We have a CFG with many production rules. Some terminal symbols are missing due to removal, making sequence impossible to produce. We're allowed to insert terminals (terminal symbols are integers from 1 to 36) into existing production rules anywhere (i.e., in any RHS we can add terminals at any position). We cannot delete or re-order symbols, nor add new productions. Our goal is to minimally insert terminals to allow the target sequence (which is a long sequence of terminals) to be derived from start symbol A1.

We need to decide where to insert terminals to match the target sequence.

First, let's parse the grammar and target sequence.

Terminals: integers 1..36.

Target sequence: given as:

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

Let's just write them with indices:

1: 13
2: 12
3: 2
4: 36
5: 1
6: 25
7: 8
8: 2
9: 35
10: 1
11: 24
12: 7
13: 2
14: 34
15: 1
16: 6
17: 5
18: 32
19: 33
20: 31
21: 4
22: 3
23: 22
24: 30
25: 21
26: 15
27: 14
28: 22
29: 29
30: 21
31: 26
32: 9
33: 2
34: 28
35: 1
36: 11
37: 10
38: 22
39: 27
40: 21
41: 23
42: 16
43: 22
44: 20
45: 21
46: 19
47: 18
48: 2
49: 17
50: 1

So target sequence length = 50.

We need to derive exactly this sequence from A1.

Given current grammar: The start rule A1 produces the sequence of non-terminals: A45 A42 A40 A38 A33 A28 A25 A19 A13 A2 (10 non-terminals). So derivation tree yields expansion of each non-terminal, possibly concatenating each of their expansions in order.

Thus the order of terminals produced will correspond to concatenation of expansions of those sub-non-terminals in this order.

Goal: The final terminal string (order) must match target sequence exactly.

We can insert missing terminals into RHS of any rule, but order cannot be changed. So we need to adapt each rule to produce exactly the needed terminal subsequence for that part.

Idea: Determine a parse tree that matches target sequence, where each rule fits. Then any missing terminals are inserted.

We need to find for each non-terminal A_i a pattern of terminal substrings; where that non-terminal appears in the production of other non-terminals; must match the target.

This is a constraint satisfaction problem: mapping each non-terminal to a substring of target sequence. According to grammar structure, we can compute expansion of each non-terminal based on its productions. Since many non-terminals have multiple productions (e.g., A2 has four alternatives). So we can choose among alternatives.

Since insertion is allowed, any rule's RHS may be extended by adding terminals. So any generation may produce extra terminals that we need to match the target in order.

Goal: Minimize total inserted terminals across all rules used.

We need to choose rules and insert minimal extra terminals.

Important: Insertion allowed anywhere in RHS, the number added (count) adds to total inserted.

We must produce exactly the target sequence; no extra terminals (i.e., any inserted must be exactly those needed). So we need to fill all missing terminal symbols.

But what does it mean that terminal symbols were removed? Probably some RHS originally contained some terminals but were removed, leaving parts incomplete. Now we need to reinsert them. We want minimal inserts.

Thus we need to find which terminals appear in some RHS originally but were removed.

Given we have grammar that appears to contain many terminals, but some terminals might be missing to produce target.

The easiest way: map each non-terminal to a known expansion pattern. Let's parse all productions with terminals.

List of productions where RHS contains terminals (coupled with possibly non-terminals). I'll list them all:

rule_1: A1 -> A45 A42 A40 A38 A33 A28 A25 A19 A13 A2

No terminals.

rule_2: A2 -> A5 A4

rule_3: A2 -> A8 A7

rule_4: A2 -> A10 A9

rule_5: A2 -> A12 A11

rule_6: A3 -> 4

rule_7: A4 -> 5 2 1

rule_8: A5 -> 6

rule_9: A6 -> 11

rule_10: A7 -> 12 1

rule_11: A8 -> 13

rule_12: A9 -> 14 2 1

rule_13: A10 -> 15

rule_14: A11 -> 18 17 1

rule_15: A12 -> 19

rule_16: A13 -> A15 A14

rule_17: A14 -> 16 22 20

rule_18: A15 -> 23

rule_19: A16 -> 24

rule_20: A17 -> 25

rule_21: A18 -> 26

rule_22: A19 -> A3 A20

rule_23: A19 -> A5 A21

rule_24: A19 -> A6 A22

rule_25: A19 -> A8 A23

rule_26: A19 -> A10 A24

rule_27: A20 -> 3 22 21

rule_28: A21 -> 5 22

rule_29: A22 -> 10 22 21

rule_30: A23 -> 12 22

rule_31: A24 -> 14 21

rule_32: A25 -> A16 A26

rule_33: A25 -> A18 A27

rule_34: A26 -> 7 2 1

rule_35: A27 -> 9 28 1

rule_36: A28 -> A3 A29

rule_37: A28 -> A5 A30

rule_38: A28 -> A6 A31

rule_39: A28 -> A10 A32

rule_40: A29 -> 3 22

rule_41: A30 -> 5 21

rule_42: A31 -> 10 22 21

rule_43: A32 -> 14

rule_44: A33 -> A3 A34

rule_45: A33 -> A5 A35

rule_46: A33 -> A6 A36

rule_47: A33 -> A8 A37

rule_48: A34 -> 3 22 30 21

rule_49: A35 -> 5 22 21

rule_50: A36 -> 10 22 21

rule_51: A37 -> 12 22 21

rule_52: A38 -> A5 A39

rule_53: A39 -> 5 32 33 31

rule_54: A40 -> A16 A41

rule_55: A41 -> 7 1

rule_56: A42 -> A15 A43

rule_57: A42 -> A17 A44

rule_58: A43 -> 16 2

rule_59: A44 -> 8 35

rule_60: A45 -> A3 A46

rule_61: A45 -> A8 A47

rule_62: A46 -> 3 2

rule_63: A47 -> 12 36

Observations: many nonterminals expand to relatively short sequences of terminal symbols (maybe 1,2, or 3 numbers) or other nonterminals.

Goal: derive target sequence (50 terminals). Let's attempt to derive step by step to see if the grammar currently can generate target (maybe it can). We need to check if it's possible without insertion; maybe some missing productions cause mismatch.

Let's try to map each part of target sequence to the expansion of each A_ in order given by A1's RHS.

Thus we have to order expansions:

1. A45
2. A42
3. A40
4. A38
5. A33
6. A28
7. A25
8. A19
9. A13
10. A2

Thus the final sequence is concatenation: expansion(A45) + expansion(A42) + expansion(A40) + expansion(A38) + expansion(A33) + expansion(A28) + expansion(A25) + expansion(A19) + expansion(A13) + expansion(A2).

Goal: match target sequence (list of 50 ints). So each sub-expansion must correspond to a contiguous segment in order.

We need to choose which production alternatives for each nonterminal to produce the specific subsegments.

Given each sub-expansion is defined by the grammar; we can also choose to insert missing terminals where needed.

Thus, the problem is to assign each non-terminal a target substring in order.

Approach: start with A1's child order and attempt to align target.

Step 1: Determine possible expansions of A45, A42, A40, etc.

We need to compute expansions for all nonterminals, some have multiple alternatives (like A45, A42, etc.). Let's list expansions (terminal sequences) for each:

Single alternatives (only one rule):

- A3 -> 4 (i.e., [4])
- A4 -> 5 2 1 (i.e., [5,2,1])
- A5 -> 6 (i.e., [6])
- A6 -> 11 (i.e., [11])
- A7 -> 12 1 (i.e., [12,1])
- A8 -> 13 (i.e., [13])
- A9 -> 14 2 1 (i.e., [14,2,1])
- A10 -> 15 (i.e., [15])
- A11 -> 18 17 1 (i.e., [18,17,1])
- A12 -> 19 (i.e., [19])
- A14 -> 16 22 20 (i.e., [16,22,20])
- A15 -> 23 ([23])
- A16 -> 24 ([24])
- A17 -> 25 ([25])
- A18 -> 26 ([26])
- A20 -> 3 22 21 ([3,22,21])
- A21 -> 5 22 ([5,22])
- A22 -> 10 22 21 ([10,22,21])
- A23 -> 12 22 ([12,22])
- A24 -> 14 21 ([14,21])
- A26 -> 7 2 1 ([7,2,1])
- A27 -> 9 28 1 ([9,28,1])
- A29 -> 3 22 ([3,22])
- A30 -> 5 21 ([5,21])
- A31 -> 10 22 21 ([10,22,21]) (same as A22)
- A32 -> 14 ([14])
- A34 -> 3 22 30 21 ([3,22,30,21])
- A35 -> 5 22 21 ([5,22,21])
- A36 -> 10 22 21 ([10,22,21])
- A37 -> 12 22 21 ([12,22,21])
- A39 -> 5 32 33 31 ([5,32,33,31])
- A41 -> 7 1 ([7,1])
- A43 -> 16 2 ([16,2])
- A44 -> 8 35 ([8,35])
- A46 -> 3 2 ([3,2])
- A47 -> 12 36 ([12,36])

