Quel modèle d'IA est le plus performant pour résoudre un problème de mathématiques de niveau CM2 ?
Le développement des modèles d'IA a progressé à un rythme étonnant ces dernières années, mais comment ces systèmes se comportent-ils lorsqu'ils sont chargés de résoudre un problème de compétition mathématique de niveau CM2 ? Dans ce test, j'examine non seulement les capacités de résolution de problèmes des modèles, mais je donne également un aperçu de leur efficacité à gérer les problèmes d'optimisation.
J'ai passé au crible les modèles d'IA actuellement largement disponibles et prometteurs pour découvrir lesquels excellent dans la résolution de cette tâche.
La Tâche et le Défi
La tâche consistait en un problème d'optimisation apparemment simple mais en réalité complexe. Il s'agissait d'une question issue d'une compétition mathématique qui nécessitait une réflexion stratégique et des algorithmes efficaces. L'objectif était de minimiser le temps total consacré au dîner selon un ensemble de règles défini.
Le problème du test admet deux solutions optimales, menant toutes deux à une durée totale d'exactement 55 minutes. L'exactitude de la solution était cruciale, et lors de l'évaluation, j'ai également pris en compte la logique et la cohérence avec lesquelles les modèles ont abordé le problème.
Pendant le test, j'ai effectué trois essais pour chaque modèle afin d'assurer la fiabilité des résultats et de donner aux modèles une chance de démontrer leurs performances.
Énoncé du problème mathématique (Original hongrois):
Van egy asztal, amelynél egyszerre három ember tud vacsorázni. Amint valaki befejezi a vacsorát, azonnal átveszi a helyét valaki más. Hét ember van: A, B, C, D, E, F és G, mindegyikük különböző idő alatt végez a vacsorázással percekben mérve: A 17 perc, B 20 perc, C 27 perc, D 26 perc, E 18 perc, F 29 perc, és G 28 perc alatt vacsorázik.
A cél az ülésrend kialakítása úgy, hogy a lehető legrövidebb idő alatt fejezzék be a vacsorát, minimalizálva a vacsorázásra fordított összes időt. A vacsora A-val és F-fel kezdődik az asztalnál.
Le même en anglais (Ceci était le prompt utilisé):
"There's a table that can seat three people at a time. As soon as someone finishes their dinner, someone else immediately takes their seat. There are seven people: A, B, C, D, E, F, and G, each with their own dining times in minutes: A takes 17 minutes, B takes 20 minutes, C takes 27 minutes, D takes 26 minutes, E takes 18 minutes, F takes 29 minutes, and G takes 28 minutes.
The goal is to arrange the seating order so that everyone finishes dinner as quickly as possible, minimizing the total time spent on dining. Dinner starts with A and F already at the table."
Ce problème est pertinent non seulement pour tester des algorithmes mathématiques, mais aussi pour modéliser les compétences en prise de décision et en gestion du temps.
L'ordre optimal des places est le suivant : initialement, A, F et G s'assoient à table. Lorsque A quitte à 17 minutes, E prend sa place. À 28 minutes, lorsque G part, C s'assoit, puis F quitte à 29 minutes, permettant à D de prendre la place. Enfin, lorsque E part à 35 minutes, B s'assoit. Avec cet arrangement, tout le monde termine en 55 minutes, ce qui est le délai le plus court possible.
Date du test : 13 janvier 2025
Résultats et Analyse
OpenAI GPT-4o - Échec
Étant l'un des modèles les plus populaires, j'attendais une performance de haut niveau de la part de GPT-4o d'OpenAI, mais il n'a pas répondu aux attentes lors du test. Lors de trois essais différents, il s'est complètement trompé de direction les deux premières fois et a donné des résultats incorrects. Lors du troisième essai, il s'est rapproché de la solution optimale, mais le résultat de 56 minutes ne répondait pas aux exigences exactes.
Anthropic Claude 3.5 Sonnet - Échec
Le deuxième modèle le plus populaire est Claude Sonnet. D'après les tests et mon opinion subjective, il a récemment mieux performé que GPT-4o d'OpenAI. Cependant, il semble qu'ils soient à un niveau similaire pour les tâches logiques, car Claude Sonnet n'a pas non plus réussi à trouver la solution optimale.
OpenAI o1-preview - Réussi
Ce modèle s'est avéré être le plus performant du test. Il a trouvé la solution optimale dès le premier essai tout en fournissant un raisonnement logique et cohérent. Le résultat soutient l'idée que dans certaines situations, il vaut la peine de supporter le coût plus élevé des modèles haut de gamme, surtout lorsqu'une précision critique est nécessaire.
OpenAI o1-mini - Échec
La performance du plus petit modèle d'OpenAI était nettement en retard par rapport à son grand frère. Aucun des essais n'a abouti à une solution optimale, et son processus de résolution n'a pas montré une cohérence logique exceptionnelle.
Deepseek Deepseek-V3 DeepThink - Réussi
Ce modèle open-source a également réussi, bien qu'il n'ait pas atteint le résultat optimal de 55 minutes lors des deux premiers essais (obtenant 56 et 61 minutes). Cependant, le troisième essai a produit la bonne solution. Il a travaillé de loin le plus dur, atteignant la solution correcte après une longue délibération et explication. Pendant le processus, il s'est engagé dans un monologue, comme s'il pensait à voix haute. Le modèle Deepseek est particulièrement remarquable en raison de son accès ouvert et de son excellente personnalisation.
