diff --git a/Exercises/Session 2 - Tensor Decomposition (with solutions).ipynb b/Exercises/Session 2 - Tensor Decomposition (with solutions).ipynb
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-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "e4f44eb7",
-   "metadata": {},
-   "source": [
-    "$\\def\\tcoreleft{\\underset{\\tiny\\mid}{\\textcolor{MidnightBlue}{⦸}}}$\n",
-    "$\\def\\tcorecenter{\\underset{\\tiny\\mid}{\\textcolor{RedOrange}{⦿}}}$\n",
-    "$\\def\\tcoreright{\\underset{\\tiny\\mid}{\\textcolor{MidnightBlue}{\\oslash}}}$\n",
-    "<h1 style=\"text-align: center;\"><b>TMQS Workshop 2024</b> @ Zuse Institute Berlin</h1>\n",
-    "<h2 style=\"text-align: center;\">Summer School on Tensor Methods for Quantum Simulation</h2>\n",
-    "<h2 style=\"text-align: center;\">June 3 - 5, 2024</h2>\n",
-    "<h1 style=\"text-align: center; background-color:#D6EAF8 ;padding:50px\">$\\tcoreleft - \\tcoreleft - \\tcoreleft - \\cdots - \\tcorecenter - \\cdots - \\tcoreright - \\tcoreright$</h1>\n",
-    "</br>"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "58a30939-f570-45cd-a736-d6f21aeb2a0c",
-   "metadata": {},
-   "source": [
-    "***"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "6c3441f3-b5a9-42c4-ba21-2bc682b0d8ac",
-   "metadata": {},
-   "source": [
-    "## **Session 2 - Tensor Decomposition**"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "6f6cc702",
-   "metadata": {},
-   "source": [
-    "***"
-   ]
-  },
-  {
-   "attachments": {},
-   "cell_type": "markdown",
-   "id": "858a33fe-e16e-4dbb-b369-511d7c71fe71",
-   "metadata": {},
-   "source": [
-    "## Exercise 2.1\n",
-    "\n",
-    "In general, the question of whether a given tensor can be represented in a low-rank form is very challenging. The following (small!) example shall illustrate this.\n",
-    "\n",
-    "**a)**$\\quad$Consider the order-3 tensor $T \\in \\mathbb{R}^{2 \\times 2 \\times 2}$ given by\n",
-    "\n",
-    "$\\hspace{1.25cm}$$T_{:, :, 1} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1\\end{pmatrix}$ and $T_{:, :, 2} = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0\\end{pmatrix}$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$Find a canonical decomposition with minimum rank!\n",
-    "\n",
-    "$\\hspace{0.75cm}$*Hint:* Both matrices are circulant. That is, they share the same eigenvectors (Fourier modes!).\n",
-    "\n",
-    "**b)**$\\quad$*extra task:* Let $T \\in \\mathbb{R}^{n \\times n \\times n}$ with $T_{:,:,k} = S^{k-1}$, where $S = \\begin{pmatrix} 0 & & & 1 \\\\ 1 & 0 & & \\\\ & \\ddots & \\ddots & \\\\ 0 & & 1 & 0 \\end{pmatrix}$. What is the canonical rank of $T$?"
-   ]
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-  {
-   "cell_type": "markdown",
-   "id": "bc6abad4-3b47-447c-9e87-86b44381164b",
-   "metadata": {},
-   "source": [
-    "***"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "11976a26-d5e7-4ca5-b02b-ce4709527c8c",
-   "metadata": {},
-   "source": [
-    "$\\textcolor{red}{\\textbf{SOLUTION:}}$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "742e700e-8485-4923-870b-d34994d2978b",
-   "metadata": {},
-   "source": [
-    "**a)**$\\quad$The (normalized) eigenvectors of $T_{:, :, 1}$ and $T_{:, :, 2}$ are given by $v_0 = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}1\\\\1\\end{pmatrix}$ and $v_1 = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}1\\\\-1\\end{pmatrix}$. \n",
-    "\n",
-    "$\\hspace{0.75cm}$The eigenvalues of $T_{:, :, 1}$ are $\\lambda_0 = \\lambda_1 =1$ and the eigenvalues of $T_{:, :, 2}$ are $\\mu_0 = 1$ and $\\mu_1 = -1$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$That is, $T_{:, :, 1} = \\sum_{i=0}^1 \\lambda_i v_i v_i^\\dagger$ and $T_{:, :, 2} = \\sum_{i=0}^1 \\mu_i v_i v_i^\\dagger$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$Since $v_i v_i^\\dagger = v_i \\otimes \\overline{v_i} = v_i \\otimes v_i$, we get\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle T = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1\\end{pmatrix} \\otimes \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} +  \\begin{pmatrix} 0 & 1 \\\\ 1 & 0\\end{pmatrix} \\otimes \\begin{pmatrix}0 \\\\ 1\\end{pmatrix}$\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle\\phantom{T} = \\sum_{i=0}^1 \\lambda_i v_i \\otimes v_i \\otimes \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} +  \\sum_{i=0}^1 \\mu_i v_i \\otimes v_i \\otimes \\begin{pmatrix}0 \\\\ 1\\end{pmatrix}$\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle\\phantom{T} = \\sum_{i=0}^1  v_i \\otimes v_i \\otimes \\begin{pmatrix} \\lambda_i \\\\ 0 \\end{pmatrix} +  \\sum_{i=0}^1 v_i \\otimes v_i \\otimes \\begin{pmatrix}0 \\\\ \\mu_i\\end{pmatrix}$\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle\\phantom{T} = \\sum_{i=0}^1  v_i \\otimes v_i \\otimes \\begin{pmatrix} \\lambda_i \\\\ \\mu_i \\end{pmatrix} $\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle\\phantom{T} = \\frac{1}{2}  \\begin{pmatrix} 1\\\\1 \\end{pmatrix}  \\otimes \\begin{pmatrix} 1\\\\1 \\end{pmatrix} \\otimes \\begin{pmatrix} 1\\\\1 \\end{pmatrix} + \\frac{1}{2} \\begin{pmatrix} 1\\\\-1 \\end{pmatrix} \\otimes \\begin{pmatrix} 1\\\\-1 \\end{pmatrix} \\otimes \\begin{pmatrix} 1\\\\-1 \\end{pmatrix}$\n",
-    "\n",
-    "Note that there exists no rank-$1$ decomposition of $T$. This means the canonical rank of $T$ is equal to $2$.\n",
-    "\n",
-    "**b)**$\\quad$Again, all \"layers\" $T_{:,:,k}$ have the same set of (complex) eigenvectors $v_0, \\dots, v_{n-1}$ and can be written as \n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle T_{:,:,k} = \\sum_{i=0}^{n-1} \\lambda_i^{(k)} v_i \\otimes \\overline{v_i}$, \n",
-    "\n",
-    "$\\hspace{0.75cm}$where $\\lambda_i^{(k)}$ is the $i$th eigenvalue of $T_{:,:,k}$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$Thus, we get \n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle T = \\sum_{i=0}^{n-1} v_i \\otimes \\overline{v_i} \\otimes \\begin{pmatrix} \\lambda_i^{(1)} \\\\ \\vdots \\\\ \\lambda_i^{(n-1)}\\end{pmatrix}$.\n"
-   ]
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-   "cell_type": "markdown",
-   "id": "57638421-d883-4f7f-ab0f-aae0505cffca",
-   "metadata": {},
-   "source": [
-    "***"
-   ]
-  },
-  {
-   "attachments": {},
-   "cell_type": "markdown",
-   "id": "c7140595-9875-4201-ad30-e186a6722ff8",
-   "metadata": {},
-   "source": [
-    "## Exercise 2.2 \n",
-    "\n",
-    "We now want to consider a very well-known model in statistical physics: the Ising model. The Ising model is a simple mathematical model used to study phase transitions in magnetic systems. It consists of a lattice arrangement of spins, each of which can point \"up\" or \"down\", and interactions between neighboring spins.\n",
-    "\n",
-    "The Hamiltonian function (i.e., the energy of a configuration) for the Ising model can be expressed as follows:\n",
-    "\n",
-    "$\\hspace{1cm}$$ \\displaystyle H(\\sigma )=- J \\sum _{\\langle ij\\rangle }\\sigma _{i}\\sigma _{j}-h \\sum _{j} \\sigma _{j}$.\n",
-    "\n",
-    "Here, $\\sigma_i$ and $\\sigma_j$ represent the spins at lattice points $i$ and $j$, respectively, where $1 \\leq i,j \\leq d$. $J$ is the interaction constant between spins, $h$ represents the external magnetic field, and the summation is over neighboring lattice points $i$ and $j$.\n",
-    "\n",
-    "Since this is a system with nearest-neighbor interactions, we can represent it using a (non-cyclic) SLIM decomposition. \n",
-    "\n",
-    "**a)**$\\quad$Find the vectors $S$, $L$, $I$, and $M$ such that the tensor train defined by\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\displaystyle \\mathbf{H} = \\begin{bmatrix} S & L & I\\end{bmatrix} \\otimes \\begin{bmatrix} I & 0& 0\\\\ M & 0& 0\\\\ S & L & I \\end{bmatrix} \\otimes \\dots \\otimes \\begin{bmatrix} I & 0& 0\\\\ M & 0& 0\\\\ S & L & I \\end{bmatrix} \\otimes \\begin{bmatrix} I\\\\ M\\\\ S\\end{bmatrix}$\n",
-    "\n",
-    "$\\hspace{0.75cm}$satisfies $\\mathbf{H}_{\\nu_1, \\dots, \\nu_d} = H(\\sigma_1, \\dots, \\sigma_d)$, where\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\nu_i = \\begin{cases} 1 & \\text{if}~\\sigma_i = +1, \\\\ 2 & \\text{if}~\\sigma_i=-1.\\end{cases}$\n",
-    "\n",
-    "$\\hspace{0.75cm}$Compare your result with Ising model representation provided by Scikit-TT (use $J=h=1$ and $d=5$):\n",
-    "\n",
-    "$\\hspace{1.25cm}$1) Import ```scikit_tt.tensor_train``` as ```tt```.\n",
-    "\n",
-    "$\\hspace{1.25cm}$2) Use tt.build_core for constructing the cores of $\\mathbf{H}$, e.g., ```cores[i] = tt.build_core([[I, 0, 0], [M, 0, 0], [S, L, I]])``` for $i=1,2,3$.\n",
