| 1 | #ifdef _CIVL
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| 2 | #include <civlc.cvh>
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| 3 | #endif
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| 4 | /* FILE: gaussJordan_elimination.c A gaussian-jordan elimination
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| 5 | * solver that converts a given matrix to a reduce row echelon form
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| 6 | * matrix
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| 7 | * RUN : mpicc gaussJordan_elimination.c; mpiexec -n 4 ./a.out numRow numCol m[0][0], m[0][1] ...
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| 8 | * VERIFY : civl verify gaussianJordan_elimination.c
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| 9 | */
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| 10 | #include <assert.h>
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| 11 | #include <mpi.h>
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| 12 | #include <stdio.h>
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| 13 | #include <stdlib.h>
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| 14 | #include <string.h>
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| 15 |
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| 16 | /* Message tag */
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| 17 | #define PRINT 0
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| 18 | /* Global parameters */
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| 19 | #ifdef _CIVL
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| 20 | $input int _mpi_nprocs_hi=3;
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| 21 | $input int _mpi_nprocs_lo=1;
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| 22 | $input int ROWB = 3; // upper bound of numRow
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| 23 | $input int numRow = 3; // number of rows in the matrix
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| 24 | $assume(0 < numRow && numRow <= ROWB);
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| 25 | $input int COLB = 4; // upper bound of numCol
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| 26 | $input int numCol = 3; // number of columns in the matrix
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| 27 | $assume(0 < numCol && numCol <= COLB && numCol >= numRow);
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| 28 | $input long double data[numRow][numCol]; // input matrix
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| 29 | long double oracle[numRow][numCol]; // results of sequential run
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| 30 | #else
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| 31 | int numRow; // number of rows in the matrix
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| 32 | int numCol; // number of columns in the matrix, the right-most
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| 33 | // column is vector B
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| 34 | #endif
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| 35 | int localRow; // number of rows owned by the process
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| 36 | int rank, nprocs;
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| 37 | int first; // the global index of the first row in original
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| 38 | // matrix
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| 39 | /* a Global Row Index -> Current Row Location table maps original
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| 40 | indices of rows to their current location in current matrix */
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| 41 | int *loc;
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| 42 | /* a Current Row Location -> Global Row Index table maps current
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| 43 | locations of current matrix to their original row indices */
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| 44 | int *idx;
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| 45 |
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| 46 | $assume(0==((data[1][1]*data[0][0] - 1*data[1][0]*data[0][1])*(data[2][2]*data[0][0] - 1*data[2][0]*data[0][2])
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| 47 | - 1*(data[1][2]*data[0][0] - 1*data[1][0]*data[0][2])*(data[2][1]*data[0][0] - 1*data[2][0]*data[0][1])));
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| 48 | $assume(0!=((data[1][1]*data[0][0])+(-1*(data[1][0]*data[0][1]))));
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| 49 | $assume(data[0][0] != 0);
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| 50 |
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| 51 |
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| 52 | /* Book keeping functions */
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| 53 | /* Return the owner of the row given by the global index of it in
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| 54 | original matrix */
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| 55 | #define OWNER(index) ((nprocs*(index+1)-1)/numRow)
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| 56 |
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| 57 | /* Returns the global index of the first row owned
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| 58 | * by the process with given rank */
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| 59 | int firstForProc(int rank) {
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| 60 | return (rank*numRow)/nprocs;
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| 61 | }
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| 62 |
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| 63 | /* Returns the number of rows the given process owns */
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| 64 | int countForProc(int rank) {
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| 65 | int a = firstForProc(rank);
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| 66 | int b = firstForProc(rank + 1);
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| 67 |
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| 68 | return b - a;
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| 69 | }
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| 70 |
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| 71 | /* Locally print a row */
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| 72 | void printRow(long double * row) {
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| 73 | for(int k=0; k < numCol; k++)
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| 74 | printf("%2.6Lf ", row[k]);
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| 75 | printf("\n");
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| 76 | }
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| 77 |
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| 78 | /* Print the given matrix. Since each process needs to send their data
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| 79 | * to root process 0, this function is collective.
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| 80 | * Parameters:
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| 81 | * a: the (part of) matrix will be printed.
