社区科普：PoD 设计成本优化公式Community guide: PoD design-cost optimization formulas
社区成员 0xDoran_NFT 发布 TapeOut 最优首创的优化思路：在 PoD 机制下，同一台处理器上每道题同时只有一个最优首创，即设计成本 C 最低的那枚电路，谁能做出严格更低的 C 就抢走席位并获得设计溢价。文中给出成本口径 b* = NAND 数量 + LATCH 数量（两者权重相同），以及面积 A = g + 6×s 的拆解，用于在流片前尽量压低设计成本。Community member 0xDoran_NFT published an optimization write-up for TapeOut's optimal first-mint: under PoD, each puzzle has only one optimal circuit per processor at a time — the one with the lowest design cost C — and whoever achieves a strictly lower C takes the slot and the design premium. The post defines cost as b* = number of NANDs + number of LATCHes (equally weighted) and breaks down area as A = g + 6×s, to push design cost down before tape-out.
