Tridiagonal matrix algorithm

In numerical linear algebra, the tridiagonal matrix algorithm, also known as the Thomas algorithm (named after Llewellyn Thomas), is a simplified form of Gaussian elimination that can be used to solve tridiagonal systems of equations. A tridiagonal system for n unknowns may be written as

${\displaystyle a_{i}x_{i-1}+b_{i}x_{i}+c_{i}x_{i+1}=d_{i},}$

where ${\displaystyle a_{1}=0}$ and ${\displaystyle c_{n}=0}$.

${\displaystyle {\begin{bmatrix}b_{1}&c_{1}&&&0\\a_{2}&b_{2}&c_{2}&&\\&a_{3}&b_{3}&\ddots &\\&&\ddots &\ddots &c_{n-1}\\0&&&a_{n}&b_{n}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\vdots \\x_{n}\end{bmatrix}}={\begin{bmatrix}d_{1}\\d_{2}\\d_{3}\\\vdots \\d_{n}\end{bmatrix}}.}$

For such systems, the solution can be obtained in ${\displaystyle O(n)}$ operations instead of ${\displaystyle O(n^{3})}$ required by Gaussian elimination. A first sweep eliminates the ${\displaystyle a_{i}}$'s, and then an (abbreviated) backward substitution produces the solution. Examples of such matrices commonly arise from the discretization of 1D

The implementation as a C function, which uses scratch space to avoid modifying its inputs for a-c, allowing them to be reused:

void thomas(const int X, double x[restrict X],
            const double a[restrict X], const double b[restrict X],
            const double c[restrict X], double scratch[restrict X]) {
    /*
     solves Ax = d, where A is a tridiagonal matrix consisting of vectors a, b, c
     X = number of equations
     x[] = initially contains the input v, and returns x. indexed from [0, ..., X - 1]
     a[] = subdiagonal, indexed from [1, ..., X - 1]
     b[] = main diagonal, indexed from [0, ..., X - 1]
     c[] = superdiagonal, indexed from [0, ..., X - 2]
     scratch[] = scratch space of length X, provided by caller, allowing a, b, c to be const
     not performed in this example: manual expensive common subexpression elimination
     */
    scratch[0] = c[0] / b[0];
    x[0] = x[0] / b[0];

    /* loop from 1 to X - 1 inclusive */
    for (int ix = 1; ix < X; ix++) {
        if (ix < X-1){
        scratch[ix] = c[ix] / (b[ix] - a[ix] * scratch[ix - 1]);
        }
        x[ix] = (x[ix] - a[ix] * x[ix - 1]) / (b[ix] - a[ix] * scratch[ix - 1]);
    }

    /* loop from X - 2 to 0 inclusive */
    for (int ix = X - 2; ix >= 0; ix--)
        x[ix] -= scratch[ix] * x[ix + 1];
}

Derivation

The derivation of the tridiagonal matrix algorithm is a special case of Gaussian elimination.

Suppose that the unknowns are ${\displaystyle x_{1},\ldots ,x_{n}}$, and that the equations to be solved are:

${\displaystyle {\begin{alignedat}{4}&&&b_{1}x_{1}&&+c_{1}x_{2}&&=d_{1}\\&a_{i}x_{i-1}&&+b_{i}x_{i}&&+c_{i}x_{i+1}&&=d_{i}\,,\quad i=2,\ldots ,n-1\\&a_{n}x_{n-1}&&+b_{n}x_{n}&&&&=d_{n}\,.\end{alignedat}}}$

Consider modifying the second (${\displaystyle i=2}$) equation with the first equation as follows:

${\displaystyle ({\mbox{equation 2}})\cdot b_{1}-({\mbox{equation 1}})\cdot a_{2}}$

which would give:

${\displaystyle (b_{2}b_{1}-c_{1}a_{2})x_{2}+c_{2}b_{1}x_{3}=d_{2}b_{1}-d_{1}a_{2}.}$

Note that ${\displaystyle x_{1}}$ has been eliminated from the second equation. Using a similar tactic with the modified second equation on the third equation yields:

${\displaystyle (b_{3}(b_{2}b_{1}-c_{1}a_{2})-c_{2}b_{1}a_{3})x_{3}+c_{3}(b_{2}b_{1}-c_{1}a_{2})x_{4}=d_{3}(b_{2}b_{1}-c_{1}a_{2})-(d_{2}b_{1}-d_{1}a_{2})a_{3}.\,}$

This time ${\displaystyle x_{2}}$ was eliminated. If this procedure is repeated until the ${\displaystyle n^{th}}$ row; the (modified) ${\displaystyle n^{th}}$ equation will involve only one unknown, ${\displaystyle x_{n}}$. This may be solved for and then used to solve the ${\displaystyle (n-1)^{th}}$ equation, and so on until all of the unknowns are solved for.

