Dragon-s-Lair-X68k/tools/analysis/23_frame_clock.py

#!/usr/bin/env python3
"""What a real frame clock can be built from, and what its cadence costs.

    python3 tools/analysis/23_frame_clock.py [--fps 12] [--vtotal 568]
                                             [--csv tmp/c68k_frames.csv]

ROADMAP P3 says "needs MFP timer or VBL", which hides the fact that ONE OF
THOSE CANNOT DO IT and the other cannot do it either without a divider.  This
file enumerates the space rather than asserting a conclusion, the way FINDINGS
47 had to be re-done once a hardware "no" turned out to be a claim about a whole
configuration space nobody had walked.

EVERY CONSTANT HERE IS SOURCED, and from a file on this machine:

  MFP timer clock   16 MHz / 4        MAME 0.277 sharp/x68k.cpp:1027-1028
  prescaler ladder  4,10,16,50,64,100,200   machine/mc68901.cpp:173
  timer data reg    8 bits, 0 means 256     machine/mc68901.cpp (TCDR/TADR)
  V-DISP -> GPIP4   x68k.cpp:1139, and it is also Timer A's event input,
                    mc68901.cpp:167 GPIO_TIMER = {GPIP_4, GPIP_3}
  line rate         31,500 Hz exactly in both 31.5 kHz modes, derived in
                    tools/bench/crtc_mode.lua from the dot clocks
  interrupt cost    MEASURED, not tabled: tools/bench/clock_run.sh

THE PART THAT IS NOT A CLOCK PROBLEM AT ALL.  12 fps on a 55.4577 Hz raster is
4.6215 refreshes per frame, so every frame is shown for 4 refreshes or 5 --
72.13 ms or 90.16 ms -- and 37.9% of them get the short one.  That is the
display's quantisation and no choice of clock changes it.  What it changes is
the BUDGET: 833,333 clocks is the mean slot, not the slot, and the short slot is
721,270.  With the per-frame decode costs in hand this file says exactly how
many frames do not fit theirs, which is a thing this project has never had to
ask because until now the tick came from a host that could not miss.
"""
import argparse
import csv
import os
import sys

MFP_HZ = 16_000_000 // 4                 # x68k.cpp:1027-1028
PRESCALER = [4, 10, 16, 50, 64, 100, 200]   # mc68901.cpp:173
HFREQ = 31500                            # lines/s, crtc_mode.lua
CPUHZ = 10_000_000                       # 40 MHz / 4, x68k.cpp:1133


def timer_space(fps):
    """Every (prescale, data) the MFP can be set to, against a target fps."""
    slowest = MFP_HZ / (PRESCALER[-1] * 256)
    print(f"\n=== 1. THE MFP TIMER, WHICH CANNOT DO IT ALONE")
    print(f"  timer clock {MFP_HZ:,} Hz, prescalers {PRESCALER}, data 1..256")
    print(f"  slowest tick any single timer can produce: "
          f"{MFP_HZ}/({PRESCALER[-1]}*256) = {slowest:.3f} Hz")
    print(f"  a {fps} fps frame needs {fps} Hz, which is {slowest/fps:.1f}x "
          f"slower than that -- so a software divider is REQUIRED whatever the "
          f"source, and 'use an MFP timer' is not by itself an answer.")
    exact = [(p, d) for p in PRESCALER for d in range(1, 257)
             if (MFP_HZ * d * p) and (MFP_HZ % (p * d) == 0)
             and (MFP_HZ // (p * d)) % fps == 0]
    print(f"  settings whose tick rate is a whole multiple of {fps} Hz, so that "
          f"a plain counter would be exact: {len(exact)}")
    if exact:
        best = min(exact, key=lambda pd: MFP_HZ / (pd[0] * pd[1]))
        p, d = best
        tick = MFP_HZ / (p * d)
        print(f"    slowest of them: prescale /{p} data {d} = {tick:.4f} Hz, "
              f"{tick/fps:.0f} ticks per frame")
        print(f"    -> {tick/fps:.0f} interrupts per frame, against 4.6215 for "
              f"the raster: {tick/fps/(HFREQ/(fps*568)):.1f}x the cost, and its "
              f"phase against the raster is arbitrary, so a frame would be "
              f"presented mid-scan.")
    else:
        print(f"    NONE.  {MFP_HZ}/{fps} = {MFP_HZ/fps:,.2f} is not an "
              f"integer, so no prescale/data pair divides to {fps} Hz at all.")


