Solve equations with substitution method: Difference between revisions
Solve equations with substitution method (view source)
Revision as of 08:08, 14 April 2024
, 1 month ago→{{header|jq}
(Solve equations with substitution method en Python) |
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=={{header|ALGOL 68}}==
{{Trans|Phix|second version (translation of Raku)}}
<syntaxhighlight lang="algol68">
BEGIN # solve equations using the substitution method - translation of Phix #
PROC solve2 = ( []REAL e1, e2 )[]REAL:
BEGIN
REAL a1 = e1[ 1 ], b1 = e1[ 2 ], c1 = e1[ 3 ];
REAL a2 = e2[ 1 ], b2 = e2[ 2 ], c2 = e2[ 3 ];
REAL x = (b2*c1 - b1*c2) / (b2*a1 - b1*a2);
REAL y = (a1*x - c1)/-b1;
( x, y )
END # solve2 # ;
OP FMT = ( REAL v )STRING: # prints v with at most 3 decimal places #
BEGIN
STRING result := fixed( ABS v, 0, 3 );
IF result[ LWB result ] = "." THEN "0" +=: result FI;
WHILE result[ UPB result ] = "0" DO result := result[ : UPB result - 1 ] OD;
IF result[ UPB result ] = "." THEN result := result[ : UPB result - 1 ] FI;
IF v < 0 THEN "-" ELSE "" FI + result
END # FMT # ;
[]REAL xy = solve2( ( 3, 1, -1 ), ( 2, -3, -19 ) );
print( ( "x = ", FMT xy[ 1 ], ", y = ", FMT xy[ 2 ], newline ) )
END
</syntaxhighlight>
{{out}}
<pre>
x = -2, y = 5
</pre>
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f SOLVE_EQUATIONS_WITH_SUBSTITUTION_METHOD.AWK
BEGIN {
main("3,1,-1","2,-3,-19")
exit(0)
}
function main(s1,s2, arr,e1,e2,result_x,result_y,r1,r2,x1,x2,y1,y2) {
split(s1,e1,",")
split(s2,e2,",")
x1 = e1[1]
y1 = e1[2]
r1 = e1[3]
x2 = e2[1]
y2 = e2[2]
r2 = e2[3]
arr[1] = x1
arr[2] = -y1
arr[3] = r1
result_y = ((arr[1]*r2) - (x2*arr[3])) / ((x2*arr[2]) + (arr[1]*y2))
result_x = (r1 - (y1*result_y)) / x1
printf("x = %g\ny = %g\n",result_x,result_y)
}
</syntaxhighlight>
{{out}}
<pre>
x = -2
y = 5
</pre>
=={{header|BASIC}}==
==={{header|BASIC256}}===
{{trans|FreeBASIC}}
<
dim firstEquation(3)
firstEquation[1] = 3
Line 45 ⟶ 108:
call getCrossingPoint(firstEquation, secondEquation)
end</
{{out}}
<pre>
Line 52 ⟶ 115:
==={{header|FreeBASIC}}===
<
Dim Shared As Integer secondEquation(1 To 3) = { 2,-3,-19}
Line 73 ⟶ 136:
getCrossingPoint(firstEquation(), secondEquation())
Sleep</
{{out}}
<pre>x = -2
Line 82 ⟶ 145:
{{works with|QuickBasic}}
{{trans|FreeBASIC}}
<
firstEquation(1) = 3
firstEquation(2) = 1
Line 109 ⟶ 172:
PRINT "x = "; resultX
PRINT "y = "; resultY
END SUB</
{{out}}
<pre>
Line 118 ⟶ 181:
{{works with|QBasic}}
{{trans|QBasic}}
<
LET firstequation(1) = 3
LET firstequation(2) = 1
Line 147 ⟶ 210:
CALL getcrossingpoint (firstequation(), secondequation())
END</
{{out}}
<pre>
Line 155 ⟶ 218:
==={{header|Yabasic}}===
{{trans|FreeBASIC}}
<
firstEquation(1) = 3
firstEquation(2) = 1
Line 182 ⟶ 245:
getCrossingPoint(firstEquation(), secondEquation())
end</
{{out}}
<pre>
Igual que la entrada de FreeBASIC.
</pre>
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
Rather than just provide a one-off solution for this one problem, I've included code that will solve any system of equations with n-equations and n-unknowns. The code revolves around a matrix object, that contains an N by N matrix. The object can perform Gausian Elimination, which reduces the matrix to "Row Eschelon" format. From Row Eschelon format, you can back substitute and get the values for the unknowns. It also includes code to do "Gauss-Jordan" elimination, which eliminates the back substitution step. The object presented here is a subset of a more elaborate object that can perform all kinds of matrix operations.
