\documentclass[11pt,draft]{article}
%
\author{Anne}
\usepackage{basic}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
   \node[shape=circle,draw,inner sep=1pt] (char) {#1};}}
%
\begin{document}
%

\section*{Kamnitzer--Knutson--Mirkovic Conjecture}

Zhijie asks about the following example.

In $A_3$ ($G=\PGL_4$) fix the usual reduced expression $\uvi = (1,2,3,1,2,1)$ and consider the Lusztig datum $n_\bullet = (1,1,0,0,1,1)$.

The smallest $\lambda$ and $\mu$ for the associated stable MV cycle satisfying Proposition XYZ in Section 2.2 of my thesis are
\[
\lambda = (5,3,2,0) \qquad \mu = (3,3,2,2)
\]
and the associated tableau is
\[
\young(11123,224,34)
\]
By applying Conjecture 4.6.2 of my thesis we can say what the free entries in each column of a matrix $A$ in the MVy slice to $\OO_\lambda$ associated to this tableau.
Let us fix notation for the coordinates of $A$.
\[
A = \left[\begin{BMAT}(e){ccc:ccc:cc:cc}{ccc:ccc:cc:cc}
    0&1&0&0&0&0&0&0&0&0\\
      0&0&1&0&0&0&0&0&0&0\\
      0&0&0&0&0& \circled{${c}_{1}$} &0&\circled{${b}_{2}$}&\circled{${a}_{3}$}&\circled{${b}_{3}$}\\
      0&0&0&0&1&0&0&0&0&0\\
      0&0&0&0&0&1&0&0&0&0\\
      0&0&0&0&0&0&0&\circled{${b}_{4}$}&{a}_{5}&\circled{${b}_{5}$}\\
      0&0&0&0&0&0&0&1&0&0\\
      0&0&0&0&0&0&0&0&0&{b}_{6}\\
      0&0&0&0&0&0&0&0&0&1\\
      0&0&0&0&0&0&0&0&0&0
\end{BMAT}\right]
\]
Our conjecture tells us that the circled entries will be the free entries, i.e. the nonzero variable entries. We went ahead and used this information to deduce and omit the entries that must be zero.

With the help of \texttt{M2} and speeding things up by substituting the first six prime numbers for the circled entries we find that the ideal of $X_\tau$ is
\[
I = \left({b}_{1},{a}_{6},{a}_{4},{a}_{2},{a}_{1},-{a}_{5}{b}_{2}+{a}_{3}{b}_{4},{b}_{2}{b}_{6}+{a}_{3},{b}_{4}{b}_{6}+{a}_{5}\right)
\]
in $R = \CC[a_1,\dots,a_6,b_1,\dots,b_6,c_1]$.
%
% [a_1,b_1,c_1,a_2,b_2,a_3,b_3,a_4,b_4,a_5,b_5,a_6,b_6,Degrees=>{{1,0,0,0},{1,0,0,1},{1,0,0,2},{1,1,0,0},{1,1,0,1},{1,1,1,0},{1,1,1,1},{0,1,0,0},{0,1,0,1},{0,1,1,0},{0,1,1,1},{0,0,1,0},{0,0,1,1}}]

We compute its multidegree
\[
\mdeg I = \left({\alpha}_{2}+{\alpha}_{1}\right)\left({\alpha}_{2}+{\alpha}_{3}\right)\left({\alpha}_{2}+{\alpha}_{3}+{\alpha}_{1}\right)\left(\hbar+{\alpha}_{1}\right)\left({\alpha}_{1}{\alpha}_{2}{\alpha}_{3}\right)
\]
and Hilbert series
{\scriptsize
\[
    \ch I=\frac{1}{\left(1-e^{\alpha_1 + \alpha_2 + \alpha_3 + \hbar}\right)\left(1-e^{\alpha_1 + \alpha_2 + \hbar}\right)\left(1-e^{\alpha_2 + \alpha_3 + \hbar}\right)\left(1-e^{\alpha_1 + 2\hbar}\right)\left(1-e^{\alpha_3 + \hbar}\right)\left(1-e^{\alpha_2 + \hbar}\right)}
\]}

