We need to analyze the problem: There is a CFG with productions given, but some terminal symbols have been removed from the production rules, making it possibly impossible to generate the target sequence. The only allowed repair operation is to insert terminal symbols (i.e., integer terminals) anywhere in existing production rules. We cannot delete symbols, reorder symbols, or create new production rules. So we can only add terminals to the right-hand side (RHS) strings of the existing rules, possibly multiple insertions in a rule, to enable generation of the target sequence. The goal is to find a minimal total number of terminal insertions across all rules, such that the target sequence can be derived (maybe with some extra inserted terminals? Actually the target sequence must be produced exactly? It says "needs to be produced by the grammar". That means we need a derivation of the exact target sequence as a string of terminals (including all inserted terminals). The inserted terminals become part of possible expansions in the grammar, so they may appear in some derived strings. Our goal is to ensure that the target sequence can be generated. Some inserted terminals may be optional in the sense they might appear in other strings, but they must not prevent the derivation of exactly target sequence. The target sequence must be possible. Because we cannot delete symbols from rules, we have to ensure that after insertion, any derivation that produces target must not generate extra terminals beyond the target. However, we can always avoid using some productions (choose ones that produce the needed terminal sequence without extra unintended symbols). Inserted terminals only add optional expansions (i.e., any new instances). But they could appear in derivations that use the rule. We can choose not to use expansions that produce inserted terminals. So inserted terminals will not appear if we never apply those rules that contain them. However, if a rule originally had a nonterminal as RHS with no terminals, and we want to produce some terminal that is missing, we might need to insert it in that rule. But then that rule will always produce that terminal whenever applied. This might cause extra terminals inserted if that rule is used more times. But it's okay if we can incorporate those terminals into the target sequence order. However, the inserted terminal must appear at some location(s) in the derived string where they appear. So we need to figure out a derivation of the target sequence using existing rules possibly with some inserted terminals to fill missing terminals in the target or produce missing nonterminals that produce needed terminals.

We need to produce the exact target: 6 3 5 4 10 9 3 8 7 3 2 1.

Let's note terminals are integers 1..10 inclusive.

Given production definitions, note that currently many rules produce only terminal 3 or specific others. We also have ambiguous ones like A2 -> 2 and A2 -> 3 (two productions). Actually rule_2: A2 -> 2, rule_3: A2 -> 3. So A2 can produce either terminal 2 or 3.

A3 -> 3 and also A3 -> 6 A6.

Thus A3 can produce terminal 3, or produce 6 followed by whatever A6 can produce.

A5 -> 3 or ->5.

A6 has many productions: 3 (six times) and also A6 -> A4. So A6 can produce only a terminal 3 (multiple choices) or expand to non-terminal A4.

A4 -> 3 or -> A3 A5.

A7 -> 3 or -> 8.

A8 -> 3 or -> 10.

Thus A1 expands to A3 A8 A3 A7 A3 A2 1 according to rule_1. That's the only rule with A1 on LHS. So the derivation must start with this expansion.

Goal: produce exactly target sequence.

Let’s compute the original RHS: A3 A8 A3 A7 A3 A2 1.

Thus the string includes the terminals produced by each nonterminal and final terminal 1.

Terminal 1 appears at the end of the string (coming from the literal "1" in rule_1). That's good because target ends with ... 2 1. Indeed target ends with "2 1". So we have trailing 1 but need also the 2 before it.

Now we need to produce the sequence: 6 3 5 4 10 9 3 8 7 3 2 1.

Let's break down the target into segments associated with each nonterminal:

Segment for A3 (first): ??? we have A3 A8 A3 A7 A3 A2 1.

Thus we have 6 nonterminal expansions (A3, A8, A3, A7, A3, A2). The ordering of terminals contributed by each nonterminal must follow the sequence order.

Thus target is [segment1 from A3][segment2 from A8][segment3 from A3][segment4 from A7][segment5 from A3][segment6 from A2][terminal 1].

We must align.

Let's denote the segments as X1 X2 X3 X4 X5 X6, then terminal 1.

We'll try to assign.

Target: 6 3 5 4 10 9 3 8 7 3 2 1

Indices: 1:6,2:3,3:5,4:4,5:10,6:9,7:3,8:8,9:7,10:3,11:2,12:1

Thus the final 1 matches last token. So X6 must produce terminal(s) that end just before 1: possibly "2". Actually after X6 there is token 2 then 1.

Thus X6 must produce the substring "2". Good because A2 can directly produce 2 by rule_2. So X6: A2->2 yields terminal 2. Good.

