,3,8,
19

Now after A53, next is A21 (since A54 -> A55 A52 A21 A51). A21 expands into A22 A23.

Look at block1 after A53 19. The remaining block1 after 19 is: [14, 17, 16, 26, 25, 11, 22, 23, 13, 7, 6, 20, 27, 31] (14 symbols). Let's verify: After 19 in block1 we have "14,17,16,26,25,11,22,23,13,7,6,20,27,31". Yes.

Thus A21 should generate that. A21 -> A22 A23. So we need A22 to generate a prefix and A23 to generate suffix.

A22 rule_48: A22 -> 14 => yields terminal 14. So A22 matches "14".

Thus A23 must generate the remaining [17,16,26,25,11,22,23,13,7,6,20,27,31] (13 symbols). Let's see A23's possible alternatives: rule_49: A23 -> 15 A24; rule_50: A23 -> 17 A25.

Thus A23 can start with either 15 or 17. Our remaining sequence starts with 17, so we must use rule_50: A23 -> 17 A25.

Thus after 17, produce A25.

Now A25 has multiple productions: rules_54-57: A25 -> 1,2,3,16 (four alternatives). Wait rule_57: A25 -> 16 (terminal 16). But also rules 54-56: produce 1,2,3.

Thus A25 should produce the continuation after 17. The rest after 17 is: [16,26,25,11,22,23,13,7,6,20,27,31] (starting with 16). So A25 must produce 16. Choose rule_57: A25 -> 16.

Thus A23 yields "17 16". Good.

Now after A23 we have still leftover: after 17,16 we have [26,25,11,22,23,13,7,6,20,27,31]. This should be generated by A51.

Recall A51 rule_102: A51 -> 26 A49. So A51 generates 26 then A49.

Thus we match 26. Good.

Now A49 must generate the remaining [25,11,22,23,13,7,6,20,27,31] (10 symbols). Let's see A49 rule_100: A49 -> A50 A44 A19 A31.

Thus A49 expands to A50, then A44, then A19, then A31.

Thus we need to produce sequence of 10 terminals: 25,11,... etc via those four nonterminals.

Let's break down:

- A50, rule_101: A50 -> 25 A13. So A50 yields "25" then A13.

Thus A50 matches 25, then A13 must generate next tokens: after 25 we have [11,22,23,13,7,6,20,27,31] (9 tokens). So A13 doit produce the prefix of that and A44 then etc.

But note we have another A13 earlier in the grammar; we must see if the A13 later can generate the required subsequence (maybe with branching). A13 alternatives: 11 OR A14 A15 (where A14=12). Since we need to produce 11 at some point, we could choose the direct 11 if we need to produce 11. Our remaining after 25 is 11, then presumably A44 must generate something like 22 A46? Let's see.

Because A44 can be either "21 A45" or "22 A46". After 25 we have 11 then "22". So likely we will use A13 -> 11 to produce 11, then A44 -> 22 A46 to produce "22" then A46 produce something that yields the rest: "23 13 7 6 20 27 31".

Thus A13 -> 11 works. Then A44's alternative rule_94: A44 -> 22 A46. Then A46 -> 9 or 23? Wait A46 has rule_96: A46 -> 9, rule_97: A46 -> 23. So we can produce either 9 or 23 after 22. However in our sequence after 22 we have 23 (immediately). So A46 should be 23. So we pick rule_97: A46 -> 23.

Thus A44 yields "22 23". That matches the sequence after 11: after 25,11 we have 22,23.

Now after A44 comes A19. Let's examine A19: rule_45: A19 -> A20 A3. So A19 yields A20 then A3.

Thus we need to generate the remainder after "22 23": after 22,23, the rest is [13,7,6,20,27,31] (six symbols). So A20 must produce a prefix and A3 produce suffix.

A20 rule_46: A20 -> 13 => yields terminal 13. So A20 matches "13". Good.

Now A3 must generate [7,6,20,27,31] (5 symbols). Let's examine A3 productions: rules_3-5: A3 -> 4 A4; A3 -> 5 A5; A3 -> 7 A6.

Thus A3 can start with 4,5,7. Our needed prefix starts with 7, so choose A3 -> 7 A6 (rule_5). So A3 yields "7" then A6.

