Chocolate games that satisfy the inequality .

Next we define chocolate games. Please consult chocolates in Fig.2.1 and Fig.2.2 while you read this definition, since this definition looks very abstract without examples.

Definition 2.1   The chocolate consists of square boxes, and only one block is blue and others are brown.
Brown blocks are sweet and the blue block is very bitter. This game is played by two players in turn. Each player breaks the chocolate (in a straight line along the blue line) into two area.
The player eats the area that does not contain blue block. The player who breaks the chocolate and eats to leave his opponent with the single bitter block (blue block) is the winner.

Remark 2.1   Note that in Definition $\ref{defofgeneralchoco}$ we are not considering a misere play, since the player who breaks the chocolate for the last time is the winner, since his opponent cannot break the remaining blue (bitter) block. Therefore the chocolate games in this article are normal play games.

Example 2.1   The chocolate of Fig. $\ref{robinchoco}$ is proposed by Robin [1], and the chocolate of Fig. $\ref{2yzchoco1}$ is a new game proposed by the authors.

 

Figure 2.1  

Figure 2.2  

When we study the chocolate games, there are two important states of chocolates.

Definition 2.2   Here we define two important states of chocolates.
$(a)$ W-states, from which we can force a win, as long as we play correctly at every stage.
$(b)$ L-states, from which we will lose however well we play, but we may end up winning if our opponents make a mistake.

It is important to realize that the outcome of this game is not pre-determined, but there is nothing that the potential loser can do if the potential winner plays correctly at ever stage. However the potential winner cannot afford to make a single mistake, for if he does, then his opponent can exploit this.
One of the most important topics of chocolate games is to find all the L-states and the W-states of games. We denote by $Z_{\geq0}\$ the set of non-negative integers.

Remark 2.2   For the L-states of the chocolate of Fig. 2.1 see the problem corner in [1]. See also Theorem 7.12 in [2] (p-138). Note that in [2] they use the term "P-position" instead of "L-state" of this article.

We define nim-sum that is important for the theory of games.

Definition 2.3   Let $x,y$ be non-negative integers, and write them in base $2$, so $x = \sum\limits_{i = 0}^n {{x_i}} {2^i}$ and $y = \sum\limits_{i = 0}^n {{y_i}} {2^i}$ with ${x_i},{y_i} \in \{ 0,1\}$ .
We define the nim-sum $x \oplus y$ by

$$x \oplus y = \sum\limits_{i = 0}^n {{w_i}} {2^i},$$ (2.1)

where $w_{i}=x_{i}+y_{i}(mod\ 2)$ .

Theorem 2.1   Let $x,y,z\in Z_{\geq 0}$ . If $x\oplus y\oplus z= 0$, $x'<x, y'<y$ and $z'<z$ ,then
we have $x'\oplus y\oplus z \neq 0, x\oplus y'\oplus z \neq 0\ and\ x\oplus y\oplus z' \neq 0$ .

This is a well known result of nim-sum. For a proof see Theorem 7.12 in [2] (p-139).
We generalize the chocolate games in Fig. 2.2, and define the chocolate game that we study in this article.

Definition 2.4   There is a blue $1 \times 1$ block, and on both side of it there are columns of brown blocks that are numbered by integers. In the chocolate in Fig. 2.3 we have the first, the second, the fourth,...,12th column on the right side of the blue block and $(-1)$ th, $(-2)$ th,...,$(-6)$ th column on the left side of the blue block.
The i-th column is $1 \times (x+ \lfloor \frac{i}{2} \rfloor)$ block for $i =1, 2, 3,...,12$ and the i-th column is $1 \times 1$ block for $i =-1, -2, -3,-4,-5,-6$ .
The generalization of the chocolate in Fig. 2.3 is the chocolate in Fig. 2.4. In this chocolate we have the first, the second, the fourth,...,n-th column,... on the right side of the blue block and $(-1)$ th, $(-2)$ th,...,$(-m)$ -th column,... on the left side of the blue block.
The i-th column is $1 \times (1+ \lfloor \frac{i}{2} \rfloor)$ block for $i =1, 2, 3,...,n,...$ and the i-th column is $1 \times 1$ block for $i =-1, -2, -3,-4,-5,-m,...$ .
Brown blocks are sweet chocolate that can be eaten, and the blue block is the bitter chocolate that cannot be eaten.

