Head to head: Kimi-K2.7-Code vs mistral-medium-3-5

Kimi-K2.7-Code vs mistral-medium-3-5

This wasn’t a contest so much as a sweep. Kimi-K2.7-Code dominated the matchup on both score and consistency, beating mistral-medium-3-5 across nearly every task type with a decisive 100% confidence verdict.

Kimi-K2.7-Code wins this head-to-head outright: **116.0 to 81.0**, with **11 task wins to 0** and **1 tie**. That’s not a narrow edge or a vibes-based preference; it’s a comprehensive result backed by a **100% confidence** statistical verdict. If you’re looking for the model that more reliably lands the assignment, this matchup answered the question emphatically. What stands out is how often Kimi won on the boring but essential stuff that separates useful models from annoying ones: following output constraints, keeping formats clean, and not smuggling in extra explanation where it wasn’t asked for. It was better on strict JSON extraction, messy contact normalization, scheduling, clinic assignment, release-note formatting, and the Python bugfix largely because it actually respected the prompt shape. Mistral-medium-3-5 repeatedly lost points by wrapping answers in markdown, adding prose, or drifting from exact structural requirements. Kimi also looked stronger on implementation-heavy work. It took the TypeScript LRU cache, the concurrency bug fix, and the Go LRU task, where correctness under edge cases and adherence to requested structure mattered more than sounding polished. Even when both models reached the right underlying answer—as in the scheduling and assignment tasks—Kimi was the one more likely to deliver it in the form the user actually requested. The lone tie, on the unit-aware math problem, says something useful: when the task is straightforward and the format is forgiving, both models can perform competently. But this benchmark wasn’t decided on easy arithmetic. It was decided on whether a model can be trusted across messy real-world asks—code repair, constrained generation, exact extraction, and format compliance—and on that measure, mistral-medium-3-5 was simply outclassed. **Final call: Kimi-K2.7-Code is the clear, decisive winner. mistral-medium-3-5 never found a lane in this matchup, while Kimi combined stronger technical execution with much better instruction discipline.**

python_fee_bugfix

Debugging & code review — Language: Python 3. The function should total invoice lines, applying each line's discount, then adding tax once to the discounted subtotal. Find the bug, fix it, and return only the corrected code. ```python def invoice_total(lines, tax_rate): total = 0 for qty, unit_price, discount in lines: line_total = qty * unit_price total += line_total * (1 - tax_rate) * (1 - discount) return round(total * (1 + tax_rate), 2) items = [ (2, 19.95, 0.10), (1, 120.00, 0.25), (3, 4.50, 0.00), ] print(invoice_total(items, 0.0825)) # expected 156.16 ```

Kimi-K2.7-Code: ```python def invoice_total(lines, tax_rate): subtotal = 0 for qty, unit_price, discount in lines: subtotal += qty * unit_price * (1 - discount) return round(subtotal * (1 + tax_rate), 2) items = [ (2, 19.95, 0.10), (1, 120.00, 0.25), (3, 4.50, 0.00), ] print(invoice_total(items, 0.0825)) ```

mistral-medium-3-5: The bug is in the application of discounts and tax. Currently, the tax is being applied to each line before discounting, and then tax is applied again to the total. The correct approach is to first calculate the line totals with their respective discounts, sum them up to get the subtotal, and then apply the tax once to this subtotal. Here is the corrected code: ```python def invoice_total(lines, tax_rate): subtotal = 0 for qty, unit_price, discount in lines: line_total = qty * unit_price * (1 - discount) subtotal += line_total total = subtotal * (1 + tax_rate) return round(total, 2) ```

Both outputs fix the tax/discount bug correctly, but Model A follows the instruction to return only corrected code, while Model B adds explanatory prose and omits the provided usage example. Model A is therefore the better fit for the task. (Order-swapped judge pass: Both outputs fix the tax/discount bug correctly, but Model B better follows the instruction to return only the corrected code. Model A adds explanatory prose and markdown, which violates the requested output format.)

proofread_policy_note

Editing & proofreading — Correct the passage for grammar, punctuation, and clarity without changing meaning. Then provide 2 brief bullet points noting the main changes. Passage: "Before anyone submits receipts for the North Quay visit, make sure they was uploaded in Concur by Friday. The finance team dont reimburse late meals unless a director approve it in writing, however taxi fares after 9pm are allowed if theres no train running. Please attached the hotel folio, not just the card slip, because thats what audit ask for last quarter."