Now Zhijie would like to know what the Plucker coordinates are.
% He asks ``What are the $a,b,c$ in terms of $\Delta_{C \in \binom {16} 8}$''

In my thesis I claim that the Plucker embedding is given by the minors $\Delta_C B$ of a certain matrix $B$ using columns $C\in\binom{S}{mp-N}$. Here
$m = 4$, $p = 5$, $N = 10$, and
$$
S = ((k,0),(k,1),(k,2),(k,3),(k,4))_{k=1}^4
$$
Recall $mp-N = \dim L/t^pL_0$ and $S$ indexes the basis $v_{(i,j)} = [e_it^j]$ of $V = L_0/t^pL_0$. The matrix $B$ is the matrix whose row vectors are the basis vectors
\[
[ge_1],[tge_1],[ge_2],[tge_2],[ge_3],[tge_3],[t^2ge_3],[ge_4],[tge_4],[t^2ge_4]
\]
of $L/t^pL_0\subset L_0/t^pL_0$.

In this case
\[
g = g_A = \begin{bmatrix}
    t^3 \\
    -c_1t^2 & t^3 \\
    -b_2t & -b_4 t & t^2 \\
    -a_3 - b_3 t & - a_5 - b_5 t & -b_6 t & t^2
\end{bmatrix}
\]
so
\begin{align*}
    [ge_1] &= [t^3 e_1] -c_1[t^2 e_2] -b_2[te_3] - a_3[e_4] - b_3[te_4] \\
    [tge_1] &= {[t^4 e_1]} -c_1[t^3 e_2] -b_2 [t^2 e_3] - a_3 [te_4] - b_3 [t^2 e_4] \\
    [ge_2] &= [t^3 e_2] - b_4 [te_3] -a_5[e_4] - b_5[t e_4] \\
    [tge_2] &= {[t^4 e_2]} - b_4 [t^2e_3] - a_5 [te_4] - b_5 [t^2 e_4] \\
    [ge_3] &= [t^2 e_3] -b_6 [te_4] \\
    [tge_3] &= {[t^3e_3]} - b_6 [t^2 e_4] \\
    [t^2 ge_3] &= [t^4e_3] - b_6 [t^3 e_4]\\
    [ge_4] &= [t^2 e_4] \\
    [tge_4] &= [t^3 e_4] \\
    [t^2 ge_4] &= [t^4 e_4]
\end{align*}
and, omitting zero columns, we get
\hfill

{
\hspace{-4cm}\begin{minipage}{1.5\textwidth}
\[
\bordermatrix{ & [t^3e_1] & [t^4e_1]  & [t^2e_2] & [t^3e_2] & [t^4 e_2] & [te_3] & [t^2e_3] & [t^3e_3] & [t^4e_3] & [e_4] & [te_4] & [t^2e_4] & [t^3e_4] & [t^4e_4] \cr
[ge_1]         & 1       &  & -c_1     &         & & -b_2   &          &          & & -a_3  & -b_3                                    \cr
[tge_1]        &         & 1  &          & -c_1  &   &        & -b_2     &         & &       & -a_3   & -b_3                           \cr
[ge_2]         &         &  &          & 1       & & -b_4   &          &          & & -a_5  & -b_5                                    \cr
[tge_2]        &         &  &          &         & 1 &        & -b_4     &          & &      & -a_5  & -b_5                            \cr
[ge_3]         &         &  &          &         & &        &           1 &    &    &   & -b_6                                    \cr
[tge_3]        &         &  &          &         & &        &          &           1 & &  &       & -b_6                            \cr
{[t^2 ge_3] }    &         &  &          &         & &        &          &  & 1             & &       & & -b_6                          \cr
[ge_4]         &         &  &          &         & &        &          &          &      & &       & 1 &                             \cr
[tge_4]        &         &  &          &         & &        &          &          &      & &       &       & 1                       \cr
[t^2 ge_4]     &         &  &          &         & &        &          &          &      & &       &       &       & 1
}
\]
    \end{minipage}
}
for $B$ so $C_0 = \{(1,3),(1,4),(2,3),(2,4),(3,2),(3,3),(3,4),(4,2),(4,3),(4,4)\}$.