Now X5 (the third A3) must produce substring that ends before the 2 token: That substring is token 10? Actually target segment preceding 2 (with token positions) is index 10: token 3. Actually X5 must produce token(s) at positions maybe index 10? Let's reconstruct:

Segments:

X1: positions 1..?
X2: ... 
X3: ...
X4: ...
X5: ...
X6: ... (should be 2). Let's assign after X5 the remaining tokens: Looking at sequence, after token 3 (position 10), we have 2 then 1. So X5 must produce token(s) ending at position 10, which is "3"? Actually position 10 is 3. So X5 could produce "3". That's plausible because A3 can produce terminal 3 directly (A3 -> 3). Good.

Thus X5 = A3 -> 3 works.

Now X4 (A7) must produce tokens position 8..9 (or maybe more) before X5's 3. Let's see: positions 8 is 8, 9 is 7. So maybe X4 must produce "8 7"? But A7 currently can produce either 3 or 8, not 7. But we can insert terminals into A7's rules. We can insert missing terminals after or before existing ones. We cannot reorder. So rule_17: A7 -> 3; rule_18: A7 -> 8. We could insert a terminal (like 7) into either rule. For example, we could change rule_18 to "A7 -> 8 7". That would produce 8 followed by 7. Order matters, we cannot reorder. Inserted after 8 yields 8 7 which matches positions 8 and 9. Good. Insert only one terminal (7). That seems plausible with minimal insert.

Alternatively, we could use rule_17 and insert "8 7" after 3? Not needed.

Thus likely we need to insert terminal 7 into rule_18: after 8.

Now X3 (second A3) must produce tokens from position maybe 5..7? Let's see after X2 then X3 then X4 (which is "8 7"), then X5 is "3". So X4 covers positions 8 and 9 (8,7). X5 covers position 10 (3). So X3 must produce tokens positions 5..7 preceding position 8. Let's see target has at positions 5:10, 6:9, 7:3. So three tokens: 10 9 3. So X3 must be able to generate "10 9 3". A3 currently can generate either "3" or "6 A6". A3 -> 6 A6 yields terminal 6 then whatever A6 expands to.

But we need "10 9 3". That's not directly possible. We need to see if we can insert terminals into A3's productions to achieve this. Possibly we could insert a terminal before or after 6 A6, or before 6 and before A6 etc.

Consider rule_5: A3 -> 6 A6. We could insert terminals in RHS of this rule. We can add terminals before 6, between 6 and A6, after A6. So we could turn this rule into something that yields a string that matches "10 9 3". For instance, we could insert "10 9" before 6? Actually "10 9 6 ..." not match. Insert after 6, before A6: "6 9 A6"? Then A6 could produce "3"? Actually A6 can produce 3 with its rule (A6->3). So "6 9 3". That's "6 9 3", but target wants "10 9 3". So not good.

Alternatively, we could insert before 6: "10 6 A6"? That yields "10 6 (? maybe 3)". Not matching.

Alternatively we could change A3's other rule "A3 -> 3" by inserting terminals to get "10 9 3"? Since we can only insert terminals anywhere, we could add "10 9" before 3, making "10 9 3". That would be perfect: rule_4: A3 -> 3, modify to "A3 -> 10 9 3". That would produce the needed sequence of three terminals. Inserting two terminals (10 and 9) yields "10 9 3". That's minimal? Let's check if we need also use other productions later for other A3 expansions that must produce the other needed tokens (X1 and X5). However, note we have multiple A3 occurrences: X1 (first), X3 (second), X5 (third). We need to allocate terminal sequences to each:

- X5: we used A3->3 (no insertion needed). However, if we change that rule to produce "10 9 3", then X5 would produce that sequence incorrectly. But we can still use the rule A3->6 A6 for X5 (maybe). Actually we have A3 rule A3 -> 3 (rule_4) and A3 -> 6 A6 (rule_5). Both exist. We can choose which rule to use for each occurrence. So we could restructure as follows:

X5 (third A3) -> use rule_4 (A3->3) unchanged.

X3 (second A3) -> use rule_4 with inserted terminals "10 9 3". So we modify rule_4 by inserting terminals before the existing "3". That would also affect any other use of rule_4 (like X5) since they share the same production rule. Changing rule_4 globally adds those inserted terminals to all uses of that rule, which will cause X5 to also expand to "10 9 3". That is not allowed because X5 must be "3". However, we could use the other rule for X5 (rule_5) to generate the needed "3" using A3 -> 6 A6 with inserted terminals? But rule_5 currently yields "6 A6". We can insert terminals to make it produce only "3"? Not possible directly.

