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update: try reverse transpositiongit status

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@@ -36,14 +36,20 @@
\newtheorem{definition}{Definition}[section]
\begin{document}
\section{Goal: Write any tensor as vector !!}
\section{Write any tensor of any order as vector or tensor of order one !!}
Let
$\mathbb{K}$ a field( for e.g $\mathbb{R}$ ), we will construct $\mathbb{K}^{d_1\times\cdots\times d_k}\simeq \mathbb{K}^{d_1\cdots d_k}$ such that $d_1,\ldots,d_k\in\mathbb{N}^*$.
\begin{definition}
A tensor is a collection of elements
its dimensions define the size of the collection
its order is the number of different dimensions
\end{definition}
In linear algebra, we have the unfolding
\begin{definition}
A dimension of a tensor $T$ is a vector $d\in \left(\mathbb{N}^*\right)^k$ $(k<\infty)$. We have $$T=(a)_d=(a)_{d_1,\ldots,d_k} =({\alpha}_{i_1,\ldots,i_k}^{\{k\}})_{0\le i_1<d_1, \ldots , 0\le i_k< d_k} $$
$$ {\alpha}_{i_1,\ldots,i_k}^{\{k\}}\in \mathbb{K} \ \ / \ \ 0\le i_j<d_j $$
\end{definition}
$\mathbb{K}$ is a field( for e.g $\mathbb{R}$ )\\
$k>0$ is the order.\\
The size of the Tensor is $d_1 \times \cdots \times d_k=\prod_{i=1}^kd_i$
@@ -55,7 +61,7 @@ We can write $$T\in \mathbb{K}^{d_1 \times d_2 \times \cdots \times d_k}$$
\section{Bijection}
For a given $d_1,\ldots,d_k\in \mathbb{N}^*$, there exists a bijective application between a set of tensor order 1 of size $n=d_1\cdot d_2 \cdots d_k$ $\mathbb{K}^{d_1\times d_2 \times \cdots \times d_k}$ and the set of tensor of order $k$ of the size $n= d_1\cdot d_2 \cdots d_k$ $\mathbb{K}^{d_1\times d_2 \times \cdots \times d_k}$.
For a given $d_1,\ldots,d_k\in \mathbb{N}^*$, there exists a bijective application between a set of tensor order 1 of size $n=d_1\cdot d_2 \cdots d_k$ $\mathbb{K}^{n}$ and the set of tensor of order $k$ of the size $n= d_1\cdot d_2 \cdots d_k$ $\mathbb{K}^{d_1\times d_2 \times \cdots \times d_k}$.
Little endian
$$
@@ -263,8 +269,380 @@ $$
The most advantage is computation, only one loop to browse all tensor elements! If we can compute in parallele, as we know the size and it is on a line, each threads or compute units available can have the same number of elements to compute!
\section{Transposition}
By definition, the transposition of a tensor $T_{\alpha}^{\{k\}}\in\mathbb{R}^{d_{1}\times\cdots\times d_{k}}$ is $T_{\beta}^{\{k\}}\in\mathbb{R}^{b_{1}\times \cdots\times b_{k}}$ and $T_{\alpha}^{\{1\}}\in\mathbb{R}^{d_{1}\cdots d_{k}}$ and $T_{\beta}^{\{1\}}\in\mathbb{R}^{b_{k}\cdots b_{1}}$ such that : $b_j=d_{k-j+1}$
$\alpha_{\left(i_1, i_2,\ldots, i_{k}\right)_k} = \beta_{\left(i_{k},\ldots,i_2,i_1\right)_k} $