def raster_space(fps, vtotal):
    hz = HFREQ / vtotal
    print(f"\n=== 2. THE RASTER, WHICH IS THE RIGHT SOURCE AND IS ALSO NOT "
          f"A WHOLE DIVIDE")
    print(f"  V-DISP is {HFREQ}/{vtotal} = {hz:.4f} Hz, and it is BOTH the "
          f"GPIP4 interrupt and Timer A's event-count input")
    print(f"  whole divides -- all the MFP can do in hardware, no software:")
    for n in (3, 4, 5, 6):
        f = hz / n
        print(f"    Timer A event count = {n}: {f:7.4f} fps  "
              f"({100*(f/fps-1):+6.2f}% from {fps})")
    print(f"  {fps} fps needs {hz/fps:.4f} refreshes per frame, which is not a "
          f"whole number, so no event-count setting is exact either.")
    print(f"\n  THE DIVIDER THAT IS EXACT: add fps*VTOTAL = {fps*vtotal} per "
          f"V-DISP, emit a tick at {HFREQ}, keep the remainder.")
    print(f"    long-run rate = {fps}*{vtotal}/{vtotal} = {fps} fps EXACTLY, "
          f"with a remainder that never accumulates")
    print(f"    the accumulator stays under {HFREQ + fps*vtotal:,}, so it is "
          f"16-bit arithmetic on a 68000 (the ceiling is fps < "
          f"{(65536-HFREQ)/vtotal:.1f} at this VTOTAL, and clk_init checks it)")


def divider_gaps(fps, vtotal, n):
    """The tick sequence src/player/clock.i emits, in refreshes per tick."""
    acc, gaps, since = 0, [], 0
    while len(gaps) < n:
        acc += fps * vtotal
        since += 1
        if acc >= HFREQ:
            acc -= HFREQ
            gaps.append(since)
            since = 0
    return gaps


def host_gaps(fps, refresh_hz, n):
    """The tick sequence tools/bench/stream.lua's HOST clock emits.

    It looks uniform in the source -- `floor((t - t_rel) * fps)` -- and is not.
    Lua only sees the machine at frame boundaries, so tick k lands on the first
    refresh at or after k/fps, and the gaps between ticks come out as the same
    two whole numbers of refreshes the divider produces.  The host-paced runs of
    FINDINGS 49 and 51 therefore already had this cadence in them; what ROADMAP
    P3 changes is who produces it, not whether it exists.
    """
    at = [-(-int(k * refresh_hz * 1000000 // fps) // 1000000) for k in range(n + 1)]
    return [at[k + 1] - at[k] for k in range(n)]


def cadence(fps, vtotal, costs, label, refresh_hz, gaps, uniform=False):
    """Charge each frame the slot it really gets, and run the pace gate."""
    rpf = 1.0 if uniform else HFREQ / (fps * vtotal)
    lo, hi = (1, 1) if uniform else (int(rpf), int(rpf) + 1)
    slot_lo = lo / refresh_hz * CPUHZ
    slot_hi = hi / refresh_hz * CPUHZ
    nominal = CPUHZ / fps

    n_lo = sum(1 for g in gaps[:len(costs)] if g == lo)
    print(f"\n  --- {label}: refresh {refresh_hz:.4f} Hz")
    print(f"      slots are {lo} refreshes = {slot_lo:,.0f} clk "
          f"({1000*lo/refresh_hz:.2f} ms) or {hi} = {slot_hi:,.0f} clk "
          f"({1000*hi/refresh_hz:.2f} ms)")
    print(f"      the nominal {fps} fps budget every figure in this project is "
          f"priced against is {nominal:,.0f} clk; the SHORT slot is "
          f"{100*(slot_lo/nominal-1):+.1f}% of it")
    over_lo = sum(1 for c in costs if c > slot_lo)
    over_hi = sum(1 for c in costs if c > slot_hi)
    over_nom = sum(1 for c in costs if c > nominal)
    print(f"      frames that do not fit: {over_lo}/{len(costs)} the short "
          f"slot, {over_hi}/{len(costs)} the long one, {over_nom}/{len(costs)} "
          f"the nominal budget")

    # The schedule, with the catch-up the pace gate actually performs: frame i
    # starts at max(finish of i-1, tick i).  A frame that overruns does not fail
    # -- it eats the next frame's idle, and the clock catches up by itself.
    t, tick, late, worst, first = 0.0, 0.0, 0, 0.0, None
    for i, c in enumerate(costs):
        if t <= tick:
            t = tick                      # idled: on time
        else:
            if i:                         # frame 0 has no predecessor
                late += 1
                if first is None:
                    first = i
            worst = max(worst, t - tick)
        t += c
        tick += gaps[i] / refresh_hz * CPUHZ
    print(f"      through the pace gate: {late}/{len(costs)} frames found "
          f"their slot already open (first at frame {first}), worst start "
          f"{worst:,.0f} clk = {1000*worst/CPUHZ:.1f} ms behind its tick")
    if not uniform:
        print(f"      {n_lo}/{len(costs)} slots were the short one "
              f"({100*n_lo/len(costs):.1f}%; the exact share is "
              f"{100*(hi-rpf):.1f}%)")
    return late