Alternatively, we could modify rule_5 to produce "3" by inserting a terminal to offset? But we can't delete the existing "6". So any rule that includes 6 will always output 6 somewhere.

Thus we must consider alternative: Use rule_5 for X5 and insert terminals such that the expansion yields "3". But rule_5 RHS "6 A6" always outputs 6 (can't delete), so the string will start with 6. Could produce "6 3" by using A6->3. That gives "6 3". But we need just "3". That's not okay. Could also insert terminals before 6 to produce "something 6 ...", but still keep 6 in the string.

Thus we cannot get X5 exactly "3" if we modify rule_4 to produce more than just "3". However we could choose to use rule_5 for X5 and change rule_5 maybe to produce "3"? We can't delete 6. So impossible.

Thus we must keep rule_4 as "3" or adjust it only by inserting terminals either before or after 3 such that we still have a derivation for X5 that results in just 3. But if we insert both 10 and 9 before 3, then any use of that rule yields 10,9,3. We can't have X5 be just 3. So we need a different solution.

Maybe we can assign the three tokens "10 9 3" across multiple nonterminals, not just a single A3. Perhaps use A8 to produce "10"? Actually A8 can produce 10 (rule_20). Also A8 could be expanded to something else maybe with insertion to produce "10"? Already has rule: A8 -> 10. That's good. We have A8 as second symbol in A1's RHS, corresponding to X2.

Given target, X2 must produce tokens after X1 and before X3. Let's allocate accordingly.

Sequence: X1 (A3) ... X2 (A8) ... X3 (A3) ... X4 (A7) ... X5 (A3) ... X6 (A2) ... 1.

We have the target: 6 3 5 4 10 9 3 8 7 3 2 1.

We need to split into 6 segments. Let's map possibilities:

Since A8 can produce 10 (or 3, but we can insert), A7 can produce 8 (or 3) etc.

Observe that target includes 10, 9, 3 in order after token 4? Actually target positions: 1:6,2:3,3:5,4:4,5:10,6:9,7:3,8:8,9:7,10:3,11:2,12:1.

Thus we see the subsequence 10,9,3 appears at positions 5,6,7. We could assign that to X2 and X3 perhaps.

Consider X2 (A8) could produce 10 (via rule_20). After X2, next X3 must produce tokens 9,3? Actually after 10, we need 9 then maybe 3 is part of X3 or X5 or something. Let's see.

If X2 = A8 -> 10 (good). Then target remaining after X2 is "9 3 8 7 3 2 1". X3 = A3 must start at next token 9.

Thus we need A3 to produce "9 3". Currently, A3 cannot produce 9 directly. A3's productions are either "3" or "6 A6". We could insert terminals into the rule to produce "9 3". For example, modify rule_4 by inserting "9 " before the 3? That would cause "9 3". However, if we modify rule_4, all uses of rule_4 produce "9 3". But we also need one of the A3s to produce maybe "6 3"? Actually maybe X1 needs to produce "6 3". Let's examine.

Potentially X1 (first A3) could produce "6 3" using rule_5: A3 -> 6 A6 and choose A6 -> 3. That gives "6 3". So X1 can be satisfied without insertions (if we use rule_5 for A3). Good. Then X5 (third A3) must produce something at the later part of target: what tokens after X4 (A7) and before X6 (A2)? The target after position 9 (7) is token 10 (3) then token 11 (2) then 1. So X5 must produce token 10: "3". So X5 can use rule_4 (A3->3) unchanged. So that works.

Thus X1 = A3 -> 6 A6 (choose A6->3) produces "6 3".

X5 = A3 -> 3 (rule_4) produces "3".

Now X2 must produce "10". Use A8 -> 10. Good.

X3 must produce "9"? Actually we need "9" then after that X4 = A7 produce "8 7"? Actually A7 will produce "8 7" after we insert 7. That yields tokens "8 7". So after X4 we need token "3"? Actually target after X4's part "8 7" we have token "3" then then X5-> "3". But wait target currently: after positions 8 and 9 (8,7), token position 10 is 3 (this could be from A7? Actually A7 gave 8 and maybe 7, but not 3). Let's examine.

Wait re-evaluate mapping:

Full target: "6 (pos1) 3 (pos2) 5 (3) 4 (4) 10 (5) 9 (6) 3 (7) 8 (8) 7 (9) 3 (10) 2 (11) 1 (12)".

We have A1's RHS: A3(1) A8 A3(2) A7 A3(3) A2 1. So we need to align six segments.

Let's define the segment positions.

Segment 1 - from A3(1) - could produce "6 3". That covers pos1-2.

Segment 2 - from A8 - could produce "5 4 10"? No we need to see.

