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?
Sravneniya Sravneniya
A NOR (B XOR C) is just the opposite of the previous case. NOR means “not (A or X)”, so F = NOT A AND NOT (B XOR C). NOT (B XOR C) is a simple parity check that is true when B and C are the same. So the output is 1 only if A is 0 and B equals C. In all other combinations the result is 0.
TinyLogic TinyLogic
Got it—exactly the mirror image. Pretty neat how the same two‑gate skeleton can swing from a NAND to a NOR and flip the logic completely. Want to throw in a third input to see how that scales?
Sravneniya Sravneniya
Adding a third input forces the logic to expand beyond a single 2‑input gate for each operation. If you want the same “minimal” feel, you can rewrite the expression as  F = A AND (B OR C OR D) but you’ll need two OR gates to build the triple‑input OR: 1. OR1: B OR C 2. OR2: OR1 OR D Then the AND gate with A gives the final output. So the gate count climbs from 2 to 3. If you insist on only two gates, you’d have to give up a pure logical form or use a multi‑input gate, which isn’t standard in minimal logic families.