Nonterminals with multiple alternatives:

- A2: four alternatives:
    - (A5 A4) = [6] + [5,2,1] = [6,5,2,1]
    - (A8 A7) = [13] + [12,1] = [13,12,1]
    - (A10 A9) = [15] + [14,2,1] = [15,14,2,1]
    - (A12 A11) = [19] + [18,17,1] = [19,18,17,1]

- A13 -> A15 A14 = [23] + [16,22,20] = [23,16,22,20]

- A19 has 5 alternatives:
    - (A3 A20) = [4] + [3,22,21] = [4,3,22,21]
    - (A5 A21) = [6] + [5,22] = [6,5,22]
    - (A6 A22) = [11] + [10,22,21] = [11,10,22,21]
    - (A8 A23) = [13] + [12,22] = [13,12,22]
    - (A10 A24) = [15] + [14,21] = [15,14,21]

- A25 has two alternatives:
    - (A16 A26) = [24] + [7,2,1] = [24,7,2,1]
    - (A18 A27) = [26] + [9,28,1] = [26,9,28,1]

- A28 has four alternatives:
    - (A3 A29) = [4] + [3,22] = [4,3,22]
    - (A5 A30) = [6] + [5,21] = [6,5,21]
    - (A6 A31) = [11] + [10,22,21] = [11,10,22,21]
    - (A10 A32) = [15] + [14] = [15,14]   (since A32 -> 14)

- A33 has four alternatives:
    - (A3 A34) = [4] + [3,22,30,21] = [4,3,22,30,21]
    - (A5 A35) = [6] + [5,22,21] = [6,5,22,21]
    - (A6 A36) = [11] + [10,22,21] = [11,10,22,21]
    - (A8 A37) = [13] + [12,22,21] = [13,12,22,21]

- A38 -> A5 A39 = [6] + [5,32,33,31] = [6,5,32,33,31]

- A40 -> A16 A41 = [24] + [7,1] = [24,7,1]

- A42 has two alternatives:
    - (A15 A43) = [23] + [16,2] = [23,16,2]
    - (A17 A44) = [25] + [8,35] = [25,8,35]

- A45 has two alternatives:
    - (A3 A46) = [4] + [3,2] = [4,3,2]
    - (A8 A47) = [13] + [12,36] = [13,12,36]

Now each nonterminal in the start rule appears with a set of possible expansions.

Goal: to assign each to a substring of target order to cover all 50 terminals.

We'll attempt to find a combination that matches exactly the target. If not possible, we need to insert missing terminals to adjust.

Since insertion is allowed, we can treat expansions as being allowed to be missing some terminals: we can add them rather than remove. But we cannot remove existing terminals. So the expansions currently define some terminals that must appear in that order.

Thus if we pick an expansion that yields some terminals we need but not enough to produce target segment, we can insert additional terminals in some positions to make up the missing ones.

Goal: Insert minimal terminals. So we want to align expansions as much as possible with target, using expansions that already match many target terminals.

Thus it's essentially a longest common subsequence type: try to match each expansion to its target substring, and any missing terminals (gaps) need to be inserted.

But we can also choose among alternative expansions for each nonterminal to maximize matches.

We have to compute minimal total inserted terminals; we can attempt to assign each expansion exactly to a target substring (consecutive) by inserting some terminals before, after, or between symbols in RHS. Insertion anywhere: we can add terminals before a symbol, between symbols, or after final symbol. So we can have expansions with extra terminals to exactly match target substring.

Thus the number of inserted terminals for a particular nonterminal equals (target substring length) - (original terminal count in expansion), assuming we can align the original terminals in order to some subset of the target substring (preserving order). But we can also insert between any two original symbols to match internal differences. As long as the original sequence appears as a subsequence of target substring, we can fill extra needed to match.

Thus the problem reduces to, for each nonterminal, we need to cover a target substring where the original expansion's terminal sequence is a subsequence; inserted terminals fill gaps.

Goal: Partition the target sequence (global) into substrings for each nonterminal according to start rule order, and for each nonterminal choose an expansion (among alternatives) that is a subsequence of its allocated substring; we then count the inserted terminals as substring length minus original length (i.e., total length of substring - #original terminals). The total insertion count across all nonterminals is sum of inserted counts (since we only insert terminals within each rule; no overlapping). We also need to consider that some nonterminals appear also within expansions, but expansions are already expanded fully to terminals. Actually, we are explicitly expanding each nonterminal to terminals as per chosen rule; that's the final derived terminals; all expansions of sub-nonterminals are considered fully (i.e., recursively). So we can precompute complete expansions for each alternative (including expansions of sub-nonterminals). Already we have for A45 etc. So those expansions are final terminal strings.

Thus we don't need to think about further recursion.

Thus: partition target sequence of length 50 into 10 parts corresponding to the 10 nonterminals in A1.

Goal: For each part, use a known expansion string (among alternatives) to match as a subsequence. Each part must contain the expansion's terminals as a subsequence, and we minimize sum of (part length - expansion length).

This is basically a segmentation + subsequence alignment problem.

Our goal: minimize total inserted characters. Equivalent to maximize total matched original terminals across all parts (sum of expansion lengths). Since total target length is fixed (50), sum of inserted = 50 - sum(original terminals used). So we need to maximize total number of original terminals that can be matched (i.e., across all expansions we pick). Because we must cover all target terminals, but original terminals can overlap across parts? No, each terminal in target must belong to exactly one nonterminal's part (since partitions are non-overlapping). And each part's original terminals must be a subsequence of that part. So total original terminals matched is sum over expansions of length of expansion.

Thus minimizing inserts = maximizing total original terminal count matched across the schedule.

Given each expansion has a fixed length L (original terminals count). For each nonterminal we choose an alternative expansion (each with L known). Our total matched original terminals sum = sum L_i. So we want to maximize sum of chosen expansions lengths, subject to them fitting as subsequences in the target sequence partition grouping.

But we also need to check that they can indeed be placed in order. The expansions must appear as subsequences within respective parts, but we can also insert anywhere within those parts. So the constraint is simply that the concatenation of expansions in order must be a subsequence of the target sequence (overall). However, because we partition the target as contiguous parts, it doesn't require that expansions appear contiguously; we can have inserted extra terminals between expansions. But also we might have overlapping positions? No separate.

Better viewpoint: We choose expansions for each of the 10 nonterminals: resulting in a total concatenated sequence of original terminals: call it S0 = E1 + E2 + ... + E10 (concatenation). This S0 must be a subsequence of the target sequence in order, because we can add extra terminals in each expansion (i.e., fill gaps) but cannot reorder the original terminals globally. If S0 is a subsequence of target, we can embed each E_i in a specific subsegment that respects ordering: because the subsequence position of the first terminal after previous ones defines boundaries. However, we also have the restriction that expansions must be assigned to contiguous blocks after insertion: can we embed S0 as subsequence but then partition that mapping into block boundaries aligned with each E_i? Yes, if we define partition points after each E_i in the target, with possibly inserted terminals in between expansions. Since we can insert extra terminals at any point within expansions, we can handle any target terminals that occur between original ones within the same part, but what about target terminals that appear between the last terminal of E_i and the first terminal of E_{i+1}? Those would have to be placed either at the end of part i (i.e., inserted after its final original terminal) or at the beginning of part i+1 (i.e., inserted before its first original terminal). Both are possible because we can insert terminals anywhere. So overall it's sufficient to ensure that concatenated expansion S0 is a subsequence of the target. That ensures we can allocate positions for each expansion within the target in order.

Thus the minimal inserted = target length - max possible matched original terminals, where matched original terminals = length of longest subsequence of target that is a concatenation of expansions from each nonterminal (one per A_i). But we cannot double count: We need exactly one expansion per start-rule nonterminal: we must choose exactly one alternative (including possible multiple options) for each of the 10 positions.

Thus the optimization is to choose expansions for each of the 10 slots to maximize total sum of lengths such that the combined sequence S0 appears as a subsequence in the target.