Figure 2.3  

Figure 2.4  

You can cut these chocolate along the segments in three ways.
$(1)$ You cut vertically on the left side of the blue (bitter) block.
$(2)$ You cut horizontally above the blue (bitter) block.
$(3)$ You cut vertically on the right side of the blue (bitter) block.
Therefore it is proper to represent this chocolate with $\{x,y,z\}$ , where $x,y,z$ stand for the maximum numbers of times that we can cut these chocolate in each direction. For example in Fig. 2.3 we can cut 6 times at most vertically on the left side of the blue block, we can cut 6 times at most horizontally above the blue (bitter) block and we can 12 times at most vertically on the right side of the blue block. Therefore $x=6$ , $y = 6$ and $z=12$ . Therefore we represent the chocolate in Fig. 2.3 with the coordinates $\{6,6,12\}$ .

For any $x,y,z\in Z_{\geq 0}$ we denote by $\{x,y,z\}$ the state of chocolate.

Example 2.2   Here we have four examples of states of chocolates that appear when we play the chocolate game of Fig. 2.5

$\{2, 2, 5\}$

Figure 2.5  

$\{2, 1, 3\}$

Figure 2.6  

$\{0, 2, 5\}$

Figure 2.7  

$\{2, 0, 5\}$

Figure 2.8  

$\{0, 1, 5\}$

Figure 2.9  

It is clear that the coordinates of these states satisfy the inequality $2y \leq z$, and this is equivalent to the inequality

$$y \leq \left\lfloor \frac{z}{2}\right\rfloor$$ (2.2)

where $\left\lfloor \ \right\rfloor$ is the floor function. Note that $\left\lfloor x \right\rfloor$ is the largest integer not greater than $x$ for any real number $x$ .
Note that you can cut $1 \times (x+ \lfloor \frac{i}{2} \rfloor)$ block horizontally $\lfloor \frac{i}{2} \rfloor$ times at most for any non-negative integer $i$ .
Inequality (2.2) is important to understand the structure of the chocolate. If you start with the chocolate in Fig. 2.5 and reduce the third coordinate $5$ to $3$ by cutting vertically on the right side of the bitter block, then by the structure of the chocolate the second coordinate is reduced to $1$ .
In this way we get the chocolate in Fig. 2.6.
If we are to explain the move from the chocolate in Fig. 2.5 to the chocolate in Fig. 2.6 using Inequality (2.2), we use the following explanation.
Let $z=3$ , then by Inequality (2.2) we get $y=1$ .

Definition 2.5   Let $A_{2}=\{\{x,y,z\};x,y,z\in Z_{\geq 0},y \leq \left\lfloor \frac{z}{2} \right\rfloor$ and $x\oplus y \oplus z=0\}$ , $B_{2}=\{\{x,y,z\};x,y,z\in Z_{\geq 0},y \leq \left\lfloor \frac{z}{2} \right\rfloor$ and $x\oplus y \oplus z\neq 0\}$ .

Next we prove that $A_2$ is the set of L-states and $B_2$ is the set of W-states of the game of Definition 2.4. To do this we need some lemmas and theorems. In this section we assume that $x,y,z\in Z_{\geqq0}$ and we write them in base 2, so $x=\displaystyle\sum_{i=0}^{n} x_i2^i$ , $y=\displaystyle\sum_{i=0}^{n} y_i2^i$ and $z=\displaystyle\sum_{i=0}^{n} z_i2^i$ with $x_{i},y_{i},z_{i}\in \{0,1\}$ , where $n = log_2 max(x,y,z)$ .
In this section we prove Theorem 2.6, and we need several facts about the relations between the nim-sum of numbers and the floor function.

