Sole Survivor: Teachers' Notes

This problem is one of a number of solitaire games. The thing that is slightly different about it is that not only do we have to remove all the stones except one but we also have to make sure that the final stone returns to its original position.

 

This problem like many others in Bright Sparks operates at two levels. First there is the level of a particular case. Can you solve the 5 x 15 array of stones, for example? Then there is the deeper level of can you solve this for all arrays or for what arrays can this not be solved?

 

This game of Sole Survivor then, gives the student a chance to analyse the game and produce a general approach. To do this the student will need to experiment until the game is understood. This provides an opportunity for working systematically and the application of logic.

It’s probably a good idea for the students to start by doing some examples to give them some idea of what Sole Survivor is all about. They can probably easily see that 1 x 4 cannot be done. In fact no 1 x n array can be reduced to a sole survivor. So we suggest that they start at the bottom by doing the 2 x 2 array. But before we get going we should say that you will get more out of what we say below if you try to do both the 2 x 2 and the 3 x 4 (that we do next) cases for yourself first. It doesn’t matter if you can’t get them out.  Even if you don’t get them out you will learn something useful.

 

In the diagram the red circle indicates the only stone that moves. It makes three jumps to remove the other three stones, and then ends up in its original cell. The first jump is to the left, then diagonally up to the right, then straight down to its starting position. We should note that as long as you only move one stone and your first jump isn’t diagonal you really can’t go wrong on this one.

 

Now what about a 3 x 4 array? Here we make more than one move at a time. To help see what stones are moving we colour them with either red or green. All other circles stay still or are removed from the board.

 

First, jump the red stone shown in the diagram below up diagonally to the left, then jump the green stone to the right, then jump the red stone back down diagonally to its original position. This removes three stones from the left of the top row.

 

Next, jump the red stone indicated below to the right, then the green stone down, then the red stone back to the left. This removes the three stones on the right hand side.

 

The last array we talk about here is the 4 x 5 one. Using techniques that we have introduced in the earlier two cases, we can reduce a 4 x 5 to a 4 x 2 array. This can be reduced by similar methods to the 2 x 3 array.

 

At this point we should say that the key to solving this problem is to find series of moves that reduce one case to a smaller case. Then we need to analyse the smaller cases.

 

For a description of how all blocks can be reduced to a sole survivor see the Sole Survivor proof page.

 

Finally it is worth noting that the computer will store series of moves so that students don’t have to repeat them laboriously in detail every time they want to use them.

Let’s look at what the game calls ‘standard sizes’; these are the 2 x 2, 2 x 3, 2 x 7, 3 x 7 and 4 x 7 arrays. Now we have already done the 2 x 2 case. You can probably do the 2 x 3 case now (if not, have a look at the second half of the explanation for the 3 x 4 case), so let’s consider the 2 x7 case. Is it possible to use any of the methods in the diagrams above to do that? And is the 3 x 7 case really any harder than the 3 x 4 case? So what can be done about the 4 x 7 case? Can you see how to reduce the 4 x 7 case to a 4 x 4 case? Where to from here?

 

With a little thought you might come up with the following conjectures.

 

Conjecture 1:  Any 2 x n array with n ≥ 2 can be reduced to a sole survivor.

Conjecture 2:  Any 3 x n array with n ≥ 3 can be reduced to a sole survivor.

Conjecture 3:  Any 4 x n array with n ≥ 4 can be reduced to a sole survivor.

Conjecture 4:  Any m x n array with n ≥ m, can be reduced to a sole survivor.

The question now is can we take the problem any further. What Sole Survivor-like problems can we build around this one? It’s very important to have this discussion with students at all levels. Extending problems is a fundamental way of working for mathematicians and students should know that. Let us suggest some related problems.