% for m2
% matrix{
    % {1,0,-c_1,0,0,-b_2,0,0,0,-a_3,-b_3,0,0,0},
    % {0,1,0,-c_1,0,0,-b_2,0,0,0,-a_3,-b_3,0,0},
    % {0,0,0,1,0,-b_4,0,0,0,-a_5,-b_5,0,0,0},
    % {0,0,0,0,1,0,-b_4,0,0,0,-a_5,-b_5,0,0},
    % {0,0,0,0,0,0,1,0,0,0,-b_6,0,0,0},
    % {0,0,0,0,0,0,0,1,0,0,0,-b_6,0,0},
    % {0,0,0,0,0,0,0,1,0,0,0,0,-b_6,0},
    % {0,0,0,0,0,0,0,0,0,0,0,1,0,0},
    % {0,0,0,0,0,0,0,0,0,0,0,0,1,0},
    % {0,0,0,0,0,0,0,0,0,0,0,0,0,1}
    % }

\section*{Finding Plucker coordinates in M2}

\begin{lstlisting}[language=Python]
R = QQ[a_1,b_1,c_1,a_2,b_2,a_3,b_3,a_4,b_4,a_5,b_5,a_6,b_6,Degrees=>{{1,0,0,0},{1,0,0,1},{1,0,0,2},{1,1,0,0},{1,1,0,1},{1,1,1,0},{1,1,1,1},{0,1,0,0},{0,1,0,1},{0,1,1,0},{0,1,1,1},{0,0,1,0},{0,0,1,1}}]

J = ideal(b_1,a_6,a_4,a_2,a_1,-a_5*b_2+a_3*b_4,b_2*b_6+a_3,b_4*b_6+a_5)

plm = matrix{{1,0,-c_1,0,0,-b_2,0,0,0,-a_3,-b_3,0,0,0}, {0,1,0,-c_1,0,0,-b_2,0,0,0,-a_3,-b_3,0,0},{0,0,0,1,0,-b_4,0,0,0,-a_5,-b_5,0,0,0},{0,0,0,0,1,0,-b_4,0,0,0,-a_5,-b_5,0,0},{0,0,0,0,0,0,1,0,0,0,-b_6,0,0,0},{0,0,0,0,0,0,0,1,0,0,0,-b_6,0,0},{0,0,0,0,0,0,0,0,1,0,0,0,-b_6,0},{0,0,0,0,0,0,0,0,0,0,0,1,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,1,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,1}}

s = subsets(14,10);

e0 = {0..9}

mins := {}

for e in s do if ( det(plm^e0_e) != 0 ) then mins = append(mins,det(plm^e0_e))

inds := {}

for e in s do if ( det(plm^e0_e) != 0 ) then inds = append(inds,e)

% check that #inds == #mins

wts := {}

for e in s do if ( det(plm^e0_e) != 0 ) then wts = append(wts,degree(det(plm^e0_e)))

pluck = QQ[apply(inds,i->p_i)]

f = map(R,pluck,mins)

Q = R/J

fbar = map(Q,pluck,mins)

K = kernel fbar

Kh = homogenize(K,p_{0,1,3,4,6,7,8,11,12,13})

indsnu := {}

for i in 0..#wts-1 do if take(wts#i,3)=={1,1,1} then indsnu = append(indsnu,inds#i)

I = ideal(apply(indsnu,i->p_i))

Knu = K + I

dim Knu

Knuh = homogenize(Knu,p_{0,1,3,4,6,7,8,11,12,13})

dim Knuh

% check that dim Kh = 7 = dim J + 1
\end{lstlisting}
%
\section*{Fixed point subscheme}
%
In this case I would like the fixed point subscheme supported at $t^{(4,3,2,1)}$.
%

Let $\nu = (4,3,2,1)$ and note that $\lambda\ge \nu\ge \mu$.