Kimi-K2.7-Code: **Corrected passage:** Before anyone submits receipts for the North Quay visit, make sure they were uploaded to Concur by Friday. The finance team does not reimburse late meals unless a director approves it in writing; however, taxi fares after 9 p.m. are allowed if there is no train running. Please attach the hotel folio, not just the card slip, because that is what the audit team asked for last quarter. **Main changes:** - Fixed subject–verb agreement and verb tenses (e.g., *was* → *were*, *approve* → *approves*, *ask* → *asked*, *attached* → *attach*). - Corrected punctuation, contractions, and spelling (e.g., *dont* → *does not*, *theres* → *there is*, *thats* → *that is*, added a semicolon before *however*, and wrote *9 p.m.*).

mistral-medium-3-5: **Corrected Passage:** "Before anyone submits receipts for the North Quay visit, make sure they *were* uploaded in Concur by Friday. The finance team *does not* reimburse late meals unless a director *approves* them in writing; however, taxi fares after 9 PM are allowed if there’s no train running. Please *attach* the hotel folio, not just the card slip, because that’s what audits *asked* for last quarter." **Main Changes:** - Grammar: Corrected verb tense ("was" → "were"), subject-verb agreement ("dont" → "does not"), and verb form ("approve" → "approves"). - Clarity: Added missing articles/positioning ("them", "after 9 PM"), adjusted punctuation (added semicolon), and fixed possessive ("that’s").

Model A is more accurate and polished overall: it fixes the grammar cleanly, improves punctuation, and its notes accurately describe the edits. Model B leaves weaker choices such as "in Concur," adds emphasis formatting and quotation marks unnecessarily, and its change note incorrectly labels "that's" as possessive. (Order-swapped judge pass: Model B provides a cleaner, more polished correction and keeps the meaning intact while improving clarity. Model A introduces a likely error by changing "audit" to "audits" and is slightly less natural in a few phrasing choices.)

Unit-aware math

A pump moves 3.5 liters every 8 seconds. A tank holds 0.9 cubic meters. Starting empty, how long to fill it, in minutes and seconds (mm:ss), rounded to the nearest second? Show the key steps, then give the final answer on its own line.

Kimi-K2.7-Code: Key steps: 1. Convert tank volume to liters: \(0.9 \text{ m}^3 \times 1000 = 900 \text{ L}\) 2. Pump rate: \(\frac{3.5 \text{ L}}{8 \text{ s}} = 0.4375 \text{ L/s}\) 3. Time in seconds: \(\frac{900}{0.4375} = 2057.14... \text{ s}\) 4. Round to nearest second: \(2057 \text{ s}\) 5. Convert to minutes and seconds: \(2057 \div 60 = 34\) minutes with \(17\) seconds remaining. 34:17

mistral-medium-3-5: **Solution Steps** 1. Convert the tank volume to liters: 0.9 m³ × 1000 = 900 liters 2. Find the pump rate in liters per second: 3.5 liters / 8 seconds = 0.4375 L/s 3. Calculate the total time in seconds: 900 L / 0.4375 L/s = 2057.142857… s 4. Round to the nearest second: 2057 s 5. Convert to minutes and seconds: 2057 s = 34 minutes 17 seconds **Answer** 34:17

Both outputs are mathematically correct, show the key steps clearly, and provide the final answer on its own line in the requested mm:ss format. The differences are purely stylistic and do not materially affect quality. (Order-swapped judge pass: Both outputs are correct, show the key steps clearly, and provide the final answer on its own line in the requested mm:ss format. The differences are purely stylistic and do not materially affect quality.)

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.

