# The ALS Algorithm

Implementing the Alternating Least Squares (ALS) algorithm (also known as single-site DMRG) for the first time was the most important step for us to understand the TT format and its intricacies. We still think that it is a good point to start so we want to provide a simple implementation of the ALS algorithm as an example. Using the xerus library this will be an efficient implementation using only about 100 lines of code (without comments).

## Introduction

The purpose of this page is not to give a full derivation of the ALS algorithm. The interested reader is instead refered to the original publications on the matter. Let us just shortly recap the general idea of the algorithm to refresh your memory though.

Solving least squares problems of the form $\operatorname{argmin}_x \|Ax - b\|^2$ for large dimensions is a difficult endeavour. Even if $x$ and $b$ are given in the TT-Tensor and $A$ in the TT-Operator format with small ranks this is far from trivial. There is a nice property of the TT format though, that we can use to construct a Gauss-Seidel-like iterative scheme: the linear dependence of the represented tensor on all of its component tensors. Due to this linearity, we can formulate a smaller subproblem: fixing all but one components of $x$, the resulting minimization problem is again of the form of a least squares problem as above but with a projected $\hat b = Pb$ and with a smaller $\hat A=PAP^T$. In practice, these projections can be obtained by simply contracting all fixed components of $x$ to $A$ and $b$ (assuming all fixed components are orthogonolized). The ALS algorithm will now simply iterate over the components of $x$ and solve these smaller subproblems.

There are a few things we should note before we start implementing this algorithm

• It is enough to restrict ourselves to the case of symmetric positive-semidefinite operators $A$. Any non-symmetric problem can be solved by setting $A’=A^TA$ and $b’ = A^Tb$. We can thus use the identity $% -2\left %]]>$
• We should always move our core of $x$ to the position currrently being optimized to make our lives easier (for several reasons…).
• Calculating the local operators $\hat A$ for components $i$ and $i+1$ is highly redundant. All components of $x$ up to the $i-1$’st have to be contracted with $A$ in both cases. Effectively this means, that we will keep stacks of $x^TAx$ contracted up to the current index (“left” of the current index) as well as contracted at all indices above the currrent one (“right” of it) and similarly for $x^T b$.

## Pseudo-Code

Let us start by writing down the algorithm in pseudo-code and then fill out the steps one by one.

// while we are not done
// for every position p = 0..degree(x) do
// local operator = left stack(A) * p'th component of A * right stack(A)
// local rhs = left stack(b) * p'th component of b * right stack(b)
// p'th component of x = solution of the local least squares problem

// remove top entry of the right stacks
// add position p to left stacks

// for every position p = degree(x)..0 do
// same as above OR simply move core and update stacks

# while we are not done
# for every position p = 0..degree(x) do
# local operator = left stack(A) * p'th component of A * right stack(A)
# local rhs = left stack(b) * p'th component of b * right stack(b)
# p'th component of x = solution of the local least squares problem

# remove top entry of the right stacks
# add position p to left stacks

# for every position p = degree(x)..0 do
# same as above OR simply move core and update stacks


## Helper Class

We want our main loop to resemble the above pseudo code as closely as possible, so we have to define some helper functions to update the stacks. To ensure that all functions work on the same data without passing along references all the time, we will define a small helper class, that holds all relevant variables: the degree d of our problem, the left and right stacks for A and b, the TT tensors A, x and b themselves and the norm of b. As a parameter of the algorithm we will also store the maximal number of iterations

class InternalSolver {
const size_t d;

std::vector<Tensor> leftAStack;
std::vector<Tensor> rightAStack;

std::vector<Tensor> leftBStack;
std::vector<Tensor> rightBStack;

TTTensor& x;
const TTOperator& A;
const TTTensor& b;
const double solutionsNorm;
public:
size_t maxIterations;

