SimClass/BlogIterativeMatrixSolversJacobi.html at master · blackencino/SimClass · GitHub

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<!DOCTYPE html>
<html>
<head>
  <title>Encino Sim Class : Iterative Matrix Solvers and the Jacobi Method</title>

  <style type="text/css" title="currentStyle">
    @import "css/SimClass.css";
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<script src='http://processingjs.org/js/processing.min.js' type='text/javascript'/></script>
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<body>
<h2>Iterative Methods for Large Linear Systems</h2>
In our
<a href="http://encinographic.blogspot.com/2013/06/simulation-class-matrix-methods-and.html">
previous post</a>,
we carefully discretized the equation of motion for a 1D wave equation into
a large set of simple linear equations, which we then converted into a sparse
matrix notation. It's important to understand that the matrix & vector notation
that we reached at the end of our derivation, after creating a list
of simultaneous linear equations, was just a change of notation, nothing more.
Matrices and the notation surrounding linear algebra can sometimes be
obfuscating and generate confusion - whenever that is the case, it's helpful
to just remember that the matrix notation is a short-hand, and that we can
always revert to the system of equations that the matrix expression represents.
<p>
So, armed with this fantastically simple linear matrix equation, we are tasked
with solving the system for the unknown solution vector, \(x\):

<div class="LaTexEquation">
  \[
  \begin{equation}
  A x = b
  \label{eq:AxEqualsB}
  \end{equation}
  \]
</div>

Surely we can just construct our matrix \(A\), invert the matrix to produce
\(A^{-1}\), and then our solution would be:

<div class="LaTexEquation">
  \[
  \begin{equation}
  x = A^{-1} b
  \label{eq:XequalsAinvB}
  \end{equation}
  \]
</div>

This simple solution is comforting, but sadly, totally unusable in any
practical setting.  Why?
<p>
The first and most important problem we run into with an explicit matrix
inversion solution is that the matrices themselves would be incredibly massive
in size for any real-world problem.  Imagine we have a fluid simulation in
2D that is \(1024 \times 1024\) in size. In any such simulation, our state
array size \(N\) is equal to the total number of points being solved for,
so in this case, \(N = 1024 \times 1024 = 1048576\).  The matrix \(A\) is,
as we've seen, size \(N \times N\), which would in that simple case be
\(N^4 = 109951162776\), or just slightly larger than \(10^{12}\) points, which
would take just over 4 <i>TERABYTES</i> of RAM to hold, for single precision
floating point.
<h3>Sparse Matrices</h3>
The obvious immediate solution to the above storage problem is to take advantage
of the fact that our matrices are <i>sparse</i>. A
<a href="http://en.wikipedia.org/wiki/Sparse_matrix">Sparse Matrix</a> is one
in which most of the matrix elements are zero. In our example, we notice that
there is at most three non-zero elements per-row, one of which is always the
diagonal element. The other non-zero elements are always one column before or
after our diagonal element. We could be clever and store our matrix \(A\)
as an array of \(N\) "Sparse Rows", where a Sparse Row has a structure something
like this:
<p>
<pre><code>
class SparseMatrixRow
{
    float indices[3];
    float values[3];
    int numIndices;
    SparseMatrixRow()
    {
        numIndices = 0;
    }
    void addElement( int i_index, float i_val )
    {
        if ( numIndices < 3 )
        {
            values[numIndices] = i_val;
            indices[numIndices] = i_index;
            ++numIndices;
        }
    }
};
</code></pre>
<p>
Indeed, structures similar to the one above are the basis of sparse-matrix
linear algebra libraries such as
<a href="http://math.nist.gov/spblas/">Sparse BLAS</a>.  With the above
representation, our matrix in the 2D height field case would take about 28
megabytes to store, a huge improvement over 4 terabytes.
<p>
However, even with the sparse representation of \(A\), how do we invert
a huge matrix like this? If you've ever worked to invert a \(3 \times 3\) or
\(4 \times 4\) matrix, you know how complex that could be - how on earth would
you go about inverting a square matrix with a million rows, and how would
you store your intermediate results, which may not enjoy the simple sparse
form that the original matrix \(A\) had?
<p>
Problems like the above are what supercomputers were essentially built to solve,
and there are excellent libraries and tools for doing this. However, such
systems are arcane in nature, usually written in FORTRAN, and in our case,
drastically more firepower than we actually need.