Gemini 2.0 Flash Thinking Experimental - Échec
Malgré le raisonnement détaillé du modèle, aucun des essais n'a abouti à une solution correcte. Fait intéressant, le modèle a fermement rejeté même mes modifications utiles du prompt. Alors que d'autres modèles prenaient en compte les suggestions utiles (appliquées uniquement lors de tentatives post-test, bien sûr), Gemini 2 s'est amusé à s'en tenir à son propre raisonnement erroné.
X Grok-2 - Échec
Malgré des tentatives de solution rapides, X Grok-2 a produit des résultats totalement incorrects et n'a montré aucune amélioration au cours des trois essais.
Google Gemini 1.5 Pro 002 - Échec
Bien que le modèle de Google soit considéré comme de génération précédente, en tant que modèle désigné "pro", il devrait théoriquement bien performer en logique. Malgré cela, sa solution s'écartait considérablement des 55 minutes optimales, donnant un résultat de 77 minutes.
Alibaba Qwen 2.5 72B Instruct - Échec
Le modèle d'Alibaba, avec 72 milliards de paramètres, n'a pas pu fournir la solution correcte, soulignant qu'avec la technologie actuelle, 72 milliards de paramètres sont probablement insuffisants pour résoudre ce type de problème.
Meta Llama 3.3 70B, 3.2 70B - Échec
Meta Llama 3.3 s'est terminé par une erreur lors des trois essais, rendant impossible l'obtention d'un résultat évaluable du dernier modèle de Meta. La version précédente 3.2 n'a pas réussi à trouver la solution correcte.
Résumé
Le tableau ci-dessous résume les modèles participants et leurs résultats :
| Fabricant | Nom du modèle | Type d'accès (Open/Fermé) | Solution optimale ? |
|---|---|---|---|
| OpenAI | GPT-4o | Fermé | Non |
| Anthropic | Claude Sonnet 3.5 | Fermé | Non |
| Deepseek | Deepseek-V3 DeepThink | Ouvert | Oui |
| OpenAI | o1-preview | Fermé | Oui |
| OpenAI | o1-mini | Fermé | Non |
| Gemini | 2.0 Flash Thinking Experimental | Fermé | Non |
| X | Grok-2 | Fermé | Non |
| Gemini 1.5 Pro 002 | Fermé | Non | |
| Alibaba | Qwen 2.5 72B instruct | Ouvert | Non |
| Meta | Llama 3.2 70B | Ouvert | Non |
| Meta | Llama 3.3 70B | Ouvert | Erreur/Non concluant |
Le test a mis en évidence des différences de performance significatives même parmi les modèles les plus avancés. La performance exceptionnelle de l'o1-preview d'OpenAI a confirmé qu'une haute précision et fiabilité peuvent être atteintes avec les modèles haut de gamme. Deepseek V3, en tant qu'alternative open-source, a été une bonne surprise dans le test ; sur la base de son résultat remarquable, nous pouvons enfin dire qu'il existe une alternative valable aux modèles fermés. En même temps, plusieurs modèles, en particulier les systèmes plus petits ou moins avancés, n'ont pas répondu aux attentes, indiquant que le développement des LLM présente encore des défis sérieux.
Résultats Détaillés
Seule la meilleure réponse parmi les trois données par chaque modèle a été copiée ci-dessous, sans modification.
OpenAI GPT 4o
This is a problem of optimizing the seating arrangement to minimize the total time taken for all seven people to finish dining. It is similar to a scheduling problem where the table can accommodate only three people at a time. Let's determine the sequence in which the people should take their seats.
Key Points:
- Initial State: A and F are already seated.
- Goal: Minimize the total dining time, which is determined by when the last person finishes their dinner.
- Constraints:
- Three people can dine simultaneously.
- A new person takes a vacated seat as soon as it becomes available.
Step-by-Step Solution:
Dining Times:
mathematica
Kód másolása
A = 17, B = 20, C = 27, D = 26, E = 18, F = 29, G = 28
Initial Setup:
- A and F start dining.
- The third seat can be occupied by someone else.
Strategy:
- Always replace the person who finishes the earliest with someone who can start dining immediately.
- This minimizes idle time at the table.
Process:
- At time t=0t=0, A (17 minutes) and F (29 minutes) are seated.
- B (20 minutes) takes the third seat.
Now, track when each person finishes and update the seating:
Iteration Table
Time Person(s) Finish Dining Available Seat(s) Remaining People New Seat(s) Occupied Total Time 0 - - B, C, D, E, G A, F, B 0 17 A 1 C, D, E, G E 17 20 B 1 C, D, G D 20 26 D 1 C, G G 26 28 G 1 C C 28 29 F 1 - - 29 56 C - - - 56 Total Time
The total time for all seven people to finish dining is 56 minutes.
