In the one-peg complement problem, we start with one hole empty and remove as many pegs as possible. Suppose that we try to do the opposite, and leave as many pegs as possible. For example, try this problem--it starts with e3 empty, and ends with 10 pegs.

Instead of doing this problem directly, try doing the anti-problem and taking the anti-solution. Here's what the anti problem looks like.

If you start with one hole, the largest number of pegs you can leave is 10, and we saw how to do that above. If you start with e1 empty, the largest number of pegs you can leave is 8. Your job is to figure out how to do this.

If you start with one hole in e5, the largest number of pegs you can leave is 7.

In most of the peg solitaire problems, we give you the start and finish positions. Here, we are also going to give you the "middle" position--what the board looks like after you have completed seven jumps.

Earlier, we saw a solution of the e1-complement where one peg made three jumps at the end. When one peg makes several jumps at the end, we call this a sweep. There are two different ways to end the e5-complement with a five-sweep.

If you can do the problem below, then you can end the e5-complement with the five-sweep a5-e5-g3-e1-c3-e5.

If you can do the problem below, then you can end the e5-complement with the sweep e1-g3-e5-c3-a5-e5.

There is no way to start with one hole empty and reduce to one peg with a final sweep of six pegs. However, it is possible to start with the c3 and i5 holes empty and end with the configuration below. Your job is to figure out how to do it. Also, can you see how to do a six-sweep from the "Finish" position below?