<h3>Iterative Methods</h3>
Our matrix has some nice properties that will allow us to solve the problem
much more simply. Firstly, it is symmetric. A
<a href="http://en.wikipedia.org/wiki/Symmetric_matrix">Symmetric Matrix</a>
is one which is identical under a transposition of the rows and columns. If you
take any element to the upper right of the diagonal, and look at the mirror
of that element in the same location on the lower left of the diagonal, you'll
see the same value. A symmetric matrix as a representation of a physical
equation of motion is an expression of
<a href="http://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton.27s_3rd_Law">
Newton's Third Law</a>, that each action has an equal but opposite reaction. It
means that the effect point \(i\) has on point \(j\) is the same as the
effect point \(j\) has on point \(i\).  Conservation of energy emerges from
this law.
<p>
Secondly, assuming that our timestep \(\Delta t\) is relatively small compared
to the ratio of wave propagation speed \(c\) to the spatial discretization
\(\Delta x\), our off-diagonal elements will be much smaller in absolute
magnitude than our diagonal elements. This means that our system is a candidate
for <a href="http://en.wikipedia.org/wiki/Jacobi_method">The Jacobi Method</a>,
which is the simplest iterative \(A x = b\) solver.
<p>
<a href="http://en.wikipedia.org/wiki/Iterative_method#Linear_systems">
Iterative methods</a>, which include Jacobi, but also include more fancy solvers
such as
<a href="http://en.wikipedia.org/wiki/Conjugate_gradient_method">
The Conjugate Gradient Method</a>, all involve some variation on making an
initial guess for the unknown solution vector \(x^{\star0}\), then using that
guess in some way to produce a better guess \(x^{\star1}\), and so on, until
the change between guesses from one iteration to the next falls below some
acceptance threshold, or a maximum number of iterations is reached.
<p>
For all iterative methods, we will require storage for the next guess and
the previous guess, and we'll swap these temporary storage arrays back and forth
as we iterate towards the solution. Our generic state-vector implementation
easily allows for this, and indeed we already used this facility when we
computed our <b>RK4</b> approximation.

<h3>Jacobi Method</h3>
The Jacobi Method is so simple, we don't even need to construct a sparse
matrix or a solution vector, we can simply solve for the new estimate of the
solution vector \(x\) in place.
<p>
Jacobi works by taking each equation and assuming that everything other
than the diagonal element is known - which we do by using values from the
previous guess of \(x\).  The new guess for the diagonal element \(x\) is then,
at any row of the matrix:

  <div class="LaTexEquation">
  \[
  \begin{equation}
  x^{\star k+1}_i = \frac{b_i - \sum_{j\neq i}A_{i,j} x^{\star k}_{j}}{A_{i,i}}
  \label{eq:JacobiIteration}
  \end{equation}
  \]
  </div>

In the above Equation \(\eqref{eq:JacobiIteration}\), the new guess iteration
\(k+1\) is computed from the values of the old guess iteration \(k\).  This
is fairly easy to code, and looks like this:
<p>
<pre><code>
// Jacobi iteration to get temp acceleration
void JacobiIterationAccel( int i_aOld, int o_aNew, int i_hStar, float i_dt )
{
    float aij = -sq( WaveSpeed ) * sq( i_dt ) / sq( DX );
    float aii = 1.0 + ( 2.0 * sq( WaveSpeed ) * sq( i_dt ) / sq( DX ) );
    float bgain = sq( WaveSpeed ) / sq( DX );

    for ( int i = 1; i < ArraySize-1; ++i )
    {
        float aLeft = State[i_aOld][i-1];
        float aRight = State[i_aOld][i+1];

        float hStarLeft = State[i_hStar][i-1];
        float hStarCen = State[i_hStar][i];
        float hStarRight = State[i_hStar][i+1];

        float bi = bgain * ( hStarLeft + hStarRight - ( 2.0 * hStarCen ) );

        float sumOffDiag = ( aij * aLeft ) + ( aij * aRight );

