07977624ae
TP5 is a 2012 Windows application supplied with the Guangzhou-Tongda LED
destination signs fitted to Yutong buses. It has no preview, so building a
destination list is guess-and-check, and it only runs on Windows.
The bus never talks to TP5 — the sign controller reads a .td5 file off an SD
card, and that is the whole interface. So this replaces the software without
touching any hardware or protocol: it just has to write byte-correct .td5.
Formats reverse-engineered from the sample files and TP5(En).exe, then verified
byte-for-byte:
.td5 the file the bus reads. Fixed-layout binary; each destination block
carries a CRC-16/ARC over block[3..len] and a rand() block id, which
together looked like one 4-byte field because RAND_MAX is 0x7fff.
.tp5 the editable project. Line-based text, UTF-16BE hex strings.
.font the sign's own bitmap fonts, each glyph row XORed with its char code.
The app is one self-contained HTML file: live LED preview at the real sign size
with real scrolling, spreadsheet/CSV import, multi-page destinations, undoable
delete, and export to both .td5 and .tp5.
Verified:
- rebuilds a real 46,080-byte TP5 export byte-for-byte with a recomputed CRC
- all 36 stored CRCs verify against the implementation
- driven through its own UI, re-exporting the real file differs in 7 bytes,
all of them the export timestamp
- running on a real bus: signs and driver's controller both correct
Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
39 lines
1.1 KiB
Python
39 lines
1.1 KiB
Python
import struct
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from parse import records
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rs = records()
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oc = [r for r in rs if r['name'].startswith('OC')]
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assert len(oc)==8
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L = set(len(r['blk']) for r in oc)
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print('lengths', L)
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b = [r['blk'] for r in oc]
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f = [r['field'] for r in oc]
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# show differing byte positions
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n = len(b[0])
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diff = [i for i in range(n) if len(set(x[i] for x in b))>1]
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print('differing byte offsets:', [hex(i) for i in diff])
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for i in diff:
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print(hex(i), [hex(x[i]) for x in b])
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# GF(2) affine test: for equal-length messages, if h is affine over GF(2),
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# then h(A)^h(B)^h(C)^h(D) == 0 whenever A^B^C^D == 0.
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# Find quadruples among the 8 whose message-XOR is 0.
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import itertools
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def bx(x,y):
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return bytes(p^q for p,q in zip(x,y))
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found=0
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for c in itertools.combinations(range(8),4):
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m = bytes(n)
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for i in c: m = bx(m,b[i])
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if m == bytes(n):
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v = 0
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for i in c: v ^= f[i]
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print('quad', c, 'msgxor=0 fieldxor=%08x' % v)
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found+=1
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print('quads with zero msg xor:', found)
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# Alternative affine test using the base-block trick:
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# Build a matrix over the differing bytes only.
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