<syntaxhighlight lang="Delphi">
{This code normally resides in a library, it is provided here for clarity}
type T2DMatrix = class(TObject)
private
FDoubleArray: array of double;
FRows,FColumns: integer;
procedure Put(Row,Col: integer; Item: double);
function Get(Row,Col: integer): double;
protected
public
constructor Create(Row,Col: integer);
property Rows: integer read FRows;
property Columns: integer read FColumns;
procedure SetDimensions(Row,Col: integer);
property Value[Row,Col: integer]: Double read Get write Put; default;
function BackSubstitute: TDoubleDynArray;
procedure GausianElimination;
function GaussBackSubsituation: TDoubleDynArray;
procedure GaussJordanElimination;
procedure ExchangeRows(R1, R2: integer);
function MatrixStr(Digits: integer): string;
end;
{ T2DMatrix }
procedure T2DMatrix.SetDimensions(Row, Col: integer);
begin
FRows:=Row; FColumns:=Col;
SetLength(FDoubleArray,Row * Col);
end;
constructor T2DMatrix.Create(Row, Col: integer);
begin
SetDimensions(Row, Col);
end;
procedure T2DMatrix.Put(Row,Col: integer; Item: double);
{Insert Double at index}
var Inx: integer;
begin
Inx:=(Col * Rows) + Row;
FDoubleArray[Inx]:=Item;
end;
function T2DMatrix.Get(Row,Col: integer): double;
{Get Double at the index}
var Inx: integer;
begin
Inx:=(Col * Rows) + Row;
Result:=FDoubleArray[Inx];
end;
{Matrix operations}
procedure T2DMatrix.ExchangeRows(R1,R2: integer);
{Exchange the specified Rows}
var Col: integer;
var T: double;
begin
for Col:=0 to Self.Columns-1 do
begin
T:=Self[R1,Col];
Self[R1,Col]:=Self[R2,Col];
Self[R2,Col]:=T;
end;
end;
procedure T2DMatrix.GausianElimination;
var I,K,J : Integer;
var S : double;
{Small value to prevent divide by zero}
const Epsilon = 5.0e-162;
begin
{Do gaussian elimination and convert Row Echelon}
K:=0;
while True do
begin
{If the pivot is zero, find another row with non-zero in the same column}
if Self[K,K]=0 then
begin
for I:=K+1 to Self.Rows-1 do
if Self[I,K]<>0 then
begin
Self.ExchangeRows(K,I);
break;
end;
end;
{Get pivot again}
S:=Self[K,K];
{if it is still zero, prevent divide by zero}
if S=0.0 then S:=Epsilon;
{Use "scaling primative row operation" to set pivot to one i.e divide each item by pivot}
for J:=K to Self.Columns-1 do Self[K,J]:=Self[K,J]/S;
{Exit if we are on the bottom row}
if K>=Self.Rows-1 then break;
{Now that the previous row has 1 in the leading coefficient (Pivot)}
{We convert all remaining columns below this item to zero }
{Using "pivot primative row operation" = Current Row - Pivot * Start Row}
for I:=K+1 to Self.Rows-1 do
{Do this to current column for all remaining rows}
begin
{Get the leading coefficient}
S:=Self[I,K];
{Use it to zero column and apply to all items in row}
for J:=K to Self.Columns-1 do Self[I,J]:=Self[I,J]-S*Self[K,J];
end;
{Point to next pivot}
K:=K+1;
end;
{Matrix is now in Row Echelon form }
end;
function T2DMatrix.BackSubstitute: TDoubleDynArray;
var Row,J: integer;
var Sum: double;
begin
SetLength(Result,Self.Rows);
for Row:=Self.Rows-1 downto 0 do
begin
Result[Row]:=Self[Row,Self.Columns-1];
Sum:=0;
for J:=Self.Rows-1 downto Row+1 do
Sum:=Sum + Result[J] * Self[Row,J];
Result[Row]:=Result[Row] - Sum;
end;
end;
function T2DMatrix.GaussBackSubsituation: TDoubleDynArray;
begin
Self.GausianElimination;
Result:=Self.BackSubstitute;
end;
procedure T2DMatrix.GaussJordanElimination;
{ Do Gauss Jordan Elimination }
var I,K,J : Integer;
var S : double;
var X,Y: integer;
var Str: string;
{Small value to prevent divide by zero}
const Epsilon = 5.0e-162;
begin
{Do Gaussian Elimination to put matrix in Row Echelon format}
Self.GausianElimination;
{Now do the Jordan part to put it in Reduced Row Echelon form}
{Point K at the both the source row and target column (left part of matrix is square)}
K:=Self.Rows-1;
while K>0 do
begin
{Point I at the target row}