Given expansions are fairly short (mostly length 2-5). So total possible sum of lengths if choose the longest expansions: look at lengths:

Compute lengths for each nonterminal:

- A45 alternatives:
    - A3 A46: expansion = [4] + [3,2] = length 3.
    - A8 A47: [13] + [12,36] = length 3.

So both length 3.

- A42 alternatives:
    - A15 A43: [23] + [16,2] = length 3.
    - A17 A44: [25] + [8,35] = length 3.

Both length 3.

- A40: only one: A16 A41 = [24] + [7,1] = [24,7,1] length3.

- A38: expansion = [6] + [5,32,33,31] = [6,5,32,33,31] length5

- A33 alternatives:
    - A3 A34: [4] + [3,22,30,21] = length5
    - A5 A35: [6] + [5,22,21] = length4? Actually A35 = [5,22,21] length3, so total length = 1+3 =4, but wait A5 -> 6 (single terminal). So length = 4.
    - A6 A36: [11] + [10,22,21] = length4? Actually A36 = [10,22,21] length3, plus 1 =4.
    - A8 A37: [13] + [12,22,21] = length4 (since A37 = 12,22,21 length3, plus 1 =4)

Thus max length =5 for A33.

- A28 alternatives:
    - A3 A29: [4] + [3,22] = length3
    - A5 A30: [6] + [5,21] = length3? Actually A30 = [5,21] length2, plus 1 =3.
    - A6 A31: [11] + [10,22,21] = length4? A31 length3 => total4.
    - A10 A32: [15] + [14] = length2.

Thus max length =4 (via A6 A31). There is also A3 A29 length3, etc.

- A25 alternatives:
    - A16 A26: [24] + [7,2,1] = length4 (actually A26 length3 => 4)
    - A18 A27: [26] + [9,28,1] = length4 (A27 length3 => 4). So both length4.

- A19 alternatives:
    - (A3 A20): [4] + [3,22,21] = length4? Wait A20 = [3,22,21] length3 => total4.
    - (A5 A21): [6] + [5,22] = length3? Actually A21 = [5,22] length2 => total3.
    - (A6 A22): [11] + [10,22,21] = length4? length4? A22 length3 => total4.
    - (A8 A23): [13] + [12,22] = length3? length3? Actually A23 = [12,22] length2 => total3.
    - (A10 A24): [15] + [14,21] = length3? A24 length2 => total3.

Thus max length =4 (through A3 A20 or A6 A22; others 3).

- A13 has only one expansion: A15 A14 = [23] + [16,22,20] = length4.

- A2 alternatives:
    - A5 A4: [6] + [5,2,1] = length4? Actually A5 6 length1 + A4 length3 => total4.
    - A8 A7: [13] + [12,1] = length3? Actually A7 = [12,1] length2 => total3.
    - A10 A9: [15] + [14,2,1] = length4? A9 length3 => total4.
    - A12 A11: [19] + [18,17,1] = length4? A11 length3 => total4.

Thus max length =4.

Thus maximum total original terminals across 10 parts:

- A45: 3
- A42: 3
- A40: 3
- A38: 5
- A33: 5
- A28: 4
- A25: 4
- A19: 4
- A13: 4
- A2: 4

Sum = 3+3+3+5+5+4+4+4+4+4 = 39.

Target length is 50, thus at least 11 inserts needed.

But that's assuming we can embed all expansions as subsequence. Might be not possible due to order constraints; might need extra inserts beyond 11. But we aim to minimize.

Thus baseline minimal inserts is target length - max total expansion length = 50 - up to 39 = 11. If we can find a mapping that matches all expansions without conflict, that's optimal. If not, may need more.

Now we need to test whether there is a subsequence assignment.

First, let's write the target sequence more clearly:

Index: Value
1:13
2:12
3:2
4:36
5:1
6:25
7:8
8:2
9:35
10:1
11:24
12:7
13:2
14:34
15:1
16:6
17:5
18:32
19:33
20:31
21:4
22:3
23:22
24:30
25:21
26:15
27:14
28:22
29:29
30:21
31:26
32:9
33:2
34:28
35:1
36:11
37:10
38:22
39:27
40:21
41:23
42:16
43:22
44:20
45:21
46:19
47:18
48:2
49:17
50:1

Now also see each expansion's sequence.

Let’s denote expansions thoroughly.

Compute all expansions with their terminal list:

- A45:
    e1: [4,3,2] (choice 1)
    e2: [13,12,36] (choice 2)

- A42:
    e1: [23,16,2] (choice 1)
    e2: [25,8,35] (choice 2)

- A40:
    e: [24,7,1] (only)

- A38:
    e: [6,5,32,33,31]

- A33 possibilities:
    o1: [4,3,22,30,21] (from A3 A34)
    o2: [6,5,22,21] (A5 A35)
    o3: [11,10,22,21] (A6 A36)
    o4: [13,12,22,21] (A8 A37)

- A28 possibilities:
    p1: [4,3,22] (A3 A29)
    p2: [6,5,21] (A5 A30)
    p3: [11,10,22,21] (A6 A31)
    p4: [15,14] (A10 A32)

- A25 possibilities:
    q1: [24,7,2,1] (A16 A26)
    q2: [26,9,28,1] (A18 A27)

- A19 possibilities:
    r1: [4,3,22,21] (A3 A20)
    r2: [6,5,22] (A5 A21)
    r3: [11,10,22,21] (A6 A22)
    r4: [13,12,22] (A8 A23)
    r5: [15,14,21] (A10 A24)

- A13 (c): [23,16,22,20] (only)

- A2 possibilities:
    s1: [6,5,2,1] (A5 A4)
    s2: [13,12,1] (A8 A7)
    s3: [15,14,2,1] (A10 A9)
    s4: [19,18,17,1] (A12 A11)

Thus target sequence length 50.

Now we need to choose one expansion per each of the 10 slots (ordered as above) to maximize total length while being a subsequence.

Let's denote the slots with indexes:

Slot1 = A45
Slot2 = A42
Slot3 = A40
Slot4 = A38
Slot5 = A33
Slot6 = A28
Slot7 = A25
Slot8 = A19
Slot9 = A13
Slot10 = A2

Goal: a concatenated list of chosen expansions (with lengths Li) of total maximum length such that the concatenated list is a subsequence of the target (preserving order). Insertions are allowed anywhere else.

We could also arguably treat inserted terminals as those matched to target positions that aren't matched to any expansion terminal.

Thus minimal insert = 50 - matched_count.

Thus find maximal matched_count using expansions.

First, consider the specific values in target. Many numbers appear:

Let's create mapping of positions for each terminal value:

- Value 1 appears at positions: 5,10,15,35,50? Actually, there is also 1 at position 35 and 50. Also maybe more? Let's list all:

Positions:
1:13
2:12
3:2
4:36
5:1
6:25
7:8
8:2
9:35
10:1
11:24
12:7
13:2
14:34
15:1
16:6
17:5
18:32
19:33
20:31
21:4
22:3
23:22
24:30
25:21
26:15
27:14
28:22
29:29
30:21
31:26
32:9
33:2
34:28
35:1
36:11
37:10
38:22
39:27
40:21
41:23
42:16
43:22
44:20
45:21
46:19
47:18
48:2
49:17
50:1

Thus frequency:

1 at 5,10,15,35,50 (5 times)
2 at 3,8,13,33,48 (5 times)
3 at 22 (once)
4 at 21 (once)
5 at 17 (once)
6 at 16 (once)
7 at 12 (once)
8 at 7 (once)
9 at 32 (once)
10 at 37 (once)
11 at 36 (once)
12 at 2 (once) and also at 2? Actually 12 appears at pos2 only, no other.
13 at pos1 (once)
14 at pos27 (once)
15 at pos26 (once)
16 at pos42 (once)
17 at pos49 (once)
18 at pos47 (once)
19 at pos46 (once)
20 at pos44 (once)
21 at 25,30,40,45 (four times)
22 at 23,28,38,43 (four times)
23 at pos41 (once)
24 at pos11 (once)
25 at pos6 (once)
26 at pos31 (once)
27 at pos39 (once)
28 at pos34 (once)
29 at pos29 (once)
30 at pos24 (once)
31 at pos20 (once)
32 at pos18 (once)
33 at pos19 (once)
34 at pos14 (once)
35 at pos9 (once)
36 at pos4 (once)

Check also 24 appears as terminal 24; that matches A40's first terminal.

Now look at expansions: many of them contain small sequences that might align.

We need to decide which expansions are best.

