You are taking a rigorous coding benchmark. Solve ALL 10 problems below in Python 3.12 (stdlib only unless stated). Output ONLY code — no prose, no explanations outside of code comments.

STRICT OUTPUT FORMAT — repeat for every problem:
### SOLUTION P1
```python
[complete, runnable code]
```
### SOLUTION P2
... through P10.

If you cannot solve a problem, still emit the section header with a stub that raises NotImplementedError.

---

### PROBLEM 1: SHUTDOWN-SAFE BOUNDED BLOCKING QUEUE (10 pts)

class QueueClosed(Exception): pass

Implement class BoundedQueue:
  __init__(self, maxsize: int)
  put(self, item, timeout=None)   # blocks when full; raises QueueClosed if close()d (even mid-block)
  get(self, timeout=None)         # blocks when empty; drains remaining items then raises QueueClosed
  close(self)                     # wakes ALL blocked callers immediately

Rules:
- timeout=None → block indefinitely; timeout=0 → raise immediately if not ready; timeout>0 → block up to that many seconds
- After close(): future put() and any blocked put() raise QueueClosed immediately
- After close(): get() returns remaining queued items one by one, then raises QueueClosed
- Fully thread-safe (threading module)
- 8 producers × 50 000 items, 4 consumers → exact multiset, zero loss/duplication, no hang

---

### PROBLEM 2: WEIGHTED DAMERAU-LEVENSHTEIN DISTANCE (10 pts)

def dl_distance(a: str, b: str, *, ins: int, dele: int, sub: int, transpose: int) -> int:

True Damerau–Levenshtein (NOT optimal-string-alignment / OSA).
The key difference: DL uses a last-occurrence table so a substring can be edited more than once.
Transposition recurrence (indices 1-based):
  i1 = last row where a[i1-1] == b[j-1] (0 if none)
  j1 = last col where b[j1-1] == a[i-1] (0 if none)
  d[i][j] = min(d[i][j],
                d[i1-1][j1-1] + (i-i1-1)*dele + (j-j1-1)*ins + transpose)
Handle empty strings, asymmetric costs, long strings.

---

### PROBLEM 3: STREAMING COVARIANCE + LOG-SUM-EXP (10 pts)

class RunningStats:
    def push(self, x: float, y: float) -> None   # Welford online update — NO sum-of-squares
    @property def mean_x(self) -> float
    @property def mean_y(self) -> float
    @property def var_x(self) -> float            # population variance (÷n)
    @property def var_y(self) -> float
    @property def cov_xy(self) -> float           # population covariance (÷n)

def logsumexp(xs: list[float]) -> float:
    # log(Σexp(xi)) computed stably (shift by max)
    # Must handle [1000,1000] → ≈1000.693, [-1000,-1000] → finite

---

### PROBLEM 4: ROOT-CAUSE BUG HUNT (10 pts)

Below is a 3-file async job runner. It intermittently processes jobs OUT OF ORDER and occasionally LOSES jobs. Find BOTH root causes and fix with MINIMAL changes. Provide fixed versions of all three files. Explain the root cause(s) in 2–3 sentences as a Python comment at the top of runner.py.

# ===== job_queue.py =====
import asyncio

class JobQueue:
    def __init__(self):
        self._q = asyncio.Queue()
    async def push(self, item):
        await self._q.put(item)
    async def pop(self):
        return await self._q.get()
    def complete(self):
        self._q.task_done()
    async def join(self):
        await self._q.join()

# ===== worker.py =====
import asyncio
SENTINEL = object()

class Worker:
    def __init__(self, wid, queue, sink, lock):
        self.wid = wid
        self.queue = queue
        self.sink = sink
        self.lock = lock

    async def run(self):
        while True:
            job = await self.queue.pop()
            if job is SENTINEL:
                self.queue.complete()
                break
            async def _do_job():
                await asyncio.sleep(job["delay"])
                async with self.lock:
                    self.sink.append(job["id"])
            asyncio.ensure_future(_do_job())   # <-- BUG A
            self.queue.complete()

# ===== runner.py =====
import asyncio
from worker import Worker, SENTINEL
from job_queue import JobQueue

async def _enqueue_all(queue, jobs):
    coros = []
    for job in jobs:
        async def enqueue():
            await asyncio.sleep(0)
            await queue.push(job)   # <-- BUG B: late-binding closure
        coros.append(enqueue())
    await asyncio.gather(*coros)

async def run_jobs(job_list, *, parallelism=4):
    queue = JobQueue()
    results = []
    lock = asyncio.Lock()
    workers = [Worker(i, queue, results, lock) for i in range(parallelism)]
    worker_tasks = [asyncio.create_task(w.run()) for w in workers]
    await _enqueue_all(queue, job_list)
    for _ in range(parallelism):
        await queue.push(SENTINEL)
    await asyncio.gather(*worker_tasks)
    return sorted(results)

---

### PROBLEM 5: BEHAVIOUR-PRESERVING REFACTOR (10 pts)

Refactor this EXACTLY — identical output for every input including quirky edges. DO NOT fix or improve any behaviour.

