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三次元形状とディープラーニング
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Tatsuya Yatagawa
December 21, 2020
Technology
0
260
三次元形状とディープラーニング
三次元形状処理とディープラーニングの初歩についてまとめたスライドです。
Tatsuya Yatagawa
December 21, 2020
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Transcript
None
• 機械学習 ニューラルネット 深層学習
• • • θ • = ; = 2 2
+ 1 + 0 ෩ = arg min =0 −1 ( , ; )
• • • 0 = 0 + 0 1 =
1 + 1 2 = 2 + 2 = 0 0 + 1 1 + 2 2 + = 0 0 + 1 1 + 2 2 +(0 0 + 1 1 + 2 2 + )
• • • 0 1 2 0 = ReLU(0 +
0 ) 1 = ReLU(1 + 1 ) 2 = ReLU(2 + 2 )
= 32 = 128 = 512 • •
• • = 512 = 1 = = 4
•
•
•
•
•
•
• • • • • • • • • •
•
(c) Velodyne (c) Andrew Tallon (c) Microsoft (c) Sony
• , , = 0
′ = − ′ = + ′
• ☺ ☺ ☺ ☺ ,
, = 0
☺ ☺ ☺ ☺ , ,
= 0
• • • • •
• • • •
•
•
• ⚫ ⚫ ⚫
• • •
• • •
• •
• • •
• • • • • • •
☺ ☺ ☺ ☺ , ,
= 0
• • • (c) Velodyne
• • • (0 , 0 , 0 ) (1
, 1 , 1 ) ( , , ) (−1 , −1 , −1 ) (+1 , +1 , +1 ) (0 , 0 , 0 ) (1 , 1 , 1 ) (−1 , −1 , −1 ) (+1 , +1 , +1 ) ( , , )
• • •
None
⚫
⚫
• •
• • •
•
•
•
• • • • • • •
☺ ☺ ☺ ☺ , ,
= 0
• • •
• • • •
• • ∗ = ℱ−1(ℱ ⋅ ℱ ) ′ =
ℱ−1(Θ ∘ ℱ )
• = 0 1 1 0 1 0 1 0
1 1 0 1 0 0 1 0 = − − 1 2− 1 2 = :
• • • = T
• () ′ = ℱ−1(Θ ∘ ℱ ) ′ =
ℱ = () ℱ−1 = ()
• • • +1 +2 +1 = ReLU ⋅
• • •
• • •
• •
• • •
• • •
• •
• • • • • •
☺ ☺ ☺ ☺ , ,
= 0
• , , =
• • •
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• • •
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•
• • • •
• • •
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• • •
• • • •
• • • • • • • •
• • • • • • • • •