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Metoda gauss


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Metoda gauss

  1. 1. Systems of linear equations Gauss method
  2. 2. Example Let the following system of linear equations is given: To solve this system we’ll remove variable x 1 from second equation: (multiply first equation with –4/2 and add to second)
  3. 3. Find x 2 from second equation: Now find x 1 from first equation (using x 2 ):
  4. 4. General Algorithm: (direct step) <ul><li>Exclude x 1 from all equations starting with 2 nd </li></ul><ul><li>Exclude x 2 from all equations starting with 3 th ............ </li></ul><ul><li>Exclude x i from all equations starting with ( i +1 ) th ............ </li></ul><ul><li>Exclude x n -1 from equation with index n </li></ul>
  5. 5. General Algorithm: (inverse step) <ul><li>Find x n from equation with index n </li></ul><ul><li>Find x n-1 from equation with index n-1 ........ </li></ul><ul><li>Find x i from equation with index i ........ </li></ul><ul><li>Find x 1 from equation with index 1 </li></ul>
  6. 6. Let the following system is given: Our task is to transform it to triangular form:
  7. 7. To transform initial system to triangular form we’ll use following algorithm: For equations from 1 to n-1 For equations from i+1 to n Multiply all coefficients of equation (i) with - a j, i / a i,i then add equation (i) to equation (j) . Notes: 1. If a i,i =0 then change equation (i) with equation (k), k > i , in which a k,i  0 2. Equation (i) of system will not be changed after multiplication
  8. 8. After transformations we have system: Our task is to find solutions x n , x n-1 , ... x 1
  9. 9. To solve this problem we’ll use following algorithm: <ul><li>From equation with index (n) find x n </li></ul><ul><li>In equation (n-1) place value of x n and find x n-1 </li></ul><ul><li>... </li></ul><ul><li>In equation (i) place values of x n , x n-1 , ... , x i+1 and find x i </li></ul><ul><li>... </li></ul><ul><li>In equation (1) place values of x n , x n-1 , ... , x 2 and find x 1 </li></ul>
  10. 10. Data structures: <ul><li>Two-dimensional array A[n,n+1] – for system coefficients and free terms of equations. Line i of array will contain coefficients of equation i of first n places and free term of this equation on n+1 place </li></ul><ul><li>One-dimensional array X[n] – for system solutions. Element X[i] of array will store solution component x i . </li></ul>
  11. 11. Sample program: program sys_equ; const max = 50; type mat = array[1..max+1, 1..max] of real; Sol = array[1..max] of real; var a: mat; x: Sol; n, i, j: integer; procedure exclude (var AA:mat;i,j:integer); var k: integer; r: real; begin r:=-AA[j,i]/AA[i,i]; for k:=i to n+1 do AA[j,k]:=AA[j,k]+AA[i,k]*r; end;
  12. 12. function find (AA:mat;i:integer):integer; var k:integer; begin find:=i; for k:=i+1 to n do if AA[k,i]<>0 then find:=k; end; procedure change (var AA:mat;i,j:integer); var r:real ; k:integer; begin for k:=1 to n do begin r:=a[i,k]; a[i,k]:=a[j,k]; a[j,k]:=r; end; end;
  13. 13. procedure transform (var AA:mat); var i,j,l:integer; begin for i:=1 to n-1 do begin if AA[i,i]= 0 then begin l:= find (AA,i); if l<>i then change (AA,i,l) else begin write ('Not unique solution'); readln; halt; end; end; for j:=i+1 to n do exclude (AA,i,j); end; end;
  14. 14. procedure readdata (var AA:mat; var n:integer); var i,j : integer; f : text; begin assign (f, ''); reset (f); readln (f,n); for i:=1 to n do for j:=1 to n+1 do read (f, AA[i,j]); close (f); end; procedure outputsol (X:sol); var i: integer; begin for i:=1 to n do writeln ('x[',i,']=',X[i]:0:3) ; readln; end ;
  15. 15. procedure calculate (AA:mat; var X:sol); var s : real; i,j : integer; begin for i:=n downto 1 do begin s:=0; for j:=i+1 to n do s:=s+AA[i,j]*x[j]; x[i]:=(AA[i,n+1]-s)/AA[i,i]; end; end; begin readdata (A,n); transform (A); calculate (A,X); outputsol (X); end .