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"""
Calculating node counts for uniform branching trees.
 
A uniform tree has the same branching factor at every internal node.
This makes node counting a straightforward geometric series calculation.
"""
 
def count_nodes_uniform_tree(branching_factor: int, depth: int) -> dict:
    """
    Calculate exact node counts for a uniform tree.
    
    Parameters:
    - branching_factor (b): Number of children per internal node
    - depth (d): Number of levels (root is level 0, leaves at level d)
    
    Returns: Dictionary with detailed node counts
    """
    b, d = branching_factor, depth
    
    # Nodes at each level
    nodes_per_level = [b**level for level in range(d + 1)]
    
    # Leaves are at the deepest level
    leaf_count = b**d
    
    # Internal nodes are at levels 0 through d-1
    internal_count = sum(nodes_per_level[:-1])
    
    # Total nodes using geometric series formula
    if b == 1:
        total = d + 1
    else:
        total = (b**(d + 1) - 1) // (b - 1)
    
    return {
        'branching_factor': b,
        'depth': d,
        'nodes_per_level': nodes_per_level,
        'leaf_count': leaf_count,
        'internal_count': internal_count,
        'total_nodes': total,
        'leaf_to_internal_ratio': leaf_count / max(1, internal_count)
    }
 
 
def print_uniform_tree_analysis():
    """
    Display analysis for common uniform tree configurations.
    """
    print("=" * 70)
    print("UNIFORM TREE NODE COUNTS")
    print("=" * 70)
    
    # Binary tree (subset generation)
    print("\nBINARY TREE (b=2) - Subset Generation Pattern")
    print("-" * 50)
    for depth in [5, 10, 15, 20]:
        result = count_nodes_uniform_tree(2, depth)
        print(f"  Depth {depth:2d}: {result['leaf_count']:,} leaves, "
              f"{result['total_nodes']:,} total")
    
    # Ternary tree
    print("\nTERNARY TREE (b=3)")
    print("-" * 50)
    for depth in [5, 10, 15]:
        result = count_nodes_uniform_tree(3, depth)
        print(f"  Depth {depth:2d}: {result['leaf_count']:,} leaves, "
              f"{result['total_nodes']:,} total")
    
    # Decisions with 4 choices (e.g., grid movement: up/down/left/right)
    print("\nQUATERNARY TREE (b=4) - Grid Navigation Pattern")
    print("-" * 50)
    for depth in [5, 10, 12]:
        result = count_nodes_uniform_tree(4, depth)
        print(f"  Depth {depth:2d}: {result['leaf_count']:,} leaves, "
              f"{result['total_nodes']:,} total")
    
    # Key insight: leaf dominance
    print("\nLEAF DOMINANCE DEMONSTRATION (b=2, d=10)")
    print("-" * 50)
    result = count_nodes_uniform_tree(2, 10)
    for level, count in enumerate(result['nodes_per_level']):
        percentage = 100 * count / result['total_nodes']
        bar = '#' * int(percentage / 2)
        print(f"  Level {level:2d}: {count:>6,} nodes ({percentage:5.2f}%) {bar}")
    print(f"\n  Leaves represent {result['leaf_count']/result['total_nodes']*100:.2f}% of total nodes")
 
 
# Run the analysis
print_uniform_tree_analysis()

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"""
Calculating node counts for non-uniform branching trees.
 
The classic case is permutation generation, where branching factor
decreases at each level: n, n-1, n-2, ..., 1.
"""
 
import math
 
def count_nodes_permutation_tree(n: int) -> dict:
    """
    Calculate exact node counts for a permutation tree of n elements.
    