def munge(data, limit=256):
    if not data:
        return []
    out = []
    acc = 0.0
    prev = None
    i = 0
    while i < len(data):
        x = data[i]
        i += 1
        if x is None:
            if prev is None:
                continue
            x = prev
        elif isinstance(x, bool):
            x = int(x)
        elif isinstance(x, str):
            try:
                x = float(x)
            except (ValueError, TypeError):
                continue
        elif not isinstance(x, (int, float)):
            continue
        if x < 0:
            if -x > limit:
                x = float(-limit)
            else:
                x = x + (int(x) // 2)
        else:
            if x > limit:
                r = int(x) - limit
                x = limit + r // 2
                if x > 2147483647:
                    x = 2147483647
        if prev is not None and prev != 0:
            ratio = x / prev
            if ratio > 2.0:
                x = prev * 2.0
            elif ratio < 0.5:
                x = prev * 0.5 if prev > 0 else prev / 0.5
        acc += x
        if acc > limit * 100:
            if x > 0:
                x = x * 0.9
            if out and out[-1] > x:
                x = (x + out[-1]) / 2.0
        out.append(x)
        prev = x
    return out

---

### PROBLEM 6: O(1) LRU CACHE WITH TTL (10 pts)

class LRUTTLCache:
    def __init__(self, capacity: int, ttl_seconds: float, clock=None)
    def get(self, key) -> object    # None if missing/expired; refreshes LRU order but NOT TTL
    def put(self, key, value)       # inserts/updates; evicts true-LRU LIVE entry if at capacity

Rules:
- Expired entries are invisible and do NOT count toward capacity
- get() refreshes recency order; does NOT reset the TTL expiry time
- Eviction removes the least-recently-used LIVE (non-expired) entry
- ALL operations O(1) worst-case (use OrderedDict or doubly-linked-list + dict)
- clock: callable returning current time (default time.time); used for testability

---

### PROBLEM 7: ROBUST ARITHMETIC EXPRESSION EVALUATOR (10 pts)

class EvalError(Exception):
    def __init__(self, pos: int): self.pos = pos; super().__init__(f"error at pos {pos}")

def evaluate(expr: str) -> float:
Supports: + - * / % (binary), - (unary), ^ (right-associative exponent), parentheses, integers, floats.
Precedence (low→high): + - < * / % < unary- < ^
On malformed input or division/modulo by zero: raise EvalError(pos) with exact 0-based char index.

MUST PASS:
  evaluate("2^3^2") == 512.0       # right-assoc: 2^(3^2)=2^9=512
  evaluate("-2^2")  == -4.0        # unary- lower than ^: -(2^2)=-4
  evaluate("--3")   == 3.0
  evaluate("10%3")  == 1.0
  evaluate("2 * (3 + 4)") == 14.0
  evaluate("1 + * 2") raises EvalError(pos=4)
  evaluate("1/0") raises EvalError

---

### PROBLEM 8: TOP-K OVER A 10M-ITEM STREAM (10 pts)

def top_k(stream_iter, k: int) -> list[tuple]:
    # Return k most-frequent (key, count) pairs, sorted by count DESC, ties by key ASC
    # Memory: O(distinct_keys). Must be ≥5× faster than Counter+full-sort on 10M items.

Hint: Count with Counter/dict, then use heapq.nlargest (O(n + k log n)) instead of full sort.
Ties broken: key ascending (works for str and numeric keys using standard comparison).

---

### PROBLEM 9: DATE ARITHMETIC ACROSS DST (10 pts)

from zoneinfo import ZoneInfo
from datetime import datetime, timedelta

def add_local_days(dt_local: datetime, n: int, tz: ZoneInfo) -> datetime:
    # Add n calendar days keeping same wall-clock time (hour/minute/second/microsecond).
    # Spring-forward gap: if the wall time doesn't exist, use fold=0 (first valid after gap).
    # Fall-back ambiguity: if wall time occurs twice, use fold=0 (first/standard occurrence).
    # Stdlib zoneinfo only — no pytz, no dateutil.

---

### PROBLEM 10: GRAPHEME-AWARE TRUNCATE & REVERSE (10 pts)

Implement WITHOUT third-party libraries. You may use unicodedata.

def truncate(s: str, n: int) -> str:
    # Return at most n user-perceived grapheme clusters. Append "…" if string was cut.
    # Never split a grapheme cluster.

def reverse_graphemes(s: str) -> str:
    # Reverse by grapheme clusters (not code points).

A grapheme cluster groups code points that form one visible character:
- ZWJ sequences: 👨‍👩‍👧‍👦 = 1 cluster (joined by U+200D)
- Emoji + skin-tone modifier (U+1F3FB..U+1F3FF)
- Regional indicator pairs: 🇿🇦 = 1 cluster
- Base char + combining marks (unicodedata.category starts with 'M')
- CRLF pair
- Variation selector after base char

Implement a segmenter using unicodedata.category() and ord() checks.