Anthropic Cloude Sonnet 3.5
Let me solve this step by step. Let me first write out the explicit constraints from your message:
- Table seats exactly 3 people at a time (quoting: "a table that can seat three people at a time")
- When someone finishes, another person immediately takes their seat (quoting: "as soon as someone finishes their dinner, someone else immediately takes their seat")
- There are 7 people total: A, B, C, D, E, F, G (quoting: "There are seven people: A, B, C, D, E, F, and G")
- Dining times in minutes:
- A: 17 minutes
- B: 20 minutes
- C: 27 minutes
- D: 26 minutes
- E: 18 minutes
- F: 29 minutes
- G: 28 minutes
- A and F must start at the table (quoting: "Dinner starts with A and F already at the table")
Let me solve this:
- First configuration (t=0):
- A (17 min), F (29 min), and we need to choose one more person
- Since A will finish first at 17 minutes, we want the third person to finish around the same time
- E takes 18 minutes, which is closest to A's time Starting seats: A, F, E
- At t=17 minutes:
- A finishes first
- F and E remain (F has 12 minutes left, E has 1 minute left)
- We should add B (20 min) since they'll finish around when F finishes New seats: B, F, E
- At t=18 minutes:
- E finishes
- F and B remain (F has 11 minutes left, B has 19 minutes left)
- We should add D (26 min) New seats: B, F, D
- At t=29 minutes:
- F finishes
- B and D remain (B has 8 minutes left, D has 15 minutes left)
- Add C (27 min) New seats: B, D, C
- At t=37 minutes:
- B finishes
- D and C remain (D has 7 minutes left, C has 19 minutes left)
- Add G (28 min) New seats: G, D, C
- At t=44 minutes:
- D finishes
- C and G remain (C has 12 minutes left, G has 21 minutes left) Both finish without needing new people.
- At t=56 minutes:
- C finishes
- Only G remains, finishing at t=65 minutes
Therefore, the optimal solution takes 65 minutes total, with this seating order:
- Start with A, F, E
- Replace A with B at t=17
- Replace E with D at t=18
- Replace F with C at t=29
- Replace B with G at t=37
OpenAI o1-mini
To minimize the total completion time (makespan) for all seven people at the table, we can strategically schedule the seating based on their eating durations. Here’s an optimal arrangement:
Initial Setup
Seated at Time 0:
A (17 minutes)
F (29 minutes)
G (28 minutes)
Scheduling Steps
- Time 0-17:
A is eating and will finish at 17 minutes.
F is eating and will finish at 29 minutes.
G is eating and will finish at 28 minutes.
- Time 17:
A finishes.
C (27 minutes) takes A’s seat.
C will finish at 44 minutes.
- Time 28:
G finishes.
D (26 minutes) takes G’s seat.
D will finish at 54 minutes.
- Time 29:
F finishes.
B (20 minutes) takes F’s seat.
B will finish at 49 minutes.
- Time 44:
C finishes.
E (18 minutes) takes C’s seat.
E will finish at 62 minutes.
Completion Times
A: 17 minutes
F: 29 minutes
G: 28 minutes
C: 44 minutes
B: 49 minutes
D: 54 minutes
E: 62 minutes
Total Makespan
All individuals finish eating by 62 minutes.
Summary of Seating Sequence
Start (0 minutes): A, F, G
At 17 minutes: Replace A with C
At 28 minutes: Replace G with D
At 29 minutes: Replace F with B
At 44 minutes: Replace C with E
This arrangement ensures that the table is utilized efficiently, minimizing the total time required for everyone to finish eating.
OpenAI o1
To minimize the total dining time, we need to schedule the longest dining times as early as possible to ensure that they are completed sooner. Here is the optimal seating arrangement:
Initial Seating (Time 0):
Seat 1: A (17 minutes)
Seat 2: F (29 minutes)
Seat 3: G (28 minutes)
Waiting List: B (20 mins), C (27 mins), D (26 mins), E (18 mins)
At Time 17 (A leaves):
- Seat 1: E (18 minutes) [from Time 17 to 35]
At Time 28 (G leaves):
- Seat 3: C (27 minutes) [from Time 28 to 55]
At Time 29 (F leaves):
- Seat 2: D (26 minutes) [from Time 29 to 55]
At Time 35 (E leaves):
- Seat 1: B (20 minutes) [from Time 35 to 55]
Completion Times:
E finishes at Time 35
B, C, D all finish at Time 55
By arranging the seating this way, we ensure that the longest dining times (F, G, C, D) start as early as possible, reducing the overall dining time. The last diners finish at 55 minutes, which is the minimal total time under the given constraints.
Answer:
The optimal seating to minimize total time is:
Start with A, F, and G at the table.
When A leaves (t = 17 min), seat E.
When G leaves (t = 28), seat C.
When F leaves (t = 29), seat D.
When E leaves (t = 35), seat B.
Everyone finishes by time 55 minutes—the earliest possible.
DeepSeek V3 (DeepThink)
Alright, so I've got this problem here. There are seven people: A, B, C, D, E, F, and G, each with their own dining times. A takes 17 minutes, B takes 20, C takes 27, D takes 26, E takes 18, F takes 29, and G takes 28 minutes. There's a table that can seat three people at a time, and as soon as someone finishes, someone else takes their seat immediately. Dinner starts with A and F already at the table.
Okay, so the goal is to arrange the seating order to minimize the total time spent on dining. I need to think about how to schedule these people so that everyone finishes as quickly as possible.