We'll need to treat each expansion as a subsequence: order needs to be preserved, but they can have gaps between its terminals. So to maximize matched length, we need to find an ordering of expansions within the target that maximize number of terminals matched.

Since we have 10 slots with relatively short expansions, maybe we can match all 39 terminals (optimal) as a subsequence. Let's see.

First, slot order: A45, A42, A40, A38, A33, A28, A25, A19, A13, A2.

Thus we need to assign expansions increasingly.

Start by analyzing target's first few terminals: 13,12,2,36,1,...

Our first slot (A45) alternatives:

- Option1 (e1): 4,3,2
- Option2 (e2): 13,12,36

Given the start of target is 13,12,2,... Option2: 13,12,36 matches 13 at pos1, 12 at pos2, 36 at pos4 (since pos3 is 2). But we need sequence 13,12,36. They appear in order: 13 (pos1), 12 (pos2), 36 (pos4). That's fine: we can skip pos3 (2) as insertion inside the expansion? Actually insertion can be anywhere inside a rule; we can insert terminals between the members. The original expansion sequence is 13,12,36. If we need to produce the target's 13,12,2,36,... we have an extra 2 at position 3 which is not in the original expansion. However, we can insert terminal 2 after the 12 maybe; but note insertion can be any terminal anywhere. So within the production rule for A45 we can insert terminal 2 between 12 and 36. That would produce [13,12,2,36] which matches target positions 1-4. So we would have insertion count = 1 (the inserted terminal 2). The original expansion length is 3, resulting substring length = 4. Perfect.

Alternatively, we could choose Option1: 4,3,2. Where would those appear? The first occurrence of 4 is at position 21. That's far later. This would require many inserted terminals before 4, skipping initial part of target entirely, which would be inefficient. So we should choose e2 = [13,12,36] for A45, with one insertion (2) between 12 and 36.

Thus A45 picks option e2.

That gives us matched terminals: 13,12,36. Including one inserted (2). So matched count = 3 from A45, insert count 1 for that slot.

Now after processing A45 we have used target positions up to 4 (including inserted 2). Actually, more formally, we considered substring positions 1-4 as A45's output: 13 (match), 12 (match), 2 (insert), 36 (match). So pointer now at 5 after position 4.

Now Slot2 A42 (alternatives: [23,16,2] or [25,8,35]).

Target next symbols from position5 onward: positions:

5:1
6:25
7:8
8:2
9:35
10:1
11:24 ...
We need to match an A42 expansion as subsequence after pointer 4.

OptionA: [23,16,2] - we look for 23 then 16 then 2.

Search in target from pos5 onward:
- 23 appears at position 41. That's far later. Could still match after many other symbols inserted, but there are many alternatives. OptionB: [25,8,35] appears earlier: 25 at pos6, 8 at pos7, 35 at pos9. That's a perfect match (25 at 6, 8 at7, 35 at9). The target at pos8 is 2, but that 2 is not part of the expansion. We can decide to insert it somewhere. The sequence [25,8,35] can produce [25,8,2,35] if we insert 2 between 8 and 35 (or anywhere). Let's examine target segment: pos5=1 is before 25. It could be inserted before the first terminal of this expansion (like inserted before the expansion). Since insertion can be anywhere in rule, we can insert terminal 1 before expansion or after or within. We want minimal insertion per rule. For this rule, maybe we can insert the 1 at beginning of A42 rule, i.e., before 25. For minimal insertion we might aim to embed as many as possible.

Alternatively, use [23,16,2] which appears later. Let's see if we can match them with minimal insertions later. 23 at pos41, 16 at pos42, 2 at pos48? Actually 2 appears later at pos48 or earlier at pos33. The order after 41: 23 at 41, 16 at42, then there is 22 at43, 20 at44,21 at45,19 at46,18 at47,2 at48,17 at49,1 at50. So the sequence 23,16,2 can be matched as 23(pos41),16(pos42),2(pos48). That would include some inserted terminals between them (22,20,21,19,18 between 16 and 2). That's 5 inserted characters. That's more insertions.

Thus better to pick [25,8,35] with insertions maybe of 1 at some point. Let's examine the segment from pos5 onward. The target substring we need to allocate to A42: we need to cover at least positions up to some point. Suppose we allocate A45 handled pos1-4. Next we start A42. The target pos5=1, pos6=25, pos7=8, pos8=2, pos9=35, pos10=1, ...

We choose expansion [25,8,35]. We can insert 1 before 25 (if we want) and insert a 2 between 8 and 35 (or after 35). The target segment includes 1 at pos5, then 25,8,2,35,1. There are two extra terminals (1 before, 2 between, 1 after). However, note after the expansion we have pos10=1, which could be considered as insertion after A42 before next slot. So perhaps we only need to insert 1 before 25 and 2 between 8 and 35. That would be 2 insertions in A42 rule.

Alternatively, we could try to incorporate the 1 after the expansion as part of a later slot's insertion; that's okay. So A42 could produce [25,8,35]; the target's 1 before and after 25.. can be inserted before/after. But insertion before the rule might be in previous slot? The prior slot's insertion part could also produce 1 before A42. However, from A45 we already inserted 2 but not 1. Inserting before A42 would be an insertion for A45? No, insertions are allowed only within rule expansions. For A45 we already inserted 2 between 12 and 36; we cannot insert preceding terminals outside A45's rule; the A45 rule's RHS can have additional terminals before its first symbol, after its last symbol, or in between any symbols. So we can insert a terminal at the beginning of A45's rule (before A8 or before 13) but we already used insertion for 2 already. We could also insert a terminal 1 before the 13 in A45 rule. Would that be allowed? Yes, we can modify any RHS: "insertion of terminal symbols anywhere in existing production rules." In a given production rule, we can add terminals anywhere. So we can add a 1 at start of A45 rule if needed. But A45 rule's current RHS is "A8 A47". (Actually rule_61 is A45 -> A8 A47). Evolution: we used Option2 e2: A8 (13) and A47 (12 36). So current RHS: A8 A47. We inserted terminal 2 between the 12 (from A47) and 36. That's inside A47's RHS between its two terminals? Actually A47 is "12 36". We could insert a 2 inside A47 RHS: after 12 before 36. That's allowed.

Now we can also insert at the start of that entire RHS: before A8. So we could insert any terminals before 13.

Thus, if we want to include the starting 1 at position5 in the sequence [13,12,...], we could have inserted it before 13. But 1 is after 36 (pos5). Actually target: 13,12,2,36,1. Since we inserted 2 inside, we have output [13,12,2,36] then we could insert a 1 after 36 (i.e., after the entire RHS). That would be an insertion at the end of rule A45. That's allowed. So we could absorb both pos5=1 as insertion for A45. Then A42 would start at position6 (25). That would reduce insertions for A42 to maybe just the 2 at position8? Actually target after the 1 (pos5) includes 25,8,2,35,... So if A45's insertion ends after 36 with an inserted 1, then A42's RHS start at 25. Good.

Thus we could incorporate the 1 after 36 as insertion within A45 rule (ending). That consumes pos5=1. So A42 would start at pos6=25.

Thus A42's expansion [25,8,35] can match, plus we need to handle the 2 at position8 (between 8 and 35). Insert this 2 within A42 rule.

Thus A42 would match positions 6:25, 7:8, 9:35 with insertion of 2 at pos8. That's one inserted terminal. Wait also we used 1 inserted after A45, not counted for A42. So A42's inserted count = 1 (the 2). Good.

Thus we have after that third rule? Let's keep track.

Thus far:

A45: matched 13,12,36; inserted 2 (pos3) and 1 (pos5). Actually A45 inserted two terminals: 2 and 1. Wait we inserted 2 between 12 and 36 (pos3). We inserted 1 after 36 (pos5). So number of insertions for A45 = 2.

But earlier we thought A45 matched [13,12,36] and had inserted 2 between 12 and 36. Now we also insert 1 after, but that insertion belongs to A45 (allowed).

Thus A45 insertion count = 2 (positions 3 and 5). Actually pos5 is not 1. The target pos5 is 1. So we inserted "1" after the entire expansion. So yes.

Now A42 will cover positions 6 to maybe 9: pos6:25 matched; pos7:8 matched; pos8:2 inserted; pos9:35 matched. So A42 insertion count = 1 (the 2). Possibly more later? We'll see.

Now after pos9=35, the next target symbol is pos10=1. However, after A42 rule (ending at 35), we could have inserted other terminals, maybe to be used for next rule. The next slot is A40 (expanded to [24,7,1]). Let's see if its expansion appears in target starting at pos10.