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| 82 | */
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| 83 | void printSystem(long double * a) {
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| 84 | long double recvbuf[numCol];
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| 85 |
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| 86 | // Every process follows the order of locations of rows to send their
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| 87 | // rows to process with rank 0
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| 88 | for(int i=0; i<numRow; i++)
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| 89 | if(OWNER(idx[i]) == rank && rank != 0)
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| 90 | MPI_Send(&a[(idx[i]-first)*numCol], numCol, MPI_LONG_DOUBLE, 0, i, MPI_COMM_WORLD);
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| 91 |
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| 92 | if(rank == 0) {
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| 93 | for(int i=0; i<numRow; i++) {
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| 94 | if(OWNER(idx[i]) != 0) {
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| 95 | MPI_Recv(recvbuf, numCol, MPI_LONG_DOUBLE, MPI_ANY_SOURCE, i,
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| 96 | MPI_COMM_WORLD, MPI_STATUS_IGNORE);
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| 97 | printRow(recvbuf);
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| 98 | #ifdef _CIVL
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| 99 | for(int j=0; j < numCol; j++) {
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| 100 | $assert((recvbuf[j] == oracle[i][j]), "Get %Lf while expecting %Lf at position [%d][%d]\n", recvbuf[j], oracle[i][j], i, j);
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| 101 | }
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| 102 | #endif
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| 103 | }
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| 104 | else {
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| 105 | printRow(&a[(idx[i]-first)*numCol]);
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| 106 | #ifdef _CIVL
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| 107 | for(int j=0; j < numCol; j++) {
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| 108 | $assert((a[(idx[i]-first)*numCol + j] == oracle[i][j]), "Get %Lf while expecting %Lf at position [%d][%d]\n",
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| 109 | a[(idx[i]-first)*numCol + j], oracle[i][j], i, j);
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| 110 | }
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| 111 | #endif
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| 112 | }
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| 113 | }
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| 114 | }
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| 115 | }
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| 116 |
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| 117 | void debugSystem(long double * a) {
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| 118 | long double recvbuf[numCol];
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| 119 |
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| 120 | // Every process follows the order of locations of rows to send their
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| 121 | // rows to process with rank 0
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| 122 | for(int i=0; i<numRow; i++)
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| 123 | if(OWNER(idx[i]) == rank && rank != 0)
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| 124 | MPI_Send(&a[(idx[i]-first)*numCol], numCol, MPI_LONG_DOUBLE, 0, i, MPI_COMM_WORLD);
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| 125 |
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| 126 | if(rank == 0) {
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| 127 | for(int i=0; i<numRow; i++) {
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| 128 | if(OWNER(idx[i]) != 0) {
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| 129 | MPI_Recv(recvbuf, numCol, MPI_LONG_DOUBLE, MPI_ANY_SOURCE, i,
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| 130 | MPI_COMM_WORLD, MPI_STATUS_IGNORE);
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| 131 | printRow(recvbuf);
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| 132 | }
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| 133 | else {
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| 134 | printRow(&a[(idx[i]-first)*numCol]);
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| 135 | }
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| 136 | }
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| 137 | }
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| 138 | }
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| 139 |
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| 140 | void specEliminationDebug(long double *a) {
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| 141 | long double denom;
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| 142 |
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| 143 | for(int i = 0; i < numRow; i++) {
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| 144 | int leadCol = numCol;
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| 145 | int rowOfLeadCol = i;
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| 146 |
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| 147 | for (int j=i; j < numCol; j++) {
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| 148 | for (int k=i; k < numRow; k++) {
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| 149 | if (a[k*numCol + j] != 0.0) {
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| 150 | leadCol = j;
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| 151 | rowOfLeadCol = k;
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| 152 | break;
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| 153 | }
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| 154 | }
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| 155 | if (leadCol < numCol)
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| 156 | break;
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| 157 | }
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| 158 | if (leadCol == numCol) return;
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| 159 |
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| 160 | denom = a[rowOfLeadCol * numCol + leadCol];
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| 161 | if (denom != 0.0) {
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| 162 | for (int j = leadCol; j < numCol; j++) {
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| 163 | long double tmp = a[rowOfLeadCol * numCol + j] / denom;
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| 164 |
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| 165 | a[rowOfLeadCol * numCol + j] = tmp;
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| 166 | }
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| 167 | }
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| 168 | if (rowOfLeadCol != i) {
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| 169 | long double tmp[numCol];
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| 170 |
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| 171 | memcpy(tmp, &a[i * numCol], numCol * sizeof(long double));
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| 172 | memcpy(&a[i * numCol], &a[rowOfLeadCol * numCol],
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| 173 | numCol * sizeof(long double));
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| 174 | memcpy(&a[rowOfLeadCol * numCol], tmp,