InternalSolver(const TTOperator& _A, TTTensor& _x, const TTTensor& _b)
: d(_x.degree()), x(_x), A(_A), b(_b), solutionsNorm(frob_norm(_b)), maxIterations(1000)
{
leftAStack.emplace_back(Tensor::ones({1,1,1}));
rightAStack.emplace_back(Tensor::ones({1,1,1}));
leftBStack.emplace_back(Tensor::ones({1,1}));
rightBStack.emplace_back(Tensor::ones({1,1}));
}

class InternalSolver :
def __init__(self, A, x, b):
self.A = A
self.b = b
self.x = x
self.d = x.degree()
self.solutionsNorm = b.frob_norm()

self.leftAStack = [ xe.Tensor.ones([1,1,1]) ]
self.leftBStack = [ xe.Tensor.ones([1,1]) ]
self.rightAStack = [ xe.Tensor.ones([1,1,1]) ]
self.rightBStack = [ xe.Tensor.ones([1,1]) ]

self.maxIterations = 1000


Note here, that we initialized the stacks with tensors of dimensions $1\times 1\times 1$ (respectively $1\times 1$). By doing this we won’t have to distinguish between the first, last or any other component that is being optimized. As the first and last component of a TT Tensor already have an additional mode of dimension 1 in xerus, we can simply contract them to these tensors containing a 1 entry as if they were any of the middle components.

## Updating the Stacks

To add the next entry to the left (right) stacks we have to contract the next components to the previous results. To increase readability of the equations, we first store references to the $p$’th components in Ai, xi and bi before we contract them using xerus’s indexed equations.

	void push_left_stack(const size_t _position) {
Index i1, i2, i3, j1 , j2, j3, k1, k2;
const Tensor &xi = x.get_component(_position);
const Tensor &Ai = A.get_component(_position);
const Tensor &bi = b.get_component(_position);

Tensor tmpA, tmpB;
tmpA(i1, i2, i3) = leftAStack.back()(j1, j2, j3)
*xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3);
leftAStack.emplace_back(std::move(tmpA));
tmpB(i1, i2) = leftBStack.back()(j1, j2)
*xi(j1, k1, i1)*bi(j2, k1, i2);
leftBStack.emplace_back(std::move(tmpB));
}

void push_right_stack(const size_t _position) {
Index i1, i2, i3, j1 , j2, j3, k1, k2;
const Tensor &xi = x.get_component(_position);
const Tensor &Ai = A.get_component(_position);
const Tensor &bi = b.get_component(_position);

Tensor tmpA, tmpB;
tmpA(i1, i2, i3) = xi(i1, k1, j1)*Ai(i2, k1, k2, j2)*xi(i3, k2, j3)
*rightAStack.back()(j1, j2, j3);
rightAStack.emplace_back(std::move(tmpA));
tmpB(i1, i2) = xi(i1, k1, j1)*bi(i2, k1, j2)
*rightBStack.back()(j1, j2);
rightBStack.emplace_back(std::move(tmpB));
}

	def push_left_stack(self, pos) :
i1,i2,i3, j1,j2,j3, k1,k2 = xe.indices(8)
Ai = self.A.get_component(pos)
xi = self.x.get_component(pos)
bi = self.b.get_component(pos)

tmpA = xe.Tensor()
tmpB = xe.Tensor()
tmpA(i1, i2, i3) << self.leftAStack[-1](j1, j2, j3)\
*xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3)
self.leftAStack.append(tmpA)
tmpB(i1, i2) << self.leftBStack[-1](j1, j2)\
*xi(j1, k1, i1)*bi(j2, k1, i2)
self.leftBStack.append(tmpB)

def push_right_stack(self, pos) :
i1,i2,i3, j1,j2,j3, k1,k2 = xe.indices(8)
Ai = self.A.get_component(pos)
xi = self.x.get_component(pos)
bi = self.b.get_component(pos)

tmpA = xe.Tensor()
tmpB = xe.Tensor()
tmpA(j1, j2, j3) << xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3) \
* self.rightAStack[-1](i1, i2, i3)
self.rightAStack.append(tmpA)
tmpB(j1, j2) << xi(j1, k1, i1)*bi(j2, k1, i2) \
* self.rightBStack[-1](i1, i2)
self.rightBStack.append(tmpB)


