Head to head: Mistral-Large-3 vs Phi-4-mini-instruct

Mistral-Large-3 vs Phi-4-mini-instruct

This one isn’t a squeaker. Mistral-Large-3 controlled the matchup across most tasks, while Phi-4-mini-instruct’s wins were real but too narrow and too isolated to change the overall picture.

Mistral-Large-3 wins this head-to-head decisively: **93.5 to 67.5 overall, 9 task wins to 2, with 1 tie, and a 98% confidence verdict**. That is not the profile of a split decision or a stylistic preference call. It’s a broad, repeatable advantage. What stands out is *where* Mistral-Large-3 won. It was better on the bread-and-butter editorial reliability tasks that matter in production: structured summarization, ticket triage, meeting-notes extraction to JSON, contradiction finding, faithful rewriting, and the warehouse SQL query. It also took the LRU cache task not because its solution was perfect—it wasn’t O(1)—but because Phi-4-mini-instruct’s implementation was simply broken. Again and again, Mistral-Large-3 was the model that more often produced something usable, correctly formatted, and closer to spec. Phi-4-mini-instruct does deserve credit for the **SQL window query**, where it was plainly better: it used the right window-function approach and correctly handled the "second-largest distinct total" requirement that Mistral-Large-3 fumbled. It also edged the nuanced classification task with sharper, more review-specific reasons. But those wins read like isolated bright spots, not evidence of a model that can hang consistently across a mixed workload. Just as important, Mistral-Large-3’s mistakes were usually the kind you can critique and still work with—extra explanation, imperfect optimality, a non-ideal algorithmic choice. Phi-4-mini-instruct more often lost on harder failure modes: invalid SQL, unsupported inferences in JSON extraction, broken data-structure logic, or missing constraints in tightly specified writing tasks. That difference matters. In real use, "mostly right and salvageable" beats "conceptually aimed in the right direction but unusable as written." **Final call: Mistral-Large-3 is the clear winner. Phi-4-mini-instruct showed flashes of technical precision, but Mistral-Large-3 was the stronger, steadier, and far more dependable model across the board.**

SQL window query

Given Postgres tables `orders(id, customer_id, created_at, total_cents)` and `customers(id, name)`, write ONE SQL query returning each customer's name and their SECOND-largest order total (in dollars, 2 decimals), excluding customers with fewer than two orders. Order by that value descending. Return ONLY the SQL.

Mistral-Large-3: ```sql SELECT c.name, ROUND(o.total_cents / 100.0, 2) AS second_largest_order_total FROM customers c JOIN orders o ON c.id = o.customer_id WHERE o.id IN ( SELECT o1.id FROM orders o1 WHERE o1.customer_id = o.customer_id ORDER BY o1.total_cents DESC LIMIT 1 OFFSET 1 ) GROUP BY c.id, c.name, second_largest_order_total HAVING COUNT(o.id) >= 2 ORDER BY second_largest_order_total DESC; ```

Phi-4-mini-instruct: ```sql WITH RankedOrders AS ( SELECT o.customer_id, c.name, o.total_cents, DENSE_RANK() OVER (PARTITION BY o.customer_id ORDER BY o.total_cents DESC) AS rank_ordered FROM orders o INNER JOIN customers c ON o.customer_id = c.id ) SELECT name, ROUND(MAX(total_cents) / 100.00, 2) AS second_largest_order_total FROM RankedOrders WHERE rank_ordered = 2 GROUP BY name ORDER BY second_largest_order_total DESC; ```

Model B is better because it uses a window function and correctly returns the second-largest distinct order total per customer while naturally excluding customers with fewer than two distinct totals. Model A can fail the "fewer than two orders" requirement because its HAVING clause is applied after filtering to a single row, and it is not a proper window-query solution for the stated task. (Second judge pass, order swapped — scores are the average of both: Model B is substantially better because it correctly uses a window function to find each customer's second-largest distinct order total and excludes customers with fewer than two distinct totals; its only notable caveat is that DENSE_RANK interprets "second-largest" as the second distinct value. Model A is not reliable in Postgres as written because the correlated subquery with ORDER BY/LIMIT can return only one row per outer order id and the HAVING COUNT(o.id) >= 2 check is ineffective after that filtering, so it does not correctly enforce the requirement.)

Ticket triage labeling

Classify each support ticket into exactly one category: Billing, Bug, Feature Request, Account Access, or How-To. For each item, give a 1-sentence justification. Return exactly 5 lines, one per item, in this format: <ID> | <Category> | <justification> Items: T1: "I was charged twice for the May invoice after updating my card." T2: "The export button spins forever in Firefox 126 on the team activity page." T3: "Can you add an option to auto-archive projects after 90 days of inactivity?" T4: "I reset my password, but the login page says my SSO account is not recognized." T5: "What steps do I use to map custom fields from HubTrail into your CRM sync?"