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| 175 | numCol * sizeof(long double));
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| 176 | }
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| 177 | for (int j = i+1; j < numRow; j++) {
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| 178 | long double factor = -a[j * numCol + leadCol];
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| 179 |
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| 180 | for (int k = leadCol; k < numCol; k++)
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| 181 | a[j * numCol + k] += factor * a[i * numCol + k];
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| 182 | }
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| 183 | }
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| 184 | }
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| 185 |
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| 186 | void specElimination(long double *a, int * rowLoc) {
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| 187 | long double denom; // a temporary variable will be used to
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| 188 | //divide other variables
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| 189 |
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| 190 | for(int i=0; i < numRow; i++) {
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| 191 | int leadCol = numCol; // the column where leading 1 be in
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| 192 | int rowOfLeadCol = i; // the row where leadCol be in
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| 193 |
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| 194 | /* step 1: Find out the leftmost nonzero column, interchange it with
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| 195 | the current iterated row. */
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| 196 | for(int j=i; j < numCol; j++) {
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| 197 | for(int k=i; k < numRow; k++) {
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| 198 | if(a[rowLoc[k]*numCol + j] != 0.0) {
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| 199 | leadCol = j;
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| 200 | rowOfLeadCol = k;
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| 201 | break;
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| 202 | }
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| 203 | }
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| 204 | if(leadCol < numCol)
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| 205 | break;
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| 206 | }
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| 207 | /* If there is no leading 1 in all unprocessed rows, elimination
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| 208 | terminates. */
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| 209 | if(leadCol == numCol)
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| 210 | return;
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| 211 | /* step 2: Reducing the leading number to one */
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| 212 | denom = a[rowLoc[rowOfLeadCol]*numCol + leadCol];
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| 213 | /* If the denominator is zero (or extremely nearing zero), do
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| 214 | * nothing. The reason is the denominator is the left-most
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| 215 | * nonzero element in all unprocessed rows, if it's zero, all
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| 216 | * numbers at that column in all unprocessed rows are zeros. For
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| 217 | * such a case, it's no need to do anything in this iteration.
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| 218 | */
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| 219 | if(denom != 0.0) {
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| 220 | for(int j=leadCol; j < numCol; j++) {
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| 221 | long double tmp = a[rowLoc[rowOfLeadCol]*numCol + j] / denom;
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| 222 |
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| 223 | a[rowLoc[rowOfLeadCol]*numCol + j] = tmp;
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| 224 | }
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| 225 | }
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| 226 | if(rowOfLeadCol != i) {
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| 227 | int tmp;
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| 228 |
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| 229 | tmp = rowLoc[i];
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| 230 | rowLoc[i] = rowLoc[rowOfLeadCol];
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| 231 | rowLoc[rowOfLeadCol] = tmp;
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| 232 | }
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| 233 | /* step 3: Add a suitable value to each row below row i so that they have zero at column i */
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| 234 | for(int j=i+1; j < numRow; j++) {
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| 235 | long double factor = -a[rowLoc[j]*numCol + leadCol];
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| 236 |
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| 237 | for(int k=leadCol; k < numCol; k++)
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| 238 | a[rowLoc[j]*numCol + k] += factor * a[rowLoc[i]*numCol + k];
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| 239 | }
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| 240 | }
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| 241 | }
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| 242 |
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| 243 | /* Working upward to make each leading one the only nonzero number in
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| 244 | * which column it be .
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| 245 | * Parameters:
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| 246 | * a: the matrix in a row echelon form.
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| 247 | * rowLoc: a look-up table for rows' locations
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| 248 | */
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| 249 | void specReduce(long double * a, int * rowLoc) {
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| 250 | int leadCol; // the column of the leading one in a row
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| 251 | int i;
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| 252 |
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| 253 | i = (numRow > (numCol - 1))?(numCol-2):(numRow-1);
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| 254 | for(; i>=0; i--) {
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| 255 | //Find the leading 1, but if it's an all-zero row, skip it.
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| 256 | leadCol = -1;
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| 257 | for(int j=i; j<numCol; j++) {
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| 258 | if(a[rowLoc[i]*numCol + j] != 0.0) {
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| 259 | leadCol = j;
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| 260 | break;
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| 261 | }
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| 262 | }
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| 263 | // if it's not an all-zero row, reducing all other numbers in all
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| 264 | // rows above at the column at where the leading 1 be.
