drinkussions(Baban and Saanjoo): TRYING TO DO CARTESIANS CONCATENATIONS ON CATALAN OPTIONS ON TREES PRESERVING THE FOREST ORDERING

i am more concerned with k are multiples of 4 and n are r times k I HAVE n numbers of symbols (obviously we cannot change the sequences of the symbols) but we can partition these symbols into k number of trees with | partitioning such that there are n/k number of symbols per tree... what are the nature of catalan numbers counting ratios in two cases where case 1 is if we take all n symbols in single tree and do bracketing Catalan (n) and case 2 is bracketing k trees seperately with n/k symbols in each tree and then (Catalan (n/k))^k numbers of possible combinations are done ... which case will have lesser count and what are the proportions????????

i am more concerned with   k are multiples of 4 and n are r times k   I HAVE n numbers of symbols (obviously we cannot change the sequences of the symbols) but we can partition these symbols into k number of trees with | partitioning such that there are n/k number of symbols per tree... what are the nature of catalan numbers counting ratios in two cases where case 1 is if we take all n symbols in single tree and do bracketing Catalan (n) and case 2 is bracketing k trees seperately with n/k symbols in each tree and then (Catalan (n/k))^k numbers of possible combinations are done ... which case will have lesser count and what are the proportions????????

Copilot said:

Copilot

Here is a visualization of the ratio between:

· Case 1: Catalan(n) — bracketing all n symbols in a single tree

· Case 2: (Catalan(n/k))^k — bracketing k trees separately, each with n/k symbols

for values where k is a multiple of 4 and n = r × k, with r ranging from 1 to 10:

 i am more concerned with   k are multiples of 4 and n are r times k   I HAVE n numbers of symbols (obviously we cannot change the sequences of the symbols) but we can partition these symbols into k number of trees with | partitioning such that there are n/k number of symbols per tree... what are the nature of catalan numbers counting ratios in two cases where case 1 is if we take all n symbols in single tree and do bracketing Catalan (n) and case 2 is bracketing k trees seperately with n/k symbols in each tree and then (Catalan (n/k))^k numbers of possible combinations are done ... which case will have lesser count and what are the proportions????????

Copilot said:

Copilot

Here is a visualization of the ratio between:

· Case 1: Catalan(n) — bracketing all n symbols in a single tree

· Case 2: (Catalan(n/k))^k — bracketing k trees separately, each with n/k symbols

for values where k is a multiple of 4 and n = r × k, with r ranging from 1 to 10:

Show code

Show thinking

Exploring Partition Counts

I'm currently focused on comparing Catalan number counts within two scenarios for partitioning 'n' symbols. I'm carefully analyzing the nuances of each scenario, aiming to identify the key differences in how Catalan numbers manifest. The goal is to articulate these differences clearly and succinctly.

Refining Partition Comparison

I'm now focusing on a more nuanced analysis. Comparing two scenarios: a single tree with 'n' symbols and 'k' separate trees, each with 'n/k' symbols, all with a fixed sequence. I'm carefully considering how Catalan numbers apply to each case, aiming for a precise formulation of the count differences. Key factors include the fixed sequence and the implication of Catalan numbers being Cn−1 not Cn.

Analyzing Partitions Further

I'm now carefully examining the relationship between the two scenarios, focusing on the implications of a fixed symbol sequence. I'm building a comparison to find the precise ratio between parenthesization possibilities. My aim is to clearly articulate the difference in bracketed structures between the two cases, emphasizing how Catalan numbers are applicable in each. The number of parenthesizations will be critical.

Evaluating Count Proportions

I'm now calculating the ratio between the parenthesization counts in the two scenarios. With n=8 and k=4, I found that Case 1 (single tree) yields C7=429, and Case 2 (k trees) gives (C1)4=14=1. The difference is significant. I'm building a generalized formula for this ratio to determine its behavior as n and k vary. Also, I must ensure precision in these Catalan number calculations and double-check my indexing.

Finalizing Count Comparisons

I've successfully computed the ratio between the parenthesization counts for the two scenarios. For n=8, k=4, the single-tree approach yields 429 possibilities, while the k-tree scenario gives only 1, confirming a significant difference. I am refining the general formula for this ratio to encompass varying values of n and k, ensuring precise Catalan number calculations throughout the process. Furthermore, the accuracy of my indexing and the underlying assumptions of Cm−1 and Cm are under review.

Quantifying Count Differences

I've been working to formalize the ratio between the counts from my examples, and I've constructed a more refined, general formula to encompass varying values of 'n' and 'k'. My calculations are precise now, and I've rigorously double-checked my application of Catalan numbers and associated indexing to guarantee accuracy.

Finalizing the Analysis

I've now fully formalized the ratio between the counts for these two scenarios, incorporating general values for n and k. I've precisely calculated and double-checked the application of Catalan numbers and their indexing.

Revisiting Count Proportions

I'm now refining the comparison for scenarios with n symbols and k groups. My initial calculations highlighted the significant count difference, but I'm generalizing the formula for the ratio: R=Cn−1(Cn/k−1)k, aiming for a more robust understanding. While the pattern of Case 2 being substantially smaller remains consistent, I'm verifying boundary conditions and ensuring Catalan number application is precise.

Synthesizing Ratio Insights

I've clarified the indexing of Catalan numbers: Cn−1 for n symbols. With this, I've constructed a Python script to compute counts and compare them. I've built a more robust formula to determine the ratio between the parenthesization counts, and tested it with numerous examples to confirm the significant difference between scenarios. Specifically, for n and k defined as such, the value of the 'k' group scenarios always results in a significantly lower count than the single, unified tree.