Kimi-K2.7-Code: ```typescript class LRUCache<K, V> { private cache: Map<K, V>; private capacity: number; constructor(capacity: number) { this.capacity = capacity; this.cache = new Map<K, V>(); } get(key: K): V | undefined { const value = this.cache.get(key); if (value === undefined) { return undefined; } this.cache.delete(key); this.cache.set(key, value); return value; } set(key: K, value: V): void { if (this.cache.has(key)) { this.cache.delete(key); } else if (this.cache.size >= this.capacity) { const lruKey = this.cache.keys().next().value as K; this.cache.delete(lruKey); } this.cache.set(key, value); } } ```

mistral-medium-3-5: ```typescript class LRUCache<K, V> { private capacity: number; private cache: Map<K, V>; private head: DoublyLinkedNode<K> | null = null; private tail: DoublyLinkedNode<K> | null = null; private nodes: Map<K, DoublyLinkedNode<K>>; constructor(capacity: number) { this.capacity = capacity; this.cache = new Map<K, V>(); this.nodes = new Map<K, DoublyLinkedNode<K>>(); } get(key: K): V | undefined { const node = this.nodes.get(key); if (!node) return undefined; this.moveToHead(node); return this.cache.get(key); } set(key: K, value: V): void { if (this.cache.has(key)) { const node = this.nodes.get(key)!; this.moveToHead(node); this.cache.set(key, value); return; } if (this.cache.size >= this.capacity) { const lruNode = this.tail!; this.cache.delete(lruNode.key); this.nodes.delete(lruNode.key); this.removeNode(lruNode); } const newNode = new DoublyLinkedNode(key); this.cache.set(key, value); this.nodes.set(key, newNode); this.addToHead(newNode); } private addToHead(node: DoublyLinkedNode<K>): void { node.prev = null; node.next = this.head; if (this.head) this.head.prev = node; this.head = node; if (!this.tail) this.tail = node; } private removeNode(node: DoublyLinkedNode<K>): void { if (node.prev) node.prev.next = node.next; else this.head = node.next; if (node.next) node.next.prev = node.prev; else this.tail = node.prev; } private moveToHead(node: DoublyLinkedNode<K>): void { this.removeNode(node); this.addToHead(node); } } class DoublyLinkedNode<K> { key: K; prev: DoublyLinkedNode<K> | null = null; next: DoublyLinkedNode<K> | null = null; constructor(key: K) { this.key = key; } } ```

A is concise, valid TypeScript, and correctly implements LRU behavior in O(1) using Map insertion order. B violates the requirement to return only the class by adding a separate top-level helper class, and its capacity-0 case can crash due to dereferencing tail! when null. (Order-swapped judge pass: Model B is simpler and correctly implements LRU behavior in O(1) average time using Map insertion order, while Model A has a serious edge-case bug when capacity is 0 and also violates the 'return ONLY the class' instruction by including an extra class. Both are readable, but B adheres better to the requested output format.)

messy_contacts_to_json

Data wrangling / structured output — Convert the messy contact notes below into valid JSON only. Schema: {"contacts":[{"name":string,"company":string,"email":string,"phones":[string],"tags":[string]}]} Rules: - Output contacts sorted by name ascending - Normalize emails to lowercase - Keep phone numbers as strings exactly as written except trim surrounding spaces - Tags must be lowercase, unique, sorted alphabetically - If a contact has no phone, use an empty array - Ignore lines marked "duplicate" Data: 1) Nia Vale | Larkspur Systems | NIA.VALE@LARKSPUR.IO | +44 20 7946 1182 | Tags: Procurement; Q3; procurement 2) duplicate -> Nia Vale | Larkspur Systems | nia.vale@larkspur.io | +44 20 7946 1182 | Tags: q3 3) Omar Sen | Brindle Health | omar.sen@brindlehealth.com | | Tags: Renewal ; VIP 4) Petra Ko | Aster Dock | PETRA.KO@ASTERDOCK.CO | (312) 555-0199, 312-555-0111 | Tags: west, Logistics, west 5) Eli Dunn | Northline Civic | eli.dunn@northline.example | 0780 123 900 | Tags: board; urgent