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| 265 | if(leadCol > -1) {
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| 266 | for(int j=i-1; j >=0; j--) {
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| 267 | long double factor = -a[rowLoc[j]*numCol + leadCol];
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| 268 |
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| 269 | for(int k=leadCol; k < numCol; k++)
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| 270 | a[rowLoc[j]*numCol + k] += a[rowLoc[i]*numCol + k] * factor;
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| 271 | }
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| 272 | }
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| 273 | }
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| 274 | }
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| 275 |
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| 276 | /* Initializing parameters and assigning input values to the original
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| 277 | * matrix.
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| 278 | * Parameter:
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| 279 | * argc: the first argument of main function
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| 280 | * argv: the second argument of main function
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| 281 | * a: the matrix owned by the process
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| 282 | * loc: the Global Row Index -> Current Row Location table
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| 283 | * idx: the Current Row Location -> Global Row Index table
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| 284 | */
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| 285 | void initialization(int argc, char * argv[],
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| 286 | long double * a, int * loc, int * idx) {
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| 287 | #ifdef _CIVL
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| 288 | long double spec[numRow*numCol];
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| 289 | int rowLoc[numRow];
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| 290 |
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| 291 | for(int i=first; i<first+localRow; i++)
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| 292 | for(int j=0; j<numCol; j++) {
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| 293 | a[(i-first)*numCol + j] = data[i][j];
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| 294 | }
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| 295 | //sequential run
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| 296 | if(rank == 0) {
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| 297 | for(int i=0; i<numRow; i++){
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| 298 | rowLoc[i] = i;
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| 299 | memcpy(&spec[i*numCol], &data[i][0], numCol * sizeof(long double));
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| 300 | }
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| 301 | specElimination(spec, rowLoc);
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| 302 | //specReduce(spec, rowLoc);
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| 303 | for(int i=0; i<numRow; i++){
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| 304 | for(int j=0; j<numCol; j++)
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| 305 | oracle[i][j] = spec[rowLoc[i]*numCol + j];
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| 306 | }
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| 307 | printf("oracle is :\n");
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| 308 | for(int i=0; i<numRow; i++)
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| 309 | printRow(&oracle[i][0]);
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| 310 | }
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| 311 | #else
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| 312 | if((argc - 3) != numRow * numCol)
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| 313 | printf("Too few arguments.\n"
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| 314 | "Usage: mpiexec -n nprocs ./a.out n m A[0,0] A[0,1] ... A[n-1,m-1]\n"
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| 315 | " n : number of rows in matrix\n"
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| 316 | " m : number of columns in matrix\n"
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| 317 | " A[0,0] .. A[n-1,m-1] : entries of matrix (doubles)\n");
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| 318 | first = firstForProc(rank);
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| 319 | localRow = countForProc(rank);
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| 320 | //initializing matrix
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| 321 | for(int i=0; i<localRow; i++)
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| 322 | for(int j=0; j<numCol; j++)
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| 323 | sscanf(argv[(first+i)*numCol + j + 3], "%Lf", &a[i*numCol + j]);
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| 324 | #endif
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| 325 | for(int i=0; i<numRow; i++){
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| 326 | loc[i] = i;
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| 327 | idx[i] = i;
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| 328 | }
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| 329 | }
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| 330 |
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| 331 | /* Set row to location loca */
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| 332 | void setLoc(int row, int loca){
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| 333 | int tmpLoc, tmpIdx;
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| 334 |
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| 335 | tmpLoc = loc[row];
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| 336 | tmpIdx = idx[loca];
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| 337 | //swap locations(update index -> location table)
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| 338 | loc[row] = loca;
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| 339 | loc[tmpIdx] = tmpLoc;
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| 340 | //update location -> index table
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| 341 | idx[loca] = row;
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| 342 | idx[tmpLoc] = tmpIdx;
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| 343 | }
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| 344 |
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| 345 | /* Performs a gaussian elimination on the given matrix, the output
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| 346 | * matrix will finally be in row echelon form .
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| 347 | */
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| 348 | void gaussianElimination(long double *a) {
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| 349 | /* Buffer for the current toppest unprocessed row. */
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| 350 | long double top[numCol];
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| 351 |
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| 352 | /* For each row of the matrix, it will be processed once. */
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| 353 | for(int i=0; i < numRow; i++) {
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| 354 | /* owner of the current unprocessed top row */
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| 355 | int owner = OWNER(idx[i]);
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| 356 | /* the column of the next leading 1, initial value is numCol
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| 357 | * because later it will pick up a minimum number.