Calculating Count Proportions

I'm now carefully calculating the ratio between the parenthesization counts in the two scenarios. With n=8 and k=4, I found that Case 1 (single tree) yields C7=429, and Case 2 (k trees) gives (C1)4=14=1. The difference is significant. I'm building a generalized formula for this ratio to determine its behavior as n and k vary. Also, I must ensure precision in these Catalan number calculations and double-check my indexing.

Analyzing Count Differences

I've formulated the ratio for the two scenarios: a single tree versus multiple trees, each with a fixed symbol sequence. Calculations show that the scenario with the k-trees count is substantially smaller, confirming initial observations. The Python script computes and verifies the relationship between n and k variations.

Synthesizing Count Ratios

The problem you're describing, where you have a fixed sequence of symbols and you want to count the unique ways to group them (parenthesize them), is directly related to Catalan numbers. The standard application of Catalan numbers for parenthesization states that the number of ways to parenthesize a sequence of m items (e.g., m matrices in a chain multiplication) is given by the (m−1)th Catalan number, Cm−1.

The nth Catalan number, Cn, can be calculated using the formula: Cn=n+11(n2n)

Let's define the two cases you've presented based on this standard interpretation:

Definitions:

Ntotal: The total number of symbols (n in your query).
Ktrees: The number of trees (k in your query).
Nper_tree: The number of symbols per tree, calculated as Ntotal/Ktrees (n/k in your query).
Your conditions state that Ktrees is a multiple of 4, and Ntotal=r×Ktrees for some integer r.

Case 1: Single Tree Parenthesization If all Ntotal symbols are in a single tree, the number of unique regroupings (parenthesizations) is given by the (Ntotal−1)th Catalan number: Count1=CNtotal−1

Case 2: K Separate Trees Parenthesization If the Ntotal symbols are partitioned into Ktrees separate trees, with Nper_tree symbols in each tree: Each individual tree, having Nper_tree symbols, will have CNper_tree−1 unique regroupings. Since there are Ktrees such independent trees, the total number of combinations is the product of the possibilities for each tree: Count2=(CNper_tree−1)Ktrees

To illustrate and determine which case has a lesser count and their proportions, let's use the Python code below to calculate these values for a few example scenarios that satisfy your conditions.

Here's the Python code to compute and compare the counts:

Python

import math

def catalan(n):

if n < 0:

return 0

# Cn = 1 / (n + 1) * C(2n, n)

try:

return math.comb(2 * n, n) // (n + 1)

except AttributeError:

# Fallback for older Python versions if math.comb is not available

if n == 0:

return 1

numerator = math.factorial(2 * n)

denominator = math.factorial(n + 1) * math.factorial(n)

return numerator // denominator

def calculate_catalan_counts(n_total, k_trees):

if k_trees <= 0 or n_total <= 0:

return "Invalid input: n and k must be positive."

if n_total % k_trees != 0:

return "Invalid input: n must be divisible by k for n/k symbols per tree."

n_per_tree = n_total // k_trees

# Case 1: All n symbols in a single tree

# For N_total symbols, the number of parenthesizations is C_{N_total-1}

count_case1 = catalan(n_total - 1)

# Case 2: k trees separately, with n/k symbols in each

# For n_per_tree symbols, the number of parenthesizations is C_{n_per_tree-1}

# This must be raised to the power of k_trees.

count_case2_per_tree = catalan(n_per_tree - 1)

count_case2 = count_case2_per_tree ** k_trees

return count_case1, count_case2

# Example values based on the conditions: k is a multiple of 4, n = r * k

examples = [

{"n": 4, "k": 4}, # r=1, n/k=1

{"n": 8, "k": 4}, # r=2, n/k=2

{"n": 12, "k": 4}, # r=3, n/k=3

{"n": 16, "k": 8}, # r=2, n/k=2

{"n": 20, "k": 4}, # r=5, n/k=5

]

results = []

for ex in examples:

n = ex["n"]

k = ex["k"]

c1, c2 = calculate_catalan_counts(n, k)

if isinstance(c1, str): # Error case

results.append(f"For n={n}, k={k}: {c1}")

continue

proportion = c2 / c1 if c1 != 0 else float('inf') # Handle division by zero if C_{n-1} is 0

results.append({

"n_total": n,

"k_trees": k,

"symbols_per_tree": n // k,

"case1_count": c1,

"case2_count": c2,

"proportion_case2_to_case1": proportion

})

# Print results in a readable format.

output_str = "Comparison of Catalan Number Counts:\n\n"

for res in results:

if isinstance(res, str):

output_str += res + "\n\n"

else:

output_str += f"For n (total symbols) = {res['n_total']}, k (number of trees) = {res['k_trees']}:\n"

output_str += f" Symbols per tree (n/k) = {res['symbols_per_tree']}\n"

output_str += f" Case 1 (Single tree: C_{res['n_total']-1}) Count: {res['case1_count']}\n"

output_str += f" Case 2 (k trees: (C_{res['symbols_per_tree']-1})^k) Count: {res['case2_count']}\n"

if res['case1_count'] > res['case2_count']:

output_str += " Which case has lesser count: Case 2\n"

elif res['case1_count'] < res['case2_count']:

output_str += " Which case has lesser count: Case 1\n"

else:

output_str += " Which case has lesser count: Both are equal\n"

output_str += f" Proportion (Case 2 / Case 1): {res['proportion_case2_to_case1']:.10f}\n\n"

print(output_str)

Code output

Comparison of Catalan Number Counts:

For n (total symbols) = 4, k (number of trees) = 4:

Symbols per tree (n/k) = 1

Case 1 (Single tree: C_3) Count: 5

Case 2 (k trees: (C_0)^k) Count: 1

Which case has lesser count: Case 2

Proportion (Case 2 / Case 1): 0.2000000000

For n (total symbols) = 8, k (number of trees) = 4:

Symbols per tree (n/k) = 2

Case 1 (Single tree: C_7) Count: 429

Case 2 (k trees: (C_1)^k) Count: 1

Which case has lesser count: Case 2

Proportion (Case 2 / Case 1): 0.0023310023

For n (total symbols) = 12, k (number of trees) = 4:

Symbols per tree (n/k) = 3

Case 1 (Single tree: C_11) Count: 58786

Case 2 (k trees: (C_2)^k) Count: 16

Which case has lesser count: Case 2

Proportion (Case 2 / Case 1): 0.0002721736

For n (total symbols) = 16, k (number of trees) = 8:

Symbols per tree (n/k) = 2

Case 1 (Single tree: C_15) Count: 9694845

Case 2 (k trees: (C_1)^k) Count: 1

Which case has lesser count: Case 2

Proportion (Case 2 / Case 1): 0.0000001031

For n (total symbols) = 20, k (number of trees) = 4:

Symbols per tree (n/k) = 5

Case 1 (Single tree: C_19) Count: 1767263190

Case 2 (k trees: (C_4)^k) Count: 38416

Which case has lesser count: Case 2

Proportion (Case 2 / Case 1): 0.0000217376

Nature of Catalan Numbers Counting Ratios:

The calculations above clearly show the nature of the counting ratios between the two cases you described.

Summary of Findings:

Which case will have a lesser count? In almost all scenarios where Nper_tree (symbols per tree, n/k) is greater than 1, Case 2 (partitioning into Ktrees separate trees) will have a significantly lesser count of possible unique regroupings compared to Case 1 (all Ntotal symbols in a single tree).

When Nper_tree=1, Case 2 count will be 1, while Case 1 count will be CKtrees−1. If Ktrees>1, Case 1 will be greater.
When Nper_tree=2, Case 2 count will be 1 (since C2−1=C1=1), while Case 1 count will be C2Ktrees−1. For Ktrees≥1, Case 1 will generally be much larger.

The only instance where Case 2 might not be strictly less is if Ntotal is very small such that CNtotal−1 is also 1. For example, if Ntotal=1, then C0=1. But this implies Nper_tree=1 and Ktrees=1, making both cases essentially the same.

What are the proportions? The proportion of Case 2 count to Case 1 count (i.e., Count2/Count1) is generally a very small number, approaching zero rapidly as Ntotal and Nper_tree increase.

Let's examine the examples:

For n = 4, k = 4 (1 symbol per tree):

Case 1 (Single tree): C3=5 ways
Case 2 (4 trees): (C0)4=(1)4=1 way
Proportion (Case 2 / Case 1): 1/5=0.20

For n = 8, k = 4 (2 symbols per tree):

Case 1 (Single tree): C7=429 ways
Case 2 (4 trees): (C1)4=(1)4=1 way
Proportion (Case 2 / Case 1): 1/429≈0.0023

For n = 12, k = 4 (3 symbols per tree):

Case 1 (Single tree): C11=58786 ways
Case 2 (4 trees): (C2)4=(2)4=16 ways
Proportion (Case 2 / Case 1): 16/58786≈0.00027

For n = 16, k = 8 (2 symbols per tree):

Case 1 (Single tree): C15=9694845 ways
Case 2 (8 trees): (C1)8=(1)8=1 way
Proportion (Case 2 / Case 1): 1/9694845≈0.0000001

For n = 20, k = 4 (5 symbols per tree):

Case 1 (Single tree): C19=1767263190 ways
Case 2 (4 trees): (C4)4=(14)4=38416 ways
Proportion (Case 2 / Case 1): 38416/1767263190≈0.0000217

Conclusion:

The results consistently demonstrate that when you partition a set of symbols into multiple trees and count the independent regroupings within each tree (Case 2), the total number of combinations is dramatically smaller than when you consider all symbols as part of a single large tree and count its regroupings (Case 1). This is because the exponential growth of Catalan numbers means that breaking a large problem into smaller, independent sub-problems (even when raised to a power) leads to a much smaller total number of combinations.

Great — you're now focusing on the structured case where:

n = r × k (i.e., total number of symbols n is a multiple of number of trees k)
Each tree will contain exactly r = n/k symbols (so uniform distribution)
You are comparing:

Case 1: All n symbols in a single tree → number of bracketings = Catalan(n−1)
Case 2: Partitioned into k trees of r symbols each → total bracketings = (Catalan(r−1))^k

🧮 Catalan Number Reminder:

Catalan(n)=1n+1(2nn)\text{Catalan}(n) = \frac{1}{n+1} \binom{2n}{n}Catalan(n)=n+11(n2n)

Let’s now compare both cases with actual numbers for different (n, k) pairs:

✅ Examples:

🔹 When n = 16, k = 4 → r = 4

Case 1: Catalan(15) = 9694845
Case 2: Catalan(3)^4 = 5^4 = 625
Ratio: 9694845 / 625 ≈ 15,511

🔹 When n = 20, k = 4 → r = 5

Case 1: Catalan(19) = 1767263190
Case 2: Catalan(4)^4 = 14^4 = 38416
Ratio: 1767263190 / 38416 ≈ 46000

🔹 When n = 24, k = 4 → r = 6

Case 1: Catalan(23) = 1289904147324
Case 2: Catalan(5)^4 = 42^4 = 3111696
Ratio: 1289904147324 / 3111696 ≈ 414,700

🔹 When n = 32, k = 4 → r = 8

Case 1: Catalan(31) = 3814986502092304
Case 2: Catalan(7)^4 = 429^4 = 33871089681
Ratio: 3814986502092304 / 33871089681 ≈ 112658

📊 Observations:

Case 1 always grows explosively faster than Case 2 as n increases.
The gap increases exponentially: the larger r = n/k, the larger Catalan(r−1), but Catalan(n−1) grows much faster.
The ratio roughly approximates:

Catalan(n−1)Catalan(r−1)kwhere r=nk\frac{Catalan(n-1)}{Catalan(r-1)^k} \quad \text{where } r = \frac{n}{k}Catalan(r−1)kCatalan(n−1)where r=kn

💡 Conclusion:

Partitioning into multiple trees (Case 2) dramatically reduces the total number of bracketings.
For performance-critical or visual clarity-sensitive applications (like dendrogram drawing, Vistaar tree synthesis), Case 2 is strongly preferable.
For mathematical richness and theoretical completeness, Case 1 is exhaustive but very expensive.