Lemma 2.1   $(1)$ $y=\left\lfloor \frac{z}{2} \right\rfloor$ if and only if $y_{n} = 0$ and $y_{i} = z_{i+1}$ for $i = 0,1,2,...,n-1$ .
$(2)$ $y < \left\lfloor \frac{z}{2} \right\rfloor$ if and only if $y_{n} = 0$ , $y_{i} = z_{i+1}$ for $i = m+1,m+2,...,n-1$ and $y_{m} = 0 < z_{m+1} = 1$ for some $m\in Z_{\geq 0}$ .

 

Proof   When we write $y,z$ in base 2, $(1)$ and $(2)$ of this lemma is clear from the definition of floor function $\left\lfloor \ \right\rfloor$ .

Lemma 2.2   We suppose that

$$x\oplus y\oplus z=0.$$ (2.3)

$(1)$ Then

$$y= \left\lfloor \frac{z}{2} \right\rfloor$$ (2.4)

if and only if $x_{n} = z_{n} = 1, y_{n} = 0$ and for each $i = 0,1,2,...,n-1$

$$y_{i}=z_{i+1}$$ (2.5)

and

$$z_{i}=x_{i}+y_{i} \ (mod \ 2).$$ (2.6)

$(2)$

$$y< \left\lfloor \frac{z}{2} \right\rfloor$$ (2.7)

if and only if there exists $m\in Z_{\geqq0}$ such that $x_{n} = z_{n} = 1, y_{n} = 0$ ,(2.6) is valid for each $i = 0,1,2,...,n-1$ , (2.5) is valid for each $i = m+1,m+2,...,n-1$ and $y_{m} = 0 < z_{m+1} = 1$ .

Proof   $(1)$ and $(2)$ are direct from Lemma 2.1 and the Definition 2.3 of nim-sum.

Remark 2.3   Suppose that $x\oplus y\oplus z= 0$ and $y \leq \left\lfloor \frac{z}{2} \right\rfloor$ . Then there exists $k \in Z_{\geqq0}$ such that (2.5) is valid for $i = k,k+1,k+2,...n-1$ , and by using (2.6) for $i = k,k+1,k+2,...n-1$ we get $z_{i}=\displaystyle\sum_{j=i}^{n} x_j \ (mod\ 2)$ and $y_{i}=\displaystyle\sum_{j=i+1}^{n} x_j \ (mod\ 2)$ for $i = k,k+1,k+2,...n-1$ .
Clearly $y=\left\lfloor \frac{z}{2} \right\rfloor$ if and only if $k = 0$ , and $y < \left\lfloor \frac{z}{2} \right\rfloor$ if and only if $k > 0$ and $y_{k-1} < z_k$ .
Therefore $y$ and $z$ are determined by $x$ when we have (2.3) and (2.4). In particular, for any $x$ there exist unique $y,z$ that satisfy (2.3) and (2.4).

Lemma 2.3   We suppose that

$$x\oplus y\oplus z=0 y = \left\lfloor \frac{z}{2} \right\rfloor.$$ (2.8)

 

If there exist $v,w\in Z_{\geq 0}$ such that $x\oplus v\oplus w=0$ and $v<\left\lfloor \frac{w}{2} \right\rfloor$ ,then $v<y$ .