Target from pos10: 1 (pos10), then pos11=24, pos12=7, pos13=2, pos14=34, pos15=1, pos16=6,...

A40 is [24,7,1].

Sequence in target: pos11=24, pos12=7, pos13=2, pos14=34, pos15=1. So we have [24,7] continuing, then we need 1 after 7, but we have 2 and 34 then 1. So we can match 24 (pos11) and 7 (pos12). For the required 1, we can match pos15=1 after skipping pos13=2 and pos14=34 (treated as insertions). However, note insertions must be in A40 rule. We can insert 2 and 34 between 7 and 1 (or just 2 and 34). Also maybe need to handle the leading 1 at pos10 before 24. That could be an insertion at the start of the rule A40.

Thus A40 can produce [?,24,7,?,1] with inserted terminals: 1 before (pos10) and 2,34 after 7 before 1. So we have inserted 3 terminals for A40. Alternatively, we could choose to treat the leading 1 as insertion before rule or after previous rule? Since we have finished A42 at pos9 (inserting 2 inside), the next target position pos10=1 could be inserted as part of A40's RHS before first symbol. That's allowed.

Thus A40 insertion count = 3 (pos10=1, pos13=2, pos14=34) maybe. So matches 24,7,1 with three insertions.

Thus far, total inserted: A45 2 + A42 1 + A40 3 = 6.

Now after A40's expansion, we've consumed up to pos15 (the 1 used for final 1 of A40). The next part of target after pos15 is pos16=6, pos17=5, pos18=32, pos19=33, pos20=31, pos21=4, pos22=3, pos23=22, pos24=30, pos25=21, etc.

Next slot is A38 (expansion [6,5,32,33,31]).

Target from pos16: 6,5,32,33,31 exactly matches first five positions (16-20). So perfect match with no insertions. That's great.

Thus A38 matched fully: [6,5,32,33,31] exactly positions 16-20, insertion count 0.

Now after pos20 (31), next target at pos21 = 4, pos22=3, pos23=22, pos24=30, pos25=21, pos26=15, pos27=14, pos28=22, pos29=29, pos30=21, and so on.

Now slot 5 is A33 (options). We need to match something starting at pos21.

Possible expansions for A33:

Option o1: [4,3,22,30,21] matches exactly pos21-25: 4,3,22,30,21. That's perfect length5, matches exactly with no insertions. Great.

Option o2: [6,5,22,21] would not start at 4.

Option o3: [11,10,22,21] none.

Option o4: [13,12,22,21] none.

Thus choose o1: A33 = [4,3,22,30,21] => matches pos21-25 exactly.

Thus A33 matched positions 21-25; insertion count 0.

Now after pos25 = 21, next target pos26 = 15, pos27=14, pos28=22, pos29=29, pos30=21, pos31=26, pos32=9, pos33=2, pos34=28, pos35=1, pos36=11, pos37=10, pos38=22, pos39=27, pos40=21, pos41=23, pos42=16, pos43=22, pos44=20, pos45=21, pos46=19, pos47=18, pos48=2, pos49=17, pos50=1.

Now slot6 = A28 (options). Let's consider possible expansions.

Option p1: [4,3,22] - does not match starting at 15.
Option p2: [6,5,21] - does not.
Option p3: [11,10,22,21] - need start at 15: mismatch.
Option p4: [15,14] - matches first two: pos26=15, pos27=14. That's exact match of two terminals. So choose p4: [15,14].

Thus A28 = [15,14] matches positions pos26-27.

Now after pos27, we have pos28=22, pos29=29, pos30=21,...

The next slot is A25 (alternatives q1 and q2). A25 expansions:

q1: [24,7,2,1] (i.e., 24,7,2,1)
q2: [26,9,28,1] (i.e., 26,9,28,1)

Looking at target after pos27 (but note, we haven't yet consumed pos28 onward). Let's see if either q1 or q2 matches.

The upcoming target sequence from pos28: 22,29,21,26,9,2,28,1,11... Wait check: Actually pos28 is 22; pos29=29; pos30=21; pos31=26; pos32=9; pos33=2; pos34=28; pos35=1; pos36=11; etc. So we could match q2: [26,9,28,1] from positions 31,32,34,35 (skipping some). We could have insertions for other numbers (22,29,21,2). However, note that q2 has no leading 24,7; but q1's first terminal is 24, which occurs earlier at pos11. So q1 is not relevant now. So we probably need to choose q2.

Thus choose q2: A25 -> [26,9,28,1].

Now the target segment from pos28 onward is: 22,29,21,26,9,2,28,1,...

We can match 26 (pos31), 9 (pos32), 28 (pos34), 1 (pos35). The numbers between: 22 (pos28), 29 (pos29), 21 (pos30) before 26; also the 2 (pos33) between 9 and 28. Those would be insertions within A25 rule. So inserted terminals for A25: 22,29,21 before first matched terminal, and 2 between 9 and 28. That's 4 insertions.

Thus A25 insertion count = 4.

Now after pos35 (1), the next target is pos36=11, pos37=10, pos38=22, pos39=27, pos40=21, pos41=23, pos42=16, pos43=22, pos44=20, pos45=21, pos46=19, pos47=18, pos48=2, pos49=17, pos50=1.

Now slot7 = A19 (alternatives). Let's look at options.

Option r1: [4,3,22,21] - not matching start at 11.
Option r2: [6,5,22] - not matching.
Option r3: [11,10,22,21] - matches [11,10,22,21] appears at pos36-39-40? Let's see: pos36=11, pos37=10, pos38=22, pos39=27 (not 21). So at pos39 we have 27 instead of 21; pos40=21. Thus we could match the 21 at pos40 after some insertions.

Thus r3 sequence: we can match 11 at pos36, 10 at pos37, 22 at pos38, 21 at pos40 (skipping pos39=27 as insertion). So we have one insertion (27). So r3 fits with one insertion.

Option r4: [13,12,22] - not matching start
Option r5: [15,14,21] - not matching.

Thus best is r3 with 1 insertion.

Thus choose r3: A19 expands to [11,10,22,21].

Thus A19 matched positions 36:11, 37:10, 38:22, and 40:21. Insertion of 27 at pos39. So inserted count = 1.

Now after finishing A19 at pos40 (which matched 21), next target symbol is pos41=23.

Now slot8 = A13 (fixed expansion [23,16,22,20]).

Target pos41=23 pos42=16 pos43=22 pos44=20 exactly matches [23,16,22,20] in order. Perfect, no insertions. Good.

Thus A13 matched pos41-44 with no insert.

Now slot9 = A2 (alternatives). After pos44=20, next target is pos45 = 21, pos46=19, pos47=18, pos48=2, pos49=17, pos50=1.

We need to choose among options:

- s1: [6,5,2,1] - does not match.
- s2: [13,12,1] - no.
- s3: [15,14,2,1] - not match.
- s4: [19,18,17,1] - matches [19,18,17,1] with a leading 21 maybe as insertion.

Target segment: 21,19,18,2,17,1.

Option s4 gives sequence [19,18,17,1]; we have 19 at pos46, 18 at pos47, then we need 17 at pos49 and 1 at pos50. There's a 2 at pos48 between 18 and 17. So we can use option s4 with insertion of 2 (and also maybe 21 before). So s4 matches with insertion of 21 before the first matched terminal, and insertion of 2 between 18 and 17.

Thus inserted count for A2 = 2 (21 at start and 2 around). That's feasible.

Alternatively, maybe we could match s4 with insertion of 21 and maybe 2. Yes.

Thus best pick s4.

Thus total inserted for A2: 2.

Now after pos50 (1) we are done.

Thus overall inserted numbers: we have:

- A45: 2 insertions (2 and 1)
- A42: 1 insertion (2)
- A40: 3 insertions (1,2,34)
- A38: 0
- A33: 0
- A28: 0
- A25: 4 insertions (22,29,21,2)
- A19: 1 insertion (27)
- A13: 0
- A2: 2 insertions (21,2)

Total inserted = 2+1+3+0+0+0+4+1+0+2 = 13.

Thus we need at least 13 insertions.

