There is some more information on this web site. Quoted from there:
"It is helpful to view this binary search in terms of the twenty bits that represent each integer. In the first pass of the algorithm we read the (at most) one million input integers and write those with a leading zero bit to one tape and those with a leading one bit to another tape. One of those two tapes contains at most 500,000 integers, so we next use that tape as the current input and repeat the probe process, but this time on the second bit. If the original input tape contains N elements, the first pass will read N integers, the second pass at most N/2, the third pass at most N/4, and so on, so the total running time is proportional to N. The missing integer could be found by sorting on tape and then scanning, but that would require time proportional to N log N."
As you can see, this is a variation on the binary search algorithm: divide the problem into two pieces and solve the problem for one of the smaller pieces.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…