Case 5

Next we study the case that $z = n = 3$ $(mod \ 4)$ and $y$ is odd.
An example of this case is Example 5.6.

Figure 5.23  

In Fig. 5.23 we suppose that Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid for $z=0,1,2,...,4k+4s-2$ and for $y = 0,1,2,...,z$ . We also suppose that these predictions are valid for $z=4k+4s-1$ and $y=0,1,...,4k-4$ . Under these assumptions we prove that $X= G1R(\{4k-3,4k+4s-1\}) =6k+4s-2$ .

Remark 5.4   In this case we assume similar conditions that are written in Remark 5.3. Note that some Grundy numbers $G1R(\{y,z\})$ with $z > 4k+4s-1$ are presented in the table. We do not assume the values of these Grundy numbers to prove this prediction, but the authors present these numbers to show the relative positions of $X= G1R(\{4k-3,4k+4s-1\})$ in the table of Grundy numbers.
Here we assume that $X= G1R(\{4k-3,4k+4s-1\})=6k-5$ , which is the last number of the sequence $\{1,4,7,...,6s-2,...,6k-8,6k-5\}$ that is an arithmetic sequence with common difference of $3$ and its length is $\lceil (4k-3)/4 \rceil$ .
Therefore the position of $X= G1R(\{4k-3,4k+4s-1\})$ is in the midst of the the sequence $\{1,4,7,...,6s-2,...,6k-8,6k-5\}$ , and this condition make Case 5 very different from Case 1, Case 2 and Case 3.

$L11(\{4k-3,4k+4s-1\})$ $=\{0,2,3,5,6,...,6s-1,6s,6s+2,6s+3,...6k-7,6k-6,6k-4 \}$

$$= \{0,2,3,5,6,...,6s-1,6s,6s+2,6s+3\}$$ (5.61)

$$\cup \{6s+5,6s+6,...,6k-7,6k-6,6k-4\}.$$ (5.62)

Clearly $L12(\{4k-3,4k+4s-1\})$

$$= \{1,4,7,...,6s-2,6s+1\}$$ (5.63)

$$\cup \{6k-2,6k,...,6k+4s-4\}.$$ (5.64)

By using Prediction 5.3 for $z=4k+4s-1$ and $y=0,1,...,4k-4$
$L13(\{4k-3,4k+4s-1\})$ $=\{4k+4s-1,4k+4s,4k+4s-2,4k+4s+1,...6k+4s-4,6s+7,6k+4s-3,6s+4\}$

$$= \{4k+4s-1,4k+4s-2,...,6s+7,6s+4\}$$ (5.65)

$$\cup \{4k+4s,4k+4s+1,...6k+4s-4,6k+4s-3\}.$$ (5.66)

By (5.61) and (5.63) we have

$$L11(\{4k-3,4k+4s-1\}) \cup L12(\{4k-3,4k+4s-1\}) \supset \{0,1,2,3,4,5,6,...,6s-1,6s,6s+1,6s+2,6s+3\}.$$ (5.67)

By (5.62) and the fact that $s < k$ we have $L11(\{4k-3,4k+4s-1\})$

$$\supset \{6s+5,6s+6,6s+8,6s+9,...,3k+3s-3,3k+3s-1,3k+3s,...,6k-6,6k-4\}.$$ (5.68)

By (5.65) and (5.66) we have $L13(\{4k-3,4k+4s-2\})$

$$\supset \{6s+4,6s+7,6s+11,...,3k+3s-2,3k+3s,3k+3s+1,3k+3s+2,...,6k+4s-4,6k+4s-3\}.$$ (5.69)

By using predictions for $z=4k+4s-1$ and $y=0,1,...,4k-4$
we know that the sequence of (5.69) starts as an arithmetic sequence with common difference of $3$ whose length is $\lceil (4k+4s-1)/4 \rceil = k+s$ , and hence the last term of this arithmetic sequence is $1+3(k+s-1)=3k+3s-2$ . The next term in (5.69) is $3k+3s$ , and after that we have an arithmetic sequence with common difference of $1$ up until it reaches $6k+4s-3$ . Therefore by (5.68), (5.67) and (5.69) we have $L11(\{4k-3,4k+4s-1\}) \cup L12(\{4k-3,4k+4s-1\}) \cup L13(\{4k-3,4k+4s-1\}) \supset \{0,1,2,3,4,5,6,...,6k+4s-3\}$ . It is clear that there is no $6k+4s-2$ in $L11(\{4k-3,4k+4s-1\}) \cup L12(\{4k-3,4k+4s-1\}) \cup L13(\{4k-3,4k+4s-1\})$ , and hence we have $X= G1R(\{4k-3,4k+4s-1\}) =6k+4s-2$ .

Lemma 5.2   Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid.

Proof   The outline of the proof is almost clear from Case 1, Case 2, Case 3, Case 4 and Case 5.

Theorem 5.1   The list of L-states of the game with inequality $y \leq z$ is the union of the following lists $(1)$ , $(2)$ , $(3)$ , $(4)$ , $(5)$ and $(6)$ .
$(1)$ . $\{\{3 n + 2 k, 2 n, 2 n + 2 k\}$ , $n \in Z_{\geq0}$    and $k \in Z_{\geq0} \}$
$(2)$ . $\{\{3 k+1, 2 n, 2 n + 2 k + 1\},$ $n \in Z_{\geq0}$    and $k =1,2,...,n \}$
$(3)$ . $\{\{3 n + 2 k+3, 2 n, 4 n + 2 k + 3\}$ , $n \in Z_{\geq0}$$\text { and } k \in Z_{\geq0} \}$
$(4)$ . $\{\{3 n + 2 k+2, 2 n + 1, 2 n + 2 k+1\}$ , $n \in Z_{\geq0}$    and $k \in Z_{\geq0} \}$
$(5)$ . $\{\{3 k+1, 2 n + 1, 2 n + 2 k + 2\}$ , $n \in Z_{\geq0}$    and $k = 0, 1, 2, ..., n\}$
$(6)$ . $\{\{3 n + 2 k+3, 2 n + 1, 4 n + 2 k+4\}$ , $n \in Z_{\geq0}$    and $k \in Z_{\geq0} \}$ .

Proof   This is direct from Prediction 5.5.