in one dim representation (little endian) we have
\begin{align*}
\alpha_{\left(i_1, i_2,\ldots, i_{k}\right)_k} &= \alpha_{\left(\sum_{j=1}^k i_j \prod_{l=1}^{j-1}d_l \right)_1} \\
\beta_{\left(t_{1},\ldots,t_{k}\right)_k} &= \beta_{\left( \sum_{j=1}^k t_j \prod_{l=1}^{j-1}b_l\right)_1}
\end{align*}
here $t_j=i_{k-j+1}$ and $b_j=d_{k-j+1}$ so we have
\begin{align*}
\alpha_{\left(i_1, i_2,\ldots, i_{k}\right)_k} &= \alpha_{\left(\sum_{j=1}^k i_j \prod_{l=1}^{j-1}d_l \right)_1} \\
&=\beta_{\left(t_{1},\ldots,t_{k}\right)_k} \\
&= \beta_{\left( \sum_{j=1}^k t_j \prod_{l=1}^{j-1}b_l\right)_1}\\
& t_j=i_{k-j+1}\text{ and }b_j=d_{k-j+1}\\
&= \beta_{\left( \sum_{j=1}^k i_{k-j+1} \prod_{l=1}^{j-1}d_{k-l+1}\right)_1}\\
&\text{we pose }t=k-j+1 \text{ so } j=k-t+1 \\
&\text{if }j=1 \Rightarrow t=k \text{ and } j=k \Rightarrow t=1 \\
&= \beta_{\left( \sum_{t=k}^{1} i_{t} \prod_{l=1}^{k-t+1-1}d_{k-l+1}\right)_1}\\
&= \beta_{\left( \sum_{t=1}^{k} i_{t} \prod_{l=1}^{k-t}d_{k-l+1}\right)_1}\\
& \text{ we pose }s=k-l+1 \text{ and }l=k-s+1 \\
&\text{if }l=1 \Rightarrow s=k \text{ and } l=k-t \Rightarrow s=k-(k-t)+1=t+1 \\
&= \beta_{\left( \sum_{t=1}^{k} i_{t} \prod_{s=k}^{t+1}d_{s}\right)_1}\\
\end{align*}
\section{Transposition}
By definition, the transposition of a tensor $T_{\alpha}^{\{k\}}\in\mathbb{R}^{d_{1}\times\cdots\times d_{k}}$ is $T_{\beta}^{\{k\}}\in\mathbb{R}^{d_{k}\times d_{k-1}\cdots\times d_{1}}$ and $T_{\alpha}^{\{1\}}\in\mathbb{R}^{d_{1}\cdots d_{k}}$ and $T_{\beta}^{\{1\}}\in\mathbb{R}^{d_{k}\cdots d_{1}}$ such that
$\alpha_{\left(i_1, i_2,\ldots, i_{k}\right)_k} = \beta_{\left(i_{k},\ldots,i_2,i_1\right)_k} $
So, if little endian
\begin{align*}
\alpha_{\left(\sum_{j=1}^{k} i_j \prod_{l=1}^{j-1}d_j \right)_1} &= \beta_{\left(\sum_{j=k}^{1} i_j \prod_{l=k}^{j+1}d_j \right)_1}\\
&= \beta_{\left(\sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j \right)_1}\\
\end{align*}
Let $l=\sum_{j=1}^{k} i_j \prod_{l=1}^{j-1}d_j$ corresponds $m=\sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j$
Suppose $i_1<d_1-1$
So $l+1 = \sum_{j=1}^{k} i_j \prod_{l=1}^{j-1}d_j +1 = i_1+1 + \sum_{j=2}^{k} i_j \prod_{l=1}^{j-1}d_j$
Corresponds to
\begin{align*}
m'&=\left(i_1+1\right)\prod_{l=2}^{k}d_j+ \sum_{j=2}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + i_1 \prod_{l=2}^{k}d_j+ \sum_{j=2}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + m\\
\end{align*}
If $i_1=d_1-1$ and $ i_2 <d_2-1$
So
\begin{align*}
l+1 &= \sum_{j=1}^{k} i_j \prod_{l=1}^{j-1}d_j +1 \\
&= i_1+1 + \sum_{j=2}^{k} i_j \prod_{l=1}^{j-1}d_j \\
&= d_1 + \sum_{j=2}^{k} i_j \prod_{l=1}^{j-1}d_j \\