Let me know if you want a chart or C# code to automatically compute this for any (n, k).

📊 Key Insights from the Graph

· The ratio grows exponentially as n increases and as k increases.

· Case 1 always yields a higher count of bracketings than Case 2.

· The logarithmic scale on the y-axis shows how dramatically the count difference increases.

· Larger k (more trees) leads to much smaller total combinations in Case 2.

✅ Conclusion

· Case 2 (splitting into multiple trees) is combinatorially much lighter.

· The proportional difference becomes astronomically large for bigger n and k.

Would you like a table of exact values or a downloadable CSV file for further analysis?

📊 Key Insights from the Graph

· The ratio grows exponentially as n increases and as k increases.

· Case 1 always yields a higher count of bracketings than Case 2.

· The logarithmic scale on the y-axis shows how dramatically the count difference increases.

· Larger k (more trees) leads to much smaller total combinations in Case 2.

✅ Conclusion

· Case 2 (splitting into multiple trees) is combinatorially much lighter.

· The proportional difference becomes astronomically large for bigger n and k.

Would you like a table of exact values or a downloadable CSV file for further analysis?

I HAVE n numbers of symbols (obviously we cannot change the sequences of the symbols) but we can partition these symbols into k number of trees with | partitioning such that there are n/k number of symbols per tree... what are the nature of catalan numbers counting ratios in two cases where case 1 is if we take all n symbols in single tree and do bracketing Catalan (n) and case 2 is bracketing k trees seperately with n/k symbols in each tree and then (Catalan (n/k))^k numbers of possible combinations are done ... which case will have lesser count and what are the proportions????????

Great combinatorial question! Let's carefully compare the growth and ratios of Catalan counts in the following two cases:

❓ Setup:

You have n symbols.
You want to compare bracketing possibilities (i.e., number of distinct full binary trees or valid parenthesizations) in two different ways:

🔹 Case 1: All n symbols in a single tree

👉 Number of bracketings:

Cn−1=1n(2(n−1)n−1)C_{n-1} = \frac{1}{n} \binom{2(n-1)}{n-1}Cn−1=n1(n−12(n−1))

🔹 Case 2: Partition n symbols into k equal-sized subtrees of size n/k

Number of trees: k
Symbols per tree: n/k
Each has C(n/k)−1C_{(n/k) - 1}C(n/k)−1 bracketing options
👉 Total bracketings:

(C(n/k)−1)k\left(C_{(n/k)-1}\right)^k(C(n/k)−1)k

🔍 Now let’s compare C_{n-1} vs. (C(n/k)−1)k\left(C_{(n/k)-1}\right)^k(C(n/k)−1)k

✅ Let’s try with real numbers:

▶️ Example: n = 12, try partitioning into k = 3 → so 4 symbols per tree

Case 1: all 12 in one tree → C₁₁ = 58786
Case 2: 3 trees of 4 symbols each → C₃ = 5, and total = 53=1255^3 = 12553=125

✅ So:

58786≫125\boxed{58786 \gg 125}58786≫125

▶️ Example: n = 16, k = 4 → 4 symbols per tree

Case 1: C₁₅ = 9694845
Case 2: C34=54=625C_3^4 = 5^4 = 625C34=54=625

✅ Again:

9694845≫625\boxed{9694845 \gg 625}9694845≫625

▶️ Example: n = 16, k = 2 → 8 symbols per tree

Case 1: C₁₅ = 9694845
Case 2: C72=4292=184041C_7^2 = 429^2 = 184041C72=4292=184041

✅ Now ratio:

C15C72=9694845184041≈52.7\frac{C_{15}}{C_7^2} = \frac{9694845}{184041} \approx 52.7C72C15=1840419694845≈52.7

✅ Conclusion

🔷 General Pattern:

Cn−1≫(C(n/k)−1)kC_{n-1} \gg \left(C_{(n/k)-1}\right)^kCn−1≫(C(n/k)−1)k

The more trees you partition into, the smaller the total number of combinations becomes (even if raised to power k), because the Catalan function grows faster than exponential:

Cn∼4nn3/2π(asymptotically)C_n \sim \frac{4^n}{n^{3/2} \sqrt{\pi}} \quad (\text{asymptotically})Cn∼n3/2π4n(asymptotically)

So for any fixed n, bracketing all symbols together (Case 1) always yields more combinations than bracketing smaller trees independently (Case 2).

🧠 Insight:

You lose combinatorial richness when you break the sequence into smaller isolated trees — because cross-tree associations aren't allowed anymore.

Let me know if you want a C# method that computes and compares these values programmatically for any n and k.

I HAVE  n numbers of symbols (obviously we cannot change the sequences of the symbols) but we can partition these symbols into  k number of trees  with | partitioning  such that there are n/k number of symbols per tree... what are the nature of catalan numbers counting ratios  in two cases where   case 1 is    if we take all n symbols in single tree and do bracketing Catalan (n) and  case 2 is    bracketing k trees seperately with n/k  symbols in each tree and then  (Catalan (n/k))^k    numbers of possible combinations are done ... which case will have lesser count and what are the proportions????????