\begin{enumerate}
    \item Take a/the chart containing the fixed point $t^\nu$ by asking that $\Delta_C\ne 0$
    for any $C$ such that $\wt (\Delta_C) = \nu$. This amounts to dehomogenizing
    (e.g. setting equal to 1) \textit{all?} such $\Delta_C$. I think this is tantamount to not homogenizing in the first place.
    \item Mod out by the generators of $K$ whose weight is not equal to $\lambda - \nu$.
\end{enumerate}
%
Where

{\scriptsize
$({p}_{\left\{4,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}+{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}+{p}_{\left\{1,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}-1,{p}_{\left\{2,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,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$\left({p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{4,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}-1,{p}_{\left\{1,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}}\right)$

$\left({p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},-{p}_{\left\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{1,\,2,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{4,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}-1\right)$

$\left({p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{4,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}-1,{p}_{\left\{0,\,1,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}}-{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},-{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},-{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}}^{2}+{p}_{\left\{0,\,1,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}}\right)$

$\left({p}_{\left\{0,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{4,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,5,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,6,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,7,\,8,\,9,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,4,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,3,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,5,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,6,\,7,\,8,\,10,\,11,\,12,\,13\right\}},{p}_{\left\{3,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,4,\,5,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{2,\,3,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,4,\,6,\,7,\,8,\,9,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{0,\,1,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}-1,{p}_{\left\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,1,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}+{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},{p}_{\left\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}+{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}},-{p}_{\left\{0,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}^{2}+{p}_{\left\{0,\,1,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}{p}_{\left\{2,\,3,\,4,\,5,\,6,\,7,\,8,\,11,\,12,\,13\right\}}\right)$

\section*{Do Zhijie's smaller example}
to make sure...

\end{document}


\subsection{PV Diagrams}
Like all things in physics, physicists really enjoy graphing thermodynamic processes. Their particular favorite is the PV graph, putting pressure on the y-axis and volume on the x-axis; from the ideal gas law, you may see why this is a natural choice of axes. Shown below is a PV graph of a thermodynamic process that takes the gas from starting temperature $T_1$ (at pressure $P_1$ and volume $V_1$) to temperature $T_2$ (at pressure $P_2$ and volume $V_2$). Of course, there's nothing stopping us from assigning numbers to $P,T,V$, as well, if the situation calls for it. The process pictured below is an expansion of the gas, but we could also very well draw the arrow the other way and depict a compression of the gas.
\begin{center}
\begin{tikzpicture}
 \draw[stealth-stealth] (0,5) node[below left]{$P$} |- (5,0) node[below left]{$V$};
 \draw[thick,->] (1,4) -- (2.5,2.5);
\draw[thick] (2.5,2.5) -- (4,1);
\draw[dashed] (4,0) -- (4,1);
\draw[dashed] (1,0) -- (1,4);
\draw[dashed] (4,1) -- (0,1);
\draw[dashed] (1,4) -- (0,4);
\filldraw (1,4) circle (2pt);
\filldraw (4,1) circle (2pt);
\node[below] at (1,0) {$V_1$};
\node[below] at (4,0) {$V_2$};
\node[left] at (0,1) {$P_2$};
\node[left] at (0,4) {$P_1$};
\node[right] at (1,4) {$T_1$};
\node[right] at (4,1) {$T_2$};
 \end{tikzpicture}
 \end{center}
 To get a feel for these kinds of graphs, I would suggest trying to make some. Write down an initial state and final state for your gas with one or two quantities remaining constant, and try to figure out what the graph between those two steps would look like. For example, if I had
\begin{itemize}
    \item Initial State: P = 10kPa, V = 1m$^3$, N = 10molecules, T = 300K
    \item Final State: P = 5kPa, V = 2$m^3$, N = 10molecules, T = 300K
\end{itemize}
What would the graph between the two states look like if temperature and number of molecules were held constant the entire time? (Hint: You might be first inclined to think it looks like a diagonal line like the diagram above, but this isn't quite correct; think about why the condition of temperature being held constant for the entire time might not hold in this case). The answer will be revealed very shortly, when we discuss the PV diagrams of the 4 basic thermodynamic processes! \\
Before we go there though, there is one very useful connection between PV diagrams and work that I would like to point out. In the previous section, we defined the work done on the gas as:
\[ W = -\int_{V_1}^{V_2} P(V)dV \]
And looking at the PV diagram, it now becomes clear that this is nothing more than the (negative) area under the PV curve! As an example, let's figure out the work done on the gas in the process shown above. All we have to do is calculate the area underneath the curve. For convenience, let us split it into two parts of the triangle and the rectangle. For the triangle, we have area $\frac{1}{2}*\left(P_1-P_2\right)*\left(V_2-V_1\right)$, and for the rectangle, we have area $P_2*\left(V_2-V_1\right)$. Therefore, the total area under the graph is:
\[ \text{Area } = \frac{1}{2}*\left(P_1-P_2\right)*\left(V_2-V_1\right) + P_2*\left(V_2-V_1\right) = \frac{1}{2}\left[\left(P_1+P_2\right)\left(V_2-V_1\right) \right]\]
This area is in fact the work done \textbf{by} the gas, and the work done on the gas is just its negative:
\[W_{on} = \frac{1}{2}\left[\left(P_1+P_2\right)\left(V_1-V_2\right) \right]\]
When using this area argument, do be careful of which direction the curve is going; for example, if the process was a compression of the gas rather than an expansion, we would have to introduce a negative sign as things would be going in the opposite direction.