Phi-4-mini-instruct: T1 | Billing | "The customer is experiencing a billing issue with being double-charged for an invoice." T2 | Bug | "A customer reports an issue where a button is non-functional due to browser compatibility." T3 | Feature Request | "The customer is requesting a new feature to automate the archiving of projects." T4 | Account Access | "The customer is having trouble accessing their account due to an SSO recognition problem after password reset." T5 | How-To | "The customer is asking for a guided explanation on how to integrate custom fields with CRM sync."

Model A fully satisfies the required format with exactly 5 lines and concise one-sentence justifications, while Model B adds unnecessary quotation marks around each justification, making its formatting less clean and slightly less compliant. Both models classify all tickets correctly, but Model A is clearer and more polished for this task. (Second judge pass, order swapped — scores are the average of both: Model A is better because it fully follows the requested output format while providing correct classifications and concise one-sentence justifications. Model B is also correct, but the added quotation marks around each justification make it slightly less compliant with the exact formatting request.)

LRU cache

Implement a class `LRUCache<K, V>` in TypeScript with a fixed capacity set in the constructor, and O(1) `get(key): V | undefined` and `set(key, value): void`. Accessing or updating a key must mark it most-recently-used; inserting beyond capacity must evict the least-recently-used entry. Return ONLY the class, no prose.

Mistral-Large-3: ```typescript class LRUCache<K, V> { private capacity: number; private cache: Map<K, V>; private accessOrder: Map<K, number>; private counter: number; constructor(capacity: number) { this.capacity = capacity; this.cache = new Map(); this.accessOrder = new Map(); this.counter = 0; } get(key: K): V | undefined { if (!this.cache.has(key)) return undefined; this.accessOrder.set(key, ++this.counter); return this.cache.get(key); } set(key: K, value: V): void { if (this.cache.has(key)) { this.cache.set(key, value); this.accessOrder.set(key, ++this.counter); return; } if (this.cache.size >= this.capacity) { const leastRecentKey = [...this.accessOrder.entries()] .reduce((a, b) => a[1] < b[1] ? a : b)[0]; this.cache.delete(leastRecentKey); this.accessOrder.delete(leastRecentKey); } this.cache.set(key, value); this.accessOrder.set(key, ++this.counter); } } ```

Phi-4-mini-instruct: ```typescript class Node<K, V> { key: K; value: V; prev?: Node<K, V>; next?: Node<K, V>; } class LRUCache<K, V> { private capacity: number; private count: number; private cache: Map<K, Node<K, V>>; private head: Node<K, V> | null; private tail: Node<K, V> | null; constructor(capacity: number) { this.capacity = capacity; this.count = 0; this.cache = new Map(); this.head = null; this.tail = null; } get(key: K): V | undefined { const node = this.cache.get(key); if (!node) return undefined; this._removeNode(node); this._addNodeToHead(node); return node.value; } set(key: K, value: V): void { let node = this.cache.get(key); if (!node) { if (this.count === this.capacity) { const tail = this._removeTail(); this.cache.delete(tail.key); this.count--; } node = { key, value, prev: null, next: null }; this._addNodeToHead(node); this.cache.set(key, node); this.count++; } else { node.value = value; this._removeNode(node); this._addNodeToHead(node); } } private _removeTail(): Node<K, V> { const tail = this.tail; if (tail) { this.tail = tail.prev; if (this.tail) this.tail.next = null; this.tail = null; return tail; } return null; } private _addNodeToHead(node: Node<K, V>): void { if (!this.head) { this.head = node; this.tail = node; } else { node.next = this.head; this.head.prev = node; this.head = node; } } private _removeNode(node: Node<K, V>): void { node.prev!.next = node.next; node.next!.prev = node.prev; } } ```

Model A is not truly O(1) because eviction scans access history, but it is mostly functional and returns only a class definition. Model B aims for the right data structure, yet its linked-list logic is broken in multiple places, includes an extra top-level class, and would fail at runtime or type-checking. (Second judge pass, order swapped — scores are the average of both: Model A is better because it is at least functionally coherent, while Model B’s linked-list implementation is broken in multiple ways, including invalid node removal logic and a faulty tail-removal method that corrupts state. Model A does not meet the O(1) requirement because eviction scans all entries, but it otherwise follows the requested interface and behavior more reliably than Model B.)

Summarize dense passage

Summarize the passage below in exactly three bullet points, each one sentence, capturing the mechanism, the tradeoff, and the caveat — no jargon a non-specialist couldn't follow. Return only the bullets. Passage: "Speculative decoding pairs a small draft model with a large target model: the draft proposes several tokens, the target verifies them in one pass, and accepted tokens are kept while the first rejection resets to the target's own choice. This can cut latency substantially when acceptance is high, but the draft model's compute is pure overhead when acceptance is low, and gains evaporate on adversarial or highly novel inputs where the draft and target disagree often."