First, I should probably list out who's already seated and who's waiting. So initially, A and F are at the table, and I need to seat one more person to make it three. So who should that be? Maybe the one with the shortest dining time to get them done quickly, but I'm not sure.
Let me see: A takes 17 minutes, F takes 29 minutes. So A will finish first. So after 17 minutes, A is done, and someone else can take their seat.
So, maybe I should seat the person with the longest dining time first, so they can start early. Wait, but I have limited seats.
Hold on, maybe I should think in terms of scheduling overlapping times. Like, when A finishes at 17 minutes, someone else can start then.
Let me try to outline a possible schedule.
Start time: 0 minutes.
Seated: A (17), F (29), and let's say B (20).
So, A finishes at 17, B at 20, F at 29.
At 17 minutes, A is done, so seat someone else. Who should that be? Maybe the next person with the shortest time.
Wait, but I need to consider who's waiting. Initially, it's A and F at the table, and I need to seat one more person. So maybe seat B first.
So, seated: A (17), F (29), B (20).
At 17 minutes, A is done, seat someone else. Who's next? Maybe E, who takes 18 minutes.
So, E starts at 17 minutes and finishes at 17 + 18 = 35 minutes.
At 20 minutes, B is done, seat someone else. Maybe C, who takes 27 minutes, starting at 20 minutes, finishing at 47 minutes.
At 29 minutes, F is done, seat someone else. Maybe D, who takes 26 minutes, starting at 29 minutes, finishing at 55 minutes.
At 35 minutes, E is done.
At 47 minutes, C is done.
At 55 minutes, D is done.
So, the last person finishes at 55 minutes. Is there a better way to arrange this?
Maybe if I seat G instead of B first.
So, seated: A (17), F (29), G (28).
A finishes at 17, seat E (18), finishes at 17 + 18 = 35.
F finishes at 29, seat D (26), finishes at 29 + 26 = 55.
G finishes at 28, seat B (20), starting at 28, finishes at 48.
At 35, E is done.
At 48, B is done.
At 55, D is done.
Still, the last finish is at 55 minutes.
Wait, same as before.
Maybe try seating C first.
Seated: A (17), F (29), C (27).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat D (26), finishes at 55.
C finishes at 27, seat B (20), starting at 27, finishes at 47.
At 35, E is done.
At 47, B is done.
At 55, D is done.
Again, 55 minutes.
Hmm.
Maybe try seating D first.
Seated: A (17), F (29), D (26).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat C (27), starting at 29, finishes at 56.
D finishes at 26 + starting at 0? Wait, no, D starts at 0, finishes at 26.
Wait, no, if D is seated at 0, he finishes at 26.
Then at 17, A is done, seat E (18), finishes at 35.
At 26, D is done, seat B (20), starting at 26, finishes at 46.
At 29, F is done, seat C (27), starting at 29, finishes at 56.
At 35, E is done.
At 46, B is done.
At 56, C is done.
So last finish at 56 minutes, which is worse than previous 55.
So seating D first is worse.
Maybe seating E first.
Seated: A (17), F (29), E (18).
A finishes at 17, seat B (20), starting at 17, finishes at 37.
F finishes at 29, seat D (26), starting at 29, finishes at 55.
E finishes at 18, seat C (27), starting at 18, finishes at 45.
At 37, B is done.
At 45, C is done.
At 55, D is done.
So last finish at 55 again.
Same as before.
Is there a way to make it finish earlier than 55 minutes?
Maybe if I seat someone else first.
Wait, maybe I need to consider who takes the longest time and seat them earlier.
For example, G takes 28 minutes, F takes 29, C takes 27, D takes 26, B takes 20, E takes 18, A takes 17.
So, the two longest are F (29) and G (28).
Since F is already seated initially, G should be seated as early as possible.
In the previous arrangements, G was seated in one of them, but still ending at 55.
Is there a better way?
Maybe if I seat F, A, and G first.
So, seated: A (17), F (29), G (28).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat D (26), finishes at 55.
G finishes at 28, seat B (20), starting at 28, finishes at 48.
At 35, E is done.
At 48, B is done.
At 55, D is done.
Still 55.
Wait, maybe seating C earlier.
If I seat C instead of B first.
Seated: A (17), F (29), C (27).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat D (26), finishes at 55.
C finishes at 27, seat B (20), starting at 27, finishes at 47.
At 35, E is done.
At 47, B is done.
At 55, D is done.
Still 55.
Is there a way to have someone finish earlier?
Maybe if I seat someone else in place of D.
Wait, what if I seat G instead of D after F is done.
Seated: A (17), F (29), B (20).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat G (28), starting at 29, finishes at 57.
B finishes at 20, seat C (27), starting at 20, finishes at 47.
At 35, E is done.
At 47, C is done.
At 57, G is done.
Now last finish is at 57 minutes, which is worse.
Not good.
Hmm.
Maybe seating D earlier.
Seated: A (17), F (29), D (26).
A finishes at 17, seat E (18), finishes at 35.
F finishes at 29, seat C (27), starting at 29, finishes at 56.
D finishes at 26, seat B (20), starting at 26, finishes at 46.
At 35, E is done.
At 46, B is done.
At 56, C is done.
