Sravneniya & TinyLogic
Sravneniya Sravneniya
TinyLogic TinyLogic
TinyLogic TinyLogic
Hey, Sravneniya, I’ve been sketching out a logic puzzle that uses the tiniest number of gates but still forces a single, clear solution—want to see if it can stand up to your efficiency test?
Sravneniya Sravneniya
Sure, send it over—I'll dissect the gate count, check for redundancy, and verify that the solution path is unique. No fluff, just the hard numbers.
TinyLogic TinyLogic
Here’s a razor‑thin puzzle: Inputs: A, B, C Output: F = A AND (B OR C) Gate count: 2 – one OR gate (B OR C) and one AND gate (A AND …) Redundancy: None – every gate is essential. Unique solution: F = 1 only when A is 1 and at least one of B or C is 1. All other input combinations give 0.
Sravneniya Sravneniya
Looks solid. Two gates, no extra logic, and the truth table is deterministic. If you wanted to trim it further you'd need to merge the operations, but with standard gates that's impossible. The function is already minimal, so it passes the efficiency test.
TinyLogic TinyLogic
Nice audit! Now, just for fun, let me tweak the wiring a bit—imagine a tiny inverter tucked inside the OR gate, so we get A NAND (B XOR C). Same gate count, but the logic dance is a little trickier. Think you can spot the hidden parity?
Sravneniya Sravneniya
You’re looking at A NAND (B XOR C). XOR is true when B and C differ – that’s the odd‑parity condition. NAND inverts the AND of A with that result. So F = 1 if either A is 0 or B and C are the same. The only time F = 0 is when A = 1 and B ≠ C. In short, the parity shows up as “A must be 1 and B XOR C must be 1 to kill the output.”
TinyLogic TinyLogic
You nailed it—exactly the flip‑flop behavior we expected. Now, if we swap the NAND for a NOR, the truth changes but the gate count stays at two. Want to see that version?