We reexamine Albert and Nowakowski's variation on
the game of Nim, called

Consider

In [

The

The positions in these games will be denoted by the corresponding sequences of coin sizes:

In Albert and Nowakowski’s entertaining paper [

In End-Nim, a position

Condition (a) corresponds to the following 8 hieroglyphs of Albert and Nowakowski.

Condition (b) corresponds to the following 4 pictures.

In these pictures, if

To give the solution to misère-End-Nim, we modify slightly the definition of

In misère-End-Nim, a position

We postpone the proof until the next section. Let us now describe the solution for Loop-End-Nim. Once again, we modify the

In Loop-End-Nim, a position

As an immediate consequence of the above three results, we have the following.

The three games, End-Nim, misère-End-Nim, and Loop-End-Nim have the same

We say that a position verifying condition (a) or (b) of Theorem

from every position

there is no move from a 0-position to a 0-position.

(1) Let

(2) Let

The following argument is modelled closely on the proof we have just given.

We say that a position verifying condition (a) or (b) of Theorem

from every position

there is no move from a 0-position to a 0-position.

(1) Let

(2) Let

Let us denote the Sprague-Grundy function for the games End-Nim, misère-End-Nim, and Loop-End-Nim by

In Loop-End-Nim,

We will prove that

The situation concerning misère-End-Nim seems to be considerably more complicated. Indeed, as far as we are aware, even for two piles, where the game is just misère Nim, the Sprague-Grundy function has not yet been determined! We have only been able to obtain very partial information. From Theorem

In the misère-End-Nim, a position

We omit the proof of the above proposition; it is simple and straightforward.

As we saw in Section

In End-Nim, a position

In misère-End-Nim, a position

In Loop-End-Nim, a position

We provide a proof of Theorem

In End-Nim, a position

We say that a position is a

there is no move from a 1-position to a 1-position,

from every position that is not a 0-position or a 1-position, there is a move to a 1-position.

To establish the first property, we suppose that

Firstly,

Now suppose that

Finally, if

To prove property 2, consider a position

In each case, we must exhibit moves to 1-positions.

We divide this further into subcases;

If

If

If

If

If

If

If

If

If

We have subcases;

If

If

If

We have subcases;

If

If

If

If

If

We have subcases;

If

If

We have subcases;

If

If

If

If

Consider the move

We have subcases;

If

If