update: try reverse transpositiongit status

onedimensiontensor.tex

@@ -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}