Now A6 productions (rules_12-15): A6 -> 1 | 2 | 3 | 6. So A6 can produce either 1,2,3,6. Our remaining after 7 is [6,20,27,31]. So we need A6 produce 6. Choose rule_15: A6 -> 6. So A6 matches the terminal 6.

Now after A6 there's nothing else (A3 ends), but we have leftover [20,27,31] that must be produced by the following parts of the A49 expansion: after A19 (which gave A20 (13) and A3 (7 6) ), we have A31 (the final component of A49). So after A19, we have A31. So at this point we need to produce [20,27,31] via A31.

Recall A31 can expand to many options: A32-A43 (each produce 8 except A36 can also produce 20). Let's explore these.

We need the sequence: 20,27,31. A31 expands to a single nonterminal which expands to perhaps multiple terminals? Actually each A32-A43 is a nonterminal with a rule that yields a single terminal (except A36 also yields 20). So A31 can produce a single terminal: either 8 or 20 via A36. Not multiple. So we need to produce three terminals but we only have one nonterminal A31 left. This is a mismatch. However note that we earlier used A31 in A47 context and that worked because it only needed a single terminal (8). But now we need to produce three terminals after A19, which appears not possible.

But maybe A31 had multiple productions that produce some sequence? Let's re-check the production rules: 

- rule_68: A31 -> A32
- rule_69: A31 -> A33
- rule_70: A31 -> A34
- rule_71: A31 -> A35
- rule_72: A31 -> A36
- rule_73: A31 -> A37
- rule_74: A31 -> A38
- rule_75: A31 -> A39
- rule_76: A31 -> A40
- rule_77: A31 -> A41
- rule_78: A31 -> A42
- rule_79: A31 -> A43

Thus exactly one option for A31; each option leads to a nonterminal that yields a single terminal (or maybe two terminals in A36? Let's check rule_85: A36 -> 20 (in addition to rule_84: A36 -> 8). So A36 can produce either 8 or 20. So A31 can produce either 8 or 20 but not multiple terminals.

Thus we cannot generate three terminals (20,27,31) from a single A31. Something must be missing: perhaps there are other productions after A31? Actually maybe the A31 in A49 is not the last, maybe there are expansions after A31? Let's examine: A49 -> A50 A44 A19 A31. So after A31 there is nothing else; A31 must produce the rest. So we need the block [20,27,31] to be produced by A31. It cannot in current grammar: we'd need to insert terminal symbols into the RHS of some productions possibly to generate extra terminals. Since we are allowed to insert terminals into any RHS of existing rules, we could possibly insert terminal symbols after A31 within the A49 RHS to produce extra terminals (like 27 and 31). But we cannot rearrange or reorder existing symbols, but we can insert terminals anywhere. So we could propose modifications:

- In rule_100 (A49 -> A50 A44 A19 A31), we could insert terminal symbols after A31, or between A31 and something else? Wait A31 is the last symbol in the RHS; we can insert terminals after A31? The grammar's RHS is a sequence. Inserting after the last symbol would simply append terminals. That might generate extra required terminals after what A31 yields.

However we need to produce 20, 27, 31. A31 currently can produce 20 (option via A36). If A31 yields 20, we could insert after it we need 27,31. So we could add terminals 27 and 31 after A31 in rule_100, thereby matching these needed terminals. Let's examine if A31 can produce 20: select A36 -> 20 (using rule_85). So A31 -> A36 -> 20.

Thus rule_100 becomes: A49 -> A50 A44 A19 A31 27 31

But we cannot reorder, so we can insert terminals after A31. That yields output: ...[A31 output]27 31.

Now need to see how to get 20,27,31. This would produce them as needed.

Thus we need to modify rule_100: "A49 -> A50 A44 A19 A31" to "A49 -> A50 A44 A19 A31 27 31". However we need to insert terminal symbols 27 and 31. That adds 2 insertions.

But also must check the previous part: A31 should output 20. We need to ensure A31 can produce 20 using A36->20. Indeed A31 must choose the production -> A36 (rule_72). A36 may produce 20 using rule_85. That's allowed.