Copilot said:

Copilot

Here’s a detailed comparison of the two cases you described using Catalan numbers:

📘 Definitions Recap

· Case 1: All n symbols are grouped into a single tree →
Total groupings = Catalan(n)

· Case 2: The n symbols are evenly split into k trees, each with n/k symbols →
Total groupings = (Catalan(n/k))^k

📊 Observations from Computation

For various values of n and valid divisors k:

n	k	Catalan(n)	(Catalan(n/k))^k	Ratio (Case1 / Case2)
6	1	132	132	1.00
6	2	132	25	5.28
6	3	132	8	16.50
6	6	132	1	132.00
10	2	16,796	1,764	9.52
10	5	16,796	32	524.88
10	10	16,796	1	16,796.00
20	4	6,564,120,420	3,111,696	2,109.50
20	10	6,564,120,420	1,024	6,410,273.85

✅ Key Insights

· Case 1 (single tree) always produces more groupings than Case 2.

· The ratio grows exponentially as k increases (i.e., more trees with fewer symbols).

· This is because Catalan numbers grow super-exponentially, so splitting into smaller trees reduces the combinatorial explosion.

📈 Conclusion

· Case 2 (splitting into k trees) is much more efficient in terms of reducing the number of groupings.

· The ratio between Case 1 and Case 2 can be thousands or even millions depending on n and k.

Would you like a graph to visualize how this ratio grows with n and k?

//USED HASHSETS TO REMOVE DUPLIKATES and it is working great

// the problem is that it is concatenating all bracketed options of one tree first and then it is generating the catalans arrangement grouping for the next tree

// this is not the good things for the vistaars To get the proper vistaring we need to concatenate all trees in order in the forest first then grouping all the arrangements of all trees in the forest again in this ways we need to get all forest with one option first and then do the re foresting ...

public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CONCATENATES_ALL_GROUP_OPTIONS_OF_ONE_TREE_THENaLL_OPTIONS_OF_NEXT_TREE

{

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.Clear();

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.Clear();

string input = input_forest_string.Replace("(", " ").Replace(")", " ").Replace("&", " ");

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var trees = ParseTrees(input);

var seenGroupings = new HashSet<string>();

foreach (var treeSymbols in trees)

{

string joined = string.Join(" ", treeSymbols);

string header = "|" + joined;

if (seenGroupings.Add(header))

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(header);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(header);

}

var groupings = GenerateGroupings(treeSymbols);

foreach (var grouping in groupings)

{

string bracketed = "|" + grouping;

if (seenGroupings.Add(bracketed))

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(bracketed);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(bracketed);

}//if (seenGroupings.Add(bracketed))

}//foreach (var grouping in groupings)

}//foreach (var treeSymbols in trees)

return PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.ToString();

}//public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

public static string ReformatAndExpandForestSyntax_Stagewise(string input)

{

var output = new StringBuilder();

cleanedInput = System.Text.RegularExpressions.Regex.Replace(cleanedInput, "\\|{2,}", "|");

var trees = cleanedInput.Split('|').Where(t => !string.IsNullOrWhiteSpace(t)).ToList();

var seen = new HashSet<string>();

foreach (var tree in trees)

{

var tokens = tree.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries);

if (tokens.Length == 0) continue;

string treeOriginal = "|" + string.Join(" ", tokens) + "|";

if (seen.Add(treeOriginal)) output.AppendLine(treeOriginal);

string treeStartDash = "|-" + string.Join(" ", tokens) + "|";

if (seen.Add(treeStartDash)) output.AppendLine(treeStartDash);

for (int j = 0; j < tokens.Length - 1; j++)

{

var stagedTokens = new List<string>();

for (int k = 0; k < tokens.Length; k++)

{

stagedTokens.Add(tokens[k]);

if (k == j) stagedTokens.Add("-");

}//for (int k = 0; k < tokens.Length; k++)

string stagedTree = "|" + string.Join(" ", stagedTokens) + "|";

if (seen.Add(stagedTree)) output.AppendLine(stagedTree);

}//for (int j = 0; j < tokens.Length - 1; j++)

string treeEndDash = "|" + string.Join(" ", tokens) + "-|";

if (seen.Add(treeEndDash)) output.AppendLine(treeEndDash);

}

return output.ToString();

}//foreach (var tree in trees)

public static List<List<string>> ParseTrees(string input)

{

List<List<string>> trees = new List<List<string>>();

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var splitTrees = input.Split('|');

foreach (var treeContent in splitTrees)

{

var symbols = treeContent.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries).ToList();

if (symbols.Count > 0)

{

if(!trees.Contains(symbols))

{

//to avoid the duplikates

trees.Add(symbols);

}//if(!trees.Contains(symbols))

}//if (symbols.Count > 0)

}//foreach (var treeContent in splitTrees)

return trees;

}//public static List<List<string>> ParseTrees(string input)

public static List<string> GenerateGroupings(List<string> symbols)

{

var results = new HashSet<string>();

Generate(symbols, results);

return results.ToList();

}//public static List<string> GenerateGroupings(List<string> symbols)

public static void Generate(List<string> symbols, HashSet<string> results)

{

if (symbols.Count == 1)

{

results.Add(symbols[0]);

return;

}// if (symbols.Count == 1)

for (int i = 1; i < symbols.Count; i++)

{

var left = symbols.Take(i).ToList();

var right = symbols.Skip(i).ToList();

var leftGroupings = GenerateGroupings(left);

var rightGroupings = GenerateGroupings(right);

foreach (var l in leftGroupings)

{

foreach (var r in rightGroupings)

{

if(!results.Contains("(" + l + " " + r + ")"))

{

//to avoid duplikates

results.Add("(" + l + " " + r + ")");

}//if(!results.Contains("(" + l + " " + r + ")"))

}//foreach (var r in rightGroupings)

}//foreach (var l in leftGroupings)

}//for (int i = 1; i < symbols.Count; i++)

}//public static void Generate(List<string> symbols, HashSet<string> results)

}//public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CONCATENATES_ALL_GROUP_OPTIONS_OF_ONE_TREE_THENaLL_OPTIONS_OF_NEXT_TREE

//STRICT NOTE THAT HASHSET WAS NOT SUFFICIENT TO REMOVE DUPLICATES AND THIS CODE IS GENERATING THE NON DUPLICATE LISTS SO YOU WILL NOT CHANGE THE BUSINESS LOGIC HERE.

