zhou yang和Claude Opus 5
efcba30a0c
- 规则：铜层 2→4，差分 0.28/0.18，间距 0.15，过孔 0.3/0.2（rules_base） - 几何：98×98 板框 + 4×M3，槽阵居中，L1 12V 大铜 + 指头（pcb_base） - 摆位：五插槽留 L1，其余 133 件翻到 L4；背面镜像规则实测；Hub 放走廊 2、 东移 2.5 mm；扇出圈 / 小件间距 1.5（去耦例外）/ 9 A 过孔阵禁区（place_base） - 叠层按用户决定改为 L1 槽面 / L2 整层 +3V3 / L3 整层 GND / L4 器件面 - 布线：D± 七对按构造布通（USB3 掉头、USB1/2 北去掉头、USB4/5 东绕、 J23 桥），+3V3 / GND / +12V 平面网每脚一孔，0.125 网格 + PathFinder 开关（route_base、remote_route.ps1）；剩 37 个网待手工补 - 铺铜：4 层重写，入口链自南向北切段 + 断面校核（pour_base）； 已下发布线 / 缝合孔 / 铜面轮廓，填铜未成功（待查） - 工具：plot_base / plot_routes / shot_base，board_check 通孔两层都算 Co-Authored-By: Claude Opus 5 <[email protected]> Claude-Session: https://claude.ai/code/session_0168Bu6SEx3cXXCJLkdnE6zB
189 行
8.1 KiB
Python
189 行
8.1 KiB
Python
# -*- coding: utf-8 -*-
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u"""下板 PCB1：**每个还要接的焊盘，能不能走得出去**。
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/c/Python314/python .claude/pcb1_escape_audit.py # 全部
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/c/Python314/python .claude/pcb1_escape_audit.py --net SDA # 只看这几个网络
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/c/Python314/python .claude/pcb1_escape_audit.py --why U8.3 # 单个焊盘，列出挡路的铜
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## 为什么要单独有这一支
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布线器报「布不通」的时候，看不出根因是**这一头出不来**还是**中间那段挤**。
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两者的处理办法完全不同：前者要动扇出过孔 / 摆位（几何上就没有出路，
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换多少种布线策略都一样），后者才是布线器的事。
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判据：在焊盘周围开一个 8 x 8 mm 的窗，按 0.05 mm 栅格算「0.2 mm 的线能走的格」，
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从出线段末端泛洪，看能不能摸到窗边。摸不到 = **这个焊盘被围死了**。
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（窗开 8 mm 是因为 0.5 脚距 QFN 的死胡同都在 2~3 mm 以内；
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再大就分不清「出不来」和「出得来但走不远」了。）
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"""
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import math
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import os
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import sys
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import numpy as np
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from scipy import ndimage
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HERE = os.path.dirname(os.path.abspath(__file__))
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sys.path.insert(0, HERE)
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import pcb1_route_sig as R # noqa: E402
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E, D = R.E, R.D
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WIN = 4.0 # 窗口半边长（mm）
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G = 0.05
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W = 0.20
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def audit(pads, obstacles, tags, verbose_tag=None):
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bytag = {p['tag']: p for p in pads}
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stubs = getattr(audit, 'stubs', {})
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alts = getattr(audit, 'alts', {})
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bad = []
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for tag in sorted(tags):
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q = bytag.get(tag)
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if q is None:
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continue
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# ⚠️ **每一个出线方向都要试。**只试首选方向会把
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# `U13.2` 这种“南边是死胡同、北边其实通”的误报进来 ——
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# `route_net()` 本来就会在失败时换方向重试。
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cands = list(alts.get(tag) or ([stubs[tag]] if tag in stubs else [(None, 0)]))
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best = None
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for st in cands:
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r = _breathe(q, st if st[0] else None, obstacles)
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if r is None:
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best = None
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break
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best = r if best is None else max(best, r)
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if best is None:
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continue
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bad.append((tag, q['net'], u'被围死：%d 个出线方向全试过，'
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u'最好的一个两层加起来只能走 %.2f mm²'
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% (len(cands), best)))
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if tag == verbose_tag:
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st = cands[0]
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sx, sy = st[0] if st[0] else (q['x'], q['y'])
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box = (q['x'] - WIN, q['y'] - WIN, q['x'] + WIN, q['y'] + WIN)
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f = E.Field(obstacles, box, skip_net=q['net'], grid=G)
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_why(f, q, sx, sy, obstacles, q['lay'] if q['lay'] != 12 else E.L_TOP)
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return bad
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def _breathe(q, st, obstacles):
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u"""从这个出线方向能不能走到窗边。能 → None；不能 → 能走的面积 mm²。否则"""
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if True:
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sx, sy = st[0] if st else (q['x'], q['y'])
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box = (q['x'] - WIN, q['y'] - WIN, q['x'] + WIN, q['y'] + WIN)
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f = E.Field(obstacles, box, skip_net=q['net'], grid=G)
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f.set_relax([E.sh_rect(cx, cy, hw, hh, R.pad_layers(q))
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for cx, cy, hw, hh in R.pad_rects(q)]
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+ ([E.sh_seg(q['x'], q['y'], sx, sy, st[1], q['lay'])] if st else []))
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lay = q['lay'] if q['lay'] != 12 else E.L_TOP
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# ⚠️ **两层一起泛洪，中间按 `via_ok()` 打通。**
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# 只在焊盘自己那一层上算的话，「出来一步就打孔翻到底层」这种最常见的逃法
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# 会被判成走不通 —— 第一版就是这样，报出 60 多个假阳性
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# （`U23.9` / `U6.*` 明明单独布都能通）。
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ok = {L: f.free(L, W) for L in (E.L_TOP, E.L_BOT)}
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vok = f.via_ok()
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i, j = f._idx(sx, sy)
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if not (0 <= i < f.nx and 0 <= j < f.ny):
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return None
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if not ok[lay][j, i]:
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return 0.0
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seed = {L: np.zeros_like(ok[L]) for L in (E.L_TOP, E.L_BOT)}
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seed[lay][j, i] = True