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| 358 | */
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| 359 | int leadCol = numCol;
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| 360 | /* the global index of the row the next leading 1 will be in */
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| 361 | int rowOfLeadCol = numRow - 1;
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| 362 | int rowOfLeadColOwner; // the owner of rowOfLeadCol
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| 363 | /* message buffer: [0]:leadCol ;[1]:rowOfLeadCol */
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| 364 | int sendbuf[2];
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| 365 | /* receive buffer: it will contain lead 1 column candidates from
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| 366 | all processes */
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| 367 | int recvbuf[2*nprocs];
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| 368 | int tmp;
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| 369 |
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| 370 | //step 1: find out the local leftmost nonzero column
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| 371 | for(int j=i; j < numCol; j++) {
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| 372 | int k, minLoc = numRow - 1;
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| 373 |
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| 374 | for(k = first; k < first + localRow; k++) {
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| 375 | // only look at unprocessed rows
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| 376 | if(loc[k] >= i && loc[k] <= minLoc) {
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| 377 | if(a[(k-first)*numCol+j] != 0.0) {
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| 378 | leadCol = j;
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| 379 | rowOfLeadCol = k;
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| 380 | minLoc = loc[k];
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| 381 | }
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| 382 | }
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| 383 | }
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| 384 | if(leadCol < numCol)
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| 385 | break;
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| 386 | }
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| 387 | sendbuf[0] = leadCol;
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| 388 | sendbuf[1] = loc[rowOfLeadCol];
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| 389 | /* All reduce the smallest column(left-most) of leading 1 to every
|
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| 390 | process */
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| 391 | MPI_Allreduce(sendbuf, recvbuf, 1, MPI_2INT, MPI_MINLOC, MPI_COMM_WORLD);
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| 392 | leadCol = recvbuf[0];
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| 393 | rowOfLeadCol = idx[recvbuf[1]];
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| 394 | //printf("rowOfLeadCol=%d, recvbuf[1]=%d\n", rowOfLeadCol, recvbuf[1]);
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| 395 | /* Now the row containing next leading 1 is decided, findout the
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| 396 | owner of it. */
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| 397 | rowOfLeadColOwner = OWNER(rowOfLeadCol);
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| 398 | // printf("rank = %d, rowOfLeadCol = %d\n", rank, rowOfLeadCol);
|
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| 399 | /* if leadCol is still initial value, it means there is no avaliable
|
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| 400 | column suitable for next leading 1. */
|
|---|
| 401 | if(leadCol == numCol)
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|---|
| 402 | return;
|
|---|
| 403 | // step 2: reducing the leading number to 1
|
|---|