We will also use a small helper function to calculate the residual norm of the current solution:

	double calc_residual_norm() {
Index i,j;
return frob_norm(A(i/2, j/2)*x(j&0) - b(i&0)) / solutionsNorm;
}

	def calc_residual_norm(self) :
i,j = xe.indices(2)
return xe.frob_norm(self.A(i/2, j/2)*self.x(j&0) - self.b(i&0)) / self.solutionsNorm


## The Main Loop

With these helper functions we can now start to write the main optimization loop. Before we start with the optimization though we have to ensrure that the core of x is at the right position and create the full right stacks - with the above helper functions this has become very easy:

	void solve() {
x.move_core(0, true);
for (size_t pos = d-1; pos > 0; --pos) {
push_right_stack(pos);
}
// ...
}

	def solve(self) :
self.x.move_core(0, True)
for pos in reversed(xrange(1, self.d)) :
self.push_right_stack(pos)

# ...


The pseudo code now starts with “while we are not done”. We have already created a maxIterations variable to decide whether we are “done”, but we typically also want an endcriterion based on the current residual. For the purpose of this example we will simply check, whether the residual decreased by at least 1% within the last 10 sweeps. To do this, we have to keep a record of the residuals after every sweep:

	void solve() {
// ...
std::vector<double> residuals(10, 1000.0);

for (size_t itr = 0; itr < maxIterations; ++itr) {
residuals.push_back(calc_residual_norm());

if (residuals.back()/residuals[residuals.size()-10] > 0.99) {
return;
}

// ...
}
}

	def solve(self) :
# ...
residuals = [1000]*10

for itr in xrange(self.maxIterations) :
residuals.append(self.calc_residual_norm())

if residuals[-1]/residuals[-10] > 0.99 :
return

# ...


The individual iterations now consist of a sweep from left to right and one back to the left. For every position we will construct the local operator and right-hand-side via indexed equations, solve this local problem and store the result in the corresponding component of x

for (size_t corePosition = 0; corePosition < d; ++corePosition) {
Tensor op, rhs;

const Tensor &Ai = A.get_component(corePosition);
const Tensor &bi = b.get_component(corePosition);

op(i1, i2, i3, j1, j2, j3) = leftAStack.back()(i1, k1, j1)
* Ai(k1, i2, j2, k2)
* rightAStack.back()(i3, k2, j3);
rhs(i1, i2, i3) = leftBStack.back()(i1, k1)
* bi(k1, i2, k2)
* rightBStack.back()(i3, k2);

xerus::solve(x.component(corePosition), op, rhs, 0);

// move core and update stacks
// ...
}


for pos in xrange(self.d):
op = xe.Tensor()
rhs = xe.Tensor()

Ai = self.A.get_component(pos)
bi = self.b.get_component(pos)

op(i1, i2, i3, j1, j2, j3) << self.leftAStack[-1](i1, k1, j1) \
* Ai(k1, i2, j2, k2) \
* self.rightAStack[-1](i3, k2, j3)
rhs(i1, i2, i3) << self.leftBStack[-1](i1, k1) \
* bi(k1, i2, k2) \
* self.rightBStack[-1](i3, k2)

tmp = xe.Tensor()
tmp(i1&0) << rhs(j1&0) / op(j1/2, i1/2)
self.x.set_component(pos, tmp)

# move core and update stacks
# ...


To prepare for the next position we simply have to move the core, pop the top element of the right stacks and add the new element to the left stacks (unless this already was the last position of course!)