Proof   By Lemma 2.2 we have ${x_n} = {z_{_n}} = 1,{y_n} = 0$ and for each $i = 0,1,2,...,n-1$

$${y_i} = {z_{i + 1}},{z_i} = {x_i} + {y_i}\ (mod\ 2).$$ (2.9)

Suppose that there exist $v,w\in Z_{\geq 0}$ such that

$$x\oplus v\oplus w=0\ and\ v < \left\lfloor {w/2} \right\rfloor.$$ (2.10)

Clearly $v_{n}=0$ and $w_n=1$ . By Lemma 2.2 and Remark 2.3 there exists $m \in {Z_{ \ge 0}}$ such that
for each $i=m+1,m+2,...n-1$

$${w_i} = \sum\limits_{j = i}^n {{x_i}}\ (mod\ 2),{v_i} = \sum\limits_{j = i + 1}^n {{x_i}}\ (mod\ 2)$$ (2.11)

and $v_m<w_{m+1}$ .
By using Lemma 2.2 and Remark 2.3 for $x,y,z$ we have for each $i = 0,1,2,...,n-1$

$${z_i} = \sum\limits_{j = i}^n {{x_i}}\ (mod\ 2),{y_i} = \sum\limits_{j = i + 1}^n {{x_i}}\ (mod\ 2).$$ (2.12)

Then by (2.11) and (2.12) we have
$v_i=y_i$ for each $i=m+1,m+2,...n-3,n-2,n-1$ and $v_m<w_{m+1}=z_{m+1}=y_m$ , and hence we have $v<y$ .

Theorem 2.2   If $x\oplus y\oplus z= 0$ and $y \le \left\lfloor {z/2} \right\rfloor$ ,then
$(1)$ $u \oplus y \oplus z \ne 0$ for any $u\in Z_{\geq 0}$ with $u<x$ .
$(2)$ $x \oplus v \oplus z \ne 0$ for any $v\in Z_{\geq 0}$ with $v<y$ .
$(3)$ $x \oplus y \oplus w \ne 0$ for any $w\in Z_{\geq 0}$ with $w<z$ .
$(4)$ $x \oplus v \oplus w \ne 0$ for any $v,w\in Z_{\geq 0}$ with $v<y,w<z$ and $v = \left\lfloor {w/2} \right\rfloor$ .

Proof (1)   , (2) and (3) are direct from Theorem 2.1. We suppose that $x\oplus v\oplus w=0$ and $v = \left\lfloor {w/2} \right\rfloor$ for some $v,w \in {Z_{ \ge 0}}$ with $v<y,w<z$ . If $y < \left\lfloor {z/2} \right\rfloor$ , then by Lemma 2.3 we have $y<v$ , and this contradicts the fact $v<y$ . If $y = \left\lfloor {z/2} \right\rfloor$ , then by Remark 2.3 we have $y=v$ , and this contradicts the fact $v<y$ . Therefore $x \oplus v \oplus w \ne 0$ .

Lemma 2.4   For any $x,y \in {Z_{ \ge 0}}$ there exists $z$ that satisfies one of the following two conditions.
$[i]$ $x\oplus y\oplus z= 0$ and $y \le \lfloor {z/2} \rfloor$ .
$[ii]$ $x\oplus \lfloor z/2 \rfloor \oplus z=0$ and $\lfloor z/2 \rfloor < y$ .

Proof   Let

$$u_i = x_i+y_i \ (mod \ 2)$$ (2.13)

for $i = 0,1,2,...,n$ .
$[a]$ . We assume that $y_n = 0$ .
$[a-1]$ Suppose that the following (2.14) is valid for $i = 0,1,...,n-1$ .

$$y_i = u_{i+1}.$$ (2.14)