Clearly, the coefficients on the modified equations get more and more complicated if stated explicitly. By examining the procedure, the modified coefficients (notated with tildes) may instead be defined recursively:

${\displaystyle {\tilde {a}}_{i}=0\,}$

${\displaystyle {\tilde {b}}_{1}=b_{1}\,}$

${\displaystyle {\tilde {b}}_{i}=b_{i}{\tilde {b}}_{i-1}-{\tilde {c}}_{i-1}a_{i}\,}$

${\displaystyle {\tilde {c}}_{1}=c_{1}\,}$

${\displaystyle {\tilde {c}}_{i}=c_{i}{\tilde {b}}_{i-1}\,}$

${\displaystyle {\tilde {d}}_{1}=d_{1}\,}$

${\displaystyle {\tilde {d}}_{i}=d_{i}{\tilde {b}}_{i-1}-{\tilde {d}}_{i-1}a_{i}.\,}$

To further hasten the solution process, ${\displaystyle {\tilde {b}}_{i}}$ may be divided out (if there's no division by zero risk), the newer modified coefficients, each notated with a prime, will be:

${\displaystyle a'_{i}=0\,}$

${\displaystyle b'_{i}=1\,}$

${\displaystyle c'_{1}={\frac {c_{1}}{b_{1}}}\,}$

${\displaystyle c'_{i}={\frac {c_{i}}{b_{i}-c'_{i-1}a_{i}}}\,}$

${\displaystyle d'_{1}={\frac {d_{1}}{b_{1}}}\,}$

${\displaystyle d'_{i}={\frac {d_{i}-d'_{i-1}a_{i}}{b_{i}-c'_{i-1}a_{i}}}.\,}$

This gives the following system with the same unknowns and coefficients defined in terms of the original ones above:

${\displaystyle {\begin{array}{lcl}x_{i}+c'_{i}x_{i+1}=d'_{i}\qquad &;&\ i=1,\ldots ,n-1\\x_{n}=d'_{n}\qquad &;&\ i=n.\\\end{array}}\,}$

The last equation involves only one unknown. Solving it in turn reduces the next last equation to one unknown, so that this backward substitution can be used to find all of the unknowns:

${\displaystyle x_{n}=d'_{n}\,}$

${\displaystyle x_{i}=d'_{i}-c'_{i}x_{i+1}\qquad ;\ i=n-1,n-2,\ldots ,1.}$

Variants

In some situations, particularly those involving periodic boundary conditions, a slightly perturbed form of the tridiagonal system may need to be solved:

${\displaystyle {\begin{alignedat}{4}&a_{1}x_{n}&&+b_{1}x_{1}&&+c_{1}x_{2}&&=d_{1}\\&a_{i}x_{i-1}&&+b_{i}x_{i}&&+c_{i}x_{i+1}&&=d_{i}\,,\quad i=2,\ldots ,n-1\\&a_{n}x_{n-1}&&+b_{n}x_{n}&&+c_{n}x_{1}&&=d_{n}\,.\end{alignedat}}}$

In this case, we can make use of the Sherman–Morrison formula to avoid the additional operations of Gaussian elimination and still use the Thomas algorithm. The method requires solving a modified non-cyclic version of the system for both the input and a sparse corrective vector, and then combining the solutions. This can be done efficiently if both solutions are computed at once, as the forward portion of the pure tridiagonal matrix algorithm can be shared.

If we indicate by:

$${\displaystyle A={\begin{bmatrix}b_{1}&c_{1}&&&a_{1}\\a_{2}&b_{2}&c_{2}&&\\&a_{3}&b_{3}&\ddots &\\&&\ddots &\ddots &c_{n-1}\\c_{n}&&&a_{n}&b_{n}\end{bmatrix}},x={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\vdots \\x_{n}\end{bmatrix}},d={\begin{bmatrix}d_{1}\\d_{2}\\d_{3}\\\vdots \\d_{n}\end{bmatrix}}}$$

Then the system to be solved is:

$${\displaystyle Ax=d}$$

In this case the coefficients ${\displaystyle a_{1}}$ and ${\displaystyle c_{n}}$ are, generally speaking, non-zero, so their presence does not allow to apply the Thomas algorithm directly. We can therefore consider ${\displaystyle B\in \mathbb {R} ^{n\times n}}$ and ${\displaystyle u,v\in \mathbb {R} ^{n}}$ as following:

$${\displaystyle B={\begin{bmatrix}b_{1}-\gamma &c_{1}&&&0\\a_{2}&b_{2}&c_{2}&&\\&a_{3}&b_{3}&\ddots &\\&&\ddots &\ddots &c_{n-1}\\0&&&a_{n}&b_{n}-{\frac {c_{n}a_{1}}{\gamma }}\end{bmatrix}},u={\begin{bmatrix}\gamma \\0\\0\\\vdots \\c_{n}\end{bmatrix}},v={\begin{bmatrix}1\\0\\0\\\vdots \\a_{1}/\gamma \end{bmatrix}}.}$$

