b8fbc0c253
`mix` over-represents L1-shaped work by 25-40x against real commit history, so this adds `tasks/real` built to the distribution actually measured: three tasks whose fix spans two source files, two single-file rewrites of real logic, one small single-file fix. Ordinary bugs — a sliding window that never prunes, a parser folding operators left to right, a loader accepting misspelled keys, interval merging that assumes sorted input, a retry loop that retries unretryable failures, a page count dropping the last partial page. The suite is sound: reference 6/6, noop 0/6. The cascade's local tiers resolve **0 of 6**, against 4/5 on `mix`. None of these bugs is a substitution, a cached repeat or a paraphrase, so L0, L1 and L2 have nothing to offer and every task falls through in under a second. That is the result the composition analysis predicted and it bounds the local claim precisely: the cheap tiers address a real but narrow class of work, and that class is not what fills a commit log. The 7.6x advantage on `mix` is a property of that suite's composition rather than a general one. What this does not touch is the cache, which is where the cascade's economics actually live: every task here runs once, so L0 never fires. An L4-enabled two-pass run over this suite would measure it — first pass pays the cloud for all six, second should close them free — and has not been run. Building a suite we score zero on is the point. A benchmark that only contains work we are good at measures nothing. Co-Authored-By: Claude Opus 5 <[email protected]>
27 lines
804 B
Python
27 lines
804 B
Python
from lexer import tokenize
|
|
|
|
|
|
def evaluate(text):
|
|
"""Evaluates with * and / binding tighter than + and -."""
|
|
tokens = tokenize(text)
|
|
# First pass: fold the high-precedence operators.
|
|
folded = [tokens[0]]
|
|
index = 1
|
|
while index < len(tokens):
|
|
op, rhs = tokens[index], tokens[index + 1]
|
|
if op == "*":
|
|
folded[-1] = folded[-1] * rhs
|
|
elif op == "/":
|
|
folded[-1] = folded[-1] // rhs
|
|
else:
|
|
folded.append(op)
|
|
folded.append(rhs)
|
|
index += 2
|
|
# Second pass: left-to-right over what is left.
|
|
total = folded[0]
|
|
index = 1
|
|
while index < len(folded):
|
|
op, rhs = folded[index], folded[index + 1]
|
|
total = total + rhs if op == "+" else total - rhs
|
|
index += 2
|
|
return total
|