I:=K-1;
repeat
begin
{Get the coefficient of target column and row, the value we want to zero }
S:=Self[I,K];
{Multiply source row by the target coefficient and subtract from each item in the target row}
{Since the target column is one in the source row, this will zero out the target coefficient}
for J:=K to Self.Columns-1 do Self[I,J]:=Self[I,J]-S*Self[K,J];
{Point to previous row}
I:=I-1;
end
until I<0;
{Point to previous column}
K:=K-1;
end;
end;
function T2DMatrix.MatrixStr(Digits: integer): string;
var Row,Col: integer;
begin
Result:=IntToStr(Self.Rows)+'X'+IntToStr(Self.Columns)+CRLF_Char;
for Row:=0 to Self.Rows-1 do
begin
Result:=Result+'[';
for Col:=0 to Self.Columns-1 do
begin
if Col<>0 then Result:=Result+' ';
Result:=Result+FloatToStrF(Self[Row,Col],ffFixed,18,Digits);
end;
Result:=Result+']'+CRLF_Char;
end;
end;
{===============================================================================}
procedure SolveEquations(Memo: TMemo);
var Mat: T2DMatrix;
var DA: TDoubleDynArray;
var I: integer;
var S: string;
begin
Mat:=T2DMatrix.Create(2,3);
try
{3x + y = -1}
{2x - 3y = -19}
Mat[0,0]:=3; Mat[0,1]:=1; Mat[0,2]:=-1;
Mat[1,0]:=2; Mat[1,1]:=-3; Mat[1,2]:=-19;
Memo.Lines.Add('Solve with Gaussian Elimination and Substitution');
Memo.Lines.Add('------------------------------------------------');
Memo.Lines.Add(Mat.MatrixStr(1));
Mat.GausianElimination;
Memo.Lines.Add('Row Echelon after Gaussian Elimination');
Memo.Lines.Add(Mat.MatrixStr(1));
DA:=Mat.BackSubstitute;
Memo.Lines.Add('Matrix after Back Substitution');
Memo.Lines.Add(Mat.MatrixStr(1));
Memo.Lines.Add(Format('Solution: X=%2.2f, Y=%2.2f',[DA[0],DA[1]]));
Mat[0,0]:=3; Mat[0,1]:=1; Mat[0,2]:=-1;
Mat[1,0]:=2; Mat[1,1]:=-3; Mat[1,2]:=-19;
Memo.Lines.Add('');
Memo.Lines.Add('Solve with Gaussian Jordan elimination');
Memo.Lines.Add('--------------------------------------');
Memo.Lines.Add(Mat.MatrixStr(1));
Mat.GaussJordanElimination;
Memo.Lines.Add('Matrix after Gauss-Jordan');
Memo.Lines.Add(Mat.MatrixStr(1));
Memo.Lines.Add(Format('Solution: X=%2.2f, Y=%2.2f',[Mat[0,2],Mat[1,2]]));
finally Mat.Free; end;
end;
</syntaxhighlight>
{{out}}
<pre>
Solve with Gaussian Elimination and Substitution
------------------------------------------------
2X3
[3.0 1.0 -1.0]
[2.0 -3.0 -19.0]
Row Echelon after Gaussian Elimination
2X3
[1.0 0.3 -0.3]
[0.0 1.0 5.0]
Matrix after Back Substitution
2X3
[1.0 0.3 -0.3]
[0.0 1.0 5.0]
Solution: X=-2.00, Y=5.00
Solve with Gaussian Jordan elimination
--------------------------------------
2X3
[3.0 1.0 -1.0]
[2.0 -3.0 -19.0]
Matrix after Gauss-Jordan
2X3
[1.0 0.0 -2.0]
[0.0 1.0 5.0]
Solution: X=-2.00, Y=5.00
Elapsed Time: 21.255 ms.
</pre>
=={{header|jq}}==
'''Works with jq, the C implementation of jq'''
'''Works with gojq, the Go implementation of jq'''
The solution presented here handles all the edge cases.
With trivial modifications, the following will also work with jaq, the Rust implementation of jq.
<syntaxhighlight lang="jq">
# The equation ax + by = c is represented by the array [a, b, c]
def solve( $e1; $e2 ):
$e1 as [$a1, $b1, $c1]
| $e2 as [$a2, $b2, $c2]
| ($b2 * $a1 - $b1 * $a2 ) as $d
| if $d == 0 then "there is no unique solution as the discriminant is 0" | error
else
{ x : (($b2 * $c1 - $b1 * $c2 ) / $d) }
| if $b1 != 0
then .y = ( $a1 * .x - $c1 ) / -$b1
else .y = ( $a2 * .x - $c2 ) / -$b2
end
end;
solve( [3,1,-1]; [2,-3,-19] )
</syntaxhighlight>
{{output}}
<pre>
{
"x": -2,
"y": 5
}
</pre>
=={{header|Julia}}==
<
ab, c = strip.(split(s, "="))
a, by = strip.(split(ab, "x"))
Line 208 ⟶ 586:
@show solvetwolinear("3x + y = -1", "2x - 3y = -19") # solvetwolinear("3x + y = -1", "2x - 3y = -19") = (-2.0, 5.0)
</syntaxhighlight>
=={{header|Nim}}==
{{trans|Python}}
<syntaxhighlight lang="Nim">type Equation = tuple[cx, cy, cr: float] # cx.x + cy.y = cr.