Kimi-K2.7-Code: { "contacts": [ { "name": "Eli Dunn", "company": "Northline Civic", "email": "eli.dunn@northline.example", "phones": [ "0780 123 900" ], "tags": [ "board", "urgent" ] }, { "name": "Nia Vale", "company": "Larkspur Systems", "email": "nia.vale@larkspur.io", "phones": [ "+44 20 7946 1182" ], "tags": [ "procurement", "q3" ] }, { "name": "Omar Sen", "company": "Brindle Health", "email": "omar.sen@brindlehealth.com", "phones": [], "tags": [ "renewal", "vip" ] }, { "name": "Petra Ko", "company": "Aster Dock", "email": "petra.ko@asterdock.co", "phones": [ "(312) 555-0199", "312-555-0111" ], "tags": [ "logistics", "west" ] } ] }

mistral-medium-3-5: ```json { "contacts": [ { "name": "Eli Dunn", "company": "Northline Civic", "email": "eli.dunn@northline.example", "phones": ["0780 123 900"], "tags": ["board", "urgent"] }, { "name": "Nia Vale", "company": "Larkspur Systems", "email": "nia.vale@larkspur.io", "phones": ["+44 20 7946 1182"], "tags": ["procurement", "q3"] }, { "name": "Omar Sen", "company": "Brindle Health", "email": "omar.sen@brindlehealth.com", "phones": [], "tags": ["renewal", "vip"] }, { "name": "Petra Ko", "company": "Aster Dock", "email": "petra.ko@asterdock.co", "phones": ["(312) 555-0199", "312-555-0111"], "tags": ["logistics", "west"] } ] } ```

Both outputs extract and normalize the contacts correctly, but Model A better follows the requirement to return valid JSON only. Model B wraps the JSON in Markdown code fences, which violates the output-format instruction. (Order-swapped judge pass: Both outputs correctly parse, normalize, deduplicate, and sort the contacts, but Model A wraps the JSON in a Markdown code fence, violating the requirement to output valid JSON only. Model B provides clean valid JSON directly and is therefore better.)

Concurrency bug fix

This TypeScript function is meant to memoize an async loader but has a race: concurrent callers can each trigger the underlying fetch. Fix it so the fetch runs at most once per key, and a rejected fetch does NOT poison the cache (a later call must retry). Return ONLY the corrected function. ```ts const cache = new Map<string, Promise<string>>(); async function load(key: string, fetcher: (k: string) => Promise<string>) { if (cache.has(key)) return cache.get(key)!; const value = await fetcher(key); cache.set(key, Promise.resolve(value)); return value; } ```

Kimi-K2.7-Code: ```ts async function load(key: string, fetcher: (k: string) => Promise<string>): Promise<string> { let promise = cache.get(key); if (!promise) { promise = fetcher(key).catch((err) => { cache.delete(key); throw err; }); cache.set(key, promise); } return promise; } ```

mistral-medium-3-5: ```ts const cache = new Map<string, { promise: Promise<string>, resolve: (v: string) => void, reject: (e: unknown) => void }>(); async function load(key: string, fetcher: (k: string) => Promise<string>) { let entry = cache.get(key); if (entry) return entry.promise; let resolve: (v: string) => void, reject: (e: unknown) => void; const promise = new Promise<string>((res, rej) => { resolve = res; reject = rej; }); cache.set(key, { promise, resolve: resolve!, reject: reject! }); try { const value = await fetcher(key); resolve!(value); cache.delete(key); return value; } catch (e) { reject!(e); cache.delete(key); throw e; } } ```

Model A correctly stores the in-flight promise before awaiting, ensuring only one fetch per key, and removes failed promises so retries work. Model B also coalesces concurrent calls and clears failures, but it unnecessarily changes the cache shape, deletes successful results so it no longer memoizes, and is more complex than needed. (Order-swapped judge pass: Model B correctly stores the in-flight promise immediately so concurrent callers share one fetch, and it removes failed promises so later calls retry. Model A avoids the race but incorrectly deletes successful results from the cache, breaking memoization semantics and adding unnecessary complexity.)

strict_release_note

Instruction following — Write a release note for an internal app update. Requirements: - Exactly 4 bullet points - Each bullet must start with "- " - Each bullet must be 7 to 10 words long - Include these exact version and date strings somewhere: "v2.8.1" and "2026-03-14" - Mention all three features exactly once each: offline drafts, SSO, audit export - Do NOT use the words: improved, seamless, bug, issue, fix - After the bullets, add a final line exactly: Owner: Mara Iqbal