Potentially we might allocate "5 4 10 9 3 8 7" across segments; but we need to allocate systematically.

Given there are no productions for terminals 5,4,9 (apart from A2->2, A5->5, etc), we will need to insert missing terminals into rules. But we also have nonterminals that can produce 5 (A5->5), and 4 perhaps via something else? Actually we have no rule that directly produces 4 or 9. The only productions that produce fixed terminals are:

- A2 -> 2 ; A2 -> 3.
- A3 -> 3 ; A3 -> 6 A6.
- A4 -> 3 ; A4 -> A3 A5.
- A5 -> 3 ; A5 -> 5.
- A6 -> 3 ; A6 -> A4.
- A7 -> 3 ; A7 -> 8.
- A8 -> 3 ; A8 -> 10.

Terminals present: 1 (explicit in start rule), 2,3,5,6,8,10.

Missing terminals 4,7,9 are not produced by any rule, so we must insert those via insertions.

Goal: to minimize insertions. So we want to reuse existing terminals as much as possible. The missing terminals are 4,7,9. So at least three insertions needed (one for each missing terminal) unless we can make a single insertion produce multiple missing terminals (perhaps by inserting "4 7 9" as a substring somewhere in a rule). Since each insertion inserts a single terminal symbol at some position, to get three distinct missing terminals, we need at least three insertions. However, if we can insert multiple terminals in same rule but counts as multiple insertions (still adds each terminal). So minimum is at least three insertions.

But also there may be the need to insert terminals like 5, which is already present via A5->5. However, where to use A5? Currently A5 is used only via A4 -> A3 A5 (or maybe indirectly via A6->A4). So we could get terminal 5 from A5. Also we need terminals 5 and 4 and 9 and 7 and maybe 8? 8 is already produced by A7->8 (present). So 8 is okay. 5 is already available. 6 is via A3->6 A6.

Thus we need to generate sequence "5 4" and "9" somewhere.

Potentially we could generate "5 4" using A3 -> 5? No A3 doesn't produce 5. Maybe we can insert 5 or 4 into A3 rule, but note 5 is already present elsewhere, but the required order is 5 after 3 (pos2), i.e., target pos3=5, pos4=4. Let's examine. After segment 1 producing "6 3", we need to produce "5 4". These could be produced by some subsequent nonterminals.

Given A1's RHS: after first A3 (which gave 6 3) we have A8. So A8 must produce the next tokens: "5 4"? Well A8 can currently produce either "3" or "10". That's not helpful. Could we insert "5 4" into A8's rule to produce "5 4"? That would require two insertions (or more). But maybe we can also use other nonterminals to produce these: The next nonterminal after A8 is second A3. So maybe A8 produce something like empty? But we cannot delete the literal "3" or "10". So changes must produce "5 4" and next tokens.

We could choose which production for each nonterminal to use: e.g., A8 -> 3 or A8 ->10. Could we insert terminals around these to generate "5 4 10"? For instance, modify rule_19 (A8 -> 3) to be "5 4 3". Then A8 would produce "5 4 3", but we need "5 4" maybe not "3". But we might then need the "3" from somewhere else: The target after "5 4" is "10". Actually target: positions: 1:6,2:3,3:5,4:4,5:10,... So after "5 4", we need a "10". So maybe A8 could produce "5 4 10"? Insert 5,4 before 10 rule. Use A8 -> 10, insert before 10: "5 4". That would produce exactly "5 4 10". Then the next A3 (second A3) must produce "9 3". This could be accomplished by inserting 9 before 3 in rule_4. Then A3 (second) would produce "9 3". Then third nonterminal A7 would produce "8 7"? We'll need to insert 7 into rule_18. Then third A3 (third) produces "3". Then A2 produces "2". Then final 1 from start rule.

Thus mapping:

- X1 (first A3): use A3 -> 6 A6, with A6 -> 3 => "6 3".

- X2 (A8): use A8 -> 10 but with insertions before 10 to output "5 4". So rule_20 (A8 -> 10) becomes "A8 -> 5 4 10". That's two insertions: terminals 5 and 4 inserted before 10.

- X3 (second A3): modify rule_4 (A3 -> 3) by inserting 9 before 3, making "A3 -> 9 3". So it outputs "9 3". That's one insertion (9).

- X4 (A7): modify rule_18 (A7 -> 8) by inserting 7 after 8, making "A7 -> 8 7". That's one insertion (7).

- X5 (third A3): maybe use A3 -> 3 (rule_4) but note that rule_4 now has insertion of 9 before 3. So any use of rule_4 will produce "9 3". That's not what we want for X5 (need just "3").