| 404 | if(rank == rowOfLeadColOwner) {
|
|---|
| 405 | long double denom = a[(rowOfLeadCol - first)*numCol + leadCol];
|
|---|
| 406 |
|
|---|
| 407 | if(denom != 0.0)
|
|---|
| 408 | for(int j=leadCol; j < numCol; j++)
|
|---|
| 409 | a[(rowOfLeadCol - first)*numCol + j] = a[(rowOfLeadCol - first)*numCol + j] / denom;
|
|---|
| 410 | memcpy(top, &a[(rowOfLeadCol - first)*numCol
|
|---|
| 411 | ], numCol*sizeof(long double));
|
|---|
| 412 | }
|
|---|
| 413 | MPI_Bcast(top, numCol, MPI_LONG_DOUBLE, rowOfLeadColOwner, MPI_COMM_WORLD);
|
|---|
| 414 | /* swap the row containing next leading 1 to the top location of
|
|---|
| 415 | current submatrix */
|
|---|
| 416 | if(loc[rowOfLeadCol] != i)
|
|---|
| 417 | setLoc(rowOfLeadCol, i);
|
|---|
| 418 | /* step 3: add a suitable value to all unprocessed rows to make
|
|---|
| 419 | all numbers at the same column as leading 1 zeros. */
|
|---|
| 420 | for(int j=0; j < localRow; j++) {
|
|---|
| 421 | if(loc[j+first] > i){
|
|---|
| 422 | long double factor = -a[j*numCol + leadCol];
|
|---|
| 423 |
|
|---|
| 424 | for(int k=leadCol; k < numCol; k++) {
|
|---|
| 425 | a[j*numCol + k] += factor * top[k];
|
|---|
| 426 | }
|
|---|
| 427 | }
|
|---|
| 428 | }
|
|---|
| 429 | }
|
|---|
| 430 | }
|
|---|
| 431 |
|
|---|
| 432 | /* Perform a backward reduction on the given matrix which transforms a
|
|---|
| 433 | row echelon form to a reduced row echelon form */
|
|---|
| 434 | void backwardReduce(long double *a) {
|
|---|
| 435 | int leadCol;
|
|---|
| 436 | int owner;
|
|---|
| 437 | int i;
|
|---|
| 438 | long double top[numCol];
|
|---|
| 439 |
|
|---|
| 440 | i = (numRow > (numCol - 1))?(numCol-2):numRow-1;
|
|---|
| 441 | for(; i>=1; i--) {
|
|---|
| 442 | leadCol = -1;
|
|---|
| 443 | owner = OWNER(idx[i]);
|
|---|
| 444 | if(rank == owner)
|
|---|
| 445 | memcpy(top, &a[(idx[i] - first)*numCol + i], (numCol-i)*sizeof(long double));
|
|---|
| 446 | MPI_Bcast(top, (numCol-i), MPI_LONG_DOUBLE, owner, MPI_COMM_WORLD);
|
|---|
| 447 | //find out the leading 1 column
|
|---|
| 448 | for(int j=0; j<(numCol-i); j++){
|
|---|
| 449 | if(top[j] != 0.0){
|
|---|
| 450 | leadCol = j+i;
|
|---|
| 451 | break;
|
|---|
| 452 | }
|
|---|
| 453 | }
|
|---|
| 454 | if(leadCol == -1)
|
|---|
| 455 | continue;
|
|---|
| 456 | else {
|
|---|
| 457 | for(int j=first; j<first+localRow; j++){
|
|---|
| 458 | if(loc[j] < i){
|
|---|
| 459 | long double factor = -a[(j-first)*numCol + leadCol];
|
|---|
| 460 |
|
|---|
| 461 | for(int k=leadCol; k<numCol; k++)
|
|---|
| 462 | a[(j-first)*numCol + k] += factor*top[k-i];
|
|---|
| 463 | }
|
|---|
| 464 | }
|
|---|
| 465 | }
|
|---|
| 466 | }
|
|---|
| 467 | }
|
|---|
| 468 |
|
|---|
| 469 | int main(int argc, char *argv[]) {
|
|---|
| 470 | long double *a;
|
|---|
| 471 |
|
|---|
| 472 | #ifndef _CIVL
|
|---|
| 473 | if(argc < 3)
|
|---|
| 474 | printf("Expecting the arguments: numberOfRows numberOfColumns\n");
|
|---|
| 475 | numRow = atoi(argv[1]);
|
|---|
| 476 | numCol = atoi(argv[2]);
|
|---|
| 477 | #else
|
|---|
| 478 | $elaborate(numRow);
|
|---|
| 479 | $elaborate(numCol);
|
|---|
| 480 | // for(int i = 0; i < numRow; i++)
|
|---|
| 481 | // for(int j = 0; j < numCol; j++)
|
|---|
| 482 | // $assume(data[i][j] != 0);
|
|---|
| 483 | #endif
|
|---|
| 484 | MPI_Init(&argc, &argv);
|
|---|
| 485 | MPI_Comm_rank(MPI_COMM_WORLD, &rank);
|
|---|
| 486 | MPI_Comm_size(MPI_COMM_WORLD, &nprocs);
|
|---|
| 487 | first = firstForProc(rank);
|
|---|
| 488 | localRow = countForProc(rank);
|
|---|
| 489 | a = (long double*)malloc(numCol*localRow*sizeof(long double));
|
|---|
| 490 | loc = (int*)malloc(numRow*sizeof(int));
|
|---|
| 491 | idx = (int*)malloc(numRow*sizeof(int));
|
|---|
| 492 | initialization(argc, argv, a, loc, idx);
|
|---|
| 493 | gaussianElimination(a);
|
|---|
| 494 | // backwardReduce(a);
|
|---|
| 495 | if(!rank)printf("After backward reduction, the matrix in reduced row echelon form is:\n");
|
|---|
| 496 | printSystem(a);
|
|---|
| 497 | MPI_Finalize();
|
|---|
| 498 | free(loc);
|
|---|
| 499 | free(idx);
|
|---|
| 500 | free(a);
|
|---|
| 501 | return 0;
|
|---|
| 502 | }
|
|---|
| 503 |
|
|---|