&= d_1 + i_2 d_1 + \sum_{j=3}^{k} i_j \prod_{l=1}^{j-1}d_j \\
&= (i_2+1) d_1 + \sum_{j=3}^{k} i_j \prod_{l=1}^{j-1}d_j \\
\end{align*}
i.e $i_1'=0$
it corresponds to
\begin{align*}
m' &=\sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + (i_2+1) \prod_{l=3}^k d_l \\
&=\sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
&=\sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
\end{align*}
\begin{align*}
m'-m &= \sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
&-\left( (d_1-1)\prod_{l=2}^k d_l +i_2\prod_{l=3}^kd_l + \sum_{j=3}i_3 \prod_{l=j+1}^kd_l \right)\\
&= \sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
&-\left( \prod_{l=1}^k d_l-\prod_{l=2}^k d_l +i_2\prod_{l=3}^kd_l + \sum_{j=3}i_3 \prod_{l=j+1}^kd_l \right) \\
&= \prod_{l=3}^k d_l+\prod_{l=2}^k d_l - \prod_{l=1}^k d_l \\
\end{align*}
Since $i_1=d_1-1$
\begin{align*}
m' &=\prod_{l=2}^{k}d_j + \sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + i_1 \prod_{l=2}^{k}d_j+ i_2 \prod_{l=3}^{k}d_j+ \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + (d_1-1) \prod_{l=2}^{k}d_j+ i_2 \prod_{l=3}^{k}d_j+ \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&= d_1 \prod_{l=2}^{k}d_j+ i_2 \prod_{l=3}^{k}d_j+ \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&= \prod_{l=1}^{k}d_j+ i_2 \prod_{l=3}^{k}d_j+ \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
& > \prod_{l=1}^{k}d_j
\end{align*}
\begin{align*}
m'-m &=\sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
&-\left( (d_1-1)\prod_{l=2}^k d_l + i_2\prod_{l=3}d_l +\sum_{j=3}i_j\prod_{l=j+1}d_l \right)\\
&=\sum_{j=3}^k i_j \prod_{l=j+1}^{k}d_l + i_2 \prod_{l=3}^k d_l + \prod_{l=3}^k d_l \\
&-\left( \prod_{l=1}^k d_l-\prod_{l=2}^k d_l + i_2\prod_{l=3}d_l +\sum_{j=3}i_j\prod_{l=j+1}d_l \right)\\
&= \prod_{l=3}^k d_l -\left( \prod_{l=1}^k d_l-\prod_{l=2}^k d_l \right)\\
&= \prod_{l=3}^k d_l + \prod_{l=2}^k d_l - \prod_{l=1}^k d_l
\end{align*}
If $i_1=d_1-1$ and $i_2=d_2-1$ in $m$ so
\begin{align*}
m'&=\left(i_1+1\right)\prod_{l=2}^{k}d_j+ \sum_{j=2}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + i_1 \prod_{l=2}^{k}d_j+ \sum_{j=2}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \sum_{j=1}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + m\\
&=\prod_{l=2}^{k}d_j + (d_1-1)\prod_{j=2}^kd_j+(d_2-1)\prod_{j=3}^kd_j + \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \prod_{j=1}^kd_j -\prod_{j=2}^kd_j+ \prod_{j=2}^kd_j -\prod_{j=3}^kd_j + \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \prod_{j=1}^kd_j -\prod_{j=3}^kd_j + \sum_{j=3}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&=\prod_{l=2}^{k}d_j + \prod_{j=1}^kd_j -\prod_{j=3}^kd_j + i_3 \prod_{l=4}^{k}d_j+ \sum_{j=4}^{k} i_j \prod_{l=j+1}^{k}d_j\\