Kimi-K2.7-Code: - v2.8.1 ships on 2026-03-14 for all teams - You can now save offline drafts anytime - SSO login is available across every workspace - Audit export supports CSV and PDF formats now Owner: Mara Iqbal

mistral-medium-3-5: - Offline drafts now available in v2.8.1 release - SSO integration added for easier access - Audit export feature enabled for all users - Release date marked as 2026-03-14 Owner: Mara Iqbal

Model A satisfies all structural constraints, includes both required strings, and mentions each required feature exactly once. Model B omits the exact version/date pairing in one bullet set and violates the 7-10 word requirement on multiple bullets. (Order-swapped judge pass: Both outputs satisfy the structural constraints and include the required version, date, features, and owner line. Model B is stronger because all bullets meet the 7-10 word requirement, while Model A has one bullet with only six words.)

Strict JSON extraction

Extract every meeting from this text into a JSON array of {"title","day","start24","durationMinutes"}. Use 24-hour "HH:MM" for start24, integer minutes for duration, and title-case titles. Return ONLY valid JSON, no commentary. Text: "Standup is daily at 9am for a quarter hour. The design review runs thursday from 2:30-4pm. Payroll sync — first monday, 11:00 to noon."

Kimi-K2.7-Code: [ {"title": "Standup", "day": "Daily", "start24": "09:00", "durationMinutes": 15}, {"title": "Design Review", "day": "Thursday", "start24": "14:30", "durationMinutes": 90}, {"title": "Payroll Sync", "day": "First Monday", "start24": "11:00", "durationMinutes": 60} ]

mistral-medium-3-5: ```json [ {"title": "Standup", "day": "Daily", "start24": "09:00", "durationMinutes": 15}, {"title": "Design Review", "day": "Thursday", "start24": "14:30", "durationMinutes": 90}, {"title": "Payroll Sync", "day": "Monday", "start24": "11:00", "durationMinutes": 60} ] ```

A is fully correct and returns only valid JSON. B incorrectly drops the ordinal in "First Monday" and wraps the JSON in a markdown code fence, violating the output constraint. (Order-swapped judge pass: Both extract the meetings and times correctly, but B better preserves the source text by keeping "First Monday" rather than collapsing it to "Monday." A also wraps its JSON in a markdown code fence, which violates the instruction to return only valid JSON.)

go_lru_cache

Algorithm & data structures — Return code only. Implement an LRU cache in Go with this API: - `type LRUCache struct { ... }` - `func NewLRU(capacity int) *LRUCache` - `func (c *LRUCache) Get(key string) (int, bool)` - `func (c *LRUCache) Put(key string, value int)` Requirements: - Both Get and Put must be O(1) average time - Evict the least recently used item when capacity is exceeded - Updating an existing key must refresh its recency - Use only the Go standard library - Include a minimal `main` demonstrating a few operations

Kimi-K2.7-Code: ```go package main import ( "container/list" "fmt" ) type lruNode struct { key string value int } type LRUCache struct { capacity int items map[string]*list.Element order *list.List } func NewLRU(capacity int) *LRUCache { return &LRUCache{ capacity: capacity, items: make(map[string]*list.Element), order: list.New(), } } func (c *LRUCache) Get(key string) (int, bool) { if elem, ok := c.items[key]; ok { c.order.MoveToFront(elem) return elem.Value.(*lruNode).value, true } return 0, false } func (c *LRUCache) Put(key string, value int) { if elem, ok := c.items[key]; ok { elem.Value.(*lruNode).value = value c.order.MoveToFront(elem) return } node := &lruNode{key: key, value: value} elem := c.order.PushFront(node) c.items[key] = elem if c.order.Len() > c.capacity { back := c.order.Back() if back != nil { c.order.Remove(back) delete(c.items, back.Value.(*lruNode).key) } } } func main() { cache := NewLRU(2) cache.Put("a", 1) cache.Put("b", 2) fmt.Println(cache.Get("a")) // 1 true cache.Put("c", 3) // evicts "b" fmt.Println(cache.Get("b")) // 0 false fmt.Println(cache.Get("c")) // 3 true cache.Put("a", 10) // updates "a" and refreshes recency fmt.Println(cache.Get("a")) // 10 true } ```