func getCrossingPoint(firstEquation, secondEquation: Equation): tuple[x, y: float] =
let (x1, y1, r1) = firstEquation
let (x2, y2, r2) = secondEquation
let temp = (x1, -y1, r1)
result.y = ((temp[0] * r2) - (x2 * temp[2])) / ((x2 * temp[1]) + (temp[0] * y2))
result.x = (r1 - (y1 * result.y)) / x1
when isMainModule:
let firstEquation: Equation = (3, 1, -1)
let secondEquation: Equation = (2, -3, -19)
let (x, y) = getCrossingPoint(firstEquation, secondEquation)
echo "x = ", x
echo "y = ", y
</syntaxhighlight>
{{out}}
<pre>x = -2.0
y = 5.0
</pre>
=={{header|Perl}}==
<
use warnings;
use feature 'say';
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}
say my $result = join ' ', solve( parse('3x + y = -1'), parse('2x - 3y = -19') );</
{{out}}
<pre>-2 5</pre>
Line 236 ⟶ 638:
=={{header|Phix}}==
Slightly modified copy of solveN() from [[Solving_coin_problems#Phix]], admittedly a tad overkill for this task, as it takes any number of rules and any number of variables.
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">procedure</span> <span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">rules</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">unknowns</span><span style="color: #0000FF;">)</span>
Line 293 ⟶ 695:
<span style="color: #000080;font-style:italic;">--for 3x + y = -1 and 2x - 3y = -19:</span>
<span style="color: #000000;">solve</span><span style="color: #0000FF;">({{-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">},{-</span><span style="color: #000000;">19</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,-</span><span style="color: #000000;">3</span><span style="color: #0000FF;">}},{</span><span style="color: #008000;">"x"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"y"</span><span style="color: #0000FF;">})</span>
<!--</
{{out}}
<pre>
Line 300 ⟶ 702:
Alternatively, since I'm staring right at it, here's a
{{trans|Raku}}
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">solve2</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">e1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">e2</span><span style="color: #0000FF;">)</span>
Line 311 ⟶ 713:
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"x = %d, y = %d\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve2</span><span style="color: #0000FF;">({</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">},{</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,-</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,-</span><span style="color: #000000;">19</span><span style="color: #0000FF;">}))</span>
<!--</
{{out}}
<pre>
Line 319 ⟶ 721:
=={{header|Python}}==
<
firstEquation = [ 3, 1, -1]
Line 342 ⟶ 744:
if __name__ == "__main__":
getCrossingPoint(firstEquation, secondEquation)</
{{out}}
<pre>x = -2.0
Line 349 ⟶ 751:
=={{header|Raku}}==
<syntaxhighlight lang="raku"
my \X = ( b2 * c1 - b1 * c2 )
/ ( b2 * a1 - b1 * a2 );
Line 357 ⟶ 759:
return X, Y;
}
say solve-system-of-two-linear-equations( (3,1,-1), (2,-3,-19) );</
{{out}}
<pre>(-2 5)</pre>
=={{header|Ring}}==
<
firstEquation = [3.0,1.0,-1.0] secondEquation = [2.0,-3.0,-19.0]
getCrossingPoint(firstEquation,secondEquation)
Line 372 ⟶ 774:
resultY = ((temp[1]* r2) - (x2 * temp[3])) / ((x2 * temp[2]) + (temp[1]*y2)) resultX = (r1 - (y1*resultY)) / x1
see "x = " + resultX + nl + "y = " + resultY + nl
</syntaxhighlight>
{{out}}
<pre>
Line 380 ⟶ 782:
=={{header|Wren}}==
<
e2 = e2.toList
for (i in 1..2) e2[i] = e2[i] * e1[0] / e2[0]
Line 391 ⟶ 793:
var e2 = [2, -3, -19]
var sol = solve.call(e1, e2)
System.print("x = %(sol[0]), y = %(sol[1])")</
{{out}}
Line 400 ⟶ 802:
=={{header|XPL0}}==
This shows the vector routines from xpllib.xpl.
<
real A, B, C; \A:= B - C
[A(0):= B(0) - C(0); \VSub(A, A, C) => A:= A - C
Line 430 ⟶ 832:
Text(0, "x = "); RlOut(0, X); CrLf(0);
Text(0, "y = "); RlOut(0, Y); CrLf(0);
]</
{{out}}
| |||