Claude Opus 5.5 (effort default (no --effort flag) · thinking on) on Undo/redo with typing coalescing for a text buffer (word boundaries, 2-second gap, backspace vs forward-delete groups, history cap)

AClaude Opus 5.5default · thinking on · Claude Code 2.1.299 (headless -p) · cloud · 2026-10-20 · by @bobsiopener

Trust 80%Fully autonomoussuccess · first tryCompare

Estimated cost: $0.25
Wall time: 54s
Human step-ins: 0
Outcome: Pass

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# Undo/redo with typing coalescing for a text buffer `editor.py` is the text buffer behind our notes app. Users expect one undo to remove a whole typed word, not one letter. Rewrite `Editor` (Python 3 stdlib only) with this exact behaviour. API: `Editor(text="")`, attribute `.text`, `insert(pos, s, t)`, `delete(pos, n, t)`, `undo() -> bool`, `redo() -> bool`. `t` is the edit time in seconds (float; compare with plain subtraction `t - last <= 2.0`; the hidden tests never use times where floating-point rounding could change that comparison). ## Edits 1. `insert`: `pos` must be an int with `0 <= pos <= len(text)`, else `ValueError`. Inserting `""` changes nothing and is not recorded (but positions and time are still validated). 2. `delete(pos, n, t)` removes `text[pos:pos+n]`; `pos` and `n` must be ints with `n >= 1`, `pos >= 0`, `pos + n <= len(text)`, else `ValueError`. 3. `t` must never be smaller than the `t` of the previous recorded edit, else `ValueError` (nothing changes). Validation errors leave everything unchanged. ## Undo groups 4. Every recorded edit either **joins** the most recent undo group or starts a new one. A new edit can only join if: the most recent group is still **open** (no `undo()`/`redo()` call happened since that group was last changed), the group was created by a **single-character** edit (a multi-character insert or delete is a group of its own that never accepts more edits), the new edit is a single character, `t - (time of the group's last edit) <= 2.0`, and the group currently holds **fewer than 20 characters**. 5. **Typing:** a 1-char insert joins an insert group if `pos` equals group start + group text length (typing right after it) and the group's last character is not a space or `\n` (a space/newline is still added to the group, but ends it for further typing). 6. **Deleting:** a 1-char delete joins a delete group either as **backspace** (`pos + 1 ==` group start; the group start moves to `pos`) or as **forward delete** (`pos ==` group start). A group's direction is fixed by its second delete; after that only the same direction joins. 7. Insert never joins a delete group and vice versa. 8. `undo()` reverts the most recent group as a whole and moves it to the redo stack; `redo()` re-applies the most recently undone group. Both return `False` (and change nothing else, except closing the open group) when their stack is empty, else `True`. Any recorded edit (including one that joins a group) clears the redo stack. 9. The undo stack keeps at most **50** groups; when a 51st is pushed (by a new group or by `redo`), the oldest is dropped. ## Scoring Hidden check groups (partial credit = fraction passed; all must pass): `validation` (1-3), `typing_groups` (4-5), `word_boundaries` (5), `time_gap` (4: exactly 2.0 s joins, 2.0001 s does not), `delete_directions` (6-7), `undo_redo_semantics` (8), `group_limits` (4: 20-char cap; 9: 50 groups), `randomized_agreement` (500 random sequences of 60 operations: `.text` and every return value must match the rules after each step).

Many small interacting rules (group closing on undo/redo, multi-character groups that never join, direction fixed by the second delete, 20-character cap, 50-group cap including redo) compared step by step on 500 random sequences. Original benchmark task written for usecaserank.com (not based on a specific public post). Headless, offline run on a reproducible fixture. Each run's notes say how its cost was measured (estimates are marked).

Success criteria: 1. validation 2. typing_groups 3. word_boundaries 4. time_gap 5. delete_directions 6. undo_redo_semantics 7. group_limits 8. randomized_agreement (500 sequences)

Setup notes: Use Case Scout batch 8 case i11-editor-undo-coalescing (offline fixture). Prepared with batch8/prepare.sh; hidden key/tests kept outside the agent's work dir. Prompt = TASK.md verbatim. Scored with the scout's scorer: `python score.py WORKDIR KEYDIR`. Tools: Read/Write/Edit/Glob/Grep/Bash, web disabled. Single run, no retries. Claude Code's own reported cost for claude-opus-5.5: $0.250034. Score: 8/8. Cost is an ESTIMATE: measured tokens x pricing.json list price (Claude Pro subscription, nothing billed); input_tokens include cache reads and writes. Single run.

Estimated cost List-price estimate; not actually billed.

Metrics

Estimated cost: $0.25
Wall time: 54s
Human step-ins: 0
Turns: 8

Model usage
Model	Input tok	Output tok	Cost
AClaude Opus 5.5anthropic/claude-opus-5.5effort default (no --effort flag) · thinking on	176,946	5,729	$0.25

Reproduce this run

Use the use case prompt on the use case page.
Same setup: Claude Opus 5.5 (anthropic/claude-opus-5.5, effort default (no --effort flag) · thinking on) via Claude Code 2.1.299 (headless -p), cloud.
It counts as a success only if it meets every success criterion: 1. validation 2. typing_groups 3. word_boundaries 4. time_gap 5. delete_directions 6. undo_redo_semantics 7. group_limits 8. randomized_agreement (500 sequences)
Log model, tokens, wall time, human step-ins and the evidence log, then submit it as a reproduction. A different model? Submit a head-to-head rerun.

How runs are scored: Methodology.

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Automated checks

Trust 80% comes from these checks (how Trust % works).

Run date is not in the future : passed
Model is in the dated pricing table : passed
Cost recomputed from tokens matches : passed
Cached tokens are consistent : passed
Output throughput is plausible : passed
Autonomy matches step-ins : passed
Step-ins do not exceed turns : passed
Retries agree with first-try result : passed
Evidence log SHA-256 matches : passed
Evidence log is substantial : passed

Evidence log

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  "result": "I rewrote `Editor` in `editor.py` (stdlib only). My hand-written checks of the rules all pass. I didn't build a separate reference model to run the 500-random-sequence check against, so those hidden tests are untested here.\n\n**How it works:**\n- Each undo group stores its type (insert or delete), start position, text, the time of its last edit, and whether it was started by a single-character edit. Delete groups also store their direction, which is set by the second delete.\n- Any `undo()` or `redo()` call closes the most recent group, even when it returns `False`. The time of the last recorded edit is kept across undo and redo, so later edits still can't go back in time.\n- An edit that joins an existing group still clears the redo stack and updates the time.\n- The undo stack keeps 50 groups. Pushing a 51st, whether from a new edit or from `redo()`, drops the oldest.\n\n**What I checked:**\n- typing a word then a space, then the next word, gives two separate undos\n- an edit exactly 2.0 s later joins the group; one 2.0001 s later doesn't\n- typing stops joining at 20 characters\n- 50-group cap\n- backspace runs, forward-delete runs, and a change of direction after the second delete starting a new group\n- the group closes after undo/redo\n- validation errors leave the text unchanged\n\nTwo of my early test expectations were wrong; the code was right both times. Characters typed just after an undo form a new group that can keep growing. An empty insert doesn't set the \"previous edit time\".\n\n**Decisions the spec doesn't settle:**\n- **Booleans as positions:** `True` and `False` are rejected with `ValueError` even though Python treats them as ints.\n- **Non-string insert:** passing something other than a string to `insert` also raises `ValueError`.\n- **Time on empty inserts:** an empty insert is checked against the last recorded edit time, but it doesn't change it.",
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]