// move core and update stacks
if (corePosition+1 < d) {
x.move_core(corePosition+1, true);
push_left_stack(corePosition);
rightAStack.pop_back();
rightBStack.pop_back();
}

# move core and update stacks
if pos+1 < self.d :
self.x.move_core(pos+1, True)
self.push_left_stack(pos)
self.rightAStack.pop()
self.rightBStack.pop()


For the sweep from right to left we could now write a similar loop, for the sake of simplicity we will just move the core and stacks back to the left though (it seems to depend on the example which of these two variants is faster). So the full iteration looks as follows:

for (size_t itr = 0; itr < maxIterations; ++itr) {
residuals.push_back(calc_residual_norm());

if (residuals.back()/residuals[residuals.size()-10] > 0.99) {
return; // We are done!
}

// Sweep Left -> Right
for (size_t corePosition = 0; corePosition < d; ++corePosition) {
Tensor op, rhs;

const Tensor &Ai = A.get_component(corePosition);
const Tensor &bi = b.get_component(corePosition);

op(i1, i2, i3, j1, j2, j3) = leftAStack.back()(i1, k1, j1)
* Ai(k1, i2, j2, k2)
* rightAStack.back()(i3, k2, j3);
rhs(i1, i2, i3) = leftBStack.back()(i1, k1)
* bi(k1, i2, k2)
* rightBStack.back()(i3, k2);

xerus::solve(x.component(corePosition), op, rhs);

if (corePosition+1 < d) {
x.move_core(corePosition+1, true);
push_left_stack(corePosition);
rightAStack.pop_back();
rightBStack.pop_back();
}
}

// Sweep Right -> Left : only move core and update stacks
x.move_core(0, true);
for (size_t corePosition = d-1; corePosition > 0; --corePosition) {
push_right_stack(corePosition);
leftAStack.pop_back();
leftBStack.pop_back();
}

}

for itr in xrange(self.maxIterations) :
residuals.append(self.calc_residual_norm())
if residuals[-1]/residuals[-10] > 0.99 :
return

# sweep left -> right
for pos in xrange(self.d):
op = xe.Tensor()
rhs = xe.Tensor()

Ai = self.A.get_component(pos)
bi = self.b.get_component(pos)

op(i1, i2, i3, j1, j2, j3) << self.leftAStack[-1](i1, k1, j1) \
* Ai(k1, i2, j2, k2) \
* self.rightAStack[-1](i3, k2, j3)
rhs(i1, i2, i3) << self.leftBStack[-1](i1, k1) \
* bi(k1, i2, k2) \
* self.rightBStack[-1](i3, k2)

tmp = xe.Tensor()
tmp(i1&0) << rhs(j1&0) / op(j1/2, i1/2)
self.x.set_component(pos, tmp)

if pos+1 < self.d :
self.x.move_core(pos+1, True)
self.push_left_stack(pos)
self.rightAStack.pop()
self.rightBStack.pop()

# right -> left, only move core and update stack
self.x.move_core(0, True)
for pos in reversed(xrange(1,self.d)) :
self.push_right_stack(pos)
self.leftAStack.pop()
self.leftBStack.pop()


## Finishing Touches

All we need now is a smal function that creates an object of our helper class and calls the .solve() method

void simpleALS(const TTOperator& _A, TTTensor& _x, const TTTensor& _b)  {
InternalSolver solver(_A, _x, _b);
return solver.solve();
}

def simpleALS(A, x, b) :
solver = InternalSolver(A, x, b)
solver.solve()


To see what is happening in the algorithm we can also add some output to the iterations along the lines of

XERUS_LOG(simpleALS, "Iteration: " << itr << " Residual: " << residuals.back());

print("Iteration:",itr, "Residual:", residuals[-1])


## Testing our Implementation

To test our freshly implemented ALS, we can use a random TT-Operator (after symmetrizing it via $A’=A^T A$). By constructing a random low-rank solution $s$ and providing the right-hand-side $b=A’ s$ to the algorithm, we can ensure that a low-rank solution actually exists. As a starting point we will furthermore provide a random initial TT tensor x of the correct rank.