mistral-medium-3-5: ```go type LRUCache struct { capacity int cache map[string]*node head *node tail *node } type node struct { key string value int prev *node next *node } func NewLRU(capacity int) *LRUCache { return &LRUCache{ capacity: capacity, cache: make(map[string]*node), head: &node{}, tail: &node{}, } } func (c *LRUCache) removeNode(n *node) { prev := n.prev next := n.next prev.next = next next.prev = prev } func (c *LRUCache) addToFront(n *node) { n.next = c.head.next n.prev = c.head c.head.next.prev = n c.head.next = n } func (c *LRUCache) Get(key string) (int, bool) { if n, ok := c.cache[key]; ok { c.removeNode(n) c.addToFront(n) return n.value, true } return 0, false } func (c *LRUCache) Put(key string, value int) { if n, ok := c.cache[key]; ok { n.value = value c.removeNode(n) c.addToFront(n) return } if len(c.cache) >= c.capacity { lru := c.tail.prev delete(c.cache, lru.key) c.removeNode(lru) } newNode := &node{key: key, value: value} c.cache[key] = newNode c.addToFront(newNode) } func main() { c := NewLRU(2) c.Put("a", 1) c.Put("b", 2) c.Get("a") c.Put("c", 3) if v, ok := c.Get("b"); !ok { println("b not found") } if v, ok := c.Get("a"); ok { println(v) } } ```

Model A provides a complete, compilable Go program that correctly implements O(1) average-time LRU behavior and includes a minimal main demonstration. Model B is incomplete and incorrect as written: it omits the package/imports, fails to initialize the sentinel links in NewLRU, and can panic during add/remove operations. (Order-swapped judge pass: Model B provides a complete, compilable Go program that correctly implements O(1) average-time LRU behavior and includes a minimal demonstration. Model A is incomplete and incorrect because it omits the package declaration/imports and fails to initialize the sentinel links, causing runtime panics in normal use.)

Localization with tone

Translate this app onboarding line into natural, friendly European Spanish suitable for a mobile toast (keep it under 60 characters, no exclamation marks): "You're all set — your first backup starts tonight." Return only the translation, then the character count in parentheses.

Kimi-K2.7-Code: Listo: tu primera copia empieza esta noche (42)

mistral-medium-3-5: Todo listo: tu primera copia de seguridad empieza hoy (58)

A is concise, natural, under the limit, and preserves the key meaning including that it starts tonight. B is also natural and within length, but it changes "tonight" to "today," which makes it less correct for the source text. (Order-swapped judge pass: B is more accurate and natural for European Spanish, especially by translating “tonight” as “esta noche” and keeping a friendly mobile-toast tone under the limit. A is understandable but mistranslates “tonight” as “hoy” and sounds slightly less natural with the longer phrasing.)

clinic_shift_assignment

Constraint satisfaction — Determine the only valid assignment of four staff to four clinic roles for Tuesday morning: Triage, Phones, Intake, Discharge. Staff: Amina, Benoit, Carla, Dev. Constraints: 1. Amina cannot do Phones. 2. Benoit must be either Triage or Discharge. 3. Carla cannot do Intake. 4. Dev must be assigned to Phones or Intake. 5. The person on Triage is not Carla. 6. If Dev is on Phones, then Amina is on Intake. 7. Benoit is not on Discharge. 8. Carla is on Phones. Return just the final mapping as four lines in the format `Role: Name`.

Kimi-K2.7-Code: Triage: Benoit Phones: Carla Intake: Dev Discharge: Amina

mistral-medium-3-5: Based on the constraints, let's determine the valid assignment: 1. **Carla is on Phones** (Constraint 8). 2. **Amina cannot do Phones** (Constraint 1) — already satisfied. 3. **Benoit must be either Triage or Discharge** (Constraint 2), but **Benoit is not on Discharge** (Constraint 7), so **Benoit is on Triage**. 4. **The person on Triage is not Carla** (Constraint 5) — already satisfied. 5. **Dev must be assigned to Phones or Intake** (Constraint 4), but Phones is taken by Carla, so **Dev is on Intake**. 6. **Carla cannot do Intake** (Constraint 3) — already satisfied. 7. **If Dev is on Phones, then Amina is on Intake** (Constraint 6) — but Dev is on Intake, so this constraint does not apply. 8. The only role left is **Discharge**, which must be assigned to **Amina**. Final mapping: ``` Triage: Benoit Phones: Carla Intake: Dev Discharge: Amina ```