Last finish at 56, which is still worse than 55.
Not better.
Is there a way to interleave these times better?
Maybe I need to consider the order in which people are seated after the initial seating.
Let me try to think differently.
Suppose I have A and F already seated. I need to seat a third person.
Option 1: Seat B (20).
So, A (17), F (29), B (20).
At 17, A is done, seat E (18), finishes at 35.
At 20, B is done, seat C (27), starting at 20, finishes at 47.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 35, E is done.
At 47, C is done.
At 55, D is done.
Total time: 55 minutes.
Option 2: Seat E (18) as the third person initially.
Seated: A (17), F (29), E (18).
At 17, A is done, seat B (20), starting at 17, finishes at 37.
At 18, E is done, seat C (27), starting at 18, finishes at 45.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 37, B is done.
At 45, C is done.
At 55, D is done.
Total time: 55 minutes.
Same as before.
Option 3: Seat C (27) initially.
Seated: A (17), F (29), C (27).
At 17, A is done, seat E (18), starting at 17, finishes at 35.
At 27, C is done, seat B (20), starting at 27, finishes at 47.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 35, E is done.
At 47, B is done.
At 55, D is done.
Again, 55 minutes.
Option 4: Seat D (26) initially.
Seated: A (17), F (29), D (26).
At 17, A is done, seat E (18), starting at 17, finishes at 35.
At 26, D is done, seat B (20), starting at 26, finishes at 46.
At 29, F is done, seat C (27), starting at 29, finishes at 56.
At 35, E is done.
At 46, B is done.
At 56, C is done.
Total time: 56 minutes.
Worse.
Option 5: Seat G (28) initially.
Seated: A (17), F (29), G (28).
At 17, A is done, seat E (18), starting at 17, finishes at 35.
At 28, G is done, seat B (20), starting at 28, finishes at 48.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 35, E is done.
At 48, B is done.
At 55, D is done.
Total time: 55 minutes.
Same as before.
Option 6: Seat someone else.
Wait, I've tried all options for the third seat initially.
It seems that no matter who I seat third initially, the total time is either 55 or 56 minutes.
Is there a way to make it faster than 55 minutes?
Maybe if I seat someone else instead of B or E at some point.
Wait, let's think about seating the person with the longest time as early as possible.
F already has the longest time at 29 minutes, seated initially.
G is next with 28, seated as early as possible.
Then C with 27, D with 26, B with 20, E with 18, and A with 17.
Maybe there's a better way to sequence them.
Let me try seating G instead of B initially.
Seated: A (17), F (29), G (28).
At 17, A is done, seat E (18), starting at 17, finishes at 35.
At 28, G is done, seat B (20), starting at 28, finishes at 48.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 35, E is done.
At 48, B is done.
At 55, D is done.
Still 55 minutes.
Same as before.
Is there a way to have D start earlier?
If I seat D earlier, but that seems to make the total time longer.
Wait, maybe if I seat D when A is done at 17.
So, seated: A (17), F (29), D (26).
At 17, A is done, seat E (18), starting at 17, finishes at 35.
At 26, D is done, seat B (20), starting at 26, finishes at 46.
At 29, F is done, seat C (27), starting at 29, finishes at 56.
At 35, E is done.
At 46, B is done.
At 56, C is done.
Total time: 56 minutes.
Still worse than 55.
Hmm.
Maybe if I seat someone else at 17 minutes.
At 17 minutes, A is done, currently seating E.
What if I seat C instead?
Seated: A (17), F (29), B (20).
At 17, A is done, seat C (27), starting at 17, finishes at 44.
At 20, B is done, seat E (18), starting at 20, finishes at 38.
At 29, F is done, seat D (26), starting at 29, finishes at 55.
At 38, E is done.
At 44, C is done.
At 55, D is done.
Total time: 55 minutes.
Same as before.
Still 55 minutes.
Is there a way to have someone finish earlier?
Maybe if I seat someone else first.
Wait, maybe I need to consider overlapping times differently.
Let me try to make a table of when each person starts and finishes.
Option: Seat B initially.
Start: 0
A: 0-17
F: 0-29
B: 0-20
At 17, seat E: 17-35
At 20, seat C: 20-47
At 29, seat D: 29-55
End at 55.
Alternative: Seat E initially.
Start: 0
A: 0-17
F: 0-29
E: 0-18
At 17, seat B: 17-37
At 18, seat C: 18-45
At 29, seat D: 29-55
End at 55.
Same.
Another option: Seat C initially.
Start: 0
A: 0-17
F: 0-29
C: 0-27
At 17, seat E: 17-35
At 27, seat B: 27-47
At 29, seat D: 29-55
End at 55.
Same as before.
Is there a way to reduce this to less than 55 minutes?
Maybe if I seat someone else at 29 minutes.
At 29 minutes, F is done, currently seating D.
What if I seat someone else, like G.
Seated: A (17), F (29), B (20).
At 17, seat E (18), 17-35.
At 20, seat C (27), 20-47.
At 29, seat G (28), 29-57.
At 35, E done.
At 47, C done.
At 57, G done.
Now, last finish at 57, which is worse.
Not better.
Alternatively, at 29, seat someone who takes less time.
But D takes 26, G takes 28, which are both longer than E and B.
