Tiling-Cost-Gradient_Descent

The problem

Programs often struggle to efficiently move through large amounts of data stored in memory, leading to slow performance. This happens because the way a computer accesses data can be disorganized and inefficient.

What it does

It calculates the most efficient path for a program to move through data by breaking it into smaller pieces. It then finds the best way to organize those pieces to minimize wasted effort.

Why it matters

It streamlines how software handles data memory to make processes run more smoothly.

Validation

It was run in the sandbox and it failed. run output shows an error/traceback — the artifact does NOT run clean.

$ python3 Tiling-Cost-Gradient_Descent_v2.py
File "/work/script_name.py", line 35
    print(f"Optimal tile: {n}x{m} (found in {grad_descents} steps)"
         ^
SyntaxError: '(' was never closed

No screenshot — there is nothing working to show. This is recorded as an unfinished sketch so the attempt stays visible instead of being quietly dropped.

The code

All of it — 59 lines, one file, standard library only.

# Tiling-Cost-Gradient_Descent.py
import math

def cost(tile_n, tile_m):
    tile_area = tile_n * tile_m
    total_cost = 0.0
    # Multi-level cache hierarchy parameters
    cache_levels = [
        {'size': 128, 'weight': 0.4},  # L1
        {'size': 512, 'weight': 0.3},  # L2
        {'size': 2048, 'weight': 0.3}  # L3
    ]
    
    for level in cache_levels:
        if tile_area <= level['size']:
            level_cost = 1.0 / (1.0 + math.exp(-tile_area))
        else:
            level_cost = 1.0
        total_cost += level['weight'] * level_cost
    return total_cost

# Parameters
N, M = 1024, 1024  # Loop bounds
initial_tile = 32     # Initial tile size
learning_rate = 0.1    # Gradient descent learning rate
precision = 0.001     # Convergence threshold

def main():
    n, m = initial_tile, initial_tile
    count = 0
    while count < 1000:  # Safety cap
        # Compute cost gradients via finite differences
        d_cost_dn = (cost(n + 1, m) - cost(n, m)) / 1
        d_cost_dm = (cost(n, m + 1) - cost(n, m)) / 1
        
        # Update tiles (project to positive space)
        n -= learning_rate * d_cost_dn
        m -= learning_rate * d_cost_dm
        n, m = max(1, math.floor(n)), max(1, math.floor(m))
        
        # Check convergence
        if abs(d_cost_dn) < precision and abs(d_cost_dm) < precision:
            break
        count += 1
    
    grad_descents = count
    optimal_tile_cost = cost(n, m)
    print(f"Optimal tile: {n}x{m} (found in {grad_descents} steps, total cost={optimal_tile_cost:.4f})")
    
    # Show cost breakdown
    print("\nCost breakdown by cache level:")
    current_area = n * m
    for i, level in enumerate(cache_levels):
        level_cost = (1.0 / (1.0 + math.exp(-current_area)) if current_area <= level['size'] else 1.0)
        weighted = level['weight'] * level_cost
        print(f"  Level {i+1} (size={level['size']}): {level_cost:.4f} * {level['weight']:.2f} = {weighted:.4f}")

if __name__ == "__main__":
    main()