Thus we need a workaround: Use a different rule for X5. Options: Use rule_5: A3 -> 6 A6. However that would output "6..." and need to produce just "3". Could we change rule_5 to output just "3"? We could insert a terminal before 6, after 6, or after A6, but we cannot delete 6. That introduces a mandatory 6. Not ok.

Alternatively, we could keep rule_4 unchanged and create a separate rule for A3 with inserted 9 only for second A3, but we cannot create new productions. All A3 expansions share same two productions. If we modify rule_4, must affect all uses of that rule. However we can also use rule_5 for X3 and modify it accordingly to produce "9 3". Let's explore: Could we make A3 -> 9 6 A6? That is insert 9 before 6. Then the production becomes "9 6 A6". With A6 -> 3, we get "9 6 3". That would be "9 6 3", not "9 3". Not match.

Could we produce "9 3" via rule_5 with insertion after 6 and before A6? E.g., rule_5: "6 9 A6". Then with A6->3, we get "6 9 3". Not match.

Could we hide the 6 by customizing A6 to produce epsilon? But we cannot delete symbols.

Alternatively, we could use a different rule for X5: rule_4 unchanged (A3 -> 3). And modify rule_4 to keep extra insertion but that would affect all uses, but perhaps we can decide to use rule_5 for X5 and modify rule_5 to produce "3"? That's not possible.

Therefore perhaps we need to assign "9" not to A3 but to some other nonterminal. Maybe we can have X3 (second A3) produce "3" (no insertion) and produce "9" via some other nonterminal inserted somewhere else like A6, or use A4 etc. Let's examine the other nonterminals that can be introduced via expansions.

Between A8 and A3 (X3), we have A8 and then A3 again. Could we have A3 produce just "3". Then we need to generate "9" somewhere else before the next "3". The next of target after "5 4 10" is "9 3". So we need "9" then "3". The "3" can come from this second A3 (producing 3). The "9" could be inserted into A8's production (i.e., we could have A8 produce "5 4 9 10". However target order is "5 4 10 9 3"? Actually check target again: after "5 4", we have "10", then "9", then "3". So we need "10" before "9". So "9" cannot be before 10. So maybe we need to insert 9 after 10, maybe inside A8's production after 10? So A8 -> 10 9 (insert after 10). Then the subsequent A3 produces 3. So that yields "5 4 10 9 3". That matches needed tokens. So we can get the "9" from insertion after 10 in rule_20. Then we don't need to modify rule_4 at all. Good! Let's verify.

Thus refine mapping:

- X1 (first A3): A3 -> 6 A6 (with A6 -> 3) => "6 3". Good.

- X2 (A8): use rule_20: A8 -> 10 (originally produced 10). Insert terminals before 10 ("5 4") and after 10 ("9") maybe. So final RHS: 5 4 10 9. That yields tokens "5 4 10 9". Good. That's three insertions (5,4,9). But note 5 is also produced by A5->5, but we need literal 5 anyway. Inserting 5 in A8's rule costs one insertion. That's fine.

- X3 (second A3): use rule_4 (A3->3) unchanged => "3". Good (position after 9 is "3").

- X4 (A7): need to produce "8 7". Already produce 8. Insert 7 after 8 in rule_18: "A7 -> 8 7". That's one insertion.

- X5 (third A3): need to produce "3" (target position 10). Use rule_4 (A3->3) unchanged => "3". Good.

- X6 (A2): need to produce "2". Use A2 -> 2 (rule_2). Good.

- final "1": literal from start rule. Good.

Thus overall target is produced as: A3(1) = "6 3", A8 = "5 4 10 9", A3(2) = "3", A7 = "8 7", A3(3) = "3", A2 = "2", "1". Concat gives: 6,3,5,4,10,9,3,8,7,3,2,1. Exactly matches target.

Now let's count insertions: we inserted 5,4,9 into rule_20 (three insertions) and 7 into rule_18 (one insertion). So total insertions = 4.

Can we reduce insertions further? Since the missing terminals are 4,7,9. But we also inserted 5. Is 5 missing from any rule? 5 is produced by A5->5, but we don't have an A5 in RHS. Could we produce "5" by a different approach with existing nonterminals without inserting a 5? Let's examine existing nonterminals and their productions for producing 5. A5 -> 5 is a direct production. However, the current derivation uses A5 maybe indirectly via A4 -> A3 A5. But A4 is not part of the start rule's RHS. However, A6 can produce A4. And A3 -> 6 A6. So we could potentially produce the "5" by using the existing nonterminal A5 through a chain, without needing to insert 5. Let's explore.