Mistral-Large-3: - **Mechanism:** A smaller model suggests words first, and a larger model quickly checks them—keeping the good ones and only redoing the rest itself. - **Tradeoff:** It speeds things up when the small model’s guesses are right, but wastes effort when they’re wrong. - **Caveat:** It doesn’t help (or can even slow things down) if the text is unusual or the two models keep disagreeing.

Phi-4-mini-instruct: - A smaller test model suggests tokens while a bigger, matching model confirms them, reducing wait times when they agree most of the time. - When the smaller model's guesses match well with the bigger model, the whole process is faster; however, if they match less often, the time saved is less and some original process work may be wasted. - This method isn't always quick; it struggles when new or unusual situations arise, and the smaller model and bigger model don't agree often.

Model A better matches the requested structure by giving exactly three one-sentence bullets that clearly cover the mechanism, tradeoff, and caveat in plain language. Model B is mostly correct, but it is less precise, repeats the tradeoff across two bullets, and uses slightly more awkward wording such as "tokens" and "original process work." (Second judge pass, order swapped — scores are the average of both: Model A better matches the requested structure by clearly covering the mechanism, tradeoff, and caveat in exactly three concise, plain-language bullets. Model B is mostly accurate, but it is more repetitive, less precise about the rejection/reset behavior, and less directly organized around the three requested points.)

Field service scheduling math

A repair coordinator is assigning 4 service calls for one technician on Tuesday. Travel starts from the depot at 08:00. Each job has a fixed service duration and a time window when work may begin. Jobs: - J1 at North Pier: 35 min service, start window 08:30-09:30 - J2 at Larch Ave: 50 min service, start window 09:45-11:00 - J3 at Brook Row: 25 min service, start window 10:30-12:00 - J4 at Elm Court: 40 min service, start window 13:00-15:00 Travel times: Depot->J1 20, Depot->J2 30, Depot->J3 25, Depot->J4 35 J1->J2 15, J1->J3 20, J1->J4 30 J2->J1 15, J2->J3 10, J2->J4 25 J3->J1 20, J3->J2 10, J3->J4 20 J4->J1 30, J4->J2 25, J4->J3 20 Find the feasible order that finishes earliest, assuming waiting is allowed if arriving before a window. Return exactly: Order: <job sequence>; Finish: <HH:MM>