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\author{Tianshuang (Ethan) Qiu}
\begin{document}
\title{Math 74, Week 6}
\maketitle

\section{Mon Lec, 1a}
Base case: $n=1, n=2, a_1=3, a_2=5$, base cases hold.
\newline
Assume that for some $n \in \N$, all $m \in \N 1, 1 \leq m \leq n$ satisfies this identity, we have
$$a_{n+1}=3a_n-2a_{n-1}=3(2^n+1)-2(2^{n-1}+1)=3\times 2^n+3-2\times 2^{n-1}-2$$
$$=2^{n-1}(6-2)+1=2^{n-1}(2^2)+1=2^{n+1}+1$$
Thus we have proven the inductive case. Q.E.D.

\section{Mon Lec, 3c}
Lemma: a sequence given by $f_n=1/\sqrt5 (\phi^{n} - \bar \phi^{n})$ where $\phi = (1+\sqrt5)/2, \bar \phi = (1-\sqrt5)/2$ has the property $f_n+f_n+1 = f_n+2$ for $n \geq 0$
\newline
Base case: $n = 0$, $f_0 = 0, f_1=1, f_2=1$, base case holds.
Assume that for some $n$, the statement hodls for all $m\leq n$.
\newline
Consider $f_{n+1}$, $$f_n = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^n-(\frac{1-\sqrt5}{2})^n)$$
We get use this property to get a similar result for $f_{n-1}$. Now we add them together:
$$f_n+f_{n-1}=\frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^n-(\frac{1-\sqrt5}{2})^n+(\frac{1+\sqrt5}{2})^{n-1}-(\frac{1-\sqrt5}{2})^{n-1})$$
$$=\frac{1}{\sqrt5}(((\frac{1+\sqrt5}{2})^{n-1}(3+\sqrt5)/2)-(\frac{1-\sqrt5}{2})^{n-1}(3+\sqrt5)/2)$$
Since $(3+\sqrt5)=(1+\sqrt5)^2/2$, the expression simplifies to
$$=\frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^{n+1}-(\frac{1-\sqrt5}{2})^{n+1})$$.
Thus we have proven the fibbonacci property from the closed form.
\newline
Base case: $n=0, m=0, f_n=1, f_m=1, f_{n+m+1}=2$, base case holds.
\newline
For any $n, m \in \N$, assume that the statement is true for all $n'\leq n, m' \leq m$. Now consider $n+1$ and $f_{n+m+2}$.
\newline
Using our inductive hypothesis, we know that $f_{n+m+1}=f_mf_n + f_{m+1}f_{n+1}$
\newline
Now consider $f_mf_{n+1}+f_{m+1}f_{n+2}$, use our lemma to get that it is equal to $f_m(f_n+f_{n-1})+f_{m+1}(f_n+f_{n+1})$
$$=f_mf_n+f_mf_{n-1}+f_{m+1}f_n+f_{m+1}f_{n+1}$$
Now we apply IH,
$$=f_{m+n}+f_{m+n+1}=f_{m+n+2}$$
The last step is using our lemma (the fibbonacci identity). Thus we have proven the inductive step. Q.E.D.

