平衡布尔函数，可以计算，因此每个输入位都不太可能被读取
Balanced Boolean functions that can be evaluated so that every input bit is unlikely to be read
摘要：
一个n比特的布尔函数是平衡的，如果它以1/2的概率取1。我们展示了一个平衡的布尔函数的随机评估过程（与概率0犯错误），使均匀随机输入，没有输入位被读取的概率大于θ（n^{-1/2} sqrt{log n}）。给出了一个平衡单调布尔函数，其对应的概率为Theta（n^{-1/3} log n）.然后，我们表明，对于任何随机算法评估平衡布尔函数，当输入位是均匀随机的，有一些输入位是读取的概率至少为Theta（n^{-1/2}）。对于平衡单调布尔函数，存在以至少Theta（n^{-1/3}）的概率读取的某个输入位。
英文摘要：
A Boolean function of n bits is balanced if it takes the value 1 with probability 1/2. We exhibit a balanced Boolean function with a randomized evaluation procedure (with probability 0 of making a mistake) so that on uniformly random inputs, no input bit is read with probability more than Theta(n^{-1/2} sqrt{log n}). We give a balanced monotone Boolean function for which the corresponding probability is Theta(n^{-1/3} log n). We then show that for any randomized algorithm for evaluating a balanced Boolean function, when the input bits are uniformly random, there is some input bit that is read with probability at least Theta(n^{-1/2}). For balanced monotone Boolean functions, there is some input bit that is read with probability at least Theta(n^{-1/3}).
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