But theoretical optimum number of insertions is at least 11 (target length 50 - max matched = 39). We achieved 50 - matched count = 13 inserted means matched count = 37. Let's compute sum of original terminals we matched:

Sum lengths of expansions used:

A45: original length 3 (13,12,36) matched 3
A42: original length 3 (25,8,35) matched 3
A40: original length 3 (24,7,1) matched 3
A38: length 5 (6,5,32,33,31) matched 5
A33: length 5 (4,3,22,30,21) matched 5
A28: length 2 (15,14) matched 2
A25: length 4 (26,9,28,1) matched 4
A19: length 4 (11,10,22,21) matched 4
A13: length 4 (23,16,22,20) matched 4
A2: length 4 (19,18,17,1) matched 4

Sum lengths = 3+3+3+5+5+2+4+4+4+4 = 37. Indeed, matches earlier.

Thus we matched 37 original terminals, leaving 13 insertions.

Upper bound optimum maybe 39 matched? Let's examine if we can improve matched count by selecting longer expansions (maybe using alternative expansions with length 5 for A28? Actually A28's max length is 4, not 5. Could we consider using A28 option p3: [11,10,22,21] which is length4; or p1 length3, p2 length3, p4 length2. So we used length2 (p4). If we could use a length4 expansion while still matching target, we could increase matched count by at most 2. Let's see: maybe we could use A28 -> [11,10,22,21] (p3) to match somewhere later. The target's segment after A33 (pos21-25) is pos26 onward: 15,14,22,29... Could we match p3? Sequence [11,10,22,21] does not appear after pos25. However, we have [11,10,22] at pos36-38 and then a 21 at pos40 (two later). But that segment is used for A19. Maybe we could incorporate A28 earlier and shift other rules. Let's consider alternative mapping.

Perhaps we could reorder via insertion, but cannot reorder nonterminals. So slots order is fixed.

Thus A28 must be before A25, A19, etc. If we use longer A28, we must match that sequence before those other sequences.

Check if after A33 (ends at pos25), we have target pos26=15, 27=14, 28=22... There is a [15,14] pair we matched with A28. But we could also match [11,10,22,21] if we could make A28 produce that pattern earlier. It does not appear at pos26-? Actually at pos26-30 we have 15,14,22,29,21. There is [15,14] but not [11,10]. So not possible.

Thus A28's longer expansions might not match earlier part. So p4 is appropriate. So length 2 is minimal; cannot do longer.

Maybe A42 could be longer? Both expansions length3, maximum length 3. So no.

A45 length 3.

A25 we used q2 length4, no longer alternatives.

A19 length 4.

Thus only place we might improve matched count is A25: we used q2 length4 matches (26,9,28,1). Could we match q1 length4 [24,7,2,1]? That would not fit later, but maybe if we reassign A40 or earlier expansions we could shift the 24 and 7 earlier. But 24 and 7 appear at positions 11 and 12, which were used for A40. We could assign A25 to match 24,7,2,1 later? No, because target positions 24,7,2? Actually after A33, doesn't have 24,7,2,1; but earlier we had them. So not possible.

Thus matched count sum of lengths seems bounded by 37 given the constraints.

But can we improve by matching A38? Already full length5, max.

A33 matched length5, max.

Thus we seems to have achieved sum lengths = 37, which may be maximal possible given ordering constraints. Let's confirm if any other slot could get a longer expansion: Only A28 maybe get length4 using A6 A31 (to get [11,10,22,21]). That length4 would increase matched count to 39 if we could fit that. Let's see if we can replace A28's expansion with p3 (length4) and adjust following slots accordingly.

Current order:

A45 (13,12,36), A42 (25,8,35), A40 (24,7,1), A38 (6,5,32,33,31), A33 (4,3,22,30,21), A28 (???), A25, A19, A13, A2.

Our target after A33 ends at pos25; the remaining target looks like:

Pos26-? : 15,14,22,29,21,26,9,2,28,1,11,10,22,27,21,23,16,22,20,21,19,18,2,17,1

Thus after pos25 we have:

26:15
27:14
28:22
29:29
30:21
31:26
32:9
33:2
34:28
35:1
36:11
37:10
38:22
39:27
40:21
41:23
42:16
43:22
44:20
45:21
46:19
47:18
48:2
49:17
50:1

Thus we need to allocate A28, A25, A19, A13, A2 across this.

We used A28 as [15,14] (length2). If we could instead use A28 as [11,10,22,21], length4, but the target segment for that would be at positions 36-38-? need to match. However A28 is before A25, which includes the 26,9,... So if A28 uses [11,10,22,21], that would consume those positions early, displacing A25 and other slots.

Alternatively, could we use A28 option p1 [4,3,22] maybe? But we don't see 4,3,22 later. Actually after pos25 we have 15,14,... So none that start with 4.

Option p2 [6,5,21] not match.

Option p3 [11,10,22,21] appears later, but if we schedule A28 to match that, then the earlier numbers (15,14, etc.) would need to be part of insertions before A28 (i.e., inserted at start of A28). That's allowed: we can insert 15,14,22,29,21,26,9,2,28,1 (all preceding) as insertions before the matched part of A28. However, the rule's insertions can be anywhere within the rule, including before the first symbol. So we could have a huge insertion of many terminals before the original terminals [11,10,22,21] appear. That would increase insertion count for A28 but may allow us to keep matched count length4 for A28. However then A25 and subsequent slots would have to start after position of A28's matched terminals, i.e., after 21 at pos40 maybe. That would shift the rest of the sequence: A25 would now start at pos41? But A25's expansion [26,9,28,1] could potentially match later? Let's examine.

If we allocate A28 with p3 [11,10,22,21] matching at pos36-40 (with insertion of 27 as before). Actually we would match 11(pos36),10(pos37),22(pos38),21(pos40) (skipping 27). That implies insertion 27 within A28 (maybe same as before). Then A28 would also need potentially insertion of earlier terminals before its matched part: i.e., we could insert the segment [15,14,22,29,21,26,9,2,28,1] before the matched part; but those could also be matched by other non-terminals (A25, etc.). However, due to order constraints, those inserted terminals belong to A28 and would not be matched by A25, and thus would not count toward matched counts for A25. That would cause overall matched count to drop (since we would use only original terminals [11,10,22,21] for A28, and then later A25 would need to match remaining target numbers after pos40, which are pos41 onward: 23,16,22,20,21,19,18,2,17,1. This tail no longer contains 26,9,28,1 that A25 wants. So A25 could not match effectively. Thus probably not.

Alternatively, we could choose A28 = p2 (6,5,21) but not possible due to mismatched later.

Thus A28 must use the small length2 expansion to keep ability for later parts.

Alternatively, A25 could be q1 [24,7,2,1] if we reorder earlier sections; but 24,7 appear earlier already, and they've been used by A40. Could we have A40 using a different expansion? A40 only has [24,7,1]; so those terminals are consumed there. Could we instead incorporate the 1 from A40 and maybe produce the 2 after as insertion? A40 we already used that insertion. Could we shift to use another alternative for A40? No alternatives.

Thus cannot.

Thus our mapping seems near optimal.

But maybe we can match more original terminals by using alternative for A45 different? A45 length3 already. That's max.

What about A42: alternative [23,16,2] length3; we used [25,8,35] length3. Could we use [23,16,2] and match later part of target where those appear? They appear near the end: 23 at pos41, 16 at pos42, 2 at pos48 (and also earlier maybe?). However, using it would shift the entire later partition drastically and may cause A13 etc. But maybe there is arrangement with less total insert.

Check possible alternative: Use A42 -> [23,16,2] and maybe A13 matches something else? But A13 is fixed [23,16,22,20]. If we used 23,16 for A42, they'd be consumed earlier, then A13 would need to match later 23,16,22,20 but those would have been taken? However, there is only one 23 and 16 each in the whole target. Actually 23 appears only at pos41, 16 appears only at pos42. So if we allocate them to A42, then A13 cannot match, because would need separate 23 and 16 not available. But maybe we could assign A42 with [23,16,2] and use A13 as something else? A13 is fixed, cannot change.

Thus we cannot use that alternative; must use [25,8,35].

Thus expansions we used are essentially forced by the target.

Thus total matched count likely max 37. Let's double-check if any other slot could have matched more original terminals than we did (i.e., using longer expansions). For A25 we used q2 length4 (max). A19 we used r3 length4 (max). A28 we used length2 (p4) but max length for A28 is 4 (p3) but p3 does not align with target in order without causing later mismatches. Could we try using p1 (length3) [4,3,22] perhaps? After A33 we have 4,3,22,30,... But we used those for A33 itself. If we use A33 as something shorter and A28 as longer, maybe we increase matched total? Let's examine alternative distribution: perhaps we can use A33 with a shorter expansion (like length4), freeing some terminals for A28 to use longer expansion, potentially increasing total matched count beyond 37.