    The tree has:
    - Root with n children
    - Level 1 nodes each have n-1 children
    - Level k nodes each have n-k children
    - Level n-1 nodes each have 1 child (leaves at level n)
    
    Parameters:
    - n: Number of elements to permute
    
    Returns: Dictionary with detailed node counts
    """
    # Nodes at each level
    nodes_per_level = []
    cumulative = 1  # Start with root
    
    for level in range(n + 1):
        if level == 0:
            nodes_per_level.append(1)
        else:
            cumulative *= (n - level + 1)
            nodes_per_level.append(cumulative)
    
    # Leaves are n!
    leaf_count = math.factorial(n)
    
    # Internal nodes are sum of levels 0 through n-1
    internal_count = sum(nodes_per_level[:-1])
    
    # Total nodes
    total = internal_count + leaf_count
    
    # Theoretical ratio: total/leaf_count should approach e
    ratio = total / leaf_count if leaf_count > 0 else 0
    
    return {
        'n': n,
        'nodes_per_level': nodes_per_level,
        'leaf_count': leaf_count,
        'internal_count': internal_count,
        'total_nodes': total,
        'total_to_factorial_ratio': ratio,
        'e_approximation': math.e
    }
 
 
def count_nodes_k_permutation_tree(n: int, k: int) -> dict:
    """
    Count nodes for k-permutation tree (arrangements of k items from n).
    
    Tree depth is k (not n), with branching factors: n, n-1, ..., n-k+1
    
    Parameters:
    - n: Total elements available
    - k: Number of positions to fill
    
    Returns: Dictionary with node counts
    """
    nodes_per_level = [1]  # Root
    cumulative = 1
    
    for level in range(1, k + 1):
        cumulative *= (n - level + 1)
        nodes_per_level.append(cumulative)
    
    leaf_count = nodes_per_level[-1]  # P(n,k) = n!/(n-k)!
    internal_count = sum(nodes_per_level[:-1])
    total = internal_count + leaf_count
    
    return {
        'n': n,
        'k': k,
        'nodes_per_level': nodes_per_level,
        'leaf_count': leaf_count,  # P(n,k)
        'internal_count': internal_count,
        'total_nodes': total
    }
 
 
def print_permutation_tree_analysis():
    """
    Display analysis of permutation tree structures.
    """
    print("=" * 70)
    print("PERMUTATION TREE NODE COUNTS")
    print("=" * 70)
    
    print("\nFULL PERMUTATION TREES")
    print("-" * 50)
    print(f"{'n':>3} | {'Leaves (n!)':>15} | {'Total Nodes':>15} | {'Ratio to n!':>10}")
    print("-" * 50)
    
    for n in range(1, 13):
        result = count_nodes_permutation_tree(n)
        print(f"{n:3d} | {result['leaf_count']:>15,} | "
              f"{result['total_nodes']:>15,} | {result['total_to_factorial_ratio']:>10.4f}")
    
    print(f"\nNote: Ratio converges to e ≈ {math.e:.4f}")
    
    # K-permutation example
    print("\n\nK-PERMUTATION TREES (selecting k from n)")
    print("-" * 50)
    n = 10
    print(f"From n={n} elements:\n")
    
    for k in range(1, n + 1):
        result = count_nodes_k_permutation_tree(n, k)
        print(f"  k={k:2d}: {result['leaf_count']:>10,} leaves (P({n},{k})), "
              f"{result['total_nodes']:>12,} total nodes")
 
 
def visualize_permutation_tree_shape(n: int):
    """
    Show the 'triangular' shape of a permutation tree.
    """
    print(f"\n\nPERMUTATION TREE SHAPE (n={n})")
    print("=" * 70)
    
    result = count_nodes_permutation_tree(n)
    max_width = 50
    max_nodes = max(result['nodes_per_level'])
    
    for level, count in enumerate(result['nodes_per_level']):
        bar_width = int((count / max_nodes) * max_width)
        bar = '█' * bar_width
        print(f"Level {level:2d} | {bar} {count:,}")
    
    print(f"\nObserve: Node count INCREASES then hits the leaf explosion at n!")
 
 
# Run all analyses
print_permutation_tree_analysis()
visualize_permutation_tree_shape(6)

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"""
Demonstrating how constraints affect tree size estimates.
 
We compare theoretical bounds with actual exploration counts
for the N-Queens problem to illustrate constraint power.
"""
 
def n_queens_estimates(n: int) -> dict:
    """
    Calculate various tree size estimates for N-Queens.
    