&> \prod_{j=1}^kd_j\\
\end{align*}
\begin{align*}
m'-m &= \prod_{l=2}^{k}d_j \\
\end{align*}
\begin{align*}
m''&= (i_3+1) \prod_{l=4}^{k}d_j + \sum_{j=4}^{k} i_j \prod_{l=j+1}^{k}d_j\\
\end{align*}
\begin{align*}
m''-m &= (i_3+1) \prod_{l=4}^{k}d_j + \sum_{j=4}^{k} i_j \prod_{l=j+1}^{k}d_j - \left((d_1-1)\prod_{l=2}^k d_l + (d_2-1)\prod_{l=3}^kd_l + i_3 \prod_{l=4}^kd_l + \sum_{j=4}^k i_j \prod_{l=j+1}^kd_l \right) \\
m''-m &= \prod_{l=4}^{k}d_j - \left((d_1-1)\prod_{l=2}^k d_l + (d_2-1)\prod_{l=3}^kd_l \right) \\
m''-m &= \prod_{l=4}^{k}d_j - \left(\prod_{l=1}^k d_l-\prod_{l=2}^k d_l + \prod_{l=2}^k d_l-\prod_{l=3}^kd_l \right) \\
m''-m &= \prod_{l=4}^{k}d_j - \left(\prod_{l=1}^k d_l-\prod_{l=3}^kd_l \right) \\
m''-m &= \prod_{l=3}^kd_l + \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l \\
\end{align*}
if $i_1=d_1-1$, $i_2=d_2-1$, $i_3=d_3-1$
\begin{align*}
m+\prod_{l=3}^kd_l + \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l &= \left( (d_1-1)\prod_{l=2}^kd_l+(d_2-1)\prod_{l=3}^kd_l+(d_3-1)\prod_{l=4}^kd_l +\sum_{j=4}^k i_j\prod_{l=j+1}d_l \right) \\& +\prod_{l=3}^kd_l+ \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l\\
\end{align*}
\begin{align*}
m+\prod_{l=3}^kd_l + \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l &= \left( (d_1-1)\prod_{l=2}^kd_l+(d_2-1)\prod_{l=3}^kd_l+(d_3-1)\prod_{l=4}^kd_l +\sum_{j=4}^k i_j\prod_{l=j+1}d_l \right) \\& +\prod_{l=3}^kd_l+ \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l\\
&= \left(\prod_{l=1}^kd_l-\prod_{l=2}^kd_l+ \prod_{l=2}^kd_l-\prod_{l=3}^kd_l+\prod_{l=3}^kd_l-\prod_{l=4}^kd_l +\sum_{j=4}^k i_j\prod_{l=j+1}d_l \right) \\& + \prod_{l=3}^kd_l+ \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l\\
&= \left(\prod_{l=1}^kd_l-\prod_{l=4}^kd_l +\sum_{j=4}^k i_j\prod_{l=j+1}d_l \right) \\& + \prod_{l=3}^kd_l+ \prod_{l=4}^{k}d_j - \prod_{l=1}^k d_l\\
&= \sum_{j=4}^k i_j\prod_{l=j+1}d_l + \prod_{l=3}^kd_l\\
&= \sum_{j=4}^k i_j\prod_{l=j+1}d_l + <\prod_{l=3}^kd_l\\
\end{align*}
\begin{align*}
l+1&=\sum_{j=1}^k i_j\prod_{l=1}^{j-1}d_l +1\\
&= 1+(d_1-1) + (d_2-1)d_1 + (d_3-1)d_1 d_2+i_4d_1d_2d_3 +\sum_{j=5}^k i_j\prod_{l=1}^{j-1}d_l\\
&= d_1 + d_2-d_1 + d_1d_2d_3-d_1 d_2+i_4d_1d_2d_3 +\sum_{j=5}^k i_j\prod_{l=1}^{j-1}d_l\\
&= (i_4+1)d_1d_2d_3 +\sum_{j=5}i_j\prod_{l=1}^{j-1}d_l\\
\end{align*}
corresponds to
\begin{align*}
m' &= (i_4+1)\prod_{l=5}^kd_l + \sum_{j=5}i_j\prod_{l=j+1}^kd_l\\
&= i_4\prod_{l=5}^kd_l +\prod_{l=5}^kd_l + \sum_{j=5}i_j\prod_{l=j+1}^kd_l\\
\end{align*}
\begin{align*}
m'-m &= i_4\prod_{l=5}^kd_l +\prod_{l=5}^kd_l + \sum_{j=5}i_j\prod_{l=j+1}^kd_l\\