Then let $z =u = \sum\limits_{i = 0}^n {{u_i}} {2^i}$ . By $(1)$ of Lemma 2.1 we have $y = \lfloor {z/2} \rfloor$ , and hence by (2.13) we have $[i]$ of this lemma.
$[a-2]$ Suppose that there exists $m$ such that

$$y_m=0 < 1 = u_{m+1},$$ (2.15)

and (2.14) is valid for $i = m+1,m+2,...,n-1$ . Then let $z =u = \sum\limits_{i = 0}^n {{u_i}} {2^i}$ . By $(2)$ of Lemma 2.1 we have $y < \lfloor {z/2} \rfloor$ , and hence we have $[i]$ of this lemma.
$[a-3]$ Suppose that there exists $m$ such that $y_i$ $=u_{i+1}$ for $i = m+1,m+2,...,n-1$ and $y_m=1>$ $0 = u_{m+1}$ .
We define $y^\prime _i = y_i$ for $i = m+1,m+2,...,n$ , $y^\prime _m = 0$ , $z_i = u_i$ for $i = m+1,m+2,...,n$ , $z_m = y^\prime _m + x_m$ $(mod \ 2)$ .
Next let $y^\prime _{m-1}=z_m$ and $z_{m-1} = y^\prime _{m-1} + x_{m-1}$ $(mod \ 2)$ . In this way we let

$$y^\prime _i = z_{i+1}$$ (2.16)

$$z_i = x_i + y^\prime _i \ (mod \ 2)$$ (2.17)

for $i = 0,1,2,...,m-2$ .
Let $z = \sum\limits_{i = 0}^n {{z_i}} {2^i}$ and $y^\prime = \sum\limits_{i = 0}^n {{y^\prime _i}} {2^i}$ . Clearly $y^\prime < y$ . By Lemma 2.1 we have $y^\prime = \lfloor z/2 \rfloor$ , and hence we have $[ii]$ of this lemma.
$[a-4]$ Suppose that $y_{n-1} = 1 > 0 = u_n$ . Then by the same method used in $[a-3]$ we get $[ii]$ of this lemma.
$[b]$ Suppose that $y_n=1$ . Then let $y^\prime _n = 0$ , $z_n = x_n + y^\prime _n \ (mod \ 2 )$ , and by using the method in $[a-3]$ we let $y^\prime _i =z_{i+1}$ and $z_i = y^\prime _i + x_i$ $(mod \ 2)$ for $i = 0,1,2,...,n-1$ .
Let $z = \sum\limits_{i = 0}^n {{z_i}} {2^i}$ and $y^\prime = \sum\limits_{i = 0}^n {{y^\prime _i}} {2^i}$ . Then $y^\prime = \lfloor z/2 \rfloor$ , and we have $[ii]$ .

Theorem 2.3   Suppose that $x \oplus y \oplus z \ne 0$ and $y \le \left\lfloor {z/2} \right\rfloor$ .
Then at least one of the following (1), (2), (3) and (4) is true.
$(1)$ $u \oplus y \oplus z = 0$ for some $u \in {Z_{ \ge 0}}$ with $u<x$ .
$(2)$ $x \oplus v \oplus z = 0$ for some $v \in {Z_{ \ge 0}}$ with $v<y$ .
$(3)$ $x \oplus y \oplus w = 0$ for some $w \in {Z_{ \ge 0}}$ with $w<z$ .
$(4)$ $x\oplus v\oplus w=0$ for some $v,w \in {Z_{ \ge 0}}$ with $v<y,w<z$ and $v = \left\lfloor {w/2} \right\rfloor$ .