Thus the remaining sequence after A19 should be 20,27,31. A31 yields 20, then we have inserted 27 and 31 after A31, achieving the needed sequence.

Now check rest of derivation to see if any other mismatches. All matches so far.

Now check total sequence: after A49->... expansions, we have matched all block1 symbols. Then after that we have block2 32 from second A54, then 30 from rule_108 (in A56), then 35 from rule_108, then we have inserted 29 before final 34 via rule_109.

Thus final string: 36 then first A54 block (block1), then 30 then second A54 block (32), then 35, then inserted 29, then 34. That yields exactly the target.

Check ordering: After the second A54 (32), we have terminal 35 (from A56), then inserted 29, then final 34 (from A57). Indeed target is ...30 32 35 29 34. So correct.

Thus we need to make the following modifications:

- Insert terminal 29 before 34 in rule_109 (A57 -> 36 A56 34). This means modify rule_109 to add 29 somewhere before 34. The simplest is to insert after A56: "A57 -> 36 A56 29 34". That's one insertion.

- Insert terminals 27 and 31 after A31 in rule_100 (A49 -> A50 A44 A19 A31). Probably at the end: "A49 -> A50 A44 A19 A31 27 31". That's two insertions.

Potential other modifications: We must ensure no other mismatches. Let's double-check the production expansions fully, step by step.

1. Starting A1 -> A2 -> A57

2. A57 -> 36 A56 29 34 (modified)

Thus output: 36 ... A56 ... 29 34

We need after the 36 to produce block1 + 30 + block2 + 35 then 29, and then 34.

3. A56 -> A54 30 A54 35

Thus after the 36 we have A54 (block1) 30 A54 (block2) 35.

Thus the string is: 36 (block1) 30 (block2) 35 29 34.

Now block2 is from second A54. Next we have insert 29 before 34, correct.

Now block1 from first A54 using rule_106: A54 -> A55 A52 A21 A51

So block1 = output of A55 + A52 + A21 + A51.

Now compute A55: rule_107: A55 -> 33 A26. Output "33" then A26.

A26: rule_58: A26 -> A27 A28, A27->18 (rule_59), A28: either 15 A29 or 17 A30. To generate 18 15 1 we need A28 -> 15 A29, A29->1 (rule_62). So block A26 yields "18 15 1". So A55 yields "33 18 15 1". Good.

Now block after A55 ends with A26 ended. Next is A52.

A52 -> 28 A47 A53 => output "28", then A47, then A53.

Thus after A55's output we need "28". Indeed block after A55 is "28 ...". Yes.

Now A47 expansion: A48 A44 A13 A31.

A48: rule_99: A48 -> 24 A7. A7 -> A8 A9. A8->10. A9->5 A11 (or 4/A10, etc). A11->2. So A48 yields "24 10 5 2". Good.

Now A44: we need to output "21 9". Use rule_93: A44 -> 21 A45, A45->9: yields "21 9". Good.

A13: we need to output "12 4 3". Use A13 -> A14 A15. A14->12; A15->4 A16 (choose rule_33). Then A16->3 (rule_38). So yields "12 4 3".

A31: we need to output "8". Choose A31->A32, A32->8. Good.

Thus A47 yields "24 10 5 2 21 9 12 4 3 8". Combined with preceding "28" yields "28 24 10 5 2 21 9 12 4 3 8". Good.

Now A53 yields "19". So after A47 we have "19". Good.

Thus after A55 + A52 we have: ["33 18 15 1"] + ["28 24 10 5 2 21 9 12 4 3 8 19"] = the block up to 19.

Now A21: rule_47: A21 -> A22 A23.

A22->14

A23: we need to output "17 16". Use rule_50: A23 -> 17 A25; A25->16.

Thus A21 yields "14 17 16".

Now A51: rule_102: A51 -> 26 A49.

Thus after "...14 17 16" we need "26 ...". Indeed block after A21 is "26 25 ...". Good.

Now A49: A50 A44 A19 A31 + insert 27 31. We must produce from here: "25 11 22 23 13 7 6 20 27 31". Let's see if the RHS of A49 with the inserted terminals yields that.

A50 -> 25 A13. A13 must output "11". So A13 -> 11 via rule_30. So A50 yields "25 11". Good.