As we know the correct solution via this construction and the random operator A is invertible with high probability, we can then calculate the residual and the actual error to verify that our algorithm works as intended. (We have no proof, that the algorithm always finds the correct solution in this setting, but empirically this is the case…)

int main() {
Index i,j,k;

auto A = TTOperator::random(std::vector<size_t>(16, 4), std::vector<size_t>(7,2));
A(i/2,j/2) = A(i/2, k/2) * A(j/2, k/2);

auto solution = TTTensor::random(std::vector<size_t>(8, 4), std::vector<size_t>(7,3));
TTTensor b;
b(i&0) = A(i/2, j/2) * solution(j&0);

auto x = TTTensor::random(std::vector<size_t>(8, 4), std::vector<size_t>(7,3));
simpleALS(A, x, b);

XERUS_LOG(info, "Residual: " << frob_norm(A(i/2, j/2) * x(j&0) - b(i&0))/frob_norm(b));
XERUS_LOG(info, "Error: " << frob_norm(solution-x)/frob_norm(x));
}

if __name__ == "__main__":
i,j,k = xe.indices(3)

A = xe.TTOperator.random([4]*16, [2]*7)
A(i/2,j/2) << A(i/2, k/2) * A(j/2, k/2)

solution = xe.TTTensor.random([4]*8, [3]*7)
b = xe.TTTensor()
b(i&0) << A(i/2, j/2) * solution(j&0)

x = xe.TTTensor.random([4]*8, [3]*7)
simpleALS(A, x, b)

print("Residual:", xe.frob_norm(A(i/2, j/2) * x(j&0) - b(i&0))/xe.frob_norm(b))
print("Error:", xe.frob_norm(solution-x)/xe.frob_norm(x))


## Complete Sourcecode

The full source code of this example looks as follows

#include <xerus.h>

using namespace xerus;

class InternalSolver {
const size_t d;

std::vector<Tensor> leftAStack;
std::vector<Tensor> rightAStack;

std::vector<Tensor> leftBStack;
std::vector<Tensor> rightBStack;

TTTensor& x;
const TTOperator& A;
const TTTensor& b;
const double solutionsNorm;
public:
size_t maxIterations;

InternalSolver(const TTOperator& _A, TTTensor& _x, const TTTensor& _b)
: d(_x.degree()), x(_x), A(_A), b(_b), solutionsNorm(frob_norm(_b)), maxIterations(1000)
{
leftAStack.emplace_back(Tensor::ones({1,1,1}));
rightAStack.emplace_back(Tensor::ones({1,1,1}));
leftBStack.emplace_back(Tensor::ones({1,1}));
rightBStack.emplace_back(Tensor::ones({1,1}));
}

void push_left_stack(const size_t _position) {
Index i1, i2, i3, j1 , j2, j3, k1, k2;
const Tensor &xi = x.get_component(_position);
const Tensor &Ai = A.get_component(_position);
const Tensor &bi = b.get_component(_position);

Tensor tmpA, tmpB;
tmpA(i1, i2, i3) = leftAStack.back()(j1, j2, j3)
*xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3);
leftAStack.emplace_back(std::move(tmpA));
tmpB(i1, i2) = leftBStack.back()(j1, j2)
*xi(j1, k1, i1)*bi(j2, k1, i2);
leftBStack.emplace_back(std::move(tmpB));
}

void push_right_stack(const size_t _position) {
Index i1, i2, i3, j1 , j2, j3, k1, k2;
const Tensor &xi = x.get_component(_position);
const Tensor &Ai = A.get_component(_position);
const Tensor &bi = b.get_component(_position);