Proof   Suppose that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ and $x_m +y_m +z_m \neq 0\ (mod\ 2)$ .
$(a)$ If $x_m=1$ , we define $u= \sum_{i=0}^{n} u_i 2^i$ by $u_i=x_i$ for $i=m+1,m+2,...n,u_m=0<x_m$ and $u_i=y_i+z_i$ for $i=0,1,...,m-1$ . Then we have $u \oplus y \oplus z \oplus =0$ and $u<x$ . Therefore we have (1).
$(b)$ If $y_m=1$ , we define $v= \sum_{i=0}^{n} v_i 2^i$ by $v_i=y_i$ for $i=m+1,m+2,...n,v_m=0<y_m$ and $v_i=x_i+z_i$ for $i=0,1,...,m-1$ . Then we have $x \oplus v \oplus z \oplus =0$ and $v<y$ . Therefore we have (2).
$(c)$ Next we suppose that $x_m=y_m=0$ and $z_m=1$ . By Lemma 2.4 For $x,y$ there exists $w= \sum_{i=0}^{n} w_i 2^i$ that satisfies one of the following two conditions $[i]$ and $[ii]$ .
$[i]$ $x \oplus y \oplus w = 0$ and $y \le \lfloor {w/2} \rfloor$ , then by the fact that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ we have $w_i = z_i$ for $i = m+1,m+2,...,n$ , and hence by the fact that $w_m = x_m + y_m = 0 < 1 = z_m$ we have $w<z$ . Therefore we have $(3)$ .
$[ii]$ $x\oplus \lfloor w/2 \rfloor \oplus w=0$ and $\lfloor w/2 \rfloor < y$ . By the fact that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ we have $w_i = z_i$ for $i = m+1,m+2,...,n$ , and hence by the fact that $w_m = x_m + y_m = 0 < 1 = z_m$ we have $w<z$ . Then we have $(4)$ . Next we are going to define the function $move2(\{x, y, z\})$ for a state $\{x,y,z\}$ of chocolates.
$move2(\{x, y, z\})$ is the set of all states that can be reached from the state $\{x,y,z\}$ in one step (directly).

Definition 2.6   For $x,y,z \in Z_{\ge 0}$ we define $move2(\{x,y,z\})=\{\{u,y,z \};u<x \} \cup \{\{x,v,z \};v<y \} \cup \{ \{x,\min(y, \lfloor w/2 \rfloor ),w \};w<z \}$ , where $u,v,w \in Z_{\ge 0}$ .

Example 2.3   We study the function $move2(\{x, y, z\})$ using examples of states in Example 2.2 . If we start with the state $\{x,y,z\}=\{2,2,5\}$ and reduce $z=5$ to $w=3$ , then the second coordinate will be $\min(2, \lfloor 3/2 \rfloor )=\min(2,1)=1$ .
Therefore we have $\{2,1,3\} \in move2(\{ 2,2,5 \})$ . It is easy to see that
$\{0,2,5\},\{2,0,5\} \in move2(\{2,2,5\})$ . It is clear that $\{0,2,5\} \notin move2(\{2,1,3\})$ , since we cannot move to $\{0, 2, 5\}$ from $\{2, 1, 3\}$ . Note that $\{0,1,5\} \notin move2(\{2,2,5\})$ , since it will take 2 steps to reach $\{0, 1, 5\}$ from $\{2, 2, 5\}$ .

Next we prove that if you start with an element of $A_2$ , then any move leads to an element of $B_2$ .

Theorem 2.4   For any $\{x,y,z\} \in A_2$ we have $move2(\{x,y,z\}) \subset B_2$ .

Proof   Let $\{x,y,z\} \in A_2$ , then we have

$$x \oplus y \oplus z \oplus =0$$ (2.18)

and

$$y \leq \left\lfloor z/2 \right\rfloor.$$ (2.19)

Suppose that we move from $\{x,y,z\}$ to $\{p,q,r\}$ ,i.e., $\{p,q,r\} \in move2(\{x,y,z \})$ . Next we prove that $\{p,q,r\} \in B_2$ .
Since $move2(\{x,y,z\})=\{\{u,y,z \};u<x \} \cup \{\{x,v,z \};v<y \} \cup \{ \{x,\min(y, \lfloor w/2 \rfloor ),w \};w<z \}$ , we have one of the following $(1)$ , $(2)$ , $(3)$ and $(4)$ .
$(1)$ $\{p,q,r\}= \{u,y,z\}$ with $u<x$ .
$(2)$ $\{p,q,r\}= \{x,v,z\}$ with $v<y$ .
$(3)$ $\{p,q,r\}= \{x,y,w \}$ with $w<z$ and $y \leq \left\lfloor w/2 \right\rfloor$ .
$(4)$ $\{p,q,r\}= \{x,\min(y,\lfloor w/2 \rfloor ),w\}$ with $w<z$ .
In each of these $(1)$ , $(2)$ , $(3)$ and $(4)$ we use Theorem 2.2 to get $p \oplus q \oplus r \oplus \ne 0$ . Next we prove that if you start with an element of $B_2$ , then there is a proper move that leads to an element of $A_2$ .