Mistral-Large-3: Alright, let's tackle this problem step by step. We have four jobs (J1, J2, J3, J4) that need to be assigned to a technician who starts from the depot at 08:00. Each job has a service duration and a time window during which the service can start. We also have travel times between the depot and each job, and between the jobs themselves. Our goal is to find the order in which the technician should visit these jobs to finish all of them as early as possible, considering that if the technician arrives before the start of a job's time window, they can wait until the window opens. ### Understanding the Problem First, let's summarize the given information: **Jobs:** - **J1 (North Pier):** - Service duration: 35 min - Start window: 08:30 - 09:30 - **J2 (Larch Ave):** - Service duration: 50 min - Start window: 09:45 - 11:00 - **J3 (Brook Row):** - Service duration: 25 min - Start window: 10:30 - 12:00 - **J4 (Elm Court):** - Service duration: 40 min - Start window: 13:00 - 15:00 **Travel Times (in minutes):** From Depot: - Depot -> J1: 20 - Depot -> J2: 30 - Depot -> J3: 25 - Depot -> J4: 35 Between Jobs: - J1 -> J2: 15 - J1 -> J3: 20 - J1 -> J4: 30 - J2 -> J1: 15 - J2 -> J3: 10 - J2 -> J4: 25 - J3 -> J1: 20 - J3 -> J2: 10 - J3 -> J4: 20 - J4 -> J1: 30 - J4 -> J2: 25 - J4 -> J3: 20 ### Approach to Solve the Problem To find the feasible order that finishes earliest, we need to consider all possible permutations of the four jobs (there are 4! = 24 possible orders) and for each, calculate the total time taken, ensuring that each job starts within its time window. Then, we'll select the order with the earliest finish time. However, evaluating all 24 permutations manually would be time-consuming. Instead, we can use some logical reasoning to narrow down the possibilities: 1. **Time Windows:** Jobs have specific time windows when they can start. If a job's window starts late (like J4 at 13:00), it's likely to be scheduled later in the sequence to avoid unnecessary waiting. 2. **Travel Times:** Minimizing travel time between consecutive jobs can help in finishing earlier. For example, if two jobs are close to each other, scheduling them consecutively might save time. 3. **Service Durations:** Jobs with longer service durations might be better scheduled earlier to prevent them from pushing the entire schedule later. Given these points, let's try to construct a sequence that seems promising. ### Constructing Possible Sequences Let's consider that J4 has a late start window (13:00), so it's likely to be last or second last in the sequence. Let's explore sequences where J4 is last. **Option 1: J1 -> J2 -> J3 -> J4** 1. Start at Depot at 08:00. 2. Travel to J1: 20 min → Arrive at 08:20. - J1's window: 08:30 - 09:30. - Arrived at 08:20, so wait until 08:30. - Start J1 at 08:30, ends at 09:05. 3. Travel to J2: 15 min → Arrive at 09:20. - J2's window: 09:45 - 11:00. - Arrived at 09:20, so wait until 09:45. - Start J2 at 09:45, ends at 10:35. 4. Travel to J3: 10 min → Arrive at 10:45. - J3's window: 10:30 - 12:00. - Arrived at 10:45, within window. - Start J3 at 10:45, ends at 11:10. 5. Travel to J4: 20 min → Arrive at 11:30. - J4's window: 13:00 - 15:00. - Arrived at 11:30, so wait until 13:00. - Start J4 at 13:00, ends at 13:40. Finish time: 13:40. **Option 2: J1 -> J3 -> J2 -> J4** 1. Start at Depot at 08:00. 2. Travel to J1: 20 min → Arrive at 08:20. - Wait until 08:30. - Start J1 at 08:30, ends at 09:05. 3. Travel to J3: 20 min → Arrive at 09:25. - J3's window: 10:30 - 12:00. - Arrived at 09:25, so wait until 10:30. - Start J3 at 10:30, ends at 10:55. 4. Travel to J2: 10 min → Arrive at 11:05. - J2's window: 09:45 - 11:00. - Window closes at 11:00, arrived at 11:05 → Too late! - This sequence is invalid. **Option 3: J2 -> J1 -> J3 -> J4** 1. Start at Depot at 08:00. 2. Travel to J2: 30 min → Arrive at 08:30. - J2's window: 09:45 - 11:00. - Arrived at 08:30, so wait until 09:45. - Start J2 at 09:45, ends at 10:35. 3. Travel to J1: 15 min → Arrive at 10:50. - J1's window: 08:30 - 09:30. - Window closed at 09:30, arrived at 10:50 → Too late! - Invalid sequence. **Option 4: J1 -> J2 -> J4 -> J3** 1. Start at Depot at 08:00. 2. Travel to J1: 20 min → Arrive at 08:20. - Wait until 08:30. - Start J1 at 08:30, ends at 09:05. 3. Travel to J2: 15 min → Arrive at 09:20. - Wait until 09:45. - Start J2 at 09:45, ends at 10:35. 4. Travel to J4: 25 min → Arrive at 11:00. - J4's window: 13:00 - 15:00. - Arrived at 11:00, so wait until 13:00. - Start J4 at 13:00, ends at 13:40. 5. Travel to J3: 20 min → Arrive at 14:00. - J3's window: 10:30 - 12:00. - Window closed at 12:00, arrived at 14:00 → Too late! - Invalid sequence. **Option 5: J2 -> J3 -> J1 -> J4** 1. Start at Depot at 08:00. 2. Travel to J2: 30 min → Arrive at 08:30. - Wait until 09:45. - Start J2 at 09:45, ends at 10:35. 3. Travel to J3: 10 min → Arrive at 10:45. - Start J3 at 10:45, ends at 11:10. 4. Travel to J1: 20 min → Arrive at 11:30. - J1's window: 08:30 - 09:30. - Window closed at 09:30, arrived at 11:30 → Too late! - Invalid sequence. **Option 6: J3 -> J2 -> J1 -> J4** 1. Start at Depot at 08:00. 2. Travel to J3: 25 min → Arrive at 08:25. - J3's window: 10:30 - 12:00. - Arrived at 08:25, so wait until 10:30. - Start J3 at 10:30, ends at 10:55. 3. Travel to J2: 10 min → Arrive at 11:05. - J2's window: 09:45 - 11:00. - Window closed at 11:00, arrived at 11:05 → Too late! - Invalid sequence. From the above, the only valid sequence so far is **J1 -> J2 -> J3 -> J4** with a finish time of 13:40. Let's check if there's a sequence that finishes earlier. **Option 7: J1 -> J3 -> J4 -> J2** 1. Start at Depot at 08:00. 2. Travel to J1: 20 min → Arrive at 08:20. - Wait until 08:30. - Start J1 at 08:30, ends at 09:05. 3. Travel to J3: 20 min → Arrive at 09:25. - Wait until 10:30. - Start J3 at 10:30, ends at 10:55. 4. Travel to J4: 20 min → Arrive at 11:15. - Wait until 13:00. - Start J4 at 13:00, ends at 13:40. 5. Travel to J2: 25 min → Arrive at 14:05. - J2's window: 09:45 - 11:00. - Window closed at 11:00, arrived at 14:05 → Too late! - Invalid sequence. **Option 8: J2 -> J1 -> J3 -> J4** We tried this earlier, and J1 was too late. **Option 9: J3 -> J1 -> J2 -> J4** 1. Start at Depot at 08:00. 2. Travel to J3: 25 min → Arrive at 08:25. - Wait until 10:30. - Start J3 at 10:30, ends at 10:55. 3. Travel to J1: 20 min → Arrive at 11:15. - J1's window: 08:30 - 09:30. - Window closed at 09:30, arrived at 11:15 → Too late! - Invalid sequence. **Option 10: J1 -> J4 -> J2 -> J3** This seems unlikely since J4 is early in the sequence, but let's try. 1. Start at Depot at 08:00. 2. Travel to J1: 20 min → Arrive at 08:20. - Wait until 08:30. - Start J1 at 08:30, ends at 09:05. 3. Travel to J4: 30 min → Arrive at 09:35. - J4's window: 13:00 - 15:00. - Wait until 13:00. - Start J4 at 13:00, ends at 13:40. 4. Travel to J2: 25 min → Arrive at 14:05. - J2's window: 09:45 - 11:00. - Window closed at 11:00, arrived at 14:05 → Too late! - Invalid sequence. After evaluating several sequences, the only valid one that doesn't violate any time windows is **J1 -> J2 -> J3 -> J4**, finishing at 13:40. Is there any other sequence that could finish earlier? Let's try one more where J4 is not last. **Option 11: J1 -> J2 -> J4 -> J3** We tried this earlier, and J3 was too late. **Option 12: J2 -> J3 -> J4 -> J1** 1. Start at Depot at 08:00. 2. Travel to J2: 30 min → Arrive at 08:30. - Wait until 09:45. - Start J2 at 09:45, ends at 10:35. 3. Travel to J3: 10 min → Arrive at 10:45. - Start J3 at 10:45, ends at 11:10. 4. Travel to J4: 20 min → Arrive at 11:30. - Wait until 13:00. - Start J4 at 13:00, ends at 13:40. 5. Travel to J1: 30 min → Arrive at 14:10. - J1's window: 08:30 - 09:30. - Window closed at 09:30, arrived at 14:10 → Too late! - Invalid sequence. It seems that any sequence where J4 is not last leads to a time window violation for the subsequent jobs. Therefore, the only feasible sequence that doesn't violate any time windows is **J1 -> J2 -> J3 -> J4**, finishing at 13:40. ### Verifying if There's a Better Sequence Is there a sequence where the technician can finish before 13:40? Let's try to see if we can rearrange the jobs to finish earlier. **Option 13: J1 -> J3 -> J2 -> J4** We tried this earlier, and J2 was too late. **Option 14: J3 -> J2 -> J1 -> J4** Also tried, J1 was too late. **Option 15: J2 -> J1 -> J4 -> J3** 1. Start at Depot at 08:00. 2. Travel to J2: 30 min → Arrive at 08:30. - Wait until 09:45. - Start J2 at 09:45, ends at 10:35. 3. Travel to J1: 15 min → Arrive at 10:50. - J1's window: 08:30 - 09:30. - Window closed at 09:30, arrived at 10:50 → Too late! - Invalid. It appears that any sequence where J1 is not first will have J1's window violated because J1's window is early (08:30 - 09:30), and other jobs take time to reach after their windows open. Similarly, J4's late window (13:00 - 15:00) makes it likely to be last to avoid waiting too long before its window opens. ### Conclusion After evaluating multiple sequences, the only feasible order that doesn't violate any time windows and finishes the earliest is: **Order: J1, J2, J3, J4; Finish: 13:40**