& -\left( (d_1-1)\prod_{l=2}^kd_l + (d_2-1)\prod_{l=3}^kd_l+(d_3-1)\prod_{l=4}^kd_l+i_4\prod_{l=5}^kd_l+\sum_{j=5}^ki_j\prod_{l=j+1}^k d_l \right)\\
&= i_4\prod_{l=5}^kd_l +\prod_{l=5}^kd_l + \sum_{j=5}i_j\prod_{l=j+1}^kd_l\\
& -\left( \prod_{l=1}^kd_l-\prod_{l=2}^kd_l + \prod_{l=2}^kd_l-\prod_{l=3}^kd_l+\prod_{l=3}^kd_l-\prod_{l=4}^kd_l+i_4\prod_{l=5}^kd_l+\sum_{j=5}^ki_j\prod_{l=j+1}^k d_l \right)\\
&= \prod_{l=5}^kd_l + \prod_{l=4}^kd_l - \prod_{l=1}^kd_l
\end{align*}
\section{Transposition backward}
$l=\sum_{j=1}^ki_j\prod_{l=1}^{j-1}d_l$ corresponds to $m=\sum_{j=1}^ki'_j\prod_{l=1}^{j-1}d'_l$ such that
\begin{align*}
d'_l&=d_{k-l+1} \\
i'_j&=d'_{j}-i_{k-j+1} -1\\
&=d_{k-j+1}-i_{k-j+1}-1 \\
\end{align*}
\begin{align*}
m&=\sum_{j=1}^ki'_j\prod_{l=1}^{j-1}d'_l\\
&=\sum_{j=1}^k\left(d_{k-j+1}-i_{k-j+1}-1\right)\prod_{l=1}^{j-1}d_{k-l+1}\\
&t=k-j+1 \Rightarrow j=k-t+1\\
&j=1 \Rightarrow t=k\\
&j=k \Rightarrow t=1\\
m&=\sum_{t=k}^1\left(d_{t}-i_{t}-1\right)\prod_{l=1}^{k-t+1-1}d_{k-l+1}\\
m&=\sum_{t=1}^k\left(d_{t}-i_{t}-1\right)\prod_{l=1}^{k-t}d_{k-l+1}\\
&s=k-l+1 \Rightarrow l=k-s+1\\
&l=1 \Rightarrow s=k\\
&l=k-t \Rightarrow s=k-(k-t)+1=t+1\\
m&=\sum_{t=1}^k\left(d_{t}-i_{t}-1\right)\prod_{s=k}^{t+1}d_{s}\\
m&=\sum_{t=1}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
\end{align*}
if $i_1<d_1-1$ then
\begin{align*}
l+1&=\sum_{j=1}^ki_j\prod_{l=1}^{j-1}d_l+1\\
&=i_1 +1 + \sum_{j=2}^ki_j\prod_{l=1}^{j-1}d_l\\
\end{align*}
corresponds to
\begin{align*}
m'&=(d_1-(i_1+1)-1)\prod_{s=2}^kd_s+ \sum_{t=2}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
&=(d_1-i_1-2)\prod_{s=2}^kd_s+ \sum_{t=2}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
\end{align*}
\begin{align*}
m'-m&=(d_1-i_1-2)\prod_{s=2}^kd_s+ \sum_{t=2}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
&-\left( (d_1-i_1-1) \prod_{s=2}^kd_s+ \sum_{t=2}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s} \right)\\
&=-\prod_{s=2}^kd_s\\
\end{align*}
if $i_1=d_1-1$ and $i_2<d_2-1$ then
\begin{align*}
l+1&=\sum_{j=1}^ki_j\prod_{l=1}^{j-1}d_l+1\\
&=d_1-1 +1 + \sum_{j=2}^ki_j\prod_{l=1}^{j-1}d_l\\
&=d_1 + i_2d_1+ \sum_{j=3}^ki_j\prod_{l=1}^{j-1}d_l\\
&=(i_2+1)d_1 +\sum_{j=3}^ki_j\prod_{l=1}^{j-1}d_l\\
\end{align*}
corresponds to
\begin{align*}
m'&=(d_1-0-1)\prod_{s=2}^kd_s+\left(d_2-(i_2+1)-1\right)\prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
&=(d_1-1)\prod_{s=2}^kd_s+(d_2-i_2-2)\prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
\end{align*}
\begin{align*}
m'-m&=(d_1-1)\prod_{s=2}^kd_s+(d_2-i_2-2)\prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
&-\left( (d_1-(d_1-1)-1) \prod_{s=2}^kd_s+ (d_2-i_2-1) \prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s} \right)\\