        State[o_aNew][i] = ( bi - sumOffDiag ) / aii;
    }

    EnforceAccelBoundaryConditions( o_aNew );
}
</code></pre>
<p>
The code above computes a single new guess for the accelerations from an
old guess for the accelerations.  We can then write our "acceleration from
position" function as we had before, as follows. We're taking a fixed number
of iterations for guessing:
<p>
<pre><code>
// Solve for acceleration.
void JacobiSolveAccel( int i_hStar, float i_dt )
{
    // Initialize acceleration to zero.
    FillArray( StateAccelStar, 0.0 );

    // Solve from StateJacobiTmp into StateAccel
    for ( int iter = 0; iter < 20; ++iter )
    {
        int tmp = StateAccelStar;
        StateAccelStar = StateJacobiTmp;
        StateJacobiTmp = tmp;

        JacobiIterationAccel( StateJacobiTmp, StateAccelStar, i_hStar, i_dt );
    }
}

void EstimateAccelStar( float i_dt )
{
    JacobiSolveAccel( StateHeightStar, i_dt );
}
</code></pre>
<p>
The rest of our implementation looks basically identical. Here's the
final timestep, which looks exactly like our previous RK4 implementation:
<p>
<pre><code>
// Time Step function.
void TimeStep( float i_dt )
{
    // Swap state
    SwapState();

    // Initialize estimate. This just amounts to copying
    // The previous values into the current values.
    CopyArray( StateHeightPrev, StateHeight );
    CopyArray( StateVelPrev, StateVel );

    // 1
    CopyArray( StateVel, StateVelStar );
    EstimateHeightStar( i_dt );
    EstimateAccelStar( i_dt );
    AccumulateEstimate( i_dt / 6.0 );

    // 2
    EstimateVelStar( i_dt / 2.0 );
    EstimateHeightStar( i_dt / 2.0 );
    EstimateAccelStar( i_dt / 2.0 );
    AccumulateEstimate( i_dt / 3.0 );

    // 3
    EstimateVelStar( i_dt / 2.0 );
    EstimateHeightStar( i_dt / 2.0 );
    EstimateAccelStar( i_dt / 2.0 );
    AccumulateEstimate( i_dt / 3.0 );

    // 4
    EstimateVelStar( i_dt );
    EstimateHeightStar( i_dt );
    EstimateAccelStar( i_dt );
    AccumulateEstimate( i_dt / 6.0 );

    // Final boundary conditions on height and vel
    EnforceHeightBoundaryConditions( StateHeight );
    EnforceVelBoundaryConditions( StateVel );

    // Update current time.
    StateCurrentTime += i_dt;
}
</code></pre>
<p>
And that's it...  Let's see how that looks in Processing!
<p>
<script type="application/processing" data-processing-target="pjsWaveEqnStableRK4">
float WaveSpeed = 0.5;

float WorldSize = 10.0;
int ArraySize = 128;
float DX = WorldSize / ArraySize;

float LX = WorldSize;
float LY = WorldSize / 2.0;

int StateSize = 8;
float[][] State = new float[StateSize][ArraySize];
int StateHeight = 0;
int StateVel = 1;
int StateHeightPrev = 2;
int StateVelPrev = 3;
int StateVelStar = 4;
int StateAccelStar = 5;
int StateHeightStar = 6;
int StateJacobiTmp = 7;

float StateCurrentTime = 0.0;

int PixelsPerCell = 4;

int WindowWidth = PixelsPerCell * ArraySize;
int WindowHeight = WindowWidth / 2;

boolean InputActive = false;
int InputIndex = 0;
float InputHeight = 0;

float snoise( float x )
{
   return ( 2.0 * noise( x )) - 1.0;
}

float anoise( float x )
{
   return ( -2.0 * abs( snoise( x )) ) + 1.0;
}

void EnforceAccelBoundaryConditions( int io_a )
{
    State[io_a][0] = 0.0;
    State[io_a][ArraySize-1] = 0.0;
}

void EnforceVelBoundaryConditions( int io_v )
{
    State[io_v][0] = State[io_v][1];
    State[io_v][ArraySize-1] = State[io_v][ArraySize-2];
}