\section{Mon Lec, 4a}
\begin{figure}[h]
    \includegraphics[width = 100mm]{GRAPH1.png}
\end{figure}
Base case: let there be three people, they form a triangle. Let the sides have length $a<b<c$. We can see that A and C will exchange pies, and B will throw at A, leaving him the survivor. Therefore base case holds.
\newline
Assume that for some odd $n$, the statement holds for all odd $m \leq n$. Now consider $n+2$.
\newline
We first remove the two people who are closest to each other. We call this the alternate game. Then the remaining $n$ people satisfies the inductive hypothesis, and there will be one survivor. Now consider the original game, since these two removed people are closest to each other, they will throw the pie at each other.
\newline
Then we break down the remaining players into two scenarios when we add the two back: the two new players are closer than their original targets, or they are further than their original targets.
\newline
In either case they will not throw their pies at the survivor in the alternate game. Therefore he will still be the survivor in the original game.
\newline
Thus we have proven the inductive case. Q.E.D.
\newpage

\section{Mon Dis, 4}
Base case: $n = 1, S = \{1,2\}, 1 \mid 2$, base case holds.
\newline
Assume that for $n \in \N$, the statement holds for $m \leq n$. Consdier $n+1$ and a set $S = \{a_1, a_2, ..., a_{n+2}\}$. We break it down into two cases.
\paragraph{If $a_{n+1}\leq 2n$}
Then we can simply take $S' = S \setminus a_{n+2}$, now we have a set that has $n+1$ elements, and the largest one is less than $2n$, so we can apply the inductive hypothesis and conclude that there is an integer that divides another in $S'$.
\paragraph{If $a_{n+1}> 2n$}
Now if $a_{n+1}>2n$, and the maximum of the set cannot exceed $2n+22$, we must have $a_{n+1}=2n+1. a_{n+2}=2n+2$. Now if $(n+1)\in S$, we will have $n+1 \mid 2n+2$. Otherwise, take the set $\{a_1, a_2, ..., a_n, n+1\}$, all of these are less than $2n$ since we know what $a_{n+1}$ and $a_{n+2}$ are. Now we can apply the inductive hypothesis to this set since it has $n+1$ items. Therefore there must be two integers that divides each other.
\newline
Thus we have proven the inductive case. Q.E.D.

\section{Mon Dis, 7}
Base case: $n=1$, we can simply take a block out, the remaining 3 form an $L$ shape. Base case holds
\newline
Assume that for some $n \in \N$, the statement holds for all $m \leq n$, now consider $2^{n+1} \times 2^{n+1}$. We first divide the square into four squares of $2^n \times 2^n$. Take the right most square and remove a single tile. By the inductive hypothesis this square is tilable with $L$'s since it has $2^n \times 2^n$ tiles. Now for the remaing three squares, we remove one tile at the center like this:
\newline
\begin{figure}[H]
    \includegraphics[width = 100mm]{GRAPH2.png}
\end{figure}
Now the remaining 3 tiles can be filled with a single $L$. Thus we have proven the inductive step. Q.E.D.

