FA-79802 / Barcode symbology encoding / Member archive

Pad randomisation uses a zero-based position · case 02

Symbols with padding fail strict verification because every randomised pad is off by one position.

Member previewVariant 2 · 3 implementations · 8 checks per implementation

Case contract

Data Matrix ASCII encodation into `capacity` data codewords: two consecutive ASCII digits become 130 + their value; ASCII 0..127 becomes ord + 1; 128..255 becomes Upper Shift 235 then ord - 127; higher code points are unencodable. Too many codewords -> overflow. Padding: the first pad is 129; each later pad at 1-based position p is 129 + ((149 * p) mod 253) + 1, minus 254 if above 254.

Why this case matters

Retail, logistics, pharmacy and document workflows depend on encoders that produce exactly the module pattern, code-set switches, separators and quiet zones scanners expect; one misplaced module or separator makes a label unreadable or, worse, scan as different data.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixture	Actual	Expected	Outcome
regression: pad randomisation position 0	`[49, 129, 175, 70, 220, 115, 11, 161, 56]`	`[49, 129, 70, 220, 115, 11, 161, 56, 206]`	Failed

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