Phi-4-mini-instruct: To solve this, let's consider the travel times, the durations of each job, and their respective time windows. We must start at the depot at 08:00 and decide an order for the jobs that minimizes the total finishing time. Organize the jobs by their possible start times and windows, considering the travel times from the depot. We can factor in each job's latest feasible arrival time (adjusted for travel and wait time). ### Jobs and Time Windows - J1: (08:30,09:30) with a 35-min service, travel (20 min) - J2: (09:45,11:00) with a 50-min service, travel (30 min) - J3: (10:30,12:00) with a 25-min service, travel (25 min) - J4: (13:00,15:00) with a 40-min service, travel (35 min) ### Travel Times - Depot to J1: 20 min - Depot to J2: 30 min - Depot to J3: 25 min - Depot to J4: 35 min - J1 to J2: 15 min - J1 to J3: 20 min - J1 to J4: 30 min - J2 to J1: 15 min - J2 to J3: 10 min - J2 to J4: 25 min - J3 to J1: 20 min - J3 to J2: 10 min - J3 to J4: 20 min - J4 to J1: 30 min - J4 to J2: 25 min - J4 to J3: 20 min ### Feasibility Check - J1: Arrive 08:00 + 20 min = 08:20, fits in (08:30,09:30) - J2: Arrive 09:02 (Depot -> J2) + 30 min -> 09:32 (no), earliest start 09:45, 25 min travel - J3: Arrive 08:38 (Depot -> J3) + 25 min -> 09:03 (no), earliest start 10:30, 25 min travel - J4: Arrive 09:56 (Depot -> J4) + 35 min -> 10:31 (no), earliest start 13:00, 35 min travel ### Optimal Order Considerations: Start with J1 (proximity) - J1: Arrive 08:00 + 20 min = 08:20 (start J1 -> 08:45) - J2: Early arrival J2 (09:30) fits (Depot -> J2: 30 min) - J3: Next fits after J2 without exceeding (Eastward) - J4: Ends with latest J4 (Eastward) ### Final Order: J1 -> J2 -> J3 -> J4 1. Job Sequence: J1 -> J2 -> J3 -> J4 2. Finish Time: - J1: 08:20 + 35 min (08:55) - J2: 09:45 + 50 min (10:35) - J3: 10:30 + 25 min (11:05) - J4: 13:00 (ends directly, 13:40) Finishing at 13:40. ### Conclusion: **Order: J1 -> J2 -> J3 -> J4** **Finish Time: 13:40**

