Efficient discrete Laplacian of a matrix

Question

I would like to compute the discrete Laplacian of a real matrix (numeric values and full), using any method and targetting efficiency (I will call the Laplacian dozens of thousands of time).

I naively defined the following function:

laplacian[Z_] := Block[{Zcenter, Ztop, Zleft, Zbottom, Zright},
  Zcenter = Z[[2 ;; -2, 2 ;; -2]];
  Ztop = Z[[;; -3, 2 ;; -2]]; 
  Zleft = Z[[2 ;; -2, ;; -3]]; 
  Zbottom = Z[[3 ;;, 2 ;; -2]];
  Zright = Z[[2 ;; -2, 3 ;;]];
  Ztop + Zleft + Zbottom + Zright - 4*Zcenter
]

It reduces the dimension of the input (because the Laplacian for the elements of the border of the array is not computed) but I am fine with that.

I also tried writing the function in a compiled way:

compileLaplacian = Compile[{{Z, _Real, 2}},
  Module[{Zcenter = Z[[2 ;; -2, 2 ;; -2]], 
          Ztop = Z[[;; -3, 2 ;; -2]],
          Zleft = Z[[2 ;; -2, ;; -3]], 
          Zbottom = Z[[3 ;;, 2 ;; -2]],
          Zright = Z[[2 ;; -2, 3 ;;]]},
    Ztop + Zleft + Zbottom + Zright - 4*Zcenter
  ]
]

but it returns the error

Compile::cpintlt: 3;;All at position 2 of Z[[3;;All,2;;-2]] should be either a nonzero integer or a vector of nonzero integers; evaluation will use the uncompiled function.

Can I improve my discrete Laplacian function in terms of computation time? (targeted matrices are $100\times 100$ to $10000\times 10000$)

---

Edit The following graph summarizes the timings for the different proposed functions. RAM is not monitored. I'll investigate Szabolcs's suggestion using packed array to see if timing can be further reduced.

Full code for the image:

laplacian[Z_] := 
  Block[{Zcenter, Ztop, Zleft, Zbottom, Zright}, 
    Zcenter = Z[[2 ;; -2, 2 ;; -2]];
    Ztop = Z[[;; -3, 2 ;; -2]];
    Zleft = Z[[2 ;; -2, ;; -3]];
    Zbottom = Z[[3 ;;, 2 ;; -2]];
    Zright = Z[[2 ;; -2, 3 ;;]];
    Ztop + Zleft + Zbottom + Zright - 4*Zcenter]
lapJM[Z_] := 
  Differences[ArrayPad[Z, {{0, 0}, {-1, -1}}], 2] + 
    Differences[ArrayPad[Z, {{-1, -1}, {0, 0}}], {0, 2}]

<< CompiledFunctionTools`
Compiler`$CCompilerOptions = {"SystemCompileOptions" -> "-fPIC -Ofast -march=native"};

lapxzczd = 
  Hold@Compile[{{z, _Real, 2}}, 
    Module[{d1, d2}, {d1, d2} = Dimensions@z;
      Table[
        z[[i + 1, j]] + z[[i, j + 1]] + z[[i - 1, j]] + 
         z[[i, j - 1]] - 4 z[[i, j]], 
        {i, 2, d1 - 1}, {j, 2, d2 - 1}]
    ], 
    CompilationTarget -> "C", 
    RuntimeOptions -> "Speed"] /. Part -> Compile`GetElement // ReleaseHold;

d2 = SparseArray@
   N@Sum[NDSolve`FiniteDifferenceDerivative[i, {#, #} &[Range[1000]], 
       "DifferenceOrder" -> 2][
      "DifferentiationMatrix"], {i, {{2, 0}, {0, 2}}}];

lapJens[values_] := Partition[d2.Flatten[values], Length[values]]

src = "
  #include \"WolframLibrary.h\"