//STRICTLY KEEP IN YOUR MIND THAT THIS IS GENERATING PERFECT BRACKETINGS (UNIQUE BRACKETING FOR THE TREES) BUT THIS CODE HAS A PROBLEM IS THAT WHEN IT IS GENERATING THE FINAL CONCATENATED REPORT IT IS CONCATENATING ALL OPTIONS OF K_TH TREE THEN CONCATENATES ALL BRACKETED OPTIONS OPTIONS OF K_PLUS_1_TH TREE IN THE FOREST WE NEED TO MODIFY THAT

//WE KNOW THAT IF THE WHOLE FOREST HAS N SYMBOLS(STRICTLY ORDERED SEQUENCE OF N SYMBOLS THEN CATALAN(N) >>>>>> (CARTESIAN PRODUCT(ALL OPTIONS OF ALL TREES IN CATALAN BRACKETED SEQUENCES ON TREES)) SO WE HAVE PARTITIONED THE SYMBOLS (WHICH ARE NODES IN THE FOREST AS STRICT LINEAR SEQUENCE OF N SYMBOLS AS A SINGLE FOREST OF LEAVES WE HAVE PARTITIONED THE STRICT ORDERED SEQUENCE INTO K NUMBER OF TREES WITH N/K NUMBERS OF SYMBOLS IN EACH TREES THEN DID THE CATALANS BRACKETING ON EACH TREE TO GET OPTIONS ON THE TREES THIS MEANS AFTER GENERATING ALL THE CATALANLY BRACKETED OPTIONS OF TREES WE NEED THE FINAL LIST AS CARTESIAN PRODUCT OF CHOOSING ALL THE I TH OPTION FROM FIRST TREE CONCATENATES AS CARTESIAN PRODUCT WITH ALL J TH OPTIONS OF SECOND TREE ... TO ALL OPTIONS OF N TH TREE SUCH THAT WHEN THE FINAL SEQUENCE (OF NODES OF FOREST )WILL COME THEN THE SYMBOLS ORDER OF INPUT FOREST REMAIN SAME CONTIGUOUSLY ...AND THIS WAYS THE ORDERED CHAIN OF BRACKETED SYMBOLS EXAMPLE TO CLARIFY THAT (IF WE REMOVE THE "(" AND ")" FROM CONCATENATED CARTESIAN PRODUCTS OF BRACKETED SEQUENCES CATALAN(N/K) ^ K NUMBER OF SYMBOLS CHAIN WILL HAVE R COPIES OF INITIAL FOREST...

//IN CURRENT CODE WE ARE NOT GETTING SEQUENCE OF FOREST REPEAT WITH BRACKETED OPTIONS INSTEAD WE ARE GETTING ALL COPIES OF ONE TREE OPTIONS FIRST THEN CONCATENATING ALL OPTIONS OF SECOND TREE ... CONTUINUING ... THEN ALL OPTIONS OF LAST TREE CONCATENATES... THAT ULTIMATELY CHANGES THE FORESTS ORIGINAL SEQUENCE OF SYMBOLS NODES

SO REVISE THE CODE (KEEP THE DUPLICACY MANAGEMENT SYSTEMS AS IT IS AND GENERATE THE OPTIONS OF TREES AS THE BELOW CODE IS DOING ... BUT WHILE GENERATING THE STRING BUILDER TORETURN THE REPORTS ... CONCATENATE CHOOSE I TH OF TREE1 OPTIONS CARTESIAN WITH CHOOSE J TH TREE2 OPTION ... IN THIS WAY ... FOR ALL TREES SUCH THAT ORIGINAL SEQUUENCE OF FOREST (AS INPUT STRING) COPIES R TIMES (DUE TO CARTESIAN PRODUCTS AND CATALANING BRACKETING... THE REPORT WILL HAVE BRACKETS BUT THE FORESTS INPUT SEQUENCE WILL REMAIN INTACT AND THE COPIES OF BRACKETED SYMBOL BUNCHES RESEMBLES R TIMES REPEAT OF THE ORIGINAL FORESTS SYMBOLS ORDERS

//USED HASHSETS TO REMOVE DUPLIKATES and it is working great

// the problem is that it is concatenating all bracketed options of one tree first and then it is generating the catalans arrangement grouping for the next tree

public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CONCATENATES_ALL_GROUP_OPTIONS_OF_ONE_TREE_THENaLL_OPTIONS_OF_NEXT_TREE

{

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.Clear();

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.Clear();

string input = input_forest_string.Replace("(", " ").Replace(")", " ").Replace("&", " ");

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var trees = ParseTrees(input);

var seenGroupings = new HashSet<string>();

foreach (var treeSymbols in trees)

{

string joined = string.Join(" ", treeSymbols);

string header = "|" + joined;

if (seenGroupings.Add(header))

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(header);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(header);

}

var groupings = GenerateGroupings(treeSymbols);

foreach (var grouping in groupings)

{

string bracketed = "|" + grouping;

if (seenGroupings.Add(bracketed))

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(bracketed);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(bracketed);

}//if (seenGroupings.Add(bracketed))

}//foreach (var grouping in groupings)

}//foreach (var treeSymbols in trees)

return PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.ToString();

}//public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

public static string ReformatAndExpandForestSyntax_Stagewise(string input)

{

var output = new StringBuilder();

cleanedInput = System.Text.RegularExpressions.Regex.Replace(cleanedInput, "\\|{2,}", "|");

var trees = cleanedInput.Split('|').Where(t => !string.IsNullOrWhiteSpace(t)).ToList();

var seen = new HashSet<string>();

foreach (var tree in trees)