A44: we need to output "22 23". Use rule_94: A44 -> 22 A46; A46 -> 23 via rule_97. So A44 yields "22 23". Good.

A19 -> A20 A3. A20 -> 13. A3 -> 7 A6 (7 from rule_5). A6 -> 6 (rule_15). So A19 yields "13 7 6". Good.

Now A31 yields "20". Choose A31 -> A36 -> 20 (rule_85). So A31 yields "20". Up to this point A49 yields "25 11 22 23 13 7 6 20". Then we have inserted terminals "27 31" after A31: that yields "27 31". So total from A49: "25 11 22 23 13 7 6 20 27 31". Exactly matches remaining block.

Thus block1 from A55, A52, A21, A51 yields the target block1 matches exactly.

Everything seems consistent.

Now block2: we need A54's second instance to produce "32". Use rule_105 (A54 -> 32). Good.

Now after that we have 30 from A56. Then second A54 yields 32. Then 35 from A56, then the inserted 29 from A57 before final 34.

Thus the full derivation matches the target.

Thus the only necessary insertions are as identified: in rule_109 and rule_100.

Now we need to also confirm there are no other missing terminals: In A31 used multiple times: we used A31 both in A47 (produces 8) and in A49 (produces 20). It may be ambiguous but fine.

Now verify that A31's expansions allow required terminal sizes. Also check A36 also yields 8 per rule_84 and 20 via rule_85. Use accordingly.

Now we must produce the answer: List only updated production rules, i.e., rule_100 and rule_109, with inserted terminals.

We must write: "Updated production rules:" then a line for each modified rule: the rule number and new definition.

Specifically:

Original rule_100: "A49 -> A50 A44 A19 A31". Modified: "A49 -> A50 A44 A19 A31 27 31". In natural format: "rule_100: A49 -> A50 A44 A19 A31 27 31". (the grammar format uses "rule_<n>:" prefix.)

Original rule_109: "A57 -> 36 A56 34". Modified: "A57 -> 36 A56 29 34". So "rule_109: A57 -> 36 A56 29 34".

These are the only modifications.

But we should check if any additional insertions might be needed elsewhere due to similar mismatches. Let's verify the end of target after 35: target includes 29 and then 34. Yes, we inserted 29. Good.

Check if any other mismatches: ensure that after 35 we did not need any other terminals before 34. The target after 35 is "29 34", done.

Now after block2 we also have to verify that 30 is in the correct position. In the target, after block1 (first 30 symbols) we have "30" at position 31, correct. Then block2: 32 at position 32. Then 35 at position 33. Good.

Thus grammar now generates exactly the target.

Now ensure no other places need insertion of terminal symbols. For completeness, verify if any component that currently eliminates some terminal values appears missing. For example, rule_85 A36 -> 20 is needed for the "20". Our derivation used A31 -> A36 selecting rule_85. That's fine, with no modifications required, because A36 already includes both 8 and 20. So we used the appropriate production.

Check if any terminal such as 27 appears elsewhere: Actually we inserted 27 explicitly. Also important: check if A1->A2->A57 leads to start sequence only 36, then expansions. Yes.

Check if any terminal appears more times than provided in the grammar; all appear accordingly.