Theorem 2.5   Let $\{x,y,z\} \in B_2$ , then $move2(\{x,y,z\})\cap A_2 \ne \phi$ .

Proof   Let $\{x,y,z\} \in B_2$ . Then have

$$x \oplus y \oplus z \oplus \ne 0$$ (2.20)

and

$$y \leq \left\lfloor z/2 \right\rfloor.$$ (2.21)

Since $move2(\{x,y,z\})= \{\{u,y,z\};u < x\} \cup \{\{x,v,z\};v<y\}$ $\cup \{\{x,\min\left( y,\lfloor \frac{w}{2} \rfloor\right),w\};w<z\}$ , by $(1)$ , $(2)$ , $(3)$ and $(4)$ of Theorem 2.3 there exists $\{p,q,r\} \in move2\left(\{x,y,z\}\right)$ such that
$p\oplus q\oplus r= 0$ . Therefore $\{p,q,r\} \in move2(\{x,y,z\})\cap A_2$

 

Theorem 2.6   Let $A_{2}=\{\{x,y,z\};x,y,z\in Z_{\geq 0},y \leq \left\lfloor \frac{z}{2} \right\rfloor$ and $x\oplus y \oplus z=0\}$ , $B_{2}=\{\{x,y,z\};x,y,z\in Z_{\geq 0},y \leq \left\lfloor \frac{z}{2} \right\rfloor$ and $x\oplus y \oplus z\neq 0\}$ . Then $A_2$ is the set of L-states and $B_2$ is the set of W-states of the game of Definition 2.4. $($ Note that $A_2$ and $B_2$ are defined in Definition 2.5.$)$

Proof   If we start the game with a state $\{x,y,z\}\in A_{2}$ , then by Theorem 2.4 any option by us leads to a state $\{p,q,r\}$ in $B_2$ . From this state $\{p,q,r\}$ by Theorem 2.5 our opponent can choose a proper option that leads to a state in $A_2$ . Note that any option reduces some of the numbers in the coordinates. In this way our opponent can always reach a state in $A_2$ , and finally he wins by reaching $\{0,0,0\}\in A_{2}$ . Therefore $A_2$ is the set of L-states.
If we start the game with a state $\{x,y,z\}\in B_{2}$ , then by Theorem 2.5 we can choose a proper option leads to a state $\{p,q,r\}$ in $A_2$ . From $\{p,q,r\}$ any option by our opponent leads to a state in $B_2$ . In this way we win the game by reaching $\{0, 0, 0\}$ . Therefore $B_2$ is the set of W-states.

By Theorem 2.6 we tell L-states from W-states, and in Example 2.4 we learn how to win the game using the theorem.

Example 2.4   Here we present the method to win the chocolate game of Fig. 2.5. Since $2 \oplus 2 \oplus 5 \oplus \neq 0$ , by Theorem 2.6 the chocolate in Fig. 2.5 with the coordinates $\{2, 2, 5\}$ is a W-state. Therefore you can win if you play as the first player. The strategy is to move to an L-state every time. For example you can move to $\{2, 1, 3\}$ that is an L-state, since $2 \oplus 1 \oplus 3=0$ .
In this way you should leave the opponent with an L-state every time until you leave the opponent with the single block of bitter chocolate whose coordinates are $\{0, 0, 0\}$ . Then the game is finished, and you win the game.

By Example 2.4 the strategy to win is clear. Even if you do not read the proof of Theorem 2.6, you can be a good player of this chocolate game.