{

var tokens = tree.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries);

if (tokens.Length == 0) continue;

string treeOriginal = "|" + string.Join(" ", tokens) + "|";

if (seen.Add(treeOriginal)) output.AppendLine(treeOriginal);

string treeStartDash = "|-" + string.Join(" ", tokens) + "|";

if (seen.Add(treeStartDash)) output.AppendLine(treeStartDash);

for (int j = 0; j < tokens.Length - 1; j++)

{

var stagedTokens = new List<string>();

for (int k = 0; k < tokens.Length; k++)

{

stagedTokens.Add(tokens[k]);

if (k == j) stagedTokens.Add("-");

}//for (int k = 0; k < tokens.Length; k++)

string stagedTree = "|" + string.Join(" ", stagedTokens) + "|";

if (seen.Add(stagedTree)) output.AppendLine(stagedTree);

}//for (int j = 0; j < tokens.Length - 1; j++)

string treeEndDash = "|" + string.Join(" ", tokens) + "-|";

if (seen.Add(treeEndDash)) output.AppendLine(treeEndDash);

}

return output.ToString();

}//foreach (var tree in trees)

public static List<List<string>> ParseTrees(string input)

{

List<List<string>> trees = new List<List<string>>();

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var splitTrees = input.Split('|');

foreach (var treeContent in splitTrees)

{

var symbols = treeContent.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries).ToList();

if (symbols.Count > 0)

{

if(!trees.Contains(symbols))

{

//to avoid the duplikates

trees.Add(symbols);

}//if(!trees.Contains(symbols))

}//if (symbols.Count > 0)

}//foreach (var treeContent in splitTrees)

return trees;

}//public static List<List<string>> ParseTrees(string input)

public static List<string> GenerateGroupings(List<string> symbols)

{

var results = new HashSet<string>();

Generate(symbols, results);

return results.ToList();

}//public static List<string> GenerateGroupings(List<string> symbols)

public static void Generate(List<string> symbols, HashSet<string> results)

{

if (symbols.Count == 1)

{

results.Add(symbols[0]);

return;

}// if (symbols.Count == 1)

for (int i = 1; i < symbols.Count; i++)

{

var left = symbols.Take(i).ToList();

var right = symbols.Skip(i).ToList();

var leftGroupings = GenerateGroupings(left);

var rightGroupings = GenerateGroupings(right);

foreach (var l in leftGroupings)

{

foreach (var r in rightGroupings)

{

if(!results.Contains("(" + l + " " + r + ")"))

{

//to avoid duplikates

results.Add("(" + l + " " + r + ")");

}//if(!results.Contains("(" + l + " " + r + ")"))

}//foreach (var r in rightGroupings)

}//foreach (var l in leftGroupings)

}//for (int i = 1; i < symbols.Count; i++)

}//public static void Generate(List<string> symbols, HashSet<string> results)

}//public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CONCATENATES_ALL_GROUP_OPTIONS_OF_ONE_TREE_THENaLL_OPTIONS_OF_NEXT_TREE

public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CROSS_JOIN_FINAL_FORESTS_ORDERED

{

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.Clear();

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.Clear();

string input = input_forest_string.Replace("(", " ").Replace(")", " ").Replace("&", " ");

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var trees = ParseTrees(input);

var allGroupingsPerTree = new List<List<string>>();

foreach (var treeSymbols in trees)

{

var treeGroupings = GenerateGroupings(treeSymbols);

treeGroupings.Insert(0, string.Join(" ", treeSymbols)); // Include the original flat tree

allGroupingsPerTree.Add(treeGroupings);

}

var cartesianProduct = CartesianConcatenation(allGroupingsPerTree);

var seenGroupings = new HashSet<string>();

foreach (var fullForest in cartesianProduct)

{

string forestStr = string.Join("|", fullForest.Select(t => t.Trim()));

forestStr = "|" + forestStr + "|";

if (seenGroupings.Add(forestStr))

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(forestStr);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(forestStr);

}

return PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.ToString();

}

public static string ReformatAndExpandForestSyntax_Stagewise(string input)

{

var output = new StringBuilder();

cleanedInput = System.Text.RegularExpressions.Regex.Replace(cleanedInput, "\\|{2,}", "|");

var trees = cleanedInput.Split('|').Where(t => !string.IsNullOrWhiteSpace(t)).ToList();

var seen = new HashSet<string>();

foreach (var tree in trees)

{

var tokens = tree.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries);

if (tokens.Length == 0) continue;

string treeOriginal = "|" + string.Join(" ", tokens) + "|";

if (seen.Add(treeOriginal)) output.AppendLine(treeOriginal);

string treeStartDash = "|-" + string.Join(" ", tokens) + "|";

if (seen.Add(treeStartDash)) output.AppendLine(treeStartDash);

for (int j = 0; j < tokens.Length - 1; j++)

{

var stagedTokens = new List<string>();

for (int k = 0; k < tokens.Length; k++)

{

stagedTokens.Add(tokens[k]);

if (k == j) stagedTokens.Add("-");

}

string stagedTree = "|" + string.Join(" ", stagedTokens) + "|";

if (seen.Add(stagedTree)) output.AppendLine(stagedTree);

}

string treeEndDash = "|" + string.Join(" ", tokens) + "-|";

if (seen.Add(treeEndDash)) output.AppendLine(treeEndDash);

}

return output.ToString();

}

public static List<List<string>> ParseTrees(string input)

{

var trees = new List<List<string>>();

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var splitTrees = input.Split('|');

foreach (var treeContent in splitTrees)

{

var symbols = treeContent.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries).ToList();

if (symbols.Count > 0)

{

trees.Add(symbols);

}

return trees;

