What keeps this engine fast — even if the semantic tree grows infinitely — is a fundamental computer science shift:
It’s the difference between O(n) and O(k).
Searching O(n) means scanning every piece of hay to find a needle.
Working in O(k) means going directly to the needle.
That’s what your Incremental Recompute (Phase 8) achieves — and why we’re seeing ~15ms recompute times.
⸻
1. The Inverted Dependency Index
In a traditional system (O(n)), if gas prices change, the system would need to scan everything to see what’s affected.
In .me, when you declare:
me.trucks["[i]"]["="]("cost", "gasoline * 20")
The kernel doesn’t just store a formula —
it builds a subscription map.
It knows:
“cost depends on gasoline.”
• n = total nodes in the system (could be millions)
• k = only the nodes directly depending on what changed
⸻
2. Surgical Updates
When you run:
me.finance.fuel_price(30)
The kernel:
• Does not scan the whole tree
• Goes straight to finance.fuel_price in its index
• Looks up its subscribers
• Recomputes only those nodes
If you have 1,000,000 nodes (n), but only 3 trucks depend on fuel price (k), the engine only touches those 3.
That’s why you went from 5 seconds (recompute everything) to 15 milliseconds (recompute the affected branch).
⸻
3. No Deep Traversal
Thanks to Proxies, paths are already resolved.
The engine doesn’t navigate:
root → fleet → trucks → 1 → cost
It already knows the exact memory reference.
It’s a desktop shortcut — not a 10-folder crawl.
⸻
The Result
Your system doesn’t slow down with volume.
It scales with immediate relational complexity, not total size.
Imagine thousands of pharmacies.
A user updates their “max budget.”
Eligible results recompute in ~15ms.