-    "\n",
-    "$\\hspace{1.25cm}$3) Create the class instance by  ```H = TT(cores)```.\n",
-    "\n",
-    "$\\hspace{1.25cm}$4) Import ```scikit_tt.models``` as ```mdl``` and load the model from Scikit-TT by ```H_SKTT = mdl.ising(d,J,h)```.\n",
-    "\n",
-    "$\\hspace{1.25cm}$5) Compute the error in the Frobenius norm: ```(H-H_SKTT).norm()```.\n",
-    "\n",
-    "**b)**$\\quad$*extra task:* The configuration probability, i.e., the probability that the system is in a state $\\sigma$ in equilibrium, is given by\n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\pi(\\sigma) = \\frac{e^{- \\beta H(\\sigma)}}{Z}$,\n",
-    "\n",
-    "$\\hspace{0.75cm}$where $Z = \\sum_{\\sigma} e^{-\\beta H(\\sigma)}$ is a normalization constant and $\\beta = 1/(k_B \\cdot T)$ is the inverse temperature with Boltzmann constant $k_B$. \n",
-    "\n",
-    "$\\hspace{0.75cm}$Suppose $\\beta=1$, can you derive a TT decomposition for $\\pi$?"
-   ]
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-    "***"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "1cb59312-1033-4dd0-886c-7a034eae661b",
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-   "source": [
-    "$\\textcolor{red}{\\textbf{SOLUTION:}}$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "81208cd9-c190-46c9-95f8-d0d23778ebce",
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-   "source": [
-    "**a)**$\\quad$The component $S$ represents the single-cell interactions, i.e., $S = -h\\begin{pmatrix}+1 \\\\-1\\end{pmatrix}$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$$L \\otimes M$ represents the two-cell (nearest-neighbor) interactions, i.e., $L = -J\\begin{pmatrix}+1 \\\\-1\\end{pmatrix}$ and $M = \\begin{pmatrix}+1 \\\\-1\\end{pmatrix}$.\n",
-    "\n",
-    "$\\hspace{0.75cm}$$I$ is simply the vector $\\begin{pmatrix}1 \\\\1\\end{pmatrix}$.\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "id": "21650b1d-183d-403f-80d5-9f27c21e3c69",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "Tensor train with order    = 10, \n",
-      "                  row_dims = [2, 2, 2, 2, 2, 2, 2, 2, 2, 2], \n",
-      "                  col_dims = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], \n",
-      "                  ranks    = [1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1] \n",
-      "\n",
-      "error: 1.9946147148160272e-13\n"
-     ]
-    }
-   ],
-   "source": [
-    "import numpy as np\n",
-    "import scikit_tt.tensor_train as tt\n",
-    "from scikit_tt.tensor_train import TT\n",
-    "import scikit_tt.models as mdl\n",
-    "\n",
-    "J = 1\n",
-    "h = 1\n",
-    "d = 10\n",
-    "\n",
-    "S = -h *np.array([1,-1])\n",
-    "L = -J *np.array([1,-1])\n",
-    "I = np.array([1,1])\n",
-    "M = np.array([1,-1])\n",
-    "\n",
-    "cores = [None]*d\n",
-    "cores[0] = tt.build_core([[S, L, I]])\n",
-    "for i in range(1,d-1):\n",
-    "    cores[i] = tt.build_core([[I, 0, 0], [M, 0, 0], [S, L, I]])\n",
-    "cores[-1] = tt.build_core([I, M, S])\n",
-    "\n",
-    "H = TT(cores)\n",
-    "print(H, '\\n')\n",
-    "\n",
-    "H_SKTT = mdl.ising(d,J,h)\n",
-    "\n",
-    "print('error:', (H-H_SKTT).norm())"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "4b3280ac-c09b-4d73-a4f7-0e46ab49cbce",
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-   "source": [
-    "**b)**$\\quad$The (unnormalized) stationary distribution can be expressed as \n",
-    "\n",
-    "$\\hspace{1.25cm}$$\\mathbf{\\Pi} = \\begin{bmatrix} \\begin{pmatrix} e^\\beta \\\\ 0 \\end{pmatrix} & \\begin{pmatrix} 0 \\\\ e^{-\\beta} \\end{pmatrix} \\end{bmatrix} \\begin{bmatrix} \\begin{pmatrix} e^{2\\beta} \\\\ 0 \\end{pmatrix} & \\begin{pmatrix} 0 \\\\ e^{-2\\beta} \\end{pmatrix} \\\\[0.4cm] \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} & \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\end{bmatrix} \\otimes \\ldots \\otimes \\begin{bmatrix} \\begin{pmatrix} e^{2\\beta} \\\\ 0 \\end{pmatrix} & \\begin{pmatrix} 0 \\\\ e^{-2\\beta} \\end{pmatrix} \\\\[0.4cm] \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} & \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\end{bmatrix} \\otimes \\begin{bmatrix} \\begin{pmatrix} e^{2\\beta} \\\\ e^{-2\\beta} \\end{pmatrix}  \\\\[0.4cm]  \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\\end{bmatrix}$."
