The chocolate problem of Example 2.1 is very difficult to study mathematically, and hence the authors have made a easier version the chocolate problem. This chocolate has a very simple formula to calculate P-positions.

Note that a list of non-negative numbers is a position of this chocolate problem when .

The position

The position

We are going to prove Theorem 3.1, and we need some corollaries and lemmas.

.

Let =** {**
,
=0 and **} **and =** {**
,
=0 and **}** =** {**
, and **}**. Let =** {**
,
=0 and **}**.

By Remark 3.1 Game 1 is a special case of the game of Example 1.1, and hence we can use Theorem 1.1 to Game 1 by making the fourth coordinate 0.

If , then at least one of the following three conditions is satisfied.

There exists u such that u x and .

There exists v such that v y and .

There exists w such that w z and .

Therefore at least one of the conditions and are satisfied.

.

*It is easy to see that ***{0, 0, 1}, {0, 0, 2}, {0, 0, 3}, {0, 0, 4}, {0, 1, 1}, {0, 1, 2}, {0, 1, 3}, {0, 2, 2}, {1, 0, 0}, {1, 1, 0}, {2, 0, 0}, {2, 1, 0}, {2, 2, 0}, {3, 0, 0},{3, 1, 0}** *and** *{4, 0, 0}*are N-positions, since you can move from each of them to {0, 0, 0}. For example, if you have {2, 1, 0}, then by subtracting 2 from the first coordinate you get ***{0, 0, 0}, ***since the position should satisfy the inequality .
*

*From ***{1, 1, 2}*** we can move to ***{0,1,2}, {1,0,2}, {1,1,1} and {1,1,0},*** and we have proved that these 4 positions are N-positions. Therefore** *{1, 1, 2}* is a P-position. Similarly** *{2, 1, 1}* **is a P-position.*

For any non-negative integers with

Note that you get

We are going to prove the following Theorem 3.2 that is essentially the same as Theorem 3.1.

Any element in

*By Lemma 3.4 any position in ***{{x, y, z}, {x, y, z}** * and ***} ***is a P-position of Game 2. *

**{{x,y,z} , **

We suppose that the conclusion of this theorem is valid for

We are going to prove of this theorem for

Suppose that we move from

Suppose that we move from

Suppose that we move from

Suppose that we move from

Suppose that we move from

Suppose that we move from

When we subtract a natural number from the first coordinate, we can use the method that we used in .

We are going to prove of this theorem for

Now we can suppose that and , and there exist non-negative numbers such that , and . Therefore we have only to show that

There exists u such that and . By Lemma 3.1 , and hence . Therefore by the hypothesis of mathematical induction we have that

There exists v such that and .

Similarly

There exists w such that and . Similarly