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oth = {E.L_TOP: E.L_BOT, E.L_BOT: E.L_TOP}
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for _it in range(8):
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grew = False
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for L in (E.L_TOP, E.L_BOT):
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r = ndimage.binary_propagation(seed[L], mask=ok[L])
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if (r != seed[L]).any():
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seed[L] = r
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grew = True
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for L in (E.L_TOP, E.L_BOT):
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cross = seed[L] & vok & ok[oth[L]]
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if (cross & ~seed[oth[L]]).any():
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seed[oth[L]] = seed[oth[L]] | cross
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grew = True
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if not grew:
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break
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# ⚠️ 环到哪一圈才算「出去了」：窗口外那圈墙本身是障碍，
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# 紧贴着它的十几格 `free()` 永远是假的（edt 不够）。
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# 拿自由区最外圈当判据会把每一个焊盘都报成“出不去”。
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nb = max(1, int(round(E.Field.RIM / G)))
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rim = nb + int(round((E.CLR + W / 2.0 + f.qerr) / G)) + 2
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edge = np.zeros(ok[lay].shape, bool)
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edge[rim, :] = True
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edge[-rim - 1, :] = True
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edge[:, rim] = True
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edge[:, -rim - 1] = True
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if any((seed[L] & edge).any() for L in (E.L_TOP, E.L_BOT)):
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return None
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return sum(int(seed[L].sum()) for L in (E.L_TOP, E.L_BOT)) * G * G
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def _why(f, q, sx, sy, obstacles, lay):
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u"""把挡在这个焊盘 3 mm 内的异网铜按距离列出来。"""
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print(u' —— %s 周围 3 mm 内的异网铜（按到出线段末端的距离）——' % q['tag'])
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rows = []
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for s, n in obstacles:
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if n == q['net'] or lay not in s[-1]:
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continue
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if s[0] == 'rect':
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d = D.seg_rect((sx, sy), (sx, sy), s[1], s[2], s[3], s[4])
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what = u'焊盘/铺铜 (%.2f,%.2f) %.2fx%.2f' % (s[1], s[2], s[3] * 2, s[4] * 2)
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elif s[0] == 'via':
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d = math.hypot(sx - s[1], sy - s[2]) - s[3]
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what = u'过孔 (%.2f,%.2f)' % (s[1], s[2])
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else:
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d = D.seg_pt(s[1], s[2], s[3], s[4], sx, sy) - s[5]
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what = u'走线 (%.2f,%.2f)-(%.2f,%.2f) w%.2f' % (s[1], s[2], s[3], s[4], s[5] * 2)
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if d < 3.0:
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rows.append((d, n, what))
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for d, n, what in sorted(rows)[:14]:
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print(u' %6.3f %-11s %s' % (d, n or u'（空）', what))
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def main():
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argv = sys.argv[1:]
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want = None
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if '--net' in argv:
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want = set(a for a in argv[argv.index('--net') + 1:] if not a.startswith('--'))
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why = None
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if '--why' in argv:
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why = argv[argv.index('--why') + 1]
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pads, vias, lines = R.board()
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pours = R.pour_rects()
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obstacles = R.obstacles_of(pads, vias, lines, pours)
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js = R.jobs(pads, want, skip=() if want else R.SKIP)
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# ⚠️ 2026-09-06 起 `route_sig` 把出线段也下发了 —— 布不通的焊盘在板上是
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# 「一个焊盘 + 一小截自己的出线段」，不再是光焊盘。只认光焊盘会报「0 个」。
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# 出线段全长 ≤ 1.2 mm、没有过孔的碎片，一样当孤立焊盘审。
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need = set()
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for net, frags in js:
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for fr in frags:
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if len(fr['pads']) != 1 or fr['vias']:
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continue
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q = fr['pads'][0]
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if all(max(math.hypot(l['x0'] - q['x'], l['y0'] - q['y']),
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math.hypot(l['x1'] - q['x'], l['y1'] - q['y'])) <= 1.2 for l in fr['lines']):
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need.add(q['tag'])
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order = [n for n, _f in js]
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stubs, alts = R.plan_stubs(order, pads, obstacles, need)
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audit.stubs = stubs
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audit.alts = alts
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if why:
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need = set([why])
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print(u'要接的孤立焊盘 %d 个，其中留到出线段的 %d 个' % (len(need), len(stubs)))
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bad = audit(pads, obstacles, need, verbose_tag=why)
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print(u'\n**被围死的焊盘 %d 个**：' % len(bad))
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for tag, net, msg in bad:
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print(u' [!!] %-10s %-12s %s' % (tag, net, msg))
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if not bad:
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print(u' （没有）')
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return 0
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if __name__ == '__main__':
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sys.exit(main())
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