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倒数第二层(2L-1)中某个node(合取规则)如果与最后一层(2L)的class node有多条连边(weight=1)时的处理

(1)你好,文中描述的这个情况:
image
我理解是否就是这样,就是在权重离散化后正向传播过程中倒数第二层的合取规则与最后一层的两个节点(如果二分类)都有连边,即权重都为1,那就选择第一个class node作为该合取规则的分类标签,但这样的话是否有些不太科学或者是降低了性能(这样可能就等于是随机分类了?),在RRL中,最后一层的weight不会被离散化,保持在[-1,1]之间,但是这样得到的解释集就没那么直观了,相当于只能得到每条规则重要程度。
image
(2)另外一个小问题就是对于网络结构的初始化,如果参数 -s 64_32_16,那么得到的网络的中间层是否应该是 [64, 64, 32, 32, 16]这样的结构(最后一层以合取层16个节点结束),但是实际建立的网络好像并不是这样的
image
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不知道是我理解还有问题吗,谢谢啦。

合取层的疑问待解决

作者你好,非常喜欢您的工作。
对于合取层,我有一个小问题,对于连续的特征来说,先对特征进行离散化然后进行编码one-hot,比如A特征,假如随机离散化划分阈值为10 20 30,那可能得到的one-hot特征就为(-inf,10)、[10,20),[30,+inf),这样训练完后,如果在合取层上的一个节点与输入的多个节点有连边,那就进行合取操作,但是这样问题是比如:(-inf,10)节点与[10,20)节点都与合取层的某一个节点有连边,那就意味着这条规则是 (-inf,10)&[10,20),这样的规则明显是不可能出现的,就意味着这条规则永远不会被激活。
想知道作者对于这种情况是如何处理考虑的呢?

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