Tensor tmpA, tmpB;
tmpA(i1, i2, i3) = xi(i1, k1, j1)*Ai(i2, k1, k2, j2)*xi(i3, k2, j3)
*rightAStack.back()(j1, j2, j3);
rightAStack.emplace_back(std::move(tmpA));
tmpB(i1, i2) = xi(i1, k1, j1)*bi(i2, k1, j2)
*rightBStack.back()(j1, j2);
rightBStack.emplace_back(std::move(tmpB));
}

double calc_residual_norm() {
Index i,j;
return frob_norm(A(i/2, j/2)*x(j&0) - b(i&0)) / solutionsNorm;
}

void solve() {
// Build right stack
x.move_core(0, true);
for (size_t pos = d-1; pos > 0; --pos) {
push_right_stack(pos);
}

Index i1, i2, i3, j1 , j2, j3, k1, k2;
std::vector<double> residuals(10, 1000.0);

for (size_t itr = 0; itr < maxIterations; ++itr) {
// Calculate residual and check end condition
residuals.push_back(calc_residual_norm());
if (residuals.back()/residuals[residuals.size()-10] > 0.99) {
XERUS_LOG(simpleALS, "Done! Residual decrease from " << std::scientific << residuals[10] << " to " << std::scientific << residuals.back() << " in " << residuals.size()-10 << " iterations.");
return; // We are done!
}
XERUS_LOG(simpleALS, "Iteration: " << itr << " Residual: " << residuals.back());

// Sweep Left -> Right
for (size_t corePosition = 0; corePosition < d; ++corePosition) {
Tensor op, rhs;

const Tensor &Ai = A.get_component(corePosition);
const Tensor &bi = b.get_component(corePosition);

op(i1, i2, i3, j1, j2, j3) = leftAStack.back()(i1, k1, j1)*Ai(k1, i2, j2, k2)*rightAStack.back()(i3, k2, j3);
rhs(i1, i2, i3) =            leftBStack.back()(i1, k1) *   bi(k1, i2, k2) *   rightBStack.back()(i3, k2);

xerus::solve(x.component(corePosition), op, rhs);

if (corePosition+1 < d) {
x.move_core(corePosition+1, true);
push_left_stack(corePosition);
rightAStack.pop_back();
rightBStack.pop_back();
}
}

// Sweep Right -> Left : only move core and update stacks
x.move_core(0, true);
for (size_t corePosition = d-1; corePosition > 0; --corePosition) {
push_right_stack(corePosition);
leftAStack.pop_back();
leftBStack.pop_back();
}

}
}

};

void simpleALS(const TTOperator& _A, TTTensor& _x, const TTTensor& _b)  {
InternalSolver solver(_A, _x, _b);
return solver.solve();
}

int main() {
Index i,j,k;

auto A = TTOperator::random(std::vector<size_t>(16, 4), std::vector<size_t>(7,2));
A(i/2,j/2) = A(i/2, k/2) * A(j/2, k/2);

auto solution = TTTensor::random(std::vector<size_t>(8, 4), std::vector<size_t>(7,3));
TTTensor b;
b(i&0) = A(i/2, j/2) * solution(j&0);

auto x = TTTensor::random(std::vector<size_t>(8, 4), std::vector<size_t>(7,3));
simpleALS(A, x, b);

XERUS_LOG(info, "Residual: " << frob_norm(A(i/2, j/2) * x(j&0) - b(i&0))/frob_norm(b));
XERUS_LOG(info, "Error: " << frob_norm(solution-x)/frob_norm(x));
}


import xerus as xe

class InternalSolver :
def __init__(self, A, x, b):
self.A = A
self.b = b
self.x = x
self.d = x.degree()
self.solutionsNorm = b.frob_norm()

self.leftAStack = [ xe.Tensor.ones([1,1,1]) ]
self.leftBStack = [ xe.Tensor.ones([1,1]) ]
self.rightAStack = [ xe.Tensor.ones([1,1,1]) ]
self.rightBStack = [ xe.Tensor.ones([1,1]) ]

self.maxIterations = 1000

def push_left_stack(self, pos) :
i1,i2,i3, j1,j2,j3, k1,k2 = xe.indices(8)
Ai = self.A.get_component(pos)
xi = self.x.get_component(pos)
bi = self.b.get_component(pos)