void EnforceHeightBoundaryConditions( int io_h )
{
    //State[io_h][0] = ( 2.0 * State[io_h][1] ) - State[io_h][2];
    //State[io_h][ArraySize-1] = ( 2.0 * State[io_h][ArraySize-2] ) - State[io_h][ArraySize-3];
    State[io_h][0] = State[io_h][1];
    State[io_h][ArraySize-1] = State[io_h][ArraySize-2];

    if ( InputActive )
    {
        State[io_h][InputIndex] = InputHeight;
    }
}

void CopyArray( int i_src, int o_dst )
{
    for ( int i = 0; i < ArraySize; ++i )
    {
        State[o_dst][i] = State[i_src][i];
    }
}

void FillArray( int o_a, float i_val )
{
    for ( int i = 0; i < ArraySize; ++i )
    {
        State[o_a][i] = i_val;
    }
}

void SetInitialState()
{
    noiseSeed( 0 );
    for ( int i = 0; i < ArraySize; ++i )
    {
        float worldX = 2341.17 + DX * ( float )i;
        State[StateHeight][i] = 0.5 * anoise( worldX * 0.0625 ) +
                         0.4 * anoise( worldX * 0.125 ) +
                         0.3 * anoise( worldX * 0.25 ) +
                         0.2 * anoise( worldX * 0.5 );
        State[StateVel][i] = 0.0;
    }
    EnforceHeightBoundaryConditions( StateHeight );
    EnforceVelBoundaryConditions( StateVel );
    StateCurrentTime = 0.0;

    CopyArray( StateHeight, StateHeightPrev );
    CopyArray( StateVel, StateVelPrev );
    InputActive = false;
}

void setup()
{
    SetInitialState();

    size( WindowWidth, WindowHeight );

    colorMode( RGB, 1.0 );
    strokeWeight( 0.5 );
    textSize( 24 );
}

void SwapHeight()
{
    int tmp = StateHeight;
    StateHeight = StateHeightPrev;
    StateHeightPrev = tmp;
}

void SwapVel()
{
    int tmp = StateVel;
    StateVel = StateVelPrev;
    StateVelPrev = tmp;
}

void SwapState()
{
    SwapHeight();
    SwapVel();
}

void GetInput()
{
    InputActive = false;

    if ( mousePressed && mouseButton == LEFT )
    {
        float mouseCellX = mouseX / ( ( float )PixelsPerCell );
        float mouseCellY = ( (height/2) - mouseY ) / ( ( float )PixelsPerCell );
        float simY = mouseCellY * DX;

        int iX = ( int )floor( mouseCellX + 0.5 );
        if ( iX > 0 && iX < ArraySize-1 )
        {
            InputIndex = iX;
            InputHeight = simY;
            InputActive = true;
        }
    }
}

// Estimate height star
void EstimateHeightStar( float i_dt )
{
    for ( int i = 0; i < ArraySize; ++i )
    {
        State[StateHeightStar][i] = State[StateHeightPrev][i] +
                ( i_dt * State[StateVelStar][i] );
    }
    EnforceHeightBoundaryConditions( StateHeightStar );
}

// Estimate vel star
void EstimateVelStar( float i_dt )
{
    for ( int i = 0; i < ArraySize; ++i )
    {
        State[StateVelStar][i] = State[StateVelPrev][i] +
                ( i_dt * State[StateAccelStar][i] );
    }
    EnforceVelBoundaryConditions( StateVelStar );
}

// Jacobi iteration to get temp acceleration
void JacobiIterationAccel( int i_aOld, int o_aNew, int i_hStar, float i_dt )
{
    float aij = -sq( WaveSpeed ) * sq( i_dt ) / sq( DX );
    float aii = 1.0 + ( 2.0 * sq( WaveSpeed ) * sq( i_dt ) / sq( DX ) );
    float bgain = sq( WaveSpeed ) / sq( DX );

    for ( int i = 1; i < ArraySize-1; ++i )
    {
        float aLeft = State[i_aOld][i-1];
        float aRight = State[i_aOld][i+1];

        float hStarLeft = State[i_hStar][i-1];
        float hStarCen = State[i_hStar][i];
        float hStarRight = State[i_hStar][i+1];

        float bi = bgain * ( hStarLeft + hStarRight - ( 2.0 * hStarCen ) );

        float sumOffDiag = ( aij * aLeft ) + ( aij * aRight );