\newpage

\section{Wed Lec, 5}
Let our 9 digit number be represented by $abcde2021$. Using the subtraction theorem we know that $abcde0000/2021$ must be an integer.
\newline
We can factor $2021 = 43 \times 47$, so it shares no factors with 10000. So $2021 \mid abcde$. Now the amount that we are seeking is equal to the amount of 5 digit numbers that is divisible by 2021
$$\floor*{\frac{99999}{2021}}-\floor*{\frac{9999}{2021}} = 49-4=45$$
\newpage

\section{Wed Dis, 6}
$$(y-\frac{1}{y})^3 = x^3$$
$$y^3 -3 \frac{1}{y} + 3 y-\frac{1}{y^3} = x^3$$
$$y^3-\frac{1}{y^3}+3(y-\frac{1}{y})=x^3$$
$$y^3-\frac{1}{y^3}+3x = x^3$$
$$y^3-\frac{1}{y^3}= x^3-3x$$
\newpage

\section{Fri Lec, 3b}
Since $a \equiv b \bmod 6$, we have $a = 6n + k$, $b = 6m + k$, where $m,n,k \in \Z$.
\newline
Let $h \in \Z$, we multiply $a, b$ by $h$: $ah = 6nh + kh$, $bh = 6mh + kh$.
\newline
Since mutiples of 6 will always be equivalent to 0 when modded by 6, the only things that are left on both sides are $kh$, and $kh \equiv kh \bmod 6$.
\newline
Q.E.D.

\section{Fri Lec, 7}
We factor 12 into $3 \times 2^2$, and 15 into $3 \times 5$. Since our perfect square needs to have these as factors, $xy^2$ must have 3 as a factor, and $xy$ must have 5 as a factor.
\newline
Furthermore, since $xy$ has 5 as a factor, $xy^2$ must have 25 as a factor since the $12xy^2$ is a perfect square and 12 cannot be divided by 5. Now we attempt to construct one such number. let $y = 5$, $x=3$, $12xy^2 = 900 = 30^2, 15xy = 225 = 15^2$. $x+y = 8$

\end{document}


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\title{\textbf{18-748 Wireless Sensor Networks Lab 1}}
\author{Emily Ruppel, Iljoo Baek, Mengwen He}

\begin{document}
\maketitle

\section{Assignment 2}

\subsection{Data Measurement}

\begin{table}[!h]
	\centering
	\caption{The Measurements of Distance and RSSI}
	\begin{tabular}{|c|c|c|c|c|}
		\hline
		\multicolumn{2}{|c|}{Distance} & \multicolumn{3}{|c|}{RSSI} \\
		\hline
		Yard	&	Meter	&	Min	&	Max	&	Ave.\\
		\hline
		0		&	0		&	39	&	51	&	45	\\
		\hline
		5		&	4.57	&	21	&	25	&	24	\\
		\hline
		10		&	9.14	&	14	&	20	&	18	\\
		\hline
		15		&	13.72	&	7	&	18	&	15	\\
		\hline
		20		&	18.29	&	11	&	14	&	12	\\
		\hline
		25		&	22.86	&	0	&	0	&	0	\\
		\hline
	\end{tabular}
\end{table}

\begin{figure}[!h]
	\centering
	\includegraphics[width=15cm]{./img/data.png}
\end{figure}

\subsection{Compute Path Loss Exponent $n$}

Path loss exponent is given by: $$L=10n\log_{10}d+C$$
\begin{itemize}
	\item $L$: path loss
	\item $n$: path loss exponent
	\item $d$: distance in meters
	\item $C$: system constant
\end{itemize}

From two different data points: $$(L2-L1)=10n\log_{10}(d2/d1) \Rightarrow n=(L2-L1)/(10\log_{10}(d2/d1))$$

As RSSI is linearly related to path loss $$(L2-L1)=(RSSI2-RSSI1)$$

Fix $d1=4.57$, $L1=24$ and get the estimation as Tab.\ref{tab:pathloss}:

\begin{table}[!h]
	\centering
	\caption{The Estimation of Path Loss Exponent}
	\label{tab:pathloss}
	\begin{tabular}{|c|c|c|}
		\hline
		$d2$	&	$L2$	&	$n$	\\
		\hline
		9.14		&	18		&	-1.993	\\
		\hline
		13.72		&	15		&	-1.885	\\
		\hline
		18.29		&	12		&	-1.992	\\
		\hline
		22.86		&	0		&	-3.433 (invalid)	\\
		\hline
	\end{tabular}
\end{table}

