Day: August 30, 2023

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202308301657 Pastime Exercise 002

The blogger claims no originality of the problem below.

The image files are licensed under the Creative Commons Attribution-Share Alike 2.5 Generic license in the public domain.

Xiangqi, commonly known as Chinese chess or elephant chess, is a strategy board game for two players.

Wikipedia on Xiangqi

Setup.

Let “帥/將” be referred to as a; “仕/士” as b, “相/象” as c; “傌/馬” as d; “俥/車” as e; and “兵/卒” as f.

Any intersection of the ten horizontal and the nine vertical lines is called a point on the board; there are ninety points.

Let t\in \{1,2,3,\dots ,\{+\infty\}\} be the number of time(s) of movement of pieces from the start (t=0). Each player moves in turn such that for one army t\in\{1,3,5,\dots ,1+2n\, |\, n\in\mathbb{N}\} and for its enemy t\in\{2,4,6,\dots ,2n\, |\, n\in\mathbb{N}\}.

And \mathbf{l}(x_{[*]}(t),y_{[*]}(t))\in\{1,2,3,\dots ,9\}\times \{1,2,3,\dots ,10\} be the location of a piece [*]\in\{a,b,c,d,e,f\} at time(s) t of movement.

And \mathbf{s}(t)=\mathbf{l}(t)-\mathbf{l}(t-2) be the displacement of the piece after one succession \Delta t=(t+2)-(t)=2, and by the inverse \mathbf{s}^{-1}(t)=\mathbf{l}(t-4)-\mathbf{l}(t-2)=-\mathbf{s}(t-2) a retreat to the original location is such that displacements for an advance and a retreat s^{-1}(t)=|\mathbf{s}^{-1}(t)|=|-\mathbf{s}(t-2)|=s(t-2) are equal in magnitudes.

And v(t)=|\mathbf{v}(t)|=\displaystyle{\bigg|\frac{\Delta\mathbf{s}(t)}{\Delta t}\bigg|=\bigg|\frac{\mathbf{s}(t)-\mathbf{s}(t-2)}{(t)-(t-2)}\bigg|} be the instantaneous speed at some point t_i in a series of movements i\in \mathbb{N}.

And \mathrm{ord}[*] called the order of a set [*] be the number of elements in that collection. For instance, in a peaceful match \mathrm{ord}(t)=+\infty the number of times t of movements could be in such countably and indefinitely infinite.

In the case of a (“帥/將”):

\mathbf{l}_a(t)=\{4,5,6\}\,\hat{\mathbf{i}}+\{1,2,3\}\,\hat{\mathbf{j}};

\mathrm{ord}(\mathbf{l}_a)=9;

\mathbf{s}_a(t)\big|_{\forall\, t\in\mathbb{N}}=\Bigg\{\begin{pmatrix}\pm 1\\0\end{pmatrix},\begin{pmatrix}0\\ \pm 1\end{pmatrix}\Bigg\}

such that \mathbf{l}_a\to \mathbf{l}_a:\mathbf{l}_a(t-2)\stackrel{\mathbf{s}_a}{\mapsto}\mathbf{l}_a(t);

\mathrm{ord}(\mathbf{s}_a)=4;

v_a(t)=\displaystyle{\frac{\sqrt{(\pm 1)^2+(0)^2}}{\Delta t}=\frac{1}{2}\,\mathrm{unit/time}};

\mathrm{ord}_a(v)=1.

In the case of b (“仕/士”):

\mathbf{l}_b(t)=\{(4,1),(4,3),(5,2),(6,1),(6,3)\};

\mathrm{ord}(\mathbf{l}_b)=5;

\mathbf{s}_b(t)\big|_{\forall\, t\in\mathbb{N}}=\Bigg\{\begin{pmatrix}1\\\pm 1\end{pmatrix},\begin{pmatrix}-1\\ \pm 1\end{pmatrix}\Bigg\}

such that \mathbf{l}_b\to \mathbf{l}_b:\mathbf{l}_b(t-2)\stackrel{\mathbf{s}_b}{\mapsto}\mathbf{l}_b(t);

\mathrm{ord}(\mathbf{s}_b)=4；

v_b(t)=\displaystyle{\frac{\sqrt{(\pm 1)^2+(\pm 1)^2}}{\Delta t}=\frac{\sqrt{2}}{2}\,\mathrm{unit/time}};

\mathrm{ord}(v_b)=1.