A33 alternatives: lengths: other options are length4 (6,5,22,21) and others length4 (11,10,22,21) and length4 (13,12,22,21). So alternative A33 could be length4 (instead of length5 we used). We used length5 giving 5 matched. If we switch to a length4, we lose one matched terminal there but maybe we can get an extra matched terminal in A28 (from length2 to length4) net +2? That could increase total by 1 overall.

We need to examine if we can assign A33 a length4 alternative that still fits the target.

After A38, we have target segment starting at pos21 = 4.

We need A33 to produce some sequence that is a subsequence of [4,3,22,30,21,15,14,...]. The A33 alternatives:

- Option o1: [4,3,22,30,21] length5 matches the first 5 exactly. Good.
- Option o2 (A5 A35): [6,5,22,21] length4. That would need to match somewhere after pos21: does target have 6,5,22,21 in order? We have 6 at pos16 earlier (but after pos20 is 31). After pos20 we have 4,3,22,30,21,... So 6 appears earlier, not after. So not possible later (maybe later later but after we pass A33). Actually later we have 6 earlier used in A38. After pos21 onward, there is no 6 again. The later occurrences of 6 are only at pos16 and no other. So we cannot fit [6,5,22,21] for A33.

- Option o3: [11,10,22,21] length4: Does target after A38 have 11? Not until pos36. So not possible at this location because there are many numbers in between; but we could treat them as insertions; we could have insertion of many symbols before matching 11 for A33. That would mean A33 would match 11 (pos36), 10 (pos37), 22 (pos38), 21 (pos40) with insertion of 27. It would skip the segment [4,3,22,30,21,15,14,22,29,21,26,9,2,28,1] etc. Those would be inserted before the first matched terminal of A33. That would heavily increase insertions for A33, but might allow A28 to match something else maybe longer. However that would not increase matched count beyond 4 for A33. Let's consider if we could assign A33 = [11,10,22,21] (len4) and then A28 possibly uses something else? Actually A28 options after that: we could then have A28 = [15,14] (len2) continues as is. But we'd have to shift A19 accordingly. The overall total matched might remain same or less.

What about A33 = [13,12,22,21] (len4). That could match target 13 at pos1 earlier but not after A38.

However, maybe we could restructure earlier slots (A45, A42, A40 etc.) to allow using different alternatives such that we can match [13,12,22,21] somewhere later. But we already used that one for A45 and A42.

Seems best is using the length5 A33.

Thus total matched = 37 is likely optimal.

But perhaps we can improve A25 to have length4 [24,7,2,1] while matching later by using A40 maybe something else? A40 is fixed. However, maybe we can insert the 24,7 from A40 into A25 instead? Actually sequence [24,7] appears at pos11-12. A40 consumed [24,7]. If we switch A40 to use a longer expansion? No alternatives.

Thus A25 cannot be [24,7,2,1] unless we generate more insertions.

Nevertheless, maybe A25 could be [24,7,2,1] and we treat A40's expansion to be just [1]? Not possible. A40 has fixed expansion 24,7,1. So A25 can't be [24,7,2,1] because that would require two 24 and 7, which we have only one.

Thus we stick.

Now does any other expansion produce longer than we used? A38 has length5, we used length5. Perfect.

Now check if we have used A2's length4 with [19,18,17,1] which matches exact later parts; we inserted 21 at start and 2 between 18 and 17 (makes 2 inserted). The original length 4, matched 4, total accounted for (including inserted). Could we use A2 variant s1 [6,5,2,1] to match later? Not plausible because 6 appears earlier (pos16), but maybe after A13 we have segment [21,19,18,2,17,1] and we want [6,5,2,1] maybe no.

Thus choice s4 is best.

Now could we improve A42 by using the other alternative but then also incorporate A13's 23 and 16 from that one? Actually using [23,16,2] for A42 would match 23,16 (later) and 2 at pos48. Then A13 would be left without 23 and 16. But A13 may be able to match something else? A13 is fixed as [23,16,22,20]; no alternative. So cannot.

Thus we need [25,8,35] for A42 as we used.

Thus matched length remains 37. So we need 13 inserted terminals.

Now we must insert these terminals into the production rules as per allowed modifications: i.e., we need to insert terminals for each rule we used.

We need to map each insertion to a particular rule's RHS.

We have identified where each insertion belongs:

- A45 rule: rule_61 (A45 -> A8 A47). We'll need to insert terminal '2' between 12 and 36 inside A47 and also insert terminal '1' after the entire expansion (i.e., after A47). However we can do both in rule_61: We can insert '2' after symbol 12 part, but note 12 is inside A47. Actually we might need to adjust rule_47 (A47 -> 12 36) by inserting '2' after 12. Also we need to insert '1' after the A47 expansion. That could be done by inserting '1' at the end of rule_61 after A47.

- A42: rule_56 (A42 -> A15 A43) OR rule_57 (A42 -> A17 A44). We chose A42 using A17 A44 (option 2) because that matches [25,8,35]? Wait A42 alternatives: rule_56: A15 A43 → 23 16 2 (not used). rule_57: A17 A44 → 25 8 35. So we need to use rule_57. The RHS of rule_57 currently A17 A44. The expansion yields 25 then 8 then 35. We need to insert a '2' between 8 and 35. Since that 2 is part of $A44$? Actually A44 -> 8 35 currently yields "8 35". But we need 2 maybe inserted between them. Actually the original expansion is [25] from A17 then [8,35] from A44. So we need to add a '2' after 8 before 35, i.e., inside rule_59 (A44 -> 8 35) we can insert '2' after 8. So modify rule_59: A44 -> 8 2 35 (inserting a terminal 2 between 8 and 35). Good.

- A40: rule_54 (A40 -> A16 A41) where A16 -> 24 (rule_19) and A41 -> 7 1 (rule_55). We need to insert three terminals: a leading 1 before 24, then 2 and 34 between 7 and 1? Actually our mapping: we had inserted 1 before first symbol (i.e., before 24). That can be inserted at the beginning of rule_54: before A16. Insert '1' before A16. Then we need to insert 2 and 34 after 7 and before 1. That's inside rule_55: currently "7 1". We can modify rule_55 to "7 2 34 1". Or we can break insertion across rules: insert 2 after 7 and 34 before 1. Both in rule_55.

Thus rule_54 gets a '1' at front, rule_55 gets "7 2 34 1" or "7 2 34 1" using insertion of 2 & 34.

Alternative: we could also insert 2 before 7 as part of rule_55; but we need order: 7,2,34,1.

Thus modifications: rule_54: A40 -> 1 A16 A41? But original RHS is "A16 A41". To insert 1 before A16, write "1 A16 A41". That's allowed.

Then rule_55: "7 2 34 1" (insertion of 2,34 before 1). So that's two insertions.

But we counted three insertions (1 at start of A40, plus 2 and 34 inside A41). Indeed total 3.

- A25: rule_33 (A25 -> A18 A27). A18 -> 26 (rule_21), A27 -> 9 28 1 (rule_35). We used q2 = [26,9,28,1] with inserted 22,29,21 before 26 and inserted 2 between 9 and 28. The inserted 22,29,21 are before 26; they could be inserted at beginning of rule_33 before the first non-terminal, i.e., before A18. So rule_33 becomes "22 29 21 A18 A27". That's three insertions.

Now for the insertion between 9 and 28: that's inside rule_35 which is "9 28 1". We need to insert a 2 between 9 and 28, i.e., modify rule_35 to "9 2 28 1". That's one insertion.

Alternatively we could also have inserted that 2 somewhere else? No, we need it between.

Thus rule_33 gets three inserts, rule_35 gets one.

- A19: rule_26? Actually we used A19 -> A6 A22 (rule_24?). Wait A19 options: rule_24 is A19 -> A6 A22. That yields [11] + [10,22,21] = [11,10,22,21]. Since we needed to insert 27 between 22 and 21? Actually we used A6 A22. Let's verify: A6 -> 11, A22 -> 10 22 21. So this yields 11,10,22,21. The target had 27 extra between 22 and 21 (i.e., after 22 we need 27 before 21). So need to insert 27 after the "22" terminal. Since "22" appears inside A22's RHS: "10 22 21". So we need to modify rule_29 (A22 -> 10 22 21) to insert 27 after 22, resulting "10 22 27 21". That's one insertion. No other insert needed.

Thus rule_29 is updated.

- A2: Using rule_5: A2 -> A12 A11 (since s4). A12 -> 19, A11 -> 18 17 1. So yields [19,18,17,1]. Insert 21 before 19 and 2 between 18 and 17.