${\displaystyle \gamma \in \mathbb {R} }$

${\displaystyle A}$

${\displaystyle A=B+uv^{\mathsf {T}}}$

$${\displaystyle By=d\qquad \qquad Bq=u}$$

Then we reconstruct the solution ${\displaystyle x}$ using the Shermann-Morrison formula:

$${\displaystyle {\begin{aligned}x&=A^{-1}d=(B+uv^{T})^{-1}d=B^{-1}d-{\frac {B^{-1}uv^{T}B^{-1}}{1+v^{T}B^{-1}u}}d=y-{\frac {qv^{T}y}{1+v^{T}q}}\end{aligned}}}$$

The implementation as a C function, which uses scratch space to avoid modifying its inputs for a-c, allowing them to be reused:

void cyclic_thomas(const int X, double x[restrict X], const double a[restrict X], const double b[restrict X], const double c[restrict X], double cmod[restrict X], double u[restrict X]) {
    /*
     solves Ax = v, where A is a cyclic tridiagonal matrix consisting of vectors a, b, c
     X = number of equations
     x[] = initially contains the input v, and returns x. indexed from [0, ..., X - 1]
     a[] = subdiagonal, regularly indexed from [1, ..., X - 1], a[0] is lower left corner
     b[] = main diagonal, indexed from [0, ..., X - 1]
     c[] = superdiagonal, regularly indexed from [0, ..., X - 2], c[X - 1] is upper right
     cmod[], u[] = scratch vectors each of length X
     */

    /* lower left and upper right corners of the cyclic tridiagonal system respectively */
    const double alpha = a[0];
    const double beta = c[X - 1];

    /* arbitrary, but chosen such that division by zero is avoided */
    const double gamma = -b[0];

    cmod[0] = alpha / (b[0] - gamma);
    u[0] = gamma / (b[0] - gamma);
    x[0] /= (b[0] - gamma);

    /* loop from 1 to X - 2 inclusive */
    for (int ix = 1; ix + 1 < X; ix++) {
        const double m = 1.0 / (b[ix] - a[ix] * cmod[ix - 1]);
        cmod[ix] = c[ix] * m;
        u[ix] = (0.0f  - a[ix] * u[ix - 1]) * m;
        x[ix] = (x[ix] - a[ix] * x[ix - 1]) * m;
    }

    /* handle X - 1 */
    const double m = 1.0 / (b[X - 1] - alpha * beta / gamma - beta * cmod[X - 2]);
    u[X - 1] = (alpha    - a[X - 1] * u[X - 2]) * m;
    x[X - 1] = (x[X - 1] - a[X - 1] * x[X - 2]) * m;

    /* loop from X - 2 to 0 inclusive */
    for (int ix = X - 2; ix >= 0; ix--) {
        u[ix] -= cmod[ix] * u[ix + 1];
        x[ix] -= cmod[ix] * x[ix + 1];
    }

    const double fact = (x[0] + x[X - 1] * beta / gamma) / (1.0 + u[0] + u[X - 1] * beta / gamma);

    /* loop from 0 to X - 1 inclusive */
    for (int ix = 0; ix < X; ix++)
        x[ix] -= fact * u[ix];
}

There is also another way to solve the slightly perturbed form of the tridiagonal system considered above.^[5] Let us consider two auxiliary linear systems of dimension ${\displaystyle (n-1)\times (n-1)}$:

$${\displaystyle {\begin{aligned}\qquad \ \ \ \ \ b_{2}u_{2}+c_{2}u_{3}&=d_{2}\\a_{3}u_{2}+b_{3}u_{3}+c_{3}u_{4}&=d_{3}\\a_{i}u_{i-1}+b_{i}u_{i}+c_{i}u_{i+1}&=d_{i}\\\dots \\a_{n}u_{n-1}+b_{n}u_{n}\qquad &=d_{n}\,.\end{aligned}}\quad i=4,\ldots ,n-1\qquad \qquad {\begin{aligned}\qquad \ \ \ \ \ b_{2}v_{2}+c_{2}v_{3}&=-a_{2}\\a_{3}v_{2}+b_{3}v_{3}+c_{3}v_{4}&=0\\a_{i}u_{i-1}+b_{i}u_{i}+c_{i}u_{i+1}&=0\\\dots \\a_{n}v_{n-1}+b_{n}v_{n}\qquad &=-c_{n}\,.\end{aligned}}\quad i=4,\ldots ,n-1}$$

For convenience, we additionally define ${\displaystyle u_{1}=0}$ and ${\displaystyle v_{1}=1}$. We can now find the solutions ${\displaystyle \{u_{2},u_{3}\dots ,u_{n}\}}$ and ${\displaystyle \{v_{2},v_{3}\dots ,v_{n}\}}$ applying Thomas algorithm to the two auxiliary tridiagonal system.