tmpA = xe.Tensor()
tmpB = xe.Tensor()
tmpA(i1, i2, i3) << self.leftAStack[-1](j1, j2, j3)\
*xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3)
self.leftAStack.append(tmpA)
tmpB(i1, i2) << self.leftBStack[-1](j1, j2)\
*xi(j1, k1, i1)*bi(j2, k1, i2)
self.leftBStack.append(tmpB)

def push_right_stack(self, pos) :
i1,i2,i3, j1,j2,j3, k1,k2 = xe.indices(8)
Ai = self.A.get_component(pos)
xi = self.x.get_component(pos)
bi = self.b.get_component(pos)

tmpA = xe.Tensor()
tmpB = xe.Tensor()
tmpA(j1, j2, j3) << xi(j1, k1, i1)*Ai(j2, k1, k2, i2)*xi(j3, k2, i3) \
* self.rightAStack[-1](i1, i2, i3)
self.rightAStack.append(tmpA)
tmpB(j1, j2) << xi(j1, k1, i1)*bi(j2, k1, i2) \
* self.rightBStack[-1](i1, i2)
self.rightBStack.append(tmpB)

def calc_residual_norm(self) :
i,j = xe.indices(2)
return xe.frob_norm(self.A(i/2, j/2)*self.x(j&0) - self.b(i&0)) / self.solutionsNorm

def solve(self) :
# build right stack
self.x.move_core(0, True)
for pos in reversed(xrange(1, self.d)) :
self.push_right_stack(pos)

i1,i2,i3, j1,j2,j3, k1,k2 = xe.indices(8)
residuals = [1000]*10

for itr in xrange(self.maxIterations) :
residuals.append(self.calc_residual_norm())
if residuals[-1]/residuals[-10] > 0.99 :
print("Done! Residual decreased from:", residuals[10], "to", residuals[-1], "in", len(residuals)-10, "sweeps")
return

print("Iteration:",itr, "Residual:", residuals[-1])

# sweep left -> right
for pos in xrange(self.d):
op = xe.Tensor()
rhs = xe.Tensor()

Ai = self.A.get_component(pos)
bi = self.b.get_component(pos)

op(i1, i2, i3, j1, j2, j3) << self.leftAStack[-1](i1, k1, j1)*Ai(k1, i2, j2, k2)*self.rightAStack[-1](i3, k2, j3)
rhs(i1, i2, i3) <<            self.leftBStack[-1](i1, k1) *   bi(k1, i2, k2) *   self.rightBStack[-1](i3, k2)

tmp = xe.Tensor()
tmp(i1&0) << rhs(j1&0) / op(j1/2, i1/2)
self.x.set_component(pos, tmp)

if pos+1 < self.d :
self.x.move_core(pos+1, True)
self.push_left_stack(pos)
self.rightAStack.pop()
self.rightBStack.pop()

# right -> left, only move core and update stack
self.x.move_core(0, True)
for pos in reversed(xrange(1,self.d)) :
self.push_right_stack(pos)
self.leftAStack.pop()
self.leftBStack.pop()

def simpleALS(A, x, b) :
solver = InternalSolver(A, x, b)
solver.solve()

if __name__ == "__main__":
i,j,k = xe.indices(3)

A = xe.TTOperator.random([4]*16, [2]*7)
A(i/2,j/2) << A(i/2, k/2) * A(j/2, k/2)

solution = xe.TTTensor.random([4]*8, [3]*7)
b = xe.TTTensor()
b(i&0) << A(i/2, j/2) * solution(j&0)

x = xe.TTTensor.random([4]*8, [3]*7)
simpleALS(A, x, b)

print("Residual:", xe.frob_norm(A(i/2, j/2) * x(j&0) - b(i&0))/xe.frob_norm(b))
print("Error:", xe.frob_norm(solution-x)/xe.frob_norm(x))