Thus minimal number of insertions: three total (one for rule_109 and two for rule_100). Could we do better? Possibly could avoid inserting 27 and 31 by using some other productions that already produce them. But we see we need to produce 27 and 31 sequentially somehow. There are no productions that generate 27 or 31 inside the existing grammar with current expansions in that location: Let's think: Could we possibly embed them inside some other nonterminal expansions? For example, A31 could produce 8 or 20, not 27 or 31. A36 could produce 20 only; can't generate 27/31. A33..A43 only 8. A31 doesn't have 27 or 31. In other nonterms, A23 etc produce only the set of numbers 14,15,17 etc. So 27 and 31 appear nowhere else as terminal productions. They only appear in the target, but not in any production. Indeed the grammar includes terminals ranges but many missing ones. To produce 27 and 31, we must insert them. Could we insert them elsewhere, e.g., in rule_52? There's no rule_52. Actually we can insert them in any rule's RHS, maybe we can embed them inside A31 expansion as e.g., A31 -> A32 27 31, but we can't because rule_68 defines A31 -> A32 precisely, and we cannot delete or reorder that, but we can insert terminals anywhere in RHS, i.e., we could change rule_68 to "A31 -> A32 27 31". However that would affect all uses of A31, including the earlier one for 8. But adding "27 31" after A32 would mean that in A47 case, A31 would produce "8 27 31". That would be extra terminals we don't want. Could we insert terminals into the other alternative productions of A31 that we don't use for the early case? However all A31 expansions are one rule each with only a single nonterminal. If we insert terminals only in one of the alternatives that is used only for the later case, we could avoid messing up early case. For instance, we can modify rule_72 (A31 -> A36) to "A31 -> A36 27 31", because in early case we used A31 -> A32 (rule_68). So the early case will not be affected. This would produce the need terminals after A36 only in the path using A31->A36. However we also need to produce 27 31 after the 20, which is exactly what A31->A36 27 31 would produce: A31->A36 27 31 yields (20) then 27 31. That would meet the requirement.

Nonetheless, we need to see that for the early A31 usage (A47 case), we used A31 -> A32 (rule_68). So rule_68 remains unchanged, producing 8 (no extra). Meanwhile rule_100 uses "A31" which could resolve to any of the expansions. In our earlier plan, we used A31->A36 (rule_72) to produce 20. If we modify rule_72 to "A31 -> A36 27 31", then we might not need to modify rule_100. Because the RHS of A49 is "A31", and A31 now expands to "A36 27 31". So A31 yields 20 27 31. So we would only need one insertion (in rule_72) instead of two in rule_100.

But we should verify that rule_72 is indeed the only rule for A31 -> A36. Yes. So we can modify rule_72: originally "A31 -> A36". Insert "27 31" after A36, making it "A31 -> A36 27 31". That adds two terminals. That's same number of insertions as adding two in rule_100. But rule_100 modifications would be two insertions. So both have same count. However perhaps there is a way to use fewer insertions: maybe we could modify A31 -> A36 (existing) to just "A31 -> A36" plus we could insert one terminal into rule_85 (A36 -> 20) after 20 to produce "20 27" then use something else to produce 31? Or you could insert both 27 and 31 after the A36's 20 via rule_85, i.e., modify rule_85 to "A36 -> 20 27 31". That would yield 20 27 31. Since A31->A36, and A31 hasn't changed. That also uses two insertions. However we could possibly produce 27 and 31 using expansions of other nonterminals such as A31 -> A37 (which we could modify to produce 20 27 31, but A37 currently => 8, cannot produce 20. So the minimal is two inserts somewhere.

Maybe we can use a combination: Insert 27 after A31, and 31 after something else, maybe earlier in a rule that already appears in both contexts, to share across. But we need to produce exactly one 27 and one 31 at that specific location (after 20). Could we amount to a single insertion if we modify rule_85 A36 -> 20 to "A36 -> 20 27 31"? That's two terminal insertions in that RHS (27,31). So it's still 2.

If we choose to modify rule_100 to insert "27 31" after A31, that's also two insertions. Could we minimize to one insertion by making a rule that uses a single terminal that includes both? No, cannot combine.

But we may consider that the 27 may be produced by some other free nonterminal accessible from A31 expansions, perhaps via A34->8? Not helpful. Could we get 27 from A31 -> A36 (20) and then by using A36 -> 20 or 20 27 and then insert 31 somewhere else later? No because after A31 there is no further symbols in that rule, unless we modify A57's RHS to insert something after 34? Wait after final 34 there is nothing else. So we would need to insert 31 before final terminal 34? Actually target after 31 is at index 30: after 27 there's 31 then 30 at index 31. Wait we need to map exactly ordering: the sequence after block1 is:

... [block1] [30] [32] [35] [29] [34].

But we have also 27 and 31 in block1? No, block1 includes 27 and 31 as part of the block after 20? Wait block1 includes 27 and 31 near its end. Let's re-evaluate block1's content: The block after 20 and before 31? Actually block1 includes "20,27,31"? Let's check: block1 after 20? The block1 includes 20 at index 28? Let's check earlier block1 list: block1 included ... after "7,6" we had 20, then 27, then 31. Indeed block1 ends with 20,27,31. So block1's tail includes 20,27,31 in that order. But we want the final part after A31->20 then we inserted 27 and 31. That is exactly within block1. So it's consistent.