def main():
    ap = argparse.ArgumentParser()
    ap.add_argument("--fps", type=int, default=12)
    ap.add_argument("--vtotal", type=int, default=568,
                    help="CRTC R04+1; 568 is the 31.5 kHz 256-line mode")
    ap.add_argument("--csv", default="tmp/c68k_frames.csv",
                    help="per-frame decode cost, from tools/bench/c68k/run.sh")
    a = ap.parse_args()

    timer_space(a.fps)
    raster_space(a.fps, a.vtotal)

    print(f"\n=== 3. WHAT THE CADENCE COSTS")
    if not os.path.exists(a.csv):
        print(f"  {a.csv} not found -- run tools/bench/c68k/run.sh first. The "
              f"cadence question CANNOT be answered from percentiles: it needs "
              f"the per-frame series, because what matters is whether an "
              f"expensive frame lands in a short slot and how long the catch-up "
              f"takes afterwards.")
        return 1
    costs = [float(r["cycles"]) for r in csv.DictReader(open(a.csv))]
    print(f"  {len(costs)} frames from {a.csv}: mean {sum(costs)/len(costs):,.0f} "
          f"clk, max {max(costs):,.0f} (frame {costs.index(max(costs))})")

    hw = HFREQ / a.vtotal
    n = len(costs)
    # MAME's screen is fast by htotal/(htotal-8): refresh_mode() builds the
    # frame period from scr.max_x*scr.max_y with scr.max_x = m_htotal - 8.  Both
    # rasters are run, because the rig measures against the fast one and the
    # player will run on the other -- reporting only one would leave the rig's
    # count and this file's differing with nobody able to say which was wrong.
    htotal = 368
    mame = hw * htotal / (htotal - 8)

    # The budget model every figure in FINDINGS assumes: a slot of exactly
    # 1/fps.  No machine has this; it is the yardstick, run through the same
    # schedule so that what the cadence ADDS can be read off.
    cadence(a.fps, a.vtotal, costs, "the NOMINAL model (a slot of exactly "
            "1/fps, which no raster produces)", float(a.fps),
            [1] * (n + 1), uniform=True)
    cadence(a.fps, a.vtotal, costs, "the HOST tick, as tools/bench/stream.lua "
            "actually emits it", mame, host_gaps(a.fps, mame, n + 1))
    cadence(a.fps, a.vtotal, costs, "the 68000's own clock on the HARDWARE "
            "raster", hw, divider_gaps(a.fps, a.vtotal, n + 1))
    cadence(a.fps, a.vtotal, costs, f"the 68000's own clock on MAME's raster "
            f"(fast by {htotal}/{htotal-8})", mame,
            divider_gaps(a.fps, a.vtotal, n + 1))

    print(f"""
Reading it.

THE CADENCE WAS ALREADY THERE.  The nominal row is the model every budget in
this project is priced against -- a slot of exactly 1/fps -- and no raster
produces it.  The host row is what tools/bench/stream.lua has been emitting all
along: `floor((t - t_rel) * fps)` looks uniform, but Lua only sees the machine at
frame boundaries, so its ticks land on refreshes and its gaps are the same two
whole numbers.  The host-paced results of FINDINGS 49 and 51 therefore already
carried a 4/5 cadence that nothing named.  ROADMAP P3 did not introduce it; it
moved who produces it onto the machine, where it belongs, and made it visible.

THE SHORT SLOT IS REAL AND IT IS NOT A FAILURE.  {sum(1 for c in costs if c > 721270)}/{len(costs)} frames do not fit
721,270 clocks.  The pace gate only says "not before tick i", so a frame that
overruns spends the next frame's idle and the clock recovers by itself; the cost
is one frame presented a refresh late, not a dropped frame.  What the counts
above measure is frames with no idle left, and the difference between the
nominal row and the raster rows -- 1 against 4 -- is the whole price of the
cadence on this container.

THE EXPENSIVE FRAME IS FRAME 0, at {max(costs)/(CPUHZ/a.fps)*100:.0f}% of the nominal budget: the first
frame of a scene has nothing to SKIP against, so it is the whole picture in one
slot.  Most of what follows it in these counts is that transient draining, which
is why the first index is printed next to the total.  It also means the cost is
paid AT A SCENE CHANGE, alongside the 18.96 ms of loader (FINDINGS 53.2) and the
seek -- not spread over the window.

SCOPE.  Decode costs are C68K's, on zero-wait-state memory, so they are a lower
bound; real DRAM moves every row here in the same direction.  The MAME rows are
the emulator's fast raster and exist to be compared with tools/bench/pace_run.sh,
not to describe hardware.""")
    return 0


if __name__ == "__main__":
    sys.exit(main())