    Provides:
    1. Naive bound (n^n): assuming any cell is valid
    2. Permutation bound (n!): one queen per column
    3. Actual exploration: with all constraint checking
    4. Valid solutions: queens that satisfy all constraints
    """
    import math
    
    # Count actual tree nodes explored with full constraints
    nodes_explored = 0
    solutions_found = 0
    
    def is_safe(board: list, row: int, col: int) -> bool:
        """Check if placing queen at (row, col) is safe."""
        for prev_row in range(row):
            prev_col = board[prev_row]
            # Same column or diagonal
            if prev_col == col or abs(prev_col - col) == abs(prev_row - row):
                return False
        return True
    
    def solve(board: list, row: int):
        nonlocal nodes_explored, solutions_found
        nodes_explored += 1
        
        if row == n:
            solutions_found += 1
            return
        
        for col in range(n):
            if is_safe(board, row, col):
                board.append(col)
                solve(board, row + 1)
                board.pop()
    
    solve([], 0)
    
    naive_bound = n ** n
    perm_bound = math.factorial(n)
    
    return {
        'n': n,
        'naive_bound_n^n': naive_bound,
        'permutation_bound_n!': perm_bound,
        'actual_nodes_explored': nodes_explored,
        'valid_solutions': solutions_found,
        'pruning_ratio_from_naive': naive_bound / nodes_explored if nodes_explored else float('inf'),
        'pruning_ratio_from_perm': perm_bound / nodes_explored if nodes_explored else float('inf')
    }
 
 
def print_constraint_analysis():
    """
    Show how constraints reduce tree exploration across N-Queens variants.
    """
    print("=" * 80)
    print("N-QUEENS: CONSTRAINT IMPACT ON TREE SIZE")
    print("=" * 80)
    print()
    print(f"{'n':>3} | {'n^n':>12} | {'n!':>10} | {'Explored':>10} | "
          f"{'Solutions':>9} | {'Pruning Factor':>14}")
    print("-" * 80)
    
    for n in range(4, 13):
        result = n_queens_estimates(n)
        print(f"{n:3d} | {result['naive_bound_n^n']:>12,} | "
              f"{result['permutation_bound_n!']:>10,} | "
              f"{result['actual_nodes_explored']:>10,} | "
              f"{result['valid_solutions']:>9} | "
              f"{result['pruning_ratio_from_naive']:>14.1f}x")
    
    print()
    print("Key insight: Actual exploration is ORDERS OF MAGNITUDE smaller than")
    print("theoretical bounds, demonstrating the power of constraint propagation.")
 
 
def visualize_constraint_layers():
    """
    Illustrate how each constraint layer reduces the search space.
    """
    print("\n\n" + "=" * 70)
    print("CONSTRAINT LAYER VISUALIZATION (8-Queens)")
    print("=" * 70)
    
    import math
    n = 8
    
    layers = [
        ("No constraints (n^n)", n**n),
        ("One queen per row (n^n)", n**n),  # Same as above with row iteration
        ("+ One per column (n!)", math.factorial(n)),
        ("+ No diagonal attacks", 2057),  # Actual 8-queens exploration
        ("Valid solutions only", 92)
    ]
    
    print()
    max_width = 50
    base = layers[0][1]
    
    for name, count in layers:
        if count > 0:
            log_ratio = math.log10(count) / math.log10(base) if count > 1 else 0
            bar_width = int(log_ratio * max_width)
            bar = '█' * bar_width
            print(f"{name:35s}: {count:>12,}  {bar}")
    
    print()
    print("Each constraint layer EXPONENTIALLY reduces the search space.")
 
 
# Run the analysis
print_constraint_analysis()
visualize_constraint_layers()

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"""
Practical techniques for estimating search tree size
when analytical formulas are unavailable or complex.
"""
 
import math
from typing import Callable, List, Tuple
 
def estimate_complexity_from_samples(
    algorithm: Callable[[int], int],
    sample_sizes: List[int]
) -> dict:
    """
    Estimate complexity class by sampling algorithm behavior.
    