&=(d_1-1)\prod_{s=2}^kd_s+(d_2-i_2-2)\prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s}\\
&-\left( 0\prod_{s=2}^kd_s+ (d_2-i_2-1) \prod_{s=3}^kd_s+ \sum_{t=3}^k\left(d_{t}-i_{t}-1\right)\prod_{s=t+1}^{k}d_{s} \right)\\
\\
&= (d_1-1)\prod_{s=2}^kd_s- \prod_{s=3}^kd_s \\
\end{align*}
\subsection{properties}
\begin{align*}
M_{(k_1,k_n)}&=\sum_{j=k_1}^{k_n}(d_j-1)\prod_{l=j+1}^{k_n}d_j\\
&=\sum_{j=k_1}^{k_n}\left(\prod_{l=j}^{k_n}d_j-\prod_{l=j+1}^{k_n}d_j\right)\\
&=\left(\prod_{l=k_1}^{k_n}d_j-\prod_{l=k_1+1}^{k_n}d_j\right)\\
&+\left(\prod_{l=k_1+1}^{k_n}d_j-\prod_{l=k_1+2}^{k_n}d_j\right)\\
& + \vdots \\
&+\left(\prod_{l=k_n-1}^{k_n}d_j-\prod_{l=k_n}^{k_n}d_j\right)\\
&+\left(\prod_{l=k_n}^{k_n}d_j-\prod_{l=k_n+1}^{k_n}d_j\right)\\
&=\prod_{l=k_1}^{k_n}d_j -1\\
\end{align*}
\begin{align*}
M_{(k_1,k_n)}&=\sum_{j=k_1}^{k_n}(d_j-1)\prod_{l=k_1}^{j-1}d_j\\
&=\sum_{j=k_1}^{k_n}\left(\prod_{l=k_1}^{j}d_j-\prod_{l=k_1}^{j-1}d_j\right)\\
&=\left(\prod_{l=k_1}^{k_1}d_j-\prod_{l=k_1}^{k_1-1}d_j\right)\\
&+\left(\prod_{l=k_1}^{k_1+1}d_j-\prod_{l=k_1}^{k_1}d_j\right)\\
& + \vdots \\
&+\left(\prod_{l=k_1}^{k_n-1}d_j-\prod_{l=k_1}^{k_n-2}d_j\right)\\
&+\left(\prod_{l=k_1}^{k_n}d_j-\prod_{l=k_1}^{k_n-1}d_j\right)\\
&=\prod_{l=k_1}^{k_n}d_j -1 \\
\end{align*}
$m=\sum_{j=1}^ki_j\prod_{l=j+1}^kd_l$ with $i_1=0$ so
\begin{align*}
m &=\sum_{j=2}^ki_j\prod_{l=j+1}^kd_l \le \prod_{l=2}^kd_l-1\\
m - \prod_{s=2}^kd_s & \le -1\\
\end{align*}
So if $ m - \prod_{s=2}^kd_s \le -1$ then
$$m'= m + (d_1-1)\prod_{s=2}^kd_s- \prod_{s=3}^kd_s $$
but $i_2<d_2-1$ then $m=(d_2-1-i_2)\prod_{l=3}^kd_l+\sum_{j=3}^ki_j\prod_{l=j+1}^kd_l$
\begin{align*}
\prod_{l=2}^kd_l -1 \ge \sum_{j=3}^ki_j\prod_{l=j+1}^kd_l & \ge 0\\
d_2-1-i_2 &\ge 1 \\
(d_2-1-i_2)\prod_{l=3}^kd_l &\ge \prod_{l=3}^kd_l \\
\prod_{l=2}^kd_l -1 \ge m &\ge \prod_{l=3}^kd_l\\
(d_1-1)\prod_{l=2}^kd_l + \prod_{l=2}^kd_l -1\ge m + (d_1-1)\prod_{l=2}^kd_l &\ge (d_1-1)\prod_{l=2}^kd_l+ \prod_{l=3}^kd_l \\
(d_1-1)\prod_{l=2}^kd_l + \prod_{l=2}^kd_l -\prod_{l=3}^kd_l -1\ge m + (d_1-1)\prod_{l=2}^kd_l -\prod_{l=3}^kd_l &\ge (d_1-1)\prod_{l=2}^kd_l \\
(d_1-1)\prod_{l=2}^kd_l + (d_2-1)\prod_{l=3}^kd_l -1\ge m + (d_1-1)\prod_{l=2}^kd_l -\prod_{l=3}^kd_l &\ge (d_1-1)\prod_{l=2}^kd_l \\
% -1\ge m-\prod_{l=2}^kd_l &\ge \prod_{l=3}^kd_l - \prod_{l=2}^kd_l \\
% -1\ge m-\prod_{l=2}^kd_l &\ge \prod_{l=3}^kd_l - \prod_{l=2}^kd_l \\
% \prod_{l=1}^kd_l-1\ge m+\prod_{l=1}^kd_l-\prod_{l=2}^kd_l &\ge \prod_{l=1}^kd_l - \prod_{l=2}^kd_l + \prod_{l=3}^kd_l = (d_1-1)\prod_{l=2}^kd_l + \prod_{l=3}^kd_l\\
% \prod_{l=1}^kd_l-1-\prod_{l=3}^kd_l\ge m+\prod_{l=1}^kd_l-\prod_{l=2}^kd_l - \prod_{l=3}^kd_l &\ge \prod_{l=1}^kd_l - \prod_{l=2}^kd_l = (d_1-1)\prod_{l=2}^kd_l \\