Thus need to insert a 21 at start of rule_5 before A12 (or before the first terminal). Actually rule_5 currently: A2 -> A12 A11. Insert 21 before A12: "21 A12 A11". That's one insertion.

Then insert 2 between 18 and 17. This is inside rule_14: A11 -> 18 17 1 (produces 18,17,1). We need to insert a 2 between 18 and 17. So modify rule_14 to "18 2 17 1". That's one insertion.

Thus rule_5 gets 21 insertion, rule_14 gets 2 insertion.

Now also A45's insertion at start: we inserted 1 after A45; we inserted '2' inside A47. Let's list needed modifications:

- A45 rule (rule_61): A45 -> A8 A47. Insert 1 after A47 (or at end). The RHS currently "A8 A47". To insert 1 after A47, we can modify to "A8 A47 1". That's an insertion.

- Then modify A47 rule (rule_63) to insert 2 between 12 and 36: originally "12 36". Change to "12 2 36". That's one insertion.

Thus rule_61 gets 1 insertion, rule_63 gets 1 insertion.

Double-check on the need for inserted 1 at the end of A45's output: does the target sequence have "1" at position5. If we insert 1 after A47, that appears after 36 (makes order 13,12,2,36,1). That matches target positions. Good.

Now ensure the inserted 2 for A45 is correctly placed: after 12 at position3 and before 36. In rule_63: "12 2 36".

Now for A42 insertion: we inserted 2 between 8 and 35. That is rule_59 change: "8 2 35". This yields sequence 25 (from A17) then 8 then 2 then 35. That matches target positions 6:25, 7:8, 8:2, 9:35. Good.

Check that for A40 we inserted leading 1 before 24: rule_54 becomes "1 A16 A41". Since A16 -> 24; A41 -> 7 1 (modified). So reading A40 -> 1 24 (via A16) then 7 2 34 1 (via A41). That yields 1 24 7 2 34 1 in order? Wait need order: rule_54: "1 A16 A41". A16 -> 24, A41 -> "7 2 34 1". So derived string: 1 24 7 2 34 1. That matches target positions 5:1 (already consumed by A45 previously?), Wait we inserted 1 as part of A45's insertion already; but after A45 we have used 1 at pos5. Then for A42 we start at pos6 25, etc. After A42 we have positions up to 9. Now A40's first inserted 1 would be at pos10 (target's 1 after 35) which matches. Good.

Thus A40 outputs 1 24 7 2 34 1. The target after pos9 (35) is 1 (pos10) then 24 (pos11) then 7 (pos12) then 2 (pos13) then 34 (pos14) then 1 (pos15). Yes matches.

Thus insertions for A40 are correct.

Now A25 insertion: rule_33 becomes "22 29 21 A18 A27". A18 -> 26; A27 -> "9 2 28 1". So A25 yields 22,29,21,26,9,2,28,1.

The target segment after A28 (including A28 produced [15,14]) is: after pos27 is 22 (pos28), 29 (29), 21 (30), 26 (31), 9 (32), 2 (33), 28 (34), 1 (35). Perfect.

Thus A25 correct.

A28: we used the rule_39 "A28 -> A10 A32". That yields 15 and 14. No insert needed.

Now A19 insertion: rule_24 is A19 -> A6 A22, and we modify A22 (rule_29) to "10 22 27 21". Then A19 yields 11 (from A6), 10,22,27,21. The target after A25 (ends at pos35=1) has pos36:11,37:10,38:22,39:27,40:21. Good.

Now A13 is fine.

A2 insertion: rule_5: "21 A12 A11". A12 ->19. A11 modified to "18 2 17 1". So A2 yields 21,19,18,2,17,1. The target after A13 at pos45:21 (pos45),19 (46),18 (47),2 (48),17 (49),1 (50). Works.

Thus all insertions placed as identified.

Now we need to produce the output with updated production rules: list only the rules that have changed.

Note: Which rules changed? Let's list all modifications:

- rule_5: originally "A2 -> A12 A11". New: "A2 -> 21 A12 A11". Inserted 21 before A12.

- rule_14: originally "A11 -> 18 17 1". New: "A11 -> 18 2 17 1". Inserted 2 after 18.

- rule_29: originally "A22 -> 10 22 21". New: "A22 -> 10 22 27 21". Inserted 27 after 22.

- rule_33: originally "A25 -> A18 A27". New: "A25 -> 22 29 21 A18 A27". Inserted 22,29,21 before A18.

- rule_35: originally "A27 -> 9 28 1". New: "A27 -> 9 2 28 1". Inserted 2 after 9.

- rule_59: originally "A44 -> 8 35". New: "A44 -> 8 2 35". Inserted 2 after 8.

- rule_63: originally "A47 -> 12 36". New: "A47 -> 12 2 36". Inserted 2 after 12.

- rule_61: originally "A45 -> A8 A47". New: "A45 -> A8 A47 1". Inserted 1 at end.

- rule_54: originally "A40 -> A16 A41". New: "A40 -> 1 A16 A41". Inserted 1 at start.

- rule_55: originally "A41 -> 7 1". New: "A41 -> 7 2 34 1". Inserted 2 and 34 between 7 and 1 (we inserted two terminals). Since rule_55 currently is "A41 -> 7 1". To insert 2 and 34, we modify to "A41 -> 7 2 34 1". That's two insertions in that rule.

Thus we have several rules changed. Also need to verify rule_57 existed for A42 as A17 A44; we inserted 2 in A44, not in rule_57. So rule_57 unchanged.

Now also rule_63 is A47; we have inserted a 2.

Check rule_60? No need.

Check rule_36? No.

Check rule_61: inserted 1 at end.

Further we might need to check A45's alternative usage; we used rule_61 (the A8 A47). So we modify that.

Now references: ensure we correctly indicate which rule numbers-- mapping: The list provided used rule_<n> labeling sequentially from rule_1 etc. We must output only the updated production rules with modifications, using same rule numbers.

Thus output lines:

rule_5: A2 -> 21 A12 A11

rule_14: A11 -> 18 2 17 1

rule_29: A22 -> 10 22 27 21

rule_33: A25 -> 22 29 21 A18 A27

rule_35: A27 -> 9 2 28 1

rule_59: A44 -> 8 2 35

rule_63: A47 -> 12 2 36

rule_61: A45 -> A8 A47 1

rule_54: A40 -> 1 A16 A41

rule_55: A41 -> 7 2 34 1

Double-check any other required inserts:

- Did we need to insert a '2' after '12' inside A47? Yes, rule_63 changed.

- Did we need to insert the leading '1' for A45 after A47? Yes, rule_61 has "A8 A47 1".

- Did we need to insert a leading '1' for A40? Yes, rule_54 becomes "1 A16 A41". Could also be placed before the non-terminal via insertion. So absolute correct.

Now confirm that there are no other insertions missed:

- For A45 we inserted also 2 inside A47 (rule_63) and 1 at end of rule_61. Did we need to insert the early 2 after 12? Yes, done.

- There's also the insertion of a '2' after 12? Actually we inserted 2 after 12; correct.

- For A42 we inserted 2 after 8 (inside A44). That's rule_59.

- For A40 we inserted 1 before A16 (rule_54) and 2 and 34 before 1 (rule_55). That matches inserted count for A40 (3). Good.

- For A25 inserted 22,29,21 before A18 A27 (rule_33). So rule_33 includes those three.

- For A27 inserted 2 between 9 and 28 (rule_35). Good.

- For A19 inserted 27 after 22 inside A22 (rule_29). Good.

- For A2 inserted 21 at beginning (rule_5) and 2 after 18 (rule_14). Good.

Thus modifications align.

Now, the solution requires we output only updated production rules, each on a line with rule number and colon. We should include only those changed.

Thus final answer: show these updates.

Potential nuance: For rule_55 we inserted two terminals (2 and 34). Ensure proper formatting: "A41 -> 7 2 34 1". That's okay.

Now final output: after listing "Updated production rules:" line, then each rule update line.

Thus final answer:Updated production rules:
rule_5: A2 -> 21 A12 A11
rule_14: A11 -> 18 2 17 1
rule_29: A22 -> 10 22 27 21
rule_33: A25 -> 22 29 21 A18 A27
rule_35: A27 -> 9 2 28 1
rule_59: A44 -> 8 2 35
rule_63: A47 -> 12 2 36
rule_61: A45 -> A8 A47 1
rule_54: A40 -> 1 A16 A41
rule_55: A41 -> 7 2 34 1