    Parameters:
    - algorithm: Function that takes input size n and returns node count
    - sample_sizes: List of n values to test
    
    Returns: Complexity analysis with predictions
    """
    # Collect samples
    samples = []
    for n in sample_sizes:
        node_count = algorithm(n)
        samples.append((n, node_count))
    
    # Analyze growth pattern
    growth_rates = []
    for i in range(1, len(samples)):
        n1, c1 = samples[i-1]
        n2, c2 = samples[i]
        
        # Ratio analysis
        ratio = c2 / c1 if c1 > 0 else 0
        
        # Check if exponential: ratio ≈ constant
        # Check if factorial: ratio ≈ n2
        growth_rates.append({
            'from_n': n1,
            'to_n': n2,
            'ratio': ratio,
            'expected_exp_ratio': 2 ** (n2 - n1),  # For O(2^n)
            'expected_fact_ratio': math.prod(range(n1 + 1, n2 + 1))  # For O(n!)
        })
    
    # Determine best-fit complexity class
    avg_ratio = sum(g['ratio'] for g in growth_rates) / len(growth_rates)
    
    # Estimate base of exponential
    log_ratio = math.log(avg_ratio) if avg_ratio > 1 else 0
    estimated_base = math.exp(log_ratio)  # Approx base if O(b^n)
    
    return {
        'samples': samples,
        'growth_rates': growth_rates,
        'average_ratio': avg_ratio,
        'estimated_exponential_base': estimated_base,
        'complexity_guess': 'O(2^n)' if 1.8 < estimated_base < 2.2 else 
                           'O(n!)' if estimated_base > 3 else
                           f'O({estimated_base:.2f}^n)'
    }
 
 
def predict_tree_size(samples: List[Tuple[int, int]], target_n: int) -> int:
    """
    Predict tree size for target_n based on observed samples.
    Uses log-linear regression for exponential extrapolation.
    """
    import statistics
    
    # Convert to log scale for linear regression
    log_counts = [(n, math.log(c)) for n, c in samples if c > 0]
    
    # Simple linear regression on log values
    n_vals = [x[0] for x in log_counts]
    log_vals = [x[1] for x in log_counts]
    
    n_mean = statistics.mean(n_vals)
    log_mean = statistics.mean(log_vals)
    
    # Calculate slope
    numerator = sum((n - n_mean) * (lv - log_mean) 
                    for n, lv in zip(n_vals, log_vals))
    denominator = sum((n - n_mean) ** 2 for n in n_vals)
    
    slope = numerator / denominator if denominator else 0
    intercept = log_mean - slope * n_mean
    
    # Predict
    predicted_log = slope * target_n + intercept
    return int(math.exp(predicted_log))
 
 
def demonstrate_estimation():
    """
    Demonstrate estimation techniques on known problems.
    """
    print("=" * 70)
    print("TREE SIZE ESTIMATION DEMONSTRATION")
    print("=" * 70)
    
    # Example: N-Queens node counting
    def n_queens_nodes(n: int) -> int:
        count = 0
        def solve(row, cols, diags, anti_diags):
            nonlocal count
            count += 1
            if row == n:
                return
            for col in range(n):
                if col in cols or (row - col) in diags or (row + col) in anti_diags:
                    continue
                cols.add(col)
                diags.add(row - col)
                anti_diags.add(row + col)
                solve(row + 1, cols, diags, anti_diags)
                cols.remove(col)
                diags.remove(row - col)
                anti_diags.remove(row + col)
        solve(0, set(), set(), set())
        return count
    
    sample_sizes = [4, 5, 6, 7, 8, 9, 10]
    samples = [(n, n_queens_nodes(n)) for n in sample_sizes]
    
    print("\nN-Queens Node Counts (actual measurements):")
    print("-" * 40)
    for n, count in samples:
        print(f"  n={n:2d}: {count:>10,} nodes")
    
    # Predict for larger n
    print("\nPredictions based on extrapolation:")
    print("-" * 40)
    for target in [11, 12, 13, 14]:
        predicted = predict_tree_size(samples, target)
        print(f"  n={target:2d}: ~{predicted:>12,} nodes (predicted)")
    
    # Verify one prediction
    print("\nVerification (computing actual):")
    actual_11 = n_queens_nodes(11)
    predicted_11 = predict_tree_size(samples, 11)
    print(f"  n=11 actual:    {actual_11:>12,}")
    print(f"  n=11 predicted: {predicted_11:>12,}")
    print(f"  Error: {abs(actual_11 - predicted_11) / actual_11 * 100:.1f}%")
 
 
# Run demonstration
demonstrate_estimation()