Not sure.
Maybe seating someone else at 17 minutes.
At 17 minutes, A is done, currently seating E (18), finishing at 35.
What if I seat D instead of E at 17.
Seated: A (17), F (29), B (20).
At 17, seat D (26), 17-43.
At 20, seat E (18), 20-38.
At 29, seat C (27), 29-56.
At 38, E done.
At 43, D done.
At 56, C done.
Last finish at 56, worse than 55.
Not better.
Hmm.
Maybe if I seat someone else at 20 minutes.
At 20 minutes, B is done, currently seating C (27), 20-47.
What if I seat G instead of C.
Seated: A (17), F (29), B (20).
At 17, seat E (18), 17-35.
At 20, seat G (28), 20-48.
At 29, seat D (26), 29-55.
At 35, E done.
At 48, G done.
At 55, D done.
Still 55 minutes.
Same as before.
Wait, maybe if I seat someone else at 29 minutes.
At 29 minutes, F is done, currently seating D (26), 29-55.
What if I seat someone who takes less time, but who's left is only D, C, G.
They all take longer than 20 minutes.
Not helpful.
Seems like no matter what I do, the total time is around 55 minutes.
Is there a way to make it faster?
Maybe if I seat someone at 35 minutes.
At 35 minutes, E is done.
Is there someone still waiting? Let's see.
Initially seated: A, F, B.
At 17, seat E.
At 20, seat C.
At 29, seat D.
At 35, E is done.
At 47, C is done.
At 55, D is done.
So at 35 minutes, E is done, but everyone else is already seated.
No one is waiting; all have been seated.
So no one else to seat at 35.
Same with other options.
It seems like 55 minutes is the minimum achievable with the given constraints.
Unless there's a smarter way to schedule.
Wait, maybe I can seat someone else earlier.
Let me try seating D at 17 minutes instead of E.
Seated: A, F, B.
At 17, seat D (26), 17-43.
At 20, seat E (18), 20-38.
At 29, seat C (27), 29-56.
At 38, E done.
At 43, D done.
At 56, C done.
Last finish at 56 minutes.
Still worse than 55.
Not better.
Alternatively, seat someone else.
Wait, maybe if I seat G at 17 minutes.
Seated: A, F, B.
At 17, seat G (28), 17-45.
At 20, seat E (18), 20-38.
At 29, seat D (26), 29-55.
At 38, E done.
At 45, G done.
At 55, D done.
Last finish at 55 minutes.
Same as before.
Hmm.
I think 55 minutes is the best I can do with these seating arrangements.
I don't see a way to make it any faster without overlapping times in a way that's not allowed.
So, the minimal total time spent on dining is 55 minutes.
The goal is to minimize the total dining time for seven people with different dining durations, given that three people can be seated at a time, starting with A and F already seated. The dining times for the individuals are:
A: 17 minutes
B: 20 minutes
C: 27 minutes
D: 26 minutes
E: 18 minutes
F: 29 minutes
G: 28 minutes
Step-by-Step Schedule:
Initial Seating (Time 0):
A (17 minutes)
F (29 minutes)
B (20 minutes)
First Rotation (Time 17 minutes):
A is done. Seat E (18 minutes), who will finish at 35 minutes.
Second Rotation (Time 20 minutes):
B is done. Seat C (27 minutes), who will finish at 47 minutes.
Third Rotation (Time 29 minutes):
F is done. Seat D (26 minutes), who will finish at 55 minutes.
Final Rotations:
At 35 minutes, E is done.
At 47 minutes, C is done.
At 55 minutes, D is done.
Conclusion:
The last person finishes at 55 minutes, which is the minimal total time achievable with this arrangement. Therefore, the optimal total dining time is 55 minutes.
Gemini 2.0 Flash Thinking Experimental
Let's strategize to minimize the total time. The key is to keep the table as full as possible.
Initial Setup (Time 0):
Table: A, F, and one more person.
Considerations:
Shortest Eaters First? Intuitively, getting the shorter eaters through might free up seats faster.
Longest Eaters Together? Putting longer eaters together might block the table for a longer period, but it also means they get their turn done.
Immediate Replacement: This is crucial. As soon as someone finishes, someone else sits down.
Let's simulate a possible scenario and track the finish times:
Scenario 1: Start with A, F, and E (shortest available)
Time 0: A (17), F (29), E (18) start.
Time 17: A finishes. Table: F (29), E (18), and someone new. Let's put B (20).
Time 18: E finishes. Table: F (29), B (20), and someone new. Let's put C (27).
Time 29: F finishes. Table: B (20), C (27), and someone new. Let's put D (26).
Time 37: B finishes (started at 17, took 20 mins). Table: C (27), D (26), and someone new. Let's put G (28).
Time 45: C finishes (started at 18, took 27 mins). Table: D (26), G (28), and the last person.
Time 55: D finishes (started at 29, took 26 mins). Table: G (28) and the last person.
This scenario is getting complex to track mentally. Let's focus on the finish times and who is available.
A More Systematic Approach:
We need to track when each person finishes their meal.
Round 1 (Time 0 - ?):
Table: A (finishes at 17), F (finishes at 29), and one more person.
