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T49: unrolling recovers the throughput without recovering the operator - #839
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Replacing an operator by k repetitions of a depth-1 step raises the clock and divides throughput by k. Unrolling into k pipeline stages restores one result per cycle at k times the step's area -- and does not restore the operator, because each stage is still the step. The closed form does restore it: the k-fold composition is a multiplication by structure constants. Measured: the pipelined golden-scale layer is 660 cells at 204.08 MHz, one element per cycle, against 1098 cells at 69.21 MHz for the multiplier -- 1.66x smaller and 2.95x faster, no multiply anywhere. This corrects T48's reading, which measured the iterative form and found it 2.15x slower per element. That was the implementation, not the lattice. Audits run locally: RU and EN PASS, 27 routes.
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T49: unrolling recovers the throughput without recovering the operator (#839) Replacing an operator by k repetitions of a depth-1 step raises the clock and divides throughput by k. Unrolling into k pipeline stages restores one result per cycle at k times the step's area -- and does not restore the operator, because each stage is still the step. The closed form does restore it: the k-fold composition is a multiplication by structure constants. Measured: the pipelined golden-scale layer is 660 cells at 204.08 MHz, one element per cycle, against 1098 cells at 69.21 MHz for the multiplier -- 1.66x smaller and 2.95x faster, no multiply anywhere. This corrects T48's reading, which measured the iterative form and found it 2.15x slower per element. That was the implementation, not the lattice. Audits run locally: RU and EN PASS, 27 routes.
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T49 — an iterative operator loses throughput; unrolling it recovers the throughput without recovering the operator.
Replacing an operator by
krepetitions of a depth-1 step raises the clock and divides throughput byk. Unrolling those repetitions intokpipeline stages restores one result per cycle atk× the step's area — and does not restore the operator, because each stage is still the step.The closed form does restore it: for a group generated by one step, the k-fold composition is a multiplication by structure constants. Here
φ^k = F(k−1) + F(k)·φmakes the scale two multiplications by Fibonacci numbers, which is the operator back.Measured by place-and-route (nextpnr-ice40, hx8k, fan-in 8), from trinity-fpga#592:
1.66× smaller and 2.95× faster, with no multiply anywhere.
This corrects the reading of T48, which measured the iterative form and found it 2.15× slower per element. That was the implementation, not the lattice — and T48's arithmetic stands, it is the conclusion drawn from it that narrows.
The refusal: unrolling costs 354 cells against 97 for the iterative scale block, 3.6×, so below about
k = 4the iterative form delivers the same rate for less. iCE40 has no DSP blocks, so the multiplier is maximally penalised; fan-in 8; no board.Ratchet 179/26 against baseline 184/27, no file gained. Audits: RU and EN PASS, 27 routes.