% &= (d_1-1)\prod_{s=2}^kd_s- \prod_{s=3}^kd_s \\
\end{align*}
if $i_1=d_1-1$ and $i_2=d_2-1$ and $i_3<d_3-1$
\begin{align*}
l+1&=\sum_{j=1}^ki_j\prod_{l=1}^{j-1}d_l+1\\
&=d_1-1 +1 + (d_2-1)d_1 + i_3d_1d_2 + \sum_{j=4}^ki_j\prod_{l=1}^{j-1}d_l\\
&=d_1 + d_1d_2-d_1+ i_3d_1d_2 + \sum_{j=4}^ki_j\prod_{l=1}^{j-1}d_l\\
&=(i_3+1)d_1d_2 +\sum_{j=4}^ki_j\prod_{l=1}^{j-1}d_l\\
\end{align*}
corresponds to
\begin{align*}
m'&=(d_1-1)\prod_{s=2}^kd_s+(d_2-1)\prod_{s=3}^kd_s + (d_3-(i_3+1)-1)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
m'&=(d_1-1)\prod_{s=2}^kd_s+(d_2-1)\prod_{s=3}^kd_s + (d_3-i_3-2)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
\end{align*}
\begin{align*}
m'-m&=(d_1-1)\prod_{s=2}^kd_s+(d_2-1)\prod_{s=3}^kd_s + (d_3-i_3-2)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
&-\left((d_1-(d_1-1)-1)\prod_{s=2}^kd_s+(d_2-(d_2-1)-1)\prod_{s=3}^kd_s \right. \\
& \left.+ (d_3-(i_3)-1)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\right)\\
&=(d_1-1)\prod_{s=2}^kd_s+(d_2-1)\prod_{s=3}^kd_s + (d_3-i_3-2)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
&-\left((0)\prod_{s=2}^kd_s+(0)\prod_{s=3}^kd_s + (d_3-(i_3)-1)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\right)\\
&=(d_1-1)\prod_{s=2}^kd_s+(d_2-1)\prod_{s=3}^kd_s -\prod_{s=4}d_s\\
&=\prod_{s=1}^kd_s-\prod_{s=2}^kd_s+\prod_{s=2}^kd_s-\prod_{s=3}^kd_s -\prod_{s=4}d_s\\
&=\prod_{s=1}^kd_s-\prod_{s=3}^kd_s -\prod_{s=4}d_s\\
\end{align*}
\begin{align*}
m + \prod_{s=1}^kd_s-\prod_{s=2}^kd_s - \prod_{s=3}^kd_s &= \left((0)\prod_{s=2}^kd_s+(0)\prod_{s=3}^kd_s + (d_3-(i_3)-1)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\right)\\
&+\prod_{s=1}^kd_s-\prod_{s=2}^kd_s - \prod_{s=3}^kd_s\\
&= \prod_{s=1}^kd_s-\prod_{s=2}^kd_s - \prod_{s=3}^kd_s + (d_3-i_3-2)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
\end{align*}
\begin{align*}
m+ (d_1-1)\prod_{s=2}^kd_s - \prod_{s=3}^kd_s &=\left((0)\prod_{s=2}^kd_s+(0)\prod_{s=3}^kd_s + (d_3-(i_3)-1)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\right)\\
& +(d_1-1)\prod_{s=2}^kd_s - \prod_{s=3}^kd_s\\
\end{align*}
as $i_1=d_1-1$ and $i_2=d_2-1$ and $i_3<d_3-1$
\begin{align*}
m&= (d_3-1-i_3)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
m-\prod_{l=2}^kd_l&= -\prod_{l=2}^kd_l+ (d_3-1-i_3)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s\\
\end{align*}
\begin{align*}
(d_3-1-i_3)&\ge 1\\
(d_3-1-i_3)\prod_{s=4}d_s &\ge \prod_{s=4}d_s \\
\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s &\ge 0 \\
\prod_{l=3}^k d_l \ge (d_3-1-i_3)\prod_{s=4}d_s+\sum_{j=4}^k(d_j-i_j-1)\prod_{s=j+1}^kd_s &\ge \prod_{s=4}d_s \\
\prod_{l=3}^k d_l -1 \ge m &\ge \prod_{s=4}d_s \\