Let's consider the impact of the third person:
If we put the shortest remaining (E):
Time 0: A, F, E
Time 17: A finishes. E (18) and F (29) are still eating. Someone else can join.
Time 18: E finishes. F (29) is still eating. Two new spots open.
If we put the longest remaining (G):
Time 0: A, F, G
Time 17: A finishes. F (29) and G (28) are still eating. Someone else can join.
It seems beneficial to get the shorter eaters through first to free up seats.
Let's try a timeline approach, making decisions at each finish time:
Timeline:
Time 0: A, F at the table. Let's add the next shortest, E.
Table: A (finishes 17), F (finishes 29), E (finishes 18)
Time 17: A finishes. Available: B, C, D, G. Table now has F, E. Let's add B (next shortest).
Table: F (finishes 29), E (finishes 18), B (starts 17, finishes 37)
Time 18: E finishes. Available: C, D, G. Table now has F, B. Let's add C.
Table: F (finishes 29), B (finishes 37), C (starts 18, finishes 45)
Time 29: F finishes. Available: D, G. Table now has B, C. Let's add D.
Table: B (finishes 37), C (finishes 45), D (starts 29, finishes 55)
Time 37: B finishes. Available: G. Table now has C, D. Let's add G.
Table: C (finishes 45), D (finishes 55), G (starts 37, finishes 65)
Time 45: C finishes. Table now has D, G. The last person needs to join.
Time 55: D finishes. Table now has G. The last person joins.
Time 65: G finishes.
This scenario seems plausible, but let's consider if we could have done better. The total time is determined by when the last person finishes.
Alternative Strategy: Prioritize filling seats immediately.
Time 0: A, F at the table. Add E.
Time 17: A finishes. Table: F, E. Add B.
Time 18: E finishes. Table: F, B. Add C.
Time 29: F finishes. Table: B, C. Add D.
Time 37: B finishes. Table: C, D. Add G.
Time 45: C finishes. Table: D, G.
Time 55: D finishes. Table: G. The last person has to eat alone now.
Let's refine the timeline with the goal of continuous seating:
Optimized Timeline:
Time 0: A (17), F (29), E (18)
Time 17: A finishes. Table: F, E. Add B (20).
Time 18: E finishes. Table: F, B. Add C (27).
Time 29: F finishes. Table: B, C. Add D (26).
Time 37: B finishes. Table: C, D. Add G (28).
Time 45: C finishes. Table: D, G.
Time 55: D finishes. Table: G.
At this point, only G is left eating. This means the last person to join the table with D and G must be the one with the longest remaining time after time 45.
Let's backtrack and think about who should be the last person to join. The goal is to have everyone finish as early as possible.
Focusing on the Last Diners:
The last few people will likely have to dine with some overlap.
Revised Timeline Attempt:
Time 0: A (17), F (29), E (18)
Time 17: A finishes. Table: F, E. Add B (20).
Time 18: E finishes. Table: F, B. Add C (27).
Time 29: F finishes. Table: B, C. Add D (26).
Time 37: B finishes. Table: C, D. Add G (28).
Time 45: C finishes. Table: D (finishes at 55), G (finishes at 65). The last person needs to join. The remaining person is the one who hasn't eaten yet.
Let's list the finish times in this scenario:
A: 17
E: 18
F: 29
B: 37
C: 45
D: 55
G: 65This assumes the order of entry is E, B, C, D, G.
Let's consider the order of people finishing:
A (17)
E (18)
F (29)
B (37)
C (45)
D (55)
G (65)
The total time is the finish time of the last person
Grok 2
To minimize the total time spent on dining, we need to consider how to cycle people in and out of the seats efficiently. Here’s a step-by-step approach:
Initial Setup:
- Start with A and F at the table. Let’s assume the third seat is taken by B initially since we need to start somewhere and B’s time is relatively average.
Track Time and Replacements:
- A finishes first at 17 minutes. Replace A with E (next shortest time after A).
- E finishes at 18 minutes (total 35 minutes from start). Replace E with C (since we want to keep the seat turning over quickly, but we also need to consider the longer dining times coming up).
- B finishes at 20 minutes (total 40 minutes from start). Replace B with D (D’s time is less than C’s remaining time).
- D finishes at 26 minutes (total 52 minutes). Replace D with G (since G’s time is less than C’s remaining time).
- C finishes at 27 minutes from when he started, which is 62 minutes total. Replace C with A again or keep the seat for someone else if A or others have eaten twice by now, but for simplicity, let’s cycle back to A if possible.
- G finishes at 28 minutes from when he started, which is 80 minutes total. Replace G with B again or another if B has eaten twice.
- F finishes last at 29 minutes from start, which is 81 minutes total.
Optimization Considerations:
We need to ensure that by the time F finishes, others who take longer should have had a chance to eat or be eating. Here’s how the seating might look:
0 min: A, F, B start
17 min: A out, E in
18 min: E out, C in (since C will take longer, we get him in early)
20 min: B out, D in
26 min: D out, G in
27 min: C finishes his first round, if we can cycle A back in or someone else who hasn’t eaten twice
28 min: G finishes first round, bring in whoever hasn’t eaten or cycle back if possible
29 min: F finishes, if we’re cycling back, bring in someone like E or B for their second round or someone new if all have eaten once.