  DLLEXPORT int laplacian(WolframLibraryData libData, mint Argc, \
MArgument *Args, MArgument Res) {
      MTensor tensor_A, tensor_B;
      mreal *a, *b;
      mint const *A_dims;
      mint n;
      int err;
      mint dims[2];
      mint i, j;
      tensor_A = MArgument_getMTensor(Args[0]);
      a = libData->MTensor_getRealData(tensor_A);
      A_dims = libData->MTensor_getDimensions(tensor_A);
      n = A_dims[0];
      dims[0] = dims[1] = n - 2;
      err = libData->MTensor_new(MType_Real, 2, dims, &tensor_B);
      b = libData->MTensor_getRealData(tensor_B);
      for (i = 1; i <= n - 2; i++) {
          for (j = 1; j <= n - 2; j++) {
              b[(n-2)*(i-1)+j-1] = a[n*(i-1)+j] + a[n*i+j-1] + \
a[n*(i+1)+j] + a[n*i+j+1]- 4*a[n*i+j];
          }
      }
      MArgument_setMTensor(Res, tensor_B);
      return LIBRARY_NO_ERROR;
  }
  ";
Needs["CCompilerDriver`"]
lib = CreateLibrary[src, "laplacian"];
lapShutao = LibraryFunctionLoad[lib, "laplacian", {{Real, 2}}, {Real, 2}];

compare[n_] := Block[{mat = RandomReal[10, {n, n}]},
  d2 = SparseArray@
    N@Sum[NDSolve`FiniteDifferenceDerivative[i, {#, #} &[Range[n]], 
        "DifferenceOrder" -> 2][
       "DifferentiationMatrix"], {i, {{2, 0}, {0, 2}}}];
  {AbsoluteTiming[Array[laplacian[mat] &, 10];], 
    If[n > 1000, {12345, 0}, 
     AbsoluteTiming[Array[lapJM[mat] &, 10];]], 
    AbsoluteTiming[Array[lapxzczd[mat] &, 10];], 
    AbsoluteTiming[Array[lapJens[mat] &, 10];], 
    AbsoluteTiming[Array[lapShutao[mat] &, 10];]}[[All, 1]]]

tab = Table[{Floor[1.3^i], #} & /@ compare[Floor[1.3^i]], {i, 6, 31}];

ListLinePlot[Transpose@tab, 
  PlotLegends -> {"original", "JM", "xzczd", "Jens", "Shutao"}, 
  AxesLabel -> {"Size", "Time"}]

Answer 1

Accodrding to your laplacian[] function, I can draw the following conclusion:

For a matrix $A_{n\times n}$

$$ \left( \begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{1,1} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \\ \end{array} \right)_{n \times n} $$

$$\mathcal{L}(a_{i,j}) \Longleftrightarrow b_{i-1,j-1}=a_{i-1,j}+a_{i+1,j}+a_{i,j-1}+a_{i,j+1}-4 \cdot a_{i,j}$$

where, the $b_{i,j}$ is the element of matrix $B_{(n-2)\times(n-2)}$, and $i=2,\cdots ,n-1, \quad j=2,\cdots,n-1$

Here, I will give a C solution with the help of LibraryLink wrapper.

src = "
#include \"WolframLibrary.h\"

DLLEXPORT int laplacian(WolframLibraryData libData, mint Argc, MArgument *Args, MArgument Res) {
    MTensor tensor_A, tensor_B;
    mreal *a, *b;
    mint const *A_dims;
    mint n;
    int err;
    mint dims[2];
    mint i, j, idx;
    tensor_A = MArgument_getMTensor(Args[0]);
    a = libData->MTensor_getRealData(tensor_A);
    A_dims = libData->MTensor_getDimensions(tensor_A);
    n = A_dims[0];
    dims[0] = dims[1] = n - 2;
    err = libData->MTensor_new(MType_Real, 2, dims, &tensor_B);
    b = libData->MTensor_getRealData(tensor_B);
    for (i = 1; i <= n - 2; i++) {
        for (j = 1; j <= n - 2; j++) {
            idx = n*i + j;
            b[idx+1-2*i-n] = a[idx-n] + a[idx-1] + a[idx+n] + a[idx+1] - 4*a[idx];
        }
    }
    MArgument_setMTensor(Res, tensor_B);
    return LIBRARY_NO_ERROR;
}
";