Model A reaches the correct schedule and finish time and shows coherent feasibility checks, though it badly violates the required output format by adding extensive explanation. Model B also identifies the same order, but its timing math is sloppy and internally inconsistent, and it likewise fails the exact-format instruction. (Second judge pass, order swapped — scores are the average of both: Model A is better because it computes the schedule more coherently and gets the stated sequence timing internally consistent, whereas Model B contains multiple arithmetic and scheduling errors. However, both models fail the prompt's required exact output format and both miss a better feasible order: J1, J3, J2, J4 finishes at 13:40 as well, so neither fully solves the optimization task.)

Meeting notes to JSON

Summarize the following meeting notes into JSON with exactly these keys: decisions, risks, owners, next_meeting. decisions must be an array of exactly 3 short strings. risks must be an array of exactly 2 short strings. owners must be an object mapping person names to one task each. next_meeting must be a date string. Notes: "Ops sync, 7 Aug. Priya said the old barcode scanners fail on about 18% of cold-room scans, so the team will replace them in Dock B first rather than all warehouses. Mateo confirmed the vendor can ship 12 new scanners by 19 Aug, but only if purchasing signs by Friday. Jen wants a one-page training sheet before rollout; Omar volunteered to draft it. They agreed not to change the inventory app this month. Main concern: if Dock B Wi-Fi keeps dropping, staff may blame the new scanners unfairly. Second concern: only two spare batteries are currently on site. Priya will track scan failure rate daily after rollout. Next check-in is 2026-08-21."

Mistral-Large-3: ```json { "decisions": [ "Replace barcode scanners in Dock B first instead of all warehouses", "Do not change the inventory app this month", "Create a one-page training sheet before rollout" ], "risks": [ "Dock B Wi-Fi drops may cause unfair blame on new scanners", "Only two spare batteries are currently available on site" ], "owners": { "Priya": "Track scan failure rate daily after rollout", "Omar": "Draft one-page training sheet" }, "next_meeting": "2026-08-21" } ```

Phi-4-mini-instruct: ```json { "decisions": [ "Replace old barcode scanners in Dock B first", "Purchase signs by Friday to get scanners by 19 Aug", "Omar will draft a one-page training sheet" ], "risks": [ "Dock B Wi-Fi may drop, staff may blame new scanners", "Only two spare batteries on site" ], "owners": { "Priya": "Track scan failure rate daily", "Mateo": "Confirm scanner shipping", "Jen": "Approve one-page training sheet", "Omar": "Draft one-page training sheet" }, "next_meeting": "2026-08-21" } ```

Model A better matches the notes and the required schema: it has exactly 3 decisions, 2 risks, appropriate owners, and the correct next meeting date. Model B includes questionable decisions, adds unsupported or unnecessary owner assignments, and is less faithful to what was actually agreed. (Second judge pass, order swapped — scores are the average of both: Model A is better because it captures three actual decisions from the notes, keeps the required schema, and avoids inventing unsupported owners/tasks. Model B includes a non-decision about purchasing timing, omits the explicit decision not to change the inventory app, and adds unsupported owner assignments for Mateo and Jen.)

Constrained outreach note

Write a follow-up email to a vendor after a delayed shipment. Constraints: - Exactly 4 lines total. - Line 1 must be: Subject: Follow-up on PO 4817 - Include these facts somewhere: 24 filter cartridges, originally due 12 June, needed for the Cedar Vale clinic. - Include exactly two bullet lines, each starting with "- ". - Final line must be exactly: Thanks, Mira - Do not use the words "urgent", "asap", or "immediately". - Keep the whole email under 80 words.

