quiz bu lëkkale ak tegtal · Diggante Niveau
Jàngale Taux Programming Quiz
Saytu tàngoor wi, cosinus ak oraaru jéego yi, tàmbaliwaat jeffin, teela ñàkka dal ak rëset bi ñuy faral di def ci tàngoor-ba noppi-yaxx.
Laaj 1 de 8
Lan mooy njariñu wàllu 'warmup' ci kalendriye tolluwaayu jàng?
Xoolaatal laaj bu nekk8 laaj
Sooy def quiz bi, tontu yi ak leeral yi dañuy feeñ.
- Lan mooy njariñu wàllu 'warmup' ci kalendriye tolluwaayu jàng?
- Tambali ci njëg bu ndaw ba noppi yokk, moytu ñàkka dal ci degrade yu teela am bruit
- Ngir wàññi njubte gi mujj
- Ngir teggi yenn done yu njëkk yi
- ngir gelé poids yépp ci ndoorte li
- Ban kalendriye mooy topp courbe demi-cosine bi wàcci dem ci njëg bi gëna ndaw?
- Oràŋ bu sax
- Cosinus
- Oraire aleatoire
- Yokkateg ligneer
- Lan mooy xew sudee dañu yéeg lool ci ni ñuy jàngee?
- Taggat yaram dafay boole ci saasi ak jaar yoon
- Model bi dafay tàggat ndànk waaye saa yu nekk dafay am ndam
- Perte bi mën na diverge wala osciller ci barabu wàññeeku
- Dara soppeekuwul; njëg bi amul benn njariñ
- Lan la step decay di def ci ni ñuy jàngee?
- Yokk ko lineairement ci lote bu nekk
- Dafay wéy di nekk
- Dafa koy def ci jéego bu nekk
- Dafa koy wàcce ci facteur fixe ci jamono yuñ tànn bu njëkk
- What does a warm restart do to a learning-rate schedule?
- It raises the learning rate again to begin another schedule cycle
- It resets the model parameters to random values
- It permanently fixes the learning rate at zero
- It deletes the optimizer’s moment estimates
- In Goyal et al.’s large-minibatch ImageNet setup, what did their warmup scheme address?
- It removed the need for a validation set
- It eased early optimization difficulties after the learning-rate scale increased with batch size
- It changed the model architecture after each epoch
- It converted the learning-rate curve into a cosine wave
- In the cosine formula, what does T denote?
- The number of model layers
- Dayo lote bi
- The schedule horizon or total number of steps used for the decay
- The optimizer’s momentum coefficient
- If the cosine schedule uses lr_min greater than zero, where does the rate approach at the end of the horizon?
- Exactly zero regardless of lr_min
- The original warmup starting rate
- An unbounded value
- The specified minimum learning rate