But maybe we can modify rule_85 A36 -> 20 27 rather than add both to A31. Then we still need 31 after that. Could we produce 31 via A31's expansion as A31 -> A36 then maybe have A36->20 27 yields "20 27", and then we could use a following nonterminal (perhaps we could insert terminal 31 into rule_100 after A31). That would be one insertion (31) plus the insertion (or modification) of rule_85 to include 27 (one insertion). That's two insertions total again. So no reduction.

Thus minimal insertions necessary likely equal 3 (one for 29, two for 27 and 31). We have to see if we could produce 27 or 31 via some existing alternative in A31 expansions that yields those terminals. At present there is no rule for any nonterminal that yields 27 or 31. Let's scan all nonterminals for productions that yield these. Look through rules: terminals range from 1 to 36. Various nonterminals produce 1-3, 4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,28,32,33,36... They haven't got 27 and 31 nor some others like 34 35? Actually lines: rule_56? Wait we saw rule_57: A25 -> 16. So 16 is there. There's also 20 from A36. 21,22,23,24,25,26,28,32,33,36. There's no 27 or 31. Indeed missing numbers: term 27 appears only in target. Term 31 also appears only in target. Also terms 30 appears in A56 (explicit). Terms 29 appears nowhere except target. So we must insert three missing terminals: 27,29,31.

Thus minimal number of insertions is at least the number of distinct missing terminals that must appear in the derived string: 27,29,31. There may be others missing but they are already present: e.g., 33 appears via A55. 18 via A27. 15 via A28. 1 via A26/A6/A10? Actually 1 appears via multiple options, etc. 28 via A52. 24 via A48. 10 via A8. 5 via A9. 2 via A11/A16/A17/A18/A29/A30. 21 via A44. 9 via A45. 12 via A14. 4 via A15. 3 via A... timing. 8 via A32. 19 via A53. 14 via A22. 17 via A23. 16 via A25.A26 is 26 via A50 etc. 25 via A50. 11 via A13. 22 via A44. 23 via A46. 13 via A20. 7 via A3. 6 via A6/A28? Actually 6 from A6. 20 via A36. 27 and 31 missing, 29 missing. So minimal number of insertions = 3 (and each insertion is one terminal per operation). We inserted 3 terminals: 27, 31, and 29. So our solution matches minimal.

Thus the answer states updated production rules for rule_109 and rule_100 (or could pick rule_100 and rule_72). Since we need to insert three terminals: we might need to split the insertions between two rules anyway. The rules we have chosen: rule_100 and rule_109 with three insertions total (1 in rule_109, 2 in rule_100). That's minimal.

Alternative approach: Could we put one of the missing terminals into rule_109 (we must) and the other two into a single rule like rule_72 (A31 -> A36) by inserting two terminals there, giving also three insertions. That would be less modifications to different rules? Not necessarily better.

But which is more "minimal"? Both require three insertions total. We can choose either. The problem wants to minimize total number of inserted terminals, not number of rules changed. So both are okay. As we already inserted all three needed, good. The answer should reflect modifications.

Better to modify rule_72 (A31 -> A36) to include both 27 and 31, leaving A49 unchanged. That results in modifications: rule_72: add "27 31". And modify rule_109: add "29". That's three insertions as well. This uses only two rules to modify. The question says "write a new line for each of the updated production rules (including rule_<n>:), no need to rewrite the production rules which haven't changed." So we can choose to output modified rule_72 and rule_109.

We need to decide which modifications to present. Let's examine heating ordering: rule_100 currently: A49 -> A50 A44 A19 A31. If we modify rule_72 to produce extra terminals after A36, then in A49's A31 expansion it yields A36 27 31, so the string after A49 expansions becomes "25 11 22 23 13 7 6 20 27 31". That's exactly correct. So we wouldn't need to modify rule_100. That's simpler to modify fewer rules. But we'd still need to modify rule_72 and rule_109.