-   ]
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-   "cell_type": "markdown",
-   "id": "179c2c04-c06d-4478-9eca-b662e7fa21ac",
-   "metadata": {},
-   "source": [
-    "***"
-   ]
-  },
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-   "source": [
-    "## Exercise 2.3\n",
-    "\n",
-    "You've already seen that any arbitrary tensor can be converted into the TT format. Let's try this out practically now. Create a random tensor with order $d$ and mode dimensions $n$:\n",
-    "\n",
-    "> T = np.random.random_sample(d*[n])\n",
-    "\n",
-    "First, we reshape $T$ into a matrix, thereby defining the \"residual tensor\" $T_{\\text{res}}$:\n",
-    "\n",
-    "> T_res = T.reshape([n, n**(d-1)])\n",
-    "\n",
-    "Now we need to write a for loop in which, at each step, one mode is separated, and the remaining matrix $T_\\text{res}$ is reshaped.\n",
-    "\n",
-    "Assume $T_\\text{res}$ is a $r*n \\times n^e$ matrix with $r \\in \\mathbb{N}$ and $e \\leq d-1$.\n",
-    "\n",
-    "To isolate the mode, we apply a (reduced) QR decomposition. The matrix $Q$ then forms the TT core for the separated mode, and the matrix $R$ represents the new residual tensor:\n",
-    "\n",
-    "> Q, R = np.linalg.qr(T_res)\n",
-    "> \n",
-    "> s = Q.shape[1]\n",
-    "> \n",
-    "> core = Q.reshape([r, n, 1, s])\n",
-    "> \n",
-    "> T_res = R.reshape([s * n, n**(e-1)])\n",
-    "\n",
-    "Try it for various values ​​of d and n, but initially don't choose them too large. What do you notice when you look at the TT ranks?"
-   ]
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-   "source": [
-    "***"
-   ]
-  },
-  {
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-   "id": "fb078531-7c5e-45f6-b9f2-631d27ad8b7d",
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-   "source": [
-    "$\\textcolor{red}{\\textbf{SOLUTION:}}$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "id": "b5325236-1967-4cae-8d61-4219d477fd2e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "Tensor train with order    = 10, \n",
-      "                  row_dims = [3, 3, 3, 3, 3, 3, 3, 3, 3, 3], \n",
-      "                  col_dims = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], \n",
-      "                  ranks    = [1, 3, 9, 27, 81, 243, 81, 27, 9, 3, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "import numpy as np\n",
-    "\n",
-    "d=10\n",
-    "n=3\n",
-    "cores = [None]*d\n",
-    "\n",
-    "T = np.random.random_sample(d*[n])\n",
-    "T_res = T.copy()\n",
-    "r = 1\n",
-    "for i in range(d-1):\n",
-    "    e = d-i-1\n",
-    "    T_res = T_res.reshape([r*n, n**e])\n",
-    "    Q, R = np.linalg.qr(T_res)\n",
-    "    s = Q.shape[1]\n",
-    "    cores[i] = Q.reshape([r, n, 1, s])\n",
-    "    T_res = R.reshape([s * n, n**(e-1)])\n",
-    "    r = s\n",
-    "cores[-1] = R.reshape([r, n, 1, 1])\n",
-    "\n",
-    "T = TT(cores)\n",
-    "\n",
-    "print(T)"
-   ]
-  }
- ],
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