        State[o_aNew][i] = ( bi - sumOffDiag ) / aii;
    }

    EnforceAccelBoundaryConditions( o_aNew );
}

// Solve for acceleration.
void JacobiSolveAccel( int i_hStar, float i_dt )
{
    // Initialize acceleration to zero.
    FillArray( StateAccelStar, 0.0 );

    // Solve from StateJacobiTmp into StateAccel
    for ( int iter = 0; iter < 20; ++iter )
    {
        int tmp = StateAccelStar;
        StateAccelStar = StateJacobiTmp;
        StateJacobiTmp = tmp;

        JacobiIterationAccel( StateJacobiTmp, StateAccelStar, i_hStar, i_dt );
    }
}

void EstimateAccelStar( float i_dt )
{
    JacobiSolveAccel( StateHeightStar, i_dt );
}

// Accumulate estimate
void AccumulateEstimate( float i_dt )
{
    for ( int i = 0; i < ArraySize; ++i )
    {
        State[StateHeight][i] += i_dt * State[StateVelStar][i];
        State[StateVel][i] += i_dt * State[StateAccelStar][i];
    }
}

// Time Step function.
void TimeStep( float i_dt )
{
    // Swap state
    SwapState();

    // Initialize estimate. This just amounts to copying
    // The previous values into the current values.
    CopyArray( StateHeightPrev, StateHeight );
    CopyArray( StateVelPrev, StateVel );

    // 1
    CopyArray( StateVel, StateVelStar );
    EstimateHeightStar( i_dt );
    EstimateAccelStar( i_dt );
    AccumulateEstimate( i_dt / 6.0 );

    // 2
    EstimateVelStar( i_dt / 2.0 );
    EstimateHeightStar( i_dt / 2.0 );
    EstimateAccelStar( i_dt / 2.0 );
    AccumulateEstimate( i_dt / 3.0 );

    // 3
    EstimateVelStar( i_dt / 2.0 );
    EstimateHeightStar( i_dt / 2.0 );
    EstimateAccelStar( i_dt / 2.0 );
    AccumulateEstimate( i_dt / 3.0 );

    // 4
    EstimateVelStar( i_dt );
    EstimateHeightStar( i_dt );
    EstimateAccelStar( i_dt );
    AccumulateEstimate( i_dt / 6.0 );

    // Final boundary conditions on height and vel
    EnforceHeightBoundaryConditions( StateHeight );
    EnforceVelBoundaryConditions( StateVel );

    // Update current time.
    StateCurrentTime += i_dt;
}

void DrawState()
{
    float OffsetY = 0.5 * ( float )WindowHeight;
    for ( int i = 0; i < ArraySize; ++i )
    {
        float SimX = DX *  ( float )i;
        float PixelsX = ( float )( i * PixelsPerCell );
        float SimY = State[StateHeight][i];
        float PixelsY = SimY * (( float )PixelsPerCell ) / DX;
        float PixelsMinY = OffsetY - PixelsY;
        float PixelsHeight = (( float )WindowHeight) - PixelsMinY;

        fill( 0.0, 0.0, 1.0 );
        rect( PixelsX, OffsetY - PixelsY, PixelsPerCell, PixelsHeight );
    }
}

void draw()
{
    background( 0.9 );

    GetInput();

    TimeStep( 1.0 / 24.0 );

    DrawState();

    // Label.
    fill( 1.0 );
    text( "Wave Equation Stable Jacobi RK4", 10, 30 );
}

// Reset function. If the key 'r' is released in the display,
// copy the initial state to the state.
void keyReleased()
{
    if ( key == 114 )
    {
        SetInitialState();
    }
}

</script>
<canvas id="pjsWaveEqnStableRK4"> </canvas>
<p>
And with that, we're finished... (for now!)

<p>
<h2>Download</h2>
All of the files associated with this class are available via a public github repository, found here:
<a href="https://github.com/blackencino/SimClass">https://github.com/blackencino/SimClass</a>


</body>