0 > \prod_{l=3}^k d_l -1 -\prod_{l=2}^k d_l \ge m-\prod_{l=2}^k d_l &\ge \prod_{s=4}d_s - \prod_{l=2}^k d_l \\
(d_1-1)\prod_{l=2}^k+\prod_{l=3}^k d_l \ge m +(d_1-1)\prod_{l=2}^k -1 &\ge (d_1-1)\prod_{l=2}^k+ \prod_{s=4}d_s \\
% (1-d_2)\prod_{l=3}^k d_l \ge m-\prod_{l=2}^k d_l &\ge \prod_{s=4}d_s - \prod_{l=2}^k d_l \\
% (1-d_2)\prod_{l=3}^k d_l \ge m-\prod_{l=2}^k d_l &\ge \prod_{s=4}d_s - \prod_{s=3}d_s + \prod_{s=3}d_s - \prod_{l=2}^k d_l \\
% (1-d_2)\prod_{l=3}^k d_l \ge m-\prod_{l=2}^k d_l &\ge (1-d_3)\prod_{s=4}d_s + (1-d_2) \prod_{s=3}d_s \\
\end{align*}
Perhaps, it's a little endian rewrite in big endian
For any $i\in \mathbb{N} , i<n=\prod_{j=1}^kd_j \exists! (i_1, \ldots, i_k) / \forall 1 \le j \le k, 0\le i_j<d_j$ and $$ i=\sum_{j=1}^{k} i_j\left( \prod_{l=1}^{j-1} d_l \right) $$
The inverse of big endian :
For any $i\in \mathbb{N} , i<n=\prod_{j=1}^kd_j \exists! (i_1, \ldots, i_k) / \forall 1 \le j \le k, 0\le i_j<d_j$ and $$ i=\sum_{s=1}^k i_s\left( \prod_{l=s+1}^k d_l \right) $$
\begin{align*}
\sum_{j=1}^{k} i_j\left( \prod_{l=1}^{j-1} d_l \right) +1
\end{align*}
\begin{align*}
l=0 &\Rightarrow m=0 \\
l=1 &\Rightarrow m=\prod_{l=2}^kd_l \\
l=2 &\Rightarrow m=2\prod_{l=2}^kd_l \\
\vdots\\
l=d_1-1 &\Rightarrow m=(d_1-1)\prod_{l=2}^kd_l = \prod_{l=1}d_l - \prod_{l=2}^k d_l \\
l=d_1 &\Rightarrow m= \prod_{l=3}^kd_l \\
l=d_1+1 &\Rightarrow m= \prod_{l=2}^kd_l + \prod_{l=3}^kd_l \\
l=d_1+2 &\Rightarrow m= 2 \prod_{l=2}^kd_l + \prod_{l=3}^kd_l \\
\vdots \\
l=d_1+d_1-1 = 2 d_1-1 &\Rightarrow m= (d_1-1) \prod_{l=2}^kd_l + \prod_{l=3}^kd_l \\
l=d_1+d_1 = 2d_1 &\Rightarrow m= 2 \prod_{l=3}^kd_l \\
l=2d_1+1 &\Rightarrow m= \prod_{l=2}^kd_l + 2\prod_{l=3}^kd_l \\
l=2d_1+2 &\Rightarrow m= 2 \prod_{l=2}^kd_l + 2\prod_{l=3}^kd_l \\
\end{align*}
\section{Examples}
In these examples, we note ${\alpha}_{(i_1,\ldots,i_k)}^{\left[n_{\left(i_1,\ldots,i_k\right)}\right]} $ such that $\left(i_1,\ldots,i_k\right)\in\mathbb{N}^k $ and $n_{(i_1,\ldots,i_k)}\in \mathbb{N} $, the index of tensor of order $k$ is $\left(i_1,\ldots,i_k \right)$ and the correspondant index in tensor of order $1$ is $\left[n_{\left(i_1,\ldots,i_k\right)}\right]$
In these examples, we note ${\alpha}_{(i_1,\ldots,i_k)}^{\left[n_{\left(i_1,\ldots,i_k\right)}\right]} $ such that $\left(i_1,\ldots,i_k\right)\in\mathbb{N}^k $ and $n_{(i_1,\ldots,i_k)}\in \mathbb{N} $, the index of tensor of order $k$ is $\left(i_1,\ldots,i_k \right)$ and the correspondant index in tensor of order $1$ is $\left[n_{\left(i_1,\ldots,i_k\right)}\right]$
\hrule
\vspace{1cm}
Big endian $k=2$
@@ -461,4 +839,6 @@ little endian $k=4$
\end{tikzpicture}
\fi
\end{document}