LinearSystem.mesa
Last edited by Maureen Stone 19-Oct-81 16:42:28
Written by Michael Plass, 8-Oct-81
LinearSystem: DEFINITIONS =
BEGIN
Real0: TYPE = REAL ← 0;
Catch the signal Real.RealException to detect singular or unstable systems, e.g.,
singular:BOOLEAN ← FALSE;
x ← LinearSystem.Solve3[A,b!
Real.RealException => {singular←TRUE;CONTINUE}];
IF NOT singular THEN ...
VecSeq: TYPE = RECORD[SEQUENCE ncols: INTEGER OF Real0];
RowN: TYPE = REF VecSeq;
MatrixSeq: TYPE = RECORD[SEQUENCE nrows: INTEGER OF RowN];
MatrixN: TYPE = REF MatrixSeq;
ColumnN: TYPE = REF VecSeq;
SolveN: PROCEDURE [A:MatrixN, b:ColumnN, n: INTEGER] RETURNS [x:ColumnN] ; -- solves Ax=b
Matrix2: TYPE = ARRAY [1..2] OF Row2;
Row2: TYPE = ARRAY [1..2] OF REAL;
Column2: TYPE = ARRAY [1..2] OF Real0;
Solve2: PROCEDURE [A:Matrix2, b:Column2] RETURNS [x:Column2] ; -- solves Ax=b
Matrix3: TYPE = ARRAY [1..3] OF Row3;
Row3: TYPE = ARRAY [1..3] OF REAL;
Column3: TYPE = ARRAY [1..3] OF Real0;
Solve3: PROCEDURE [A:Matrix3, b:Column3] RETURNS [x:Column3] ; -- solves Ax=b
Matrix4: TYPE = ARRAY [1..4] OF Row4;
Row4: TYPE = ARRAY [1..4] OF REAL;
Column4: TYPE = ARRAY [1..4] OF Real0;
Solve4: PROCEDURE [A:Matrix4, b:Column4] RETURNS [x:Column4] ; -- solves Ax=b
Matrix5: TYPE = ARRAY [1..5] OF Row5;
Row5: TYPE = ARRAY [1..5] OF REAL;
Column5: TYPE = ARRAY [1..5] OF Real0;
Solve5: PROCEDURE [A:Matrix5, b:Column5] RETURNS [x:Column5] ; -- solves Ax=b
Matrix6: TYPE = ARRAY [1..6] OF Row6;
Row6: TYPE = ARRAY [1..6] OF REAL;
Column6: TYPE = ARRAY [1..6] OF Real0;
Solve6: PROCEDURE [A:Matrix6, b:Column6] RETURNS [x:Column6] ; -- solves Ax=b
END.