Mistral-Large-3: **Subject: Follow-up on PO 4817** I’m checking on the status of our order for 24 filter cartridges, originally due 12 June for the Cedar Vale clinic. - Could you confirm the revised delivery timeline? - Any updates on the cause of the delay would be helpful. Thanks, Mira

Phi-4-mini-instruct: Subject: Follow-up on PO 4817 Dear [Vendor], The 24 filter cartridges originally due 12 June for Cedar Vale clinic remains undelivered. Please confirm receipt status. - Remaining order: 24 filter cartridges - Urgent necessity for clinic operations Thanks, Mira

Model A is better because it includes the required facts and avoids the banned words, though it still violates the format by using more than 4 lines, bolding the subject, and splitting the sign-off across two lines. Model B also exceeds 4 lines and splits the sign-off, but additionally omits one required bullet line, includes a banned word, and has weaker phrasing. (Second judge pass, order swapped — scores are the average of both: Model A is better because it is clearer and more professional, but both outputs fail key constraints: neither has exactly 4 lines, neither ends with the exact final line, and Model A also alters the required subject line formatting while Model B uses a prohibited word. Model B additionally includes extra blank lines and awkward grammar, making it less compliant overall.)

Python discount bugfix

The following Python function is supposed to apply a percentage discount, then cap the final price at a minimum of 0, rounded to 2 decimals. It fails on normal inputs. Find the bug and return only the corrected code. ```python def apply_discount(price, percent): discounted = price * (percent / 100) final = price - discounted / 100 if final < 0: final = 0 return round(final, 2) ```

Mistral-Large-3: ```python def apply_discount(price, percent): discounted = price * (percent / 100) final = price - discounted if final < 0: final = 0 return round(final, 2) ```

Phi-4-mini-instruct: ```python def apply_discount(price, percent): discounted = price * (percent / 100) final = price - discounted if final < 0: final = 0 return round(final, 2) ```

Model A and Model B provide the same corrected function, fixing the bug by subtracting the full discounted amount instead of dividing it by 100 again. Both follow the instruction to return only corrected code and are equally clear. (Second judge pass, order swapped — scores are the average of both: Model B and Model A provide the same corrected function, fixing the bug by subtracting the full discounted amount instead of dividing it by 100 again. Both follow the instruction to return only corrected code and are equally clear.)

Nuanced classification

Classify each review's sentiment as "positive", "negative", or "mixed", and give a 6-word-max reason. Return ONLY a JSON array of {"text","label","reason"} in input order. Reviews: ["Fast shipping but the fabric feels cheap.", "Absolutely love it, wearing it daily!", "It broke after a week. Refund was quick and painless though."]

Mistral-Large-3: ```json [ {"text": "Fast shipping but the fabric feels cheap.", "label": "mixed", "reason": "Pros and cons"}, {"text": "Absolutely love it, wearing it daily!", "label": "positive", "reason": "Highly enthusiastic praise"}, {"text": "It broke after a week. Refund was quick and painless though.", "label": "mixed", "reason": "Bad product, good service"} ] ```

Phi-4-mini-instruct: ```json [ {"text": "Fast shipping but the fabric feels cheap.", "label": "mixed", "reason": "Quick delivery, cheap feel."}, {"text": "Absolutely love it, wearing it daily!", "label": "positive", "reason": "Loves it, daily wear."}, {"text": "It broke after a week. Refund was quick and painless though.", "label": "mixed", "reason": "Product broke, refund good."} ] ```

Model B is better because it matches the requested labels while giving concise, review-specific reasons within the six-word limit. Model A is also correct, but its reasons are more generic and slightly less informative. (Second judge pass, order swapped — scores are the average of both: Model B and Model A both classify all three reviews correctly and follow the JSON-only format. Model B is better because its reasons are more specific to each review while still staying within the six-word limit, whereas Model A uses vaguer phrases like "Pros and cons.")

Find the contradiction

The following spec contains exactly one internal contradiction. Quote the two conflicting sentences verbatim and explain the conflict in one sentence. Do not fix it. Spec: "Free accounts may create up to three projects. Every account, regardless of tier, may archive unlimited projects. Archiving a project does not count against the project limit. Free accounts are limited to three projects total, including archived ones."

Mistral-Large-3: **Conflicting sentences:** 1. *"Archiving a project does not count against the project limit."* 2. *"Free accounts are limited to three projects total, including archived ones."* **Conflict:** The first sentence states archived projects don’t count toward the limit, while the second says they *do* count.

Phi-4-mini-instruct: The conflicting sentences are: "Free accounts may create up to three projects." and "Every account, regardless of tier, may archive unlimited projects." The conflict arises because if free accounts are strictly limited to three projects total and free accounts can archive unlimited projects, then creating three projects and archiving an unlimited number of them suggests that a free account can exceed the three-project limit by archiving projects, which contradicts the initial project limit per account.

Model A identifies the actual contradictory pair verbatim and explains the conflict accurately in one sentence. Model B selects a non-contradictory pair and bases its explanation on an inferred contradiction rather than the spec’s direct conflict. (Second judge pass, order swapped — scores are the average of both: Model A identifies the actual contradictory pair verbatim and explains the conflict directly in one sentence. Model B quotes the wrong pair of sentences and its explanation relies on an inferred scenario rather than the spec’s explicit contradiction.)

