مسابقة نماذج فك التشفير التأملية

تظهر الإجابات والتفسيرات أثناء إجراء الاختبار.

1. ما هو الدور الأساسي لنموذج "المسودة" في فك التشفير التأملي؟
   • لضغط النموذج الكبير إلى عدد أقل من المعلمات بشكل دائم
   • لترجمة المطالبة إلى لغة مختلفة
   • لاقتراح العديد من الرموز المرشحة بسرعة للنموذج الكبير للتحقق منها
   • لتخزين المحادثات السابقة لاسترجاعها
2. Why can verifying a drafted token block accelerate target-model generation in a suitable workload?
   • Verification has no compute cost at any block size
   • A verification pass can process several positions together and amortize target work across accepted tokens
   • The draft model is guaranteed more accurate than the target
   • The target model is skipped for every accepted token
3. ماذا يحدث للرموز المميزة التي تمت صياغتها بعد الرمز المميز الأول الذي يرفضه النموذج المستهدف؟
   • يتم تجاهل كل شيء بعد عدم التطابق الأول
   • يتم قبولهم جميعًا على أي حال
   • يتم إرسالها إلى النموذج الثالث
   • يتم تخزينها مؤقتًا وإعادة استخدامها في الدور التالي
4. What correctness property does the standard exact speculative-sampling rejection procedure preserve under its assumptions?
   • An identical random sequence for every possible random seed
   • Guaranteed truth of every generated statement
   • The draft model's distribution regardless of the target
   • The target model's configured sampling distribution
5. Which mechanism distinguishes EAGLE's drafting approach from Medusa's additional decoding heads?
   • EAGLE performs only word-count-based lexical retrieval
   • EAGLE drafts autoregressively at the target model's feature level
   • EAGLE removes target verification from every algorithm
   • EAGLE and Medusa are the same architecture with different names
6. Why should a speculative-decoding benchmark report the draft-target pair and acceptance rate?
   • Every model pair has the same guaranteed speedup
   • Acceptance rate measures whether generated facts are true
   • Speedup depends on drafting and verification costs and how many proposed tokens are accepted
   • The target model's hardware and context length cannot affect performance
7. In exact speculative sampling, a proposed token has q equal to 0.2 and p equal to 0.1. What is its acceptance probability?
   • 0.1, because only p is used
   • 0.2, because only q is used
   • 1.0, because every proposed token is accepted
   • 0.5, from min(1, p/q)
8. What can the standard speculative-sampling algorithm emit when every token in a drafted block is accepted?
   • The accepted block plus an additional token sampled from the target
   • Only the last drafted token, discarding the accepted prefix
   • An arbitrary extra draft token without target sampling
   • No output until the draft model is retrained