The segment after A3(1) is A8 (where we need "5 4 10 9"). Could we restructure to incorporate the "5" via an existing A5? For instance, we could make A8 produce "4 10 9" and then produce "5" from some other nonterminal, maybe A4 -> A3 A5. But we have a fixed ordering of nonterminals: after A3(1) we have A8, then A3(2), A7, A3(3), A2, 1. There's no place to insert A4. We cannot reorder production or add new non-terminal expansions unless we insert them into the RHS of existing productions. The only insertion we can do is adding terminal symbols anywhere within existing RHS strings. We cannot add new nonterminals to the RHS, only terminals. So we cannot introduce an A5 at that point. So to generate "5", we must either use an existing rule that yields 5 (like A5->5) but there is no A5 in the start rule's RHS. However, we can perhaps modify A8's rule to produce "5" via indirect means? We cannot add nonterminals. So we need to insert terminal "5" explicitly. So we must insert 5 somewhere.

But maybe we can produce "5" as part of A8's rule by using the rule A8 -> 3, and injection of a nonterminal "A5"? Not allowed. Only terminals can be inserted. So we must insert 5.

Thus minimal insertions likely include one for each missing terminal 4,7,9, plus maybe 5 if we can't get it elsewhere. But 5 is missing from the target? It's present and present in grammar (A5->5). However we cannot get A5's production in this derivation because there is no A5 in the start rule or any of the nonterminal expansions we can use (except through A4->A3 A5 etc). But maybe we can restructure the derivation to incorporate A5 by using the fact that A6 can go to A4 via rule_16. A3(1) uses rule_5: A3 -> 6 A6. Then A6 -> A4 (instead of 3) yields A4. Then A4 -> A3 A5. Then that could produce A5 (terminal 5). That would shift the order though.

Let's examine if we can produce the "5" via A3(1) expansions, not via A8. Maybe we could modify the first A3 to produce "6 3 5"? Actually we need "6 3 5 4 ..." in target. After "6 3" we have "5". So first A3 could produce "6 3 5"? But the rules for A3: either "3" or "6 A6". From "6 A6" we could have A6 -> A4 -> A3 A5. That yields "6"+"(expansion of A4)". If A4 -> A3 A5, then A5 -> 5. And A3 again can produce "3" or "6 A6". We could shape it to produce "6 3 5". Let's try: A3(1) -> 6 A6. Then A6 -> A4. A4 -> A3 A5. In this, the inner A3 could produce "3". A5 -> 5. So overall: 6 + (then A4 expands) => A3 (->3) A5 (->5) => yields "6 3 5". Indeed that yields the sequence "6 3 5". This matches the first three tokens: 6 (pos1), 3 (pos2), 5 (pos3). Actually target has "6 3 5 4...". So we need also "4". So maybe we can generate "4" after that using another insertion in one of the productions.

So we could perhaps generate the "5" embedded using existing nonterminals and thus avoid inserting a 5.

But note that to obtain "6 3 5", we used the second A3 inside A4, and that A3 must use rule_4 (->3). That's fine. So the first A3 (outer) uses rule_5 (6 A6). That's okay.

Now we also need "4" after this "5". The next nonterminal in the start rule is A8, so after we finish expanding A3(outer) we get to A8, which must produce "4 10 9 ..."? Actually target after "5" is "4". So we could insert 4 before the output of A8 (i.e., in rule_20 or rule_19). So we may need to insert 4 anyway. So at least one insertion for 4. However we might be able to generate the "4" using a similar trick with A6/A4 expansions that produce maybe A5->something else? There is no rule to produce terminal 4. So we have to insert a 4 somewhere. So that's one insertion.

Now the "9" appears after "10". So maybe we have to insert 9 as before? Let's see if we can get "9" via some other nonterminal expansions: maybe via A2? A2 can generate 2 or 3. No.

A5->5, A5->3 as alternative. Not 9. A6->3. Not 9. A3->6 A6 (could generate 9 if inserted). So we need to insert 9 somewhere.

Thus minimal insertions needed: 4,7,9. That's three. Possibly we can also get 5 without insertion if we use A5 as described, thus reducing total to 3. Let's test if we can generate target using 3 insertions: 4 inserted somewhere (maybe in A8), 7 inserted after 8 in A7, and 9 inserted in A8 after 10. No insertion of 5 needed as we embed a "5" using A5. Let's see if we can embed 5 seamlessly in the derivation.

We need to produce the exact sequence:

Positions:
1: 6
2: 3
3: 5
4: 4
5: 10
6: 9
7: 3
8: 8
9: 7
10:3
11:2
12:1

We need to assign production breakdown across start rule's RHS: A3 A8 A3 A7 A3 A2 1.

