To do this asynchronously, the net traverses the distorted pattern, and at each node, n, it asks whether n's value should be set to -1 or + 1.
如果是异步完成，网络会遍历失真的图案，在每一个节点n，它会询问n的值应该设置为-1还是+ 1。
Once we multiply it by a very large number n, it becomes very much greater than kT.
一旦我用一个很大的n与它相乘，它就会变得比kT大得多。
But you mark me, cap 'n, it won't do twice, by thunder!
但是你听着，船长，决不会有第二次了，我赌咒！
But you mark me, cap 'n, it won't do twice, by thunder!
但是你听着，船长，决不会有第二次了，我赌咒！
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