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倍。\n\n可见位置0、3、9、14进入编码器，12个隐藏块只在目标端保留。 查看对应图解状态\n\n- MAE 教学网格 · 16 块中保留 0、3、9、14\n- 0\n- 1\n- 2\n- 3\n- 4\n- 5\n- 6\n- 7\n- 8\n- 9\n- 10\n- 11\n- 12\n- 13\n- 14\n- 15\n- 4 块可见，12 块隐藏\n- 块 0\n- 位置 0\n- 块 3\n- 位置 3\n- 块 9\n- 位置 9\n- 块 14\n- 位置 14\n- 编码器只处理这 4 块\n- 隐藏位置不送入 mask token\n- 4 / 16 是输入块数量；并不等于整个系统必然加速 4 倍。可见位置0、3、9、14进入编码器，12个隐藏块只在目标端保留。 [查看对应图解状态](/#/l3-r2-p20/1)[查看完整图解](/#/l3-r2-p20/1)","sha256":"e086c681d8ce3b82138a3cf0321e0ed99e635d2d8b8384b1b5f3532ba1b77dca","chapter":{"id":"I","title":"学习信号与表示"},"unit":{"id":"i-masked-prediction","title":"限制可见信息，怎样构成可验证的重建任务？"},"page":{"id":"reading-opening","title":"16 块中，编码器真的只看见 4 块","number":1},"sectionTitle":"16 块中，编码器真的只看见 4 块","currentQuestion":"16 块中，编码器真的只看见 4 块","sourceSlides":[],"knowledgeBlock":null,"formulas":[],"figure":{"caption":"可见位置0、3、9、14进入编码器，12个隐藏块只在目标端保留。 查看对应图解状态","labels":["MAE 教学网格 · 16 块中保留 0、3、9、14","0","1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","4 块可见，12 块隐藏","块 0","位置 0","块 3","位置 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