The solution ${\displaystyle \{x_{1},x_{2}\dots ,x_{n}\}}$ can be then represented in the form:

$${\displaystyle x_{i}=u_{i}+x_{1}v_{i}\qquad i=1,2,\dots ,n}$$

Indeed, multiplying each equation of the second auxiliary system by ${\displaystyle x_{1}}$, adding with the corresponding equation of the first auxiliary system and using the representation ${\displaystyle x_{i}=u_{i}+x_{1}v_{i}}$, we immediately see that equations number ${\displaystyle 2}$ through ${\displaystyle n}$ of the original system are satisfied; it only remains to satisfy equation number ${\displaystyle 1}$. To do so, consider formula for ${\displaystyle i=2}$ and ${\displaystyle i=n}$ and substitute ${\displaystyle x_{2}=u_{2}+x_{1}v_{2}}$and ${\displaystyle x_{n}=u_{n}+x_{1}v_{n}}$ into the first equation of the original system. This yields one scalar equation for ${\displaystyle x_{1}}$:

$${\displaystyle b_{1}x_{1}+c_{1}(u_{2}+x_{1}v_{2})+a_{1}(u_{n}+x_{1}v_{n})=d_{1}}$$

As such, we find:

$${\displaystyle x_{1}={\frac {d_{1}-a_{1}u_{n}-c_{1}u_{2}}{b_{1}+a_{1}v_{n}+c_{1}v_{2}}}}$$

The implementation as a C function, which uses scratch space to avoid modifying its inputs for a-c, allowing them to be reused:

void cyclic_thomas(const int X, double x[restrict X], const double a[restrict X], const double b[restrict X], const double c[restrict X], double cmod[restrict X], double v[restrict X]) {
    /* first solve a system of length X - 1 for two right hand sides, ignoring ix == 0 */
    cmod[1] = c[1] / b[1];
    v[1] = -a[1] / b[1];
    x[1] = x[1] / b[1];

    /* loop from 2 to X - 1 inclusive */
    for (int ix = 2; ix < X - 1; ix++) {
        const double m = 1.0 / (b[ix] - a[ix] * cmod[ix - 1]);
        cmod[ix] = c[ix] * m;
        v[ix] = (0.0f  - a[ix] * v[ix - 1]) * m;
        x[ix] = (x[ix] - a[ix] * x[ix - 1]) * m;
    }

    /* handle X - 1 */
    const double m = 1.0 / (b[X - 1] - a[X - 1] * cmod[X - 2]);
    cmod[X - 1] = c[X - 1] * m;
    v[X - 1] = (-c[0]    - a[X - 1] * v[X - 2]) * m;
    x[X - 1] = (x[X - 1] - a[X - 1] * x[X - 2]) * m;

    /* loop from X - 2 to 1 inclusive */
    for (int ix = X - 2; ix >= 1; ix--) {
        v[ix] -= cmod[ix] * v[ix + 1];
        x[ix] -= cmod[ix] * x[ix + 1];
    }

    x[0] = (x[0] - a[0] * x[X - 1] - c[0] * x[1]) / (b[0] + a[0] * v[X - 1] + c[0] * v[1]);

    /* loop from 1 to X - 1 inclusive */
    for (int ix = 1; ix < X; ix++)
        x[ix] += x[0] * v[ix];
}

In both cases the auxiliary systems to be solved are genuinely tri-diagonal, so the overall computational complexity of solving system ${\displaystyle Ax=d}$ remains linear with the respect to the dimension of the system ${\displaystyle n}$, that is ${\displaystyle O(n)}$ arithmetic operations.

In other situations, the system of equations may be block tridiagonal (see

The textbook Numerical Mathematics by Alfio Quarteroni, Sacco and Saleri, lists a modified version of the algorithm which avoids some of the divisions (using instead multiplications), which is beneficial on some computer architectures.

Parallel tridiagonal solvers have been published for many vector and parallel architectures, including GPUs^[7][8]

For an extensive treatment of parallel tridiagonal and block tridiagonal solvers see ^[9]

References

The Wikibook Algorithm Implementation has a page on the topic of: Tridiagonal matrix algorithm