---

Needs["CCompilerDriver`"]
lib = CreateLibrary[src, "laplacian"];

lapShutao = LibraryFunctionLoad[lib, "laplacian", {{Real, 2}}, {Real, 2}]

OK, let's test it

mat = RandomReal[10, {1000, 1000}];
lapShutao[mat]; // AbsoluteTiming

Remark:

In my laptop that with 4GB RAM, I discovered that when $n = 15000$, the lapShutao[] and cLa[] will lead to system halted.

Update

For the following code:

b[(n-2)*(i-1)+j-1] = a[n*(i-1)+j] + a[n*i+j-1] + a[n*(i+1)+j] + a[n*i+j+1] - 4*a[n*i+j];

let idx = n*i+j, then the above code could be refactored as below:

b[idx+1-2*i-n] = a[idx-n] + a[idx-1] + a[idx+n] + a[idx+1] - 4*a[idx];

Answer 2

ListCorrelate is a very general function to handle operations of this type. The Laplacian is a correlation with a specific kernel.

ker = {{0, 1, 0}, {1, -4, 1}, {0, 1, 0}};
ListCorrelate[ker, Z]

which gives the same answer as your laplacian function. As for speed, this appears to be faster than laplacian for small sizes, slower for medium sizes, and faster for larger sizes. Consider the comparison:

n = 2^14;
mat = RandomReal[10, {n, n}];
AbsoluteTiming[laplacian[mat]][[1]]
ker = {{0, 1, 0}, {1, -4, 1}, {0, 1, 0}};
AbsoluteTiming[ListCorrelate[ker, mat]][[1]]
(*
  81.0357
  35.7189 
*)

Answer 3

Your compiling fails because 3;; can't be compiled, see here for more information. The easiest fix is to modify it to 3 ;; -1, but this leads to no improvement for speed. To write a faster laplacian, we can make use of the experience obtained in this post:

cLa = Hold@Compile[{{z, _Real, 2}}, 
     Module[{d1, d2}, {d1, d2} = Dimensions@z; 
      Table[z[[i + 1, j]] + z[[i, j + 1]] + z[[i - 1, j]] + z[[i, j - 1]] - 
        4 z[[i, j]], {i, 2, d1 - 1}, {j, 2, d2 - 1}]], CompilationTarget -> C, 
     RuntimeOptions -> "Speed"] /. Part -> Compile`GetElement // ReleaseHold;

mat = RandomReal[10, {1000, 1000}];
Array[laplacian[mat] &, 10]; // AbsoluteTiming
(* {0.447978, Null} *)
Array[cLa[mat] &, 10]; // AbsoluteTiming
(* {0.084375, Null} *)

Answer 4

You can get very good speed and additional flexibility by using the built-in functionality described in "The Numerical Method of Lines". For example, here I assume that the dimensions of the matrix you call Z are known a priori, so that they can be hard-coded. That allows me to define a matrix d2 that performs the Laplacian on the flattened version of Z:

d2 = 
  SparseArray@
   N@Sum[NDSolve`FiniteDifferenceDerivative[i, {#, #} &[Range[1000]], 
       "DifferenceOrder" -> 2]["DifferentiationMatrix"], {i, {{2, 0}, {0, 2}}}];

lap2[values_] := Partition[d2.Flatten[values], Length[values]]

Z = RandomReal[10, {1000, 1000}];
Array[lap2[Z] &, 10]; // AbsoluteTiming

(* ==> {0.079342, Null} *)

In the function FiniteDifferenceDerivative, I could now also specify higher differentiation orders than 2. The generalization to three dimensions would also be straightforward, by changing the sum to run over {{2,0,0},{0,2,0},{0,0,2}} instead of {{2, 0}, {0, 2}}.

Another option is to add ,PeriodicInterpolation->True after the "DifferenceOrder" option. So this lets you do lots of variations on the Laplacian very efficiently.