Because we cannot get RSSI value when $d2=22.86$, we cannot trust this measurement that just represents the margin of the signal coverage. Therefore, the estimated Path Loss Exponent is:
$$\mathbb{E}[n]=\frac{-1.993-1.885-1.992}{3}=-1.957$$
$$Var[n]=\frac{(1.957-1.993)^2+(1.957-1.885)^2+(1.957-1.992)^2}{3}=0.0039$$
$$\sigma=\sqrt{Var[n]}=0.0621$$

\newpage
\section{Assignment 3}

\subsection{Data Collection}

\begin{figure}[!h]
	\centering
	\subfloat[Case1: Task1 has lower priority.]{\includegraphics[width=3.5cm]{./img/case1_t1.png}\includegraphics[width=3.5cm]{./img/case1_t2.png}}
	~~
	\subfloat[Case2: Task2 has lower priority.]{\includegraphics[width=3.5cm]{./img/case2_t1.png}\includegraphics[width=3.5cm]{./img/case2_t2.png}}
\end{figure}

\subsection{Worst-case Response Time Estimation}

Because the task1 and task2 are released every 1s and 2s, we can directly get their response time from the nanosecond part, and thus easily get the worst-case response time as below:

\begin{itemize}
	\item Case 1:
	\begin{itemize}
		\item Task1: 601562808
		\item Task2: 299804841
	\end{itemize}
	\item Case 2:
	\begin{itemize}
		\item Task1: 298828278
		\item Task2: 601562808
	\end{itemize}
\end{itemize}

\subsection{Explanation of The Differences}

\subsubsection{Response Time}

For case1, because task1 has lower priority, it will be preempted by task2 every 2 seconds; therefore, the response time of task1 is not stable. For case2, because task1 has higher priority, it will not be preempted by task2; therefore, the response time of task1 is stable.

For task2, because its period interval is longer than task1's, and they form a harmonized taskset; therefore, task2's response time is stable for both cases.

\subsubsection{Worst-case Response Time}

Comparing the worst-case response time of task1 and task2 from two cases, we found that the task with higher priority will have smaller worst-case response time.

\subsection{Bonus1: Find a tight CPU reserve budget for each task.}

From the experiment, we know that if task1 with shorter period interval has higher priority, both tasks can have stable response time; therefore, our method is first to use the RMS(Rate-monotonic scheduling) to assign the priority to each task, and then assign each task's reserve budget equals to its worst-case response time.

\newpage
\subsection{Bonus2: Plot graphs for both basks to show their response times.}

\begin{figure}[!h]
	\centering
	\includegraphics[width=15cm]{./img/Task2_prio2_updated.eps}
	\includegraphics[width=15cm]{./img/Task1_prio2_updated.eps}
\end{figure}

\end{document}\grid
\grid


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\begin{document}

\section*{Problem} Moe has built an awesome atomic clock.  It has one hand that
moves at a perfectly constant speed around a 24 hour dial. The problem is he has
no way to calibrate it.  He asks Joe for advice and Joe asks him what his
objective is.  He says he wants it to be exactly right as often as possible.
Joe thinks for a minute and says, ``That’s going to take a lot of electricity.''
Explain what he means.

\section*{Solution} Let $s$ be the rotational speed that Moe chooses expressed
in revolutions per day. If Moe can find \emph{exactly} 1 as the rotational
speed, he will be right all the time.  Unfortunately, the probability of that is
$0$. Now consider what happens when Moe's chosen speed is off by $\delta$
revolutions per day.  If $\delta < 0$ and Moe manages to start the clock at
exactly the right time, it will drift by $\delta$ each 24 hours and it will take
$1/\delta$ days for it to be exactly right again (when it has fallen so far
behind that it is showing exactly the right time). Somewhat paradoxically, the
smaller $\delta$ is, the longer this will take and hence the smaller the number
of times per any time interval that the clock will be exactly right. The same
analysis obviously applies for $\delta > 0$. Now suppose that Moe just lets the
hand move as fast as possible. Then it will pass the correct time once every
revolution it makes (plus whatever time it takes to complete a revolution). So
the faster he can make it go, the more frequently the reading is exactly right.
Hence Joe's comment ``that will take a lot of electricity.'' \end{document}