We need to map the token sequence into those segments.

Segment 1 (from first A3) can generate "6 3 5"? That's possible via A3 -> 6 A6 -> 6 A4 -> 6 A3 A5 -> Expand inner A3->3, A5->5 gives "6 3 5". That yields first three tokens. That uses no insertions. Then we need next token "4" to be from A8's expansion. So A8 must generate "4 10 9"? Actually after "5" we have 4, then 10, then 9, then 3 (which is from second A3). So A8 should produce the sequence "4 10 9". That would be three terminals. Currently A8 can produce "10" (rule_20) or "3" (rule_19). We can insert terminals before or after 10 to get "4 10 9". We need to insert "4" before 10 and "9" after 10. Two insertions. So now we've inserted 4 and 9.

Segment 2 (second A3) must produce "3". That's fine via rule_4 unchanged.

Segment 3 (A7) must produce "8 7". Insert 7 after 8 in rule_18. That's one insertion.

Segment 4 (third A3) must produce "3". That's unchanged.

Segment 5 (A2) produces "2". That's unchanged.

All good.

Thus total insertions are 4 (4,9,7). Wait we inserted 4 & 9 (2 insertions) + an insertion of 7 (1). So total 3 insertions? No, we also inserted 4 and 9 = two, plus 7 = one, total 3? Let's count precisely:

- Insert 4 before the 10: one insertion.
- Insert 9 after the 10: second insertion.
- Insert 7 after 8: third insertion.

Thus total of 3 insertions. Good. So we improved earlier solution that had 4 insertions (5,4,9,7). Now we have inserted only 3. Good.

Now we need to verify that with the above changes we can generate the target.

Let's propose modifications:

- rule_5: A3 -> 6 A6 (unchanged). This will be used for outer A3 (first segment). No change needed.

- rule_16: A6 -> A4 (unchanged) to route to A4.

- rule_4: A3 -> 3 (unchanged) to produce inner A3 for "3" inside the first segment.

- rule_8: A4 -> A3 A5 (we need to produce "3 5"? Actually we want the inner A3 to produce "3" then A5 produce "5". So rule_8 unchanged.

- rule_9: A5 -> 3 (unused) and rule_10: A5 -> 5 (which we need). Actually rule_9: A5 -> 3, rule_10: A5 -> 5 (makes sense). So inner A5 should use rule_10. No change required.

Thus first segment generation: A3 (outer) -> 6 A6 -> 6 A4 -> 6 (A3 A5) -> A3 -> 3 [rule_4], A5 -> 5 [rule_10] => "6 3 5". Good.

Now second segment (A8) using rule_20: A8 -> 10. Need to insert 4 before 10 and 9 after 10. So new RHS is "4 10 9"? Or "4 10 9"? Yes.

We cannot change the literal ordering of 10, but can add terminals before or after. So we change rule_20 to: A8 -> 4 10 9. That's two insertions.

Now third segment (A3) after A8: this must generate "3". Use rule_4 unchanged.

Next A7 -> we need "8 7". Currently rule_18: A7 -> 8. Insert 7 after 8: new RHS "8 7". That's one insertion.

Next A3 (third) must generate "3": rule_4 unchanged.

A2 -> must generate "2": rule_2 unchanged (A2 -> 2).

Then final 1 literal from rule_1.

Thus full derivation:

Start: A1 -> A3 A8 A3 A7 A3 A2 1.

- Expand first A3 (call it A3_1) via rule_5: 6 A6.
- Expand A6 via rule_16: A4.
- Expand A4 via rule_8: A3_2 A5.
- Expand A3_2 via rule_4: 3.
- Expand A5 via rule_10: 5.

Now we have "6 3 5" generated, correspond to tokens 1-3.

Now expand A8 via modified rule_20: 4 10 9.

Now tokens "4 10 9" appended. So tokens 4-6.

Now second A3 (A3_3) via rule_4: 3. Token 7.

Now A7 via modified rule_18: 8 7. Tokens 8-9.

Now third A3 (A3_4) via rule_4: 3. Token 10.

Now A2 via rule_2: 2. Token 11.

Then final terminal "1". Token 12.

Thus final string: 6 3 5 4 10 9 3 8 7 3 2 1, matching target exactly. No extra tokens. Great.

Thus we have a solution with 3 insertions total: a 4, a 9, a 7 (three terminals). Inserted terminals: 4, 9, and 7. All other needed tokens are generated by existing productions.