Thus total insertions: Insert 27 and 31 after A36 in rule_72; Insert 29 after A56 in rule_109 (or before 34). That yields 3 insertions.

Thus final answer: rule_72 and rule_109.

We need to ensure that rule_72: originally "A31 -> A36". We'll insert after A36: "27 31". So new RHS: "A36 27 31". So new rule: "rule_72: A31 -> A36 27 31". This is allowed: insert two terminals anywhere in RHS (here after A36). That implies new productions: A31 yields A36 then 27 then 31. So in the case where we need 20 (via A36), then we get "20 27 31". That matches the required suffix.

Now, we must ensure that this modification does not affect other uses of rule_72, i.e., any other place where A31 is used, which is only in rule_100 and also many other rules (68-79) produce A31 alternatives; if there is any other direct reference to A31 besides those rules (like from other expansions), they could also use rule_72 now and produce undesired extra terminals. Let's see: rule_68: A31 -> A32, rule_69: A31 -> A33, etc. Many alternatives also produce single terminals. The one we modified is rule_72. So there is no other direct derivation of A31 via rule_72 unless that rule is used. So we only affect those expansions that choose rule_72. The A49 uses A31, but may also choose other expansions for A31 (like A32 etc). But in our derivation, we need to produce "20 27 31", which is only possible via rule_72 (A31->A36 27 31). If we mistakenly used rule_68 or others for that occurrence, they'd just produce "8". So we need to ensure A49 selects rule_72 for A31. That's permissible because grammar is nondeterministic; we can choose that production.

Now check if rule_72 conflict with earlier usage of A31 in A47: there we used rule_68: A31 -> A32 to produce 8. That's unaffected because the rule we selected is different. So fine.

Thus modification of rule_72 will not break other parts.

Thus final modifications: rule_72 and rule_109.

Now, check if any other missing terminals still needed. After modifications, A31 (in A49) yields A36 (20) then terminals 27 and 31. Is 27 and 31 inserted in correct order? Need sequence: after block1's 20, then 27, then 31 (which matches). Good.

Now check the final part: after A31 expansion includes "20 27 31". Then following A57's "29 34". So final segment is "20 27 31 29 34". The target after block2 35 is "29 34". Wait queue: Actually after A56 there is 35 then A57's inserted 29 then 34. But we had block2 assigned A54 yields 32, then 35, then inserted 29, then final 34. Let's write string fully:

- 36 (A57)
- block1: [33 18 15 1 28 24 10 5 2 21 9 12 4 3 8 19 14 17 16 26 25 11 22 23 13 7 6 20 27 31]
- 30 (A56)
- block2: [32] (second A54)
- 35 (A56)
- 29 (inserted in A57)
- 34 (end)

Sequence order: let's write them sequentially:

[36,
33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31,
30,
32,
35,
29,
34]

Now compare with target:

target = [36,33,18,15,1,28,24,10,5,2,21,9,12,4,3,8,19,14,17,16,26,25,11,22,23,13,7,6,20,27,31,30,32,35,29,34]

Our derived has 30 and 32 swapped? Actually derived: 30 appears after block1, then 32, then 35, then 29, then 34. That matches target's ordering: after ...31 we have 30, then 32, then 35, then 29, then 34. Indeed check target: after ...31 (position 30) we have 30 (position 31), then 32 (position 32), then 35 (position 33), then 29 (position 34), then 34 (position 35). So ordering matches.

Thus our modifications produce exactly target.

Thus final answer: updated production rules for rule_72 and rule_109.

We need to output as:

Updated production rules:
rule_72: A31 -> A36 27 31
rule_109: A57 -> 36 A56 29 34

Now we need to ensure the formatting exactly uses "rule_<n>:" prefix. Yes.

Thus answer can be those two rules, no extra text.

Be careful to include newlines after each.

Thus final output should be:

Updated production rules:
rule_72: A31 -> A36 27 31
rule_109: A57 -> 36 A56 29 34

Double-check that rule_72 appears originally as rule_72: A31 -> A36. Yes.

Check rule_109 originally: A57 -> 36 A56 34. We'll modify.

Yes, all good.

Thus answer.

Updated production rules:
rule_72: A31 -> A36 27 31
rule_109: A57 -> 36 A56 29 34