Total Time Calculation:
- The last person to finish will be whoever takes the seat after F, assuming optimal cycling. If we manage to get everyone through once by the time F finishes, the total time would ideally be just over F’s time since we’re continuously seating people:
- F finishes at 81 minutes, but we need to account for the last person’s time who might take over from F or shortly after. If we manage to get D or G in again (as they are close in time), we might push towards 85-90 minutes for everyone to have had dinner assuming optimal replacements.
This solution assumes perfect timing and immediate replacements, which might not be practical, but in an ideal scenario, this arrangement would minimize idle time at the table. However, real-world scenarios might extend this time due to transition periods between diners.
Google Gemini 1.5 Pro 002
Let’s figure out the optimal seating arrangement to minimize the total dining time. Since we want to finish as quickly as possible, we should prioritize getting the people with longer dining times to the table early. This is because their time is “on the clock” as soon as they sit down. Delaying them only delays the overall finish time.
Here’s how we can approach this problem:
Initial State: A (17 mins) and F (29 mins) are already seated. We need to choose the third person for the first round.
Strategy: We want the longest diners at the table first. From the remaining people (B, C, D, E, G), C and G have the longest dining times (27 and 28 minutes respectively).
First Round: A (17), F (29), and either C (27) or G (28) should start. Let’s consider both scenarios:
Scenario 1: A, F, and C: The first round will take 29 minutes (the longest of the three).
Scenario 2: A, F, and G: The first round will take 29 minutes (the longest of the three).
Since both scenarios lead to the same first-round time, let’s pick A, F, and C for the first round (it doesn’t matter in this case).
Subsequent Rounds: After the first round (29 minutes), A and C are finished, and F still has 0 minutes left. We now have B (20), D (26), E (18), and G (28) remaining. Again, prioritize the longest diners. Let’s seat F (0 remaining), G (28), and D (26). This round takes 28 minutes.
Third Round: Now, F, G, and D are finished. We have B (20) and E (18) left. We can seat them with anyone who has already finished. Let’s seat them with A. This round takes 20 minutes.
Total Time: 29 minutes (round 1) + 28 minutes (round 2) + 20 minutes (round 3) = 77 minutes.
Therefore, one optimal seating arrangement is:
- Round 1: A, F, C (29 minutes)
- Round 2: F, G, D (28 minutes)
- Round 3: A, B, E (20 minutes)
Total time: 77 minutes.
It’s important to note that other optimal solutions might exist with the same total time, but this method ensures we find an optimal solution by prioritizing the longest dining times.
Qwen 2.5 72B Instruct
To minimize the total time spent on dining for all seven people, we need to carefully consider the seating arrangement and the order in which people take their seats. The key is to ensure that the person who takes the longest to eat is seated as early as possible, while also making sure that the table is always occupied by the next available person.
Given:
- A takes 17 minutes
- B takes 20 minutes
- C takes 27 minutes
- D takes 26 minutes
- E takes 18 minutes
- F takes 29 minutes
- G takes 28 minutes
Initial state:
- A and F are already seated.
Step-by-Step Solution:
- Initial Seating:
- A (17 minutes)
- F (29 minutes)
- First Replacement:
- After 17 minutes, A finishes eating.
- The next person to sit should be the one with the longest remaining dining time to minimize the total time.
- Remaining times: B (20), C (27), D (26), E (18), G (28).
- Choose G (28 minutes) to sit next.
- Current table: F (29 - 17 = 12 minutes left), G (28 minutes).
- Second Replacement:
- After 12 more minutes, F finishes eating.
- The next person to sit should be the one with the longest remaining dining time.
- Remaining times: B (20), C (27), D (26), E (18).
- Choose C (27 minutes) to sit next.
- Current table: G (28 - 12 = 16 minutes left), C (27 minutes).
- Third Replacement:
- After 16 more minutes, G finishes eating.
- The next person to sit should be the one with the longest remaining dining time.
- Remaining times: B (20), D (26), E (18).
- Choose D (26 minutes) to sit next.
- Current table: C (27 - 16 = 11 minutes left), D (26 minutes).
- Fourth Replacement:
- After 11 more minutes, C finishes eating.
- The next person to sit should be the one with the longest remaining dining time.
- Remaining times: B (20), E (18).
- Choose B (20 minutes) to sit next.
- Current table: D (26 - 11 = 15 minutes left), B (20 minutes).
- Fifth Replacement:
- After 15 more minutes, D finishes eating.
- The next person to sit should be the one with the longest remaining dining time.
- Remaining times: E (18).
- Choose E (18 minutes) to sit next.
- Current table: B (20 - 15 = 5 minutes left), E (18 minutes).
- Final Replacement:
- After 5 more minutes, B finishes eating.
- The last person to sit is E (18 - 5 = 13 minutes left).
Total Time Calculation:
- A finishes after 17 minutes.
- F finishes after 17 + 12 = 29 minutes.
- G finishes after 29 + 16 = 45 minutes.
- C finishes after 45 + 11 = 56 minutes.
- D finishes after 56 + 15 = 71 minutes.
- B finishes after 71 + 5 = 76 minutes.
- E finishes after 76 + 13 = 89 minutes.
The total time for all seven people to finish dining is 89 minutes. This is the minimum time possible given the constraints.