In the case of c (“相/象”):

\mathbf{l}_c(t)=\{(3,1),(7,1),(1,3),(5,3),(9,3),(3,5),(7,5)\};

\mathrm{ord}(\mathbf{l}_c)=7;

\mathbf{s}_c(t)\big|_{\forall\, t\in\mathbb{N}}=\Bigg\{\begin{pmatrix}2\\\pm 2\end{pmatrix},\begin{pmatrix}-2\\ \pm 2\end{pmatrix}\Bigg\}

such that \mathbf{l}_c\to \mathbf{l}_c:\mathbf{l}_c(t-2)\stackrel{\mathbf{s}_c}{\mapsto}\mathbf{l}_c(t);

\mathrm{ord}(\mathbf{s}_c)=4；

v_c(t)=\displaystyle{\frac{\sqrt{(\pm 2)^2+(\pm 2)^2}}{\Delta t}=\frac{2\sqrt{2}}{2}=\sqrt{2}\,\mathrm{unit/time}};

\mathrm{ord}(v_c)=1.

In the case of d (“傌/馬”):

\mathbf{l}_d(t)=\{ 1,2,3,\dots ,9\}\,\hat{\mathbf{i}}+\{1,2,3,\dots ,10\}\,\hat{\mathbf{j}};

\mathrm{ord}(\mathbf{l}_d)=9\times 10=90;

\mathbf{s}_d(t)\big|_{\forall\, t\in\mathbb{N}}=\Bigg\{\begin{pmatrix}-1\\\pm 2\end{pmatrix},\begin{pmatrix}2\\ \pm 1\end{pmatrix},\begin{pmatrix}1\\ \pm 2\end{pmatrix},\begin{pmatrix}-2\\ \pm 1\end{pmatrix}\Bigg\}

such that \mathbf{l}_d\to \mathbf{l}_d:\mathbf{l}_d(t-2)\stackrel{\mathbf{s}_d}{\mapsto}\mathbf{l}_d(t);

\mathrm{ord}(\mathbf{s}_d)=8；

v_d(t)=\displaystyle{\frac{\sqrt{(\pm 1)^2+(\pm 2)^2}}{\Delta t}=\frac{\sqrt{5}}{2}\,\mathrm{unit/time}};

\mathrm{ord}(v_d)=1.

In the case of e (“俥/車”):

\mathbf{l}_e(t)=\{ 1,2,3,\dots ,9\}\,\hat{\mathbf{i}}+\{1,2,3,\dots ,10\}\,\hat{\mathbf{j}};

\mathrm{ord}(\mathbf{l}_e)=9\times 10=90;

\mathbf{s}_e(t)\big|_{\forall\, t\in\mathbb{N}}=\Bigg\{\pm\begin{pmatrix}\{1,2,3,\dots ,8\}\\0\end{pmatrix},\pm\begin{pmatrix}0\\ \{1,2,3,\dots , 9\}\end{pmatrix}\Bigg\}

such that \mathbf{l}_e\to \mathbf{l}_e:\mathbf{l}_e(t-2)\stackrel{\mathbf{s}_e}{\mapsto}\mathbf{l}_e(t);

\mathrm{ord}(\mathbf{s}_e)=8\times 2+9\times 2=34；

\begin{aligned} v_e(t) & =[\min v_e(t),\max v_e(t)]\big|_{\in n/2\textrm{ for } n\in\mathbb{N}\backslash\{ 0 \} } \\ & = \bigg[ \frac{\sqrt{(1)^2+(0)^2}}{2},\frac{\sqrt{(0)^2+(9)^2}}{2}\bigg] \\ &= \{ 1/2,1,3/2,\dots , \, 9/2\}\,\mathrm{unit/time} \\ \end{aligned};

\mathrm{ord}(v_e)=9.

In the case of f (“兵/卒”):

Exercise.