Reading the damage backwards#
Five witnesses. No original. Which tree does the evidence support — and what does it leave unresolved?
Three short witnesses#
…he shall set a point above it and pass on. (A)
…he shall rest again at the twelfth. (B)
…he shall rest again at the twelfth. (C)
Witness A disagrees with B and C at one line. B and C agree with each other at a different line, against the reading every other witness carries.
Text: A. Halloway, Ordinance of the Copy-House (authored teaching scenario).
The observed differences#
| Point | Line | Type | Carriers |
|---|---|---|---|
| 1 | 4 | substitution | A |
| 2 | 7 | haplography | B, C, D, E |
| 3 | 12 | homoeoteleuton | C, D, E |
| 4 | 6 | normalisation | A, C |
| 5 | 2 | dittography | B |
| 6 | 5 | omission | C |
| 7 | 8 | substitution | D, E |
Full table: collation.csv.
The hypothesis#
A shared error may be evidence of a shared ancestor.
Not: a shared reading proves one. Not: any two witnesses that agree are related.
The question every point above has to answer: could two copyists have made this same change independently?
The assumption it needs#
A shared error supports shared ancestry only if there is no ordinary way for it to happen twice, independently.
- a haplography (eye skips a repeated word) — no
- an omission that breaks the sense of its own line — no
- a spelling normalisation — yes, easily
Get this wrong and the tree is built on agreements that mean nothing.
Build: start with the strongest point#
Point 2 — a haplography shared by B, C, D, E, absent from A.
No mechanism makes four copyists skip the same repeated phrase by chance.
Edge: A sits outside a group the other four share an ancestor within.
Build: split the group#
Within the group B, C, D, E, point 3 — a homoeoteleuton shared by C, D, E, absent from B.
Same logic: an eye-skip between two occurrences of “shall keep” is not a coincidence four times over, and B does not have it.
Edge: B branches off; C, D, E share a further common ancestor.
Build: the innermost pair#
Points from lines 6 and 8 leave D and E agreeing with each other at every single point where the table has any variation at all.
Edge: C branches off from D and E. D and E share the closest ancestor the evidence can locate — but see the next slide before you call that settled.
An unresolved edge#
D and E’s text is identical at every variant point in this table.
That evidence supports D copied from E, E copied from D, and both copied from a lost intermediate equally well. Nothing here chooses between them.
Marking the edge “unresolved” is not a gap in the analysis. It is the analysis.
A competing tree#
Point 4 (line 6): a normalisation, shared by A and C only.
Taken at face value, that reading groups A with C — directly against the tree just built, which puts A outside the whole B, C, D, E group.
Which tree does the full table support?
The trap, and why it is a trap#
A normalisation is exactly the kind of change point 2’s rule says not to trust: two copyists can tidy the same archaic phrasing independently, with no shared ancestor at all.
Point 2’s haplography has no such independent mechanism. When the two points disagree, the table backs the one with no ordinary way to repeat by chance.
Contamination and multiple sources#
The same caution applies to a witness copied from more than one source at once — a copyist checking two exemplars and blending their readings.
A tree assumes one parent per witness. Where the evidence instead shows a witness agreeing with one group on some points and a different group on others, for reasons other than independent error, that assumption itself needs stating and testing, not assuming.
The reveal#
reveal/production-log.md (ask your tutor for the link once the tree is
built) names the true archetype and every variant’s real point of entry.
Compare it against the tree just built. Where they agree, say why the evidence supported it. Where they don’t, ask whether the tree was still the best call available from what was given — not whether it was “right”.
Same method, one family over#
Weeks 2 and 3’s audio dubs — copies W, X, Y, Z — fit the same reasoning: a shared feature with no independent mechanism supports shared ancestry; a shared feature two dubbing decks could produce separately does not.
Copying mechanisms differ across collections. The logic transfers; the specific list of error types does not.
Practice now, then The Stemma#
Workshop: Reading the damage backwards — build this tree yourself from your own week 5 witnesses.
Assessment: The Stemma — pack releases when the workshop ends. Same method, a separate set of witnesses, no reveal until after submission.
Text walkthrough#
For a reader who cannot use the deck view: five witnesses disagree at seven
points (collation.csv). Two points (a haplography, then a homoeoteleuton
inside it) build a nested grouping of B, C, D, E inside all five, then
C, D, E inside that. Two more points show D and E agreeing at everything,
leaving their order unresolved. One point (a normalisation shared by A and
C) looks like a fourth edge and is not, because normalisation has no
independent-repeat protection the way an eye-skip does. The reveal names the
true archetype and each variant’s real origin.