Check missing terminals: target includes 5 (generated via A5->5), 6 ( via rule_5 ), 3's (via many rules), 8 (via A7->8 originally), 10 (via A8->10), 2 (via A2->2), 1 (via start rule), all covered.

Thus minimal insertions appear to be 3: insert 4 and 9 into rule_20, and insert 7 into rule_18. Could there be a solution with only 2 insertions? Possibly combine 4 and 9 into the same rule insertion making a single insertion? Each insertion inserts a single terminal symbol at a position; you cannot insert multiple terminals in a single insertion. So at least we need to insert 4 and 9 (two separate insertions) because they are distinct terminals. There is also a 7 missing, requiring another insertion; thus at least 3 insertions total. So this is minimal.

Thus final updates:

- rule_20: original right side "10". After insertion, becomes "4 10 9". Inserted two terminals: 4 before, 9 after.

But we need to be careful about representation: we must show the updated rule with the terminals in appropriate order. So rule_20: A8 -> 4 10 9. Or "A8 -> 4 10 9". I think we should keep original rule indices unchanged but with modifications.

- rule_18: original A7 -> 8. Change to "A7 -> 8 7".

Thus we must output:

Updated production rules:
rule_18: A7 -> 8 7
rule_20: A8 -> 4 10 9

Check that rule_20 is indeed original rule_20, which in the given grammar is:

rule_20: A8 -> 10

Thus we modify that to "A8 -> 4 10 9". Insert 4 before 10 and 9 after.

Make sure we didn't also need to adjust any other rule? No. All other rules remain same and are sufficient for derivation.

But need to consider that rule_5: A3 -> 6 A6 may need to produce the outer 6 and then A6 to produce A4, which we have; A6 produces A4 via rule_16. That's fine. No insertion needed there.

We should double-check that the grammar can indeed generate "6 3 5". Let's simulate step by step clearly.

We start derivation as mentioned. Let's do details:

1. A1 -> A3 A8 A3 A7 A3 A2 1

Take first A3 => use rule_5: A3 -> 6 A6, yields 6 A6.

So far string: 6 A6 A8 A3 A7 A3 A2 1.

Now expand A6 with rule_16: A6 -> A4.

String: 6 A4 A8 A3 A7 A3 A2 1.

Now expand A4 with rule_8: A4 -> A3 A5.

String: 6 A3 A5 A8 A3 A7 A3 A2 1.

Now we need to decide which expansions for these A3 and A5.

- Expand the inner A3 (the one from A4) with rule_4: A3 -> 3.

String: 6 3 A5 A8 A3 A7 A3 A2 1.

- Expand A5 with rule_10: A5 -> 5.

String: 6 3 5 A8 A3 A7 A3 A2 1.

Now we have 6 3 5 as required.

Now A8 (which is after those three). Expand A8 via rule_20 modified: 4 10 9.

String: 6 3 5 4 10 9 A3 A7 A3 A2 1.

Now next A3 (second in start) expand using rule_4: 3.

String: 6 3 5 4 10 9 3 A7 A3 A2 1.

Now A7 with rule_18 modified: 8 7.

String: 6 3 5 4 10 9 3 8 7 A3 A2 1.

Now third A3: rule_4: 3.

String: 6 3 5 4 10 9 3 8 7 3 A2 1.

Now A2: rule_2: 2.

String: 6 3 5 4 10 9 3 8 7 3 2 1.

Thus final string matches target.

So the solution works.

Now minimality: we inserted three terminals (4, 9, 7). Could we embed the 9 somewhere else using existing nonterminals? Possibly using A6 expansions: A6 -> A4 -> ... but then would require terminal 9 somewhere else. However, there's no natural 9 generation. So we need to insert 9 somewhere. So 3 is minimal.

Thus final answer: we list updated production rules for rule_18 and rule_20, with insertions accordingly.

Potentially also note that rule_5 and rule_16 are used. They remain as originally.

Thus final output:

Updated production rules:
rule_18: A7 -> 8 7
rule_20: A8 -> 4 10 9

Make sure to settle on correct formatting: each line with rule id and colon then space then left-hand side -> RHS.

Thus answer.

But check if any other rule must be updated because we inserted two terminals before and after 10. Insert 4 before 10, 9 after. No other modifications.

Make sure no other rule inadvertently requires insertion: we also inserted "4" before "10", but "4" does not appear elsewhere, but that's fine.

Thus output as required.

One more nuance: In rule_20 we have to ensure that it reflects the original integer tokens and preserve order. So "A8 -> 4 10 9" is okay.

Thus final answer.

Now final message.

Updated production rules:
rule_18: A7 -> 8 7
rule_20: A8 -> 4 10 9