Both models give the same correct assignment, but Model A follows the instruction to return just the final mapping, while Model B adds unnecessary explanation and formatting beyond the requested output. Model B’s reasoning is clear, but it is less compliant with the prompt. (Order-swapped judge pass: Both models give the correct unique assignment, but Model B follows the instruction exactly by returning just the final mapping. Model A adds unnecessary reasoning and formatting beyond what was requested.)

Constraint scheduling

Four talks (A, B, C, D) fill four 1-hour slots 9,10,11,12. Constraints: A is before D; C is not first; B is immediately after A; D is not at 12. Give the ONE valid schedule as 'slot: talk' lines, then a one-line justification. If impossible, say so and explain.

Kimi-K2.7-Code: 9: A 10: B 11: D 12: C The AB block must start at 9, 10, or 11; only starting at 9 leaves a pre-12 slot for D while keeping C out of first place, so D takes 11 and C takes 12.

mistral-medium-3-5: Let's analyze the constraints step by step to find the valid schedule. ### Given: - **Slots**: 9, 10, 11, 12 - **Talks**: A, B, C, D - **Constraints**: 1. A is before D. 2. C is not first (i.e., C cannot be at 9). 3. B is immediately after A (i.e., B is in the next slot after A). 4. D is not at 12. ### Step-by-Step Reasoning: 1. **B is immediately after A**: - This means A and B must occupy two consecutive slots. The possible pairs are: - A at 9, B at 10 - A at 10, B at 11 - A at 11, B at 12 - A at 12 is not possible because there is no slot after 12 for B. 2. **A is before D**: - D must come after A in the schedule. So if A is at 9, D can be at 10, 11, or 12 (but D cannot be at 12 due to constraint 4). - If A is at 10, D can be at 11 or 12 (but D cannot be at 12). - If A is at 11, D can only be at 12 (but D cannot be at 12). So A cannot be at 11. Thus, A cannot be at 11 because that would force D to be at 12, which violates constraint 4. So possible A positions are 9 or 10. 3. **Possible A and B placements**: - Case 1: A at 9, B at 10 - Remaining slots: 11, 12 - Remaining talks: C, D - Constraints: - C is not first (already satisfied, since first is A at 9). - D is not at 12, so D must be at 11, and C at 12. - Check if this satisfies all constraints: - A (9) is before D (11): Yes. - C is not first: Yes (C is at 12). - B is immediately after A: Yes (A at 9, B at 10). - D is not at 12: Yes (D is at 11). - This is a valid schedule: 9:A, 10:B, 11:D, 12:C. - Case 2: A at 10, B at 11 - Remaining slots: 9, 12 - Remaining talks: C, D - Constraints: - C is not first, so C cannot be at 9. Thus, C must be at 12, and D at 9. - Check constraints: - A (10) is before D (9): No, because 10 is after 9. This violates "A is before D". - So this case is invalid. - Case 3: A at 11, B at 12 - This was already ruled out because D cannot be at 12 (and A at 11 would force D to be at 12). 4. **Only valid schedule**: - The only valid schedule is from Case 1: - 9: A - 10: B - 11: D - 12: C ### Verification: - A (9) is before D (11): Yes. - C is not first: C is at 12, so yes. - B is immediately after A: A at 9, B at 10, so yes. - D is not at 12: D is at 11, so yes. ### Final Answer: ``` 9: A 10: B 11: D 12: C ``` Justification: This is the only schedule satisfying all constraints (A before D, C not first, B immediately after A, D not at 12).

Both outputs give the correct unique schedule, but Model A follows the requested format exactly with slot lines plus a one-line justification. Model B is correct and clear, but it ignores the instruction to provide only the schedule and a one-line justification by adding extensive step-by-step analysis. (Order-swapped judge pass: Both models give the correct unique schedule, but Model B follows the requested format exactly by providing only the slot lines and a one-line justification. Model A is correct but adds unnecessary step-by-step reasoning instead of the concise requested output.)