}

public static List<string> GenerateGroupings(List<string> symbols)

{

var results = new HashSet<string>();

Generate(symbols, results);

return results.ToList();

}

public static void Generate(List<string> symbols, HashSet<string> results)

{

if (symbols.Count == 1)

{

results.Add(symbols[0]);

return;

}

for (int i = 1; i < symbols.Count; i++)

{

var left = symbols.Take(i).ToList();

var right = symbols.Skip(i).ToList();

var leftGroupings = GenerateGroupings(left);

var rightGroupings = GenerateGroupings(right);

foreach (var l in leftGroupings)

{

foreach (var r in rightGroupings)

{

results.Add("(" + l + " " + r + ")");

}

public static List<List<string>> CartesianConcatenation(List<List<string>> lists)

{

var result = new List<List<string>> { new List<string>() };

foreach (var currentList in lists)

{

var temp = new List<List<string>>();

foreach (var prefix in result)

{

foreach (var item in currentList)

{

var newList = new List<string>(prefix) { item };

temp.Add(newList);

}

result = temp;

}

return result;

}

public class TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS___CARTESIANED_VISTAARS

{

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static StringBuilder PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS = new StringBuilder();

public static string TreeGroupingGenerator___CATALANING______BRACKETING_REGROUPING_COPILOTS_Main(string input_forest_string)

{

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.Clear();

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.Clear();

string input = input_forest_string.Replace("(", " ").Replace(")", " ").Replace("&", " ");

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var trees = ParseTrees(input);

var allOptionsPerTree = new List<List<string>>();

foreach (var treeSymbols in trees)

{

var options = GenerateGroupings(treeSymbols);

options.Insert(0, string.Join(" ", treeSymbols)); // Include ungrouped

allOptionsPerTree.Add(options);

}

var cartesian = CartesianProduct(allOptionsPerTree);

foreach (var forestOption in cartesian)

{

string full = "|" + string.Join("||", forestOption.Select(opt => opt)) + "|";

PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.AppendLine(full);

PUBLIC_STATIC_STRINGBUILDER___ONLY_INDIANNOTES_GROUPINGS___CATALANED_TREES______BRACKETING_REGROUPING_COPILOTS.AppendLine(full);

}

return PUBLIC_STATIC_STRINGBUILDER____RAW_CATALANING___CATALANING______BRACKETING_REGROUPING_COPILOTS.ToString();

}

public static string ReformatAndExpandForestSyntax_Stagewise(string input)

{

var output = new StringBuilder();

cleanedInput = System.Text.RegularExpressions.Regex.Replace(cleanedInput, "\\|{2,}", "|");

var trees = cleanedInput.Split('|').Where(t => !string.IsNullOrWhiteSpace(t)).ToList();

var seen = new HashSet<string>();

foreach (var tree in trees)

{

var tokens = tree.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries);

if (tokens.Length == 0) continue;

string treeOriginal = "|" + string.Join(" ", tokens) + "|";

if (seen.Add(treeOriginal)) output.AppendLine(treeOriginal);

string treeStartDash = "|-" + string.Join(" ", tokens) + "|";

if (seen.Add(treeStartDash)) output.AppendLine(treeStartDash);

for (int j = 0; j < tokens.Length - 1; j++)

{

var stagedTokens = new List<string>();

for (int k = 0; k < tokens.Length; k++)

{

stagedTokens.Add(tokens[k]);

if (k == j) stagedTokens.Add("-");

}

string stagedTree = "|" + string.Join(" ", stagedTokens) + "|";

if (seen.Add(stagedTree)) output.AppendLine(stagedTree);

}

string treeEndDash = "|" + string.Join(" ", tokens) + "-|";

if (seen.Add(treeEndDash)) output.AppendLine(treeEndDash);

}

return output.ToString();

}

public static List<List<string>> ParseTrees(string input)

{

List<List<string>> trees = new List<List<string>>();

input = System.Text.RegularExpressions.Regex.Replace(input, "\\|{2,}", "|");

var splitTrees = input.Split('|');

foreach (var treeContent in splitTrees)

{

var symbols = treeContent.Split(new[] { ' ', ',', ';' }, StringSplitOptions.RemoveEmptyEntries).ToList();

if (symbols.Count > 0)

{

trees.Add(symbols);

}

return trees;

}

public static List<string> GenerateGroupings(List<string> symbols)

{

var results = new HashSet<string>();

Generate(symbols, results);

return results.ToList();

}

public static void Generate(List<string> symbols, HashSet<string> results)

{

if (symbols.Count == 1)

{

results.Add(symbols[0]);

return;

}

for (int i = 1; i < symbols.Count; i++)

{

var left = symbols.Take(i).ToList();

var right = symbols.Skip(i).ToList();

var leftGroupings = GenerateGroupings(left);

var rightGroupings = GenerateGroupings(right);

foreach (var l in leftGroupings)

{

foreach (var r in rightGroupings)

{

results.Add("(" + l + " " + r + ")");

}

public static List<List<string>> CartesianProduct(List<List<string>> sequences)

{

var result = new List<List<string>> { new List<string>() };

foreach (var sequence in sequences)

{

var temp = new List<List<string>>();

foreach (var acc in result)

{

foreach (var item in sequence)

{

var newAcc = new List<string>(acc) { item };

temp.Add(newAcc);

}

result = temp;

}

return result;

}

drinkussions(Baban and Saanjoo)

Wednesday, June 18, 2025

TRYING TO DO CARTESIANS CONCATENATIONS ON CATALAN OPTIONS ON TREES PRESERVING THE FOREST ORDERING

Copilot said:

Copilot said:

📊 Key Insights from the Graph

✅ Conclusion

📊 Key Insights from the Graph

✅ Conclusion

Copilot said:

📘 Definitions Recap

📊 Observations from Computation

✅ Key Insights

📈 Conclusion

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