Faithful rewrite

Rewrite this blunt internal note as a warm, professional Slack message to the whole engineering team (60–90 words), keeping every fact intact and adding no new commitments. Return only the message. Note: "Deploy is frozen until the memory leak in the image service is fixed. Priya is on it. Do not merge to main. ETA tomorrow noon."

Mistral-Large-3: **Hey team!** 👋 Just a quick heads-up: we’ve hit a temporary deploy freeze while Priya tackles a memory leak in the image service. To keep things stable, please hold off on merging to `main` for now. We’re aiming to resolve this by **tomorrow noon**—thanks for your patience and teamwork! Let us know if you have questions. 🚀

Phi-4-mini-instruct: Hey Team, Just a heads-up: Deployment is on hold until we resolve the memory leak in the image service. Priya is tackling it and we'll proceed after noon tomorrow. Please refrain from merging to main. Thanks for understanding! Best, [Your Name]

Model A is warmer and more polished while preserving the key facts more faithfully, though it adds minor extra language. Model B changes the timing by implying action will resume after noon tomorrow rather than preserving the original ETA of tomorrow noon, and it is less suited to a Slack message because of the sign-off. (Second judge pass, order swapped — scores are the average of both: Model A is warmer and more polished while preserving the key facts more faithfully, especially the ETA as tomorrow noon. Model B adds a sign-off and changes the timing to "after noon tomorrow," which is less precise and slightly alters the original note.)

Warehouse SQL query

Write a single SQL query to answer this question. Schema: customers(customer_id INT, name TEXT, region TEXT) orders(order_id INT, customer_id INT, order_date DATE, status TEXT) order_items(order_id INT, sku TEXT, quantity INT, unit_price DECIMAL(10,2)) Question: For completed orders only (status = 'completed') placed in Q1 2025 (2025-01-01 through 2025-03-31 inclusive), return the top 3 customers in the 'Midwest' region by total revenue, where revenue is SUM(quantity * unit_price). Output columns: customer_id, name, total_revenue. Sort by total_revenue descending, then customer_id ascending.

Mistral-Large-3: Here's the SQL query to answer your question: ```sql SELECT c.customer_id, c.name, SUM(oi.quantity * oi.unit_price) AS total_revenue FROM customers c JOIN orders o ON c.customer_id = o.customer_id JOIN order_items oi ON o.order_id = oi.order_id WHERE c.region = 'Midwest' AND o.status = 'completed' AND o.order_date BETWEEN '2025-01-01' AND '2025-03-31' GROUP BY c.customer_id, c.name ORDER BY total_revenue DESC, c.customer_id ASC LIMIT 3; ``` This query: 1. Joins the three tables to connect customers with their orders and order items 2. Filters for Midwest region customers with completed orders in Q1 2025 3. Calculates total revenue for each customer by summing quantity × unit_price 4. Groups by customer_id and name 5. Orders by total revenue descending, then customer_id ascending 6. Returns only the top 3 customers

Phi-4-mini-instruct: ```sql WITH CompletedOrders AS ( SELECT o.order_id, o.customer_id, o.order_date FROM orders o WHERE o.status = 'completed' AND o.order_date BETWEEN '2025-01-01' AND '2025-03-31' ), OrderDetails AS ( SELECT ci.customer_id, SUM(oi.quantity * oi.unit_price) AS revenue FROM CompletedOrders co JOIN order_items oi ON co.order_id = oi.order_id GROUP BY ci.customer_id ), MidwestCustomers AS ( SELECT c.customer_id, c.name, od.revenue FROM customers c JOIN OrderDetails od ON c.customer_id = od.customer_id WHERE c.region = 'Midwest' ) SELECT customer_id, name, total_revenue FROM ( SELECT customer_id, name, total_revenue, RANK() OVER (ORDER BY total_revenue DESC, customer_id) AS rank FROM MidwestCustomers ) AS RankedCustomers WHERE rank <= 3 ORDER BY total_revenue DESC, customer_id ASC; ```

Model A provides a correct single query that filters the right date range and status, aggregates revenue per Midwest customer, sorts correctly, and limits to the top 3. Model B contains multiple correctness issues, including invalid aliases/column references (using ci.customer_id, selecting total_revenue where only revenue exists) and it does not reliably satisfy the requested top-3 behavior as cleanly as Model A. (Second judge pass, order swapped — scores are the average of both: Model A provides a correct single query that joins the right tables, filters completed Q1 2025 Midwest orders, aggregates revenue per customer, and returns the top 3 with the requested ordering. Model B has multiple fatal errors, including undefined aliases, inconsistent column names, and selecting a non-existent total_revenue field, so it would not run as written.)