How to simplify output from solve

조회 수: 31 (최근 30일)
Simon
Simon 2024년 12월 13일 21:12
댓글: Star Strider 2024년 12월 14일 2:00
I execute the following commands:
% Define symbolic variables
syms a b c d e f g h i
syms x1 y1 x2 y2 x3 y3 x4 y4
syms x y
% Define the formula
denom = g*x + h*y + i;
f1 = (a*x + b*y + c)/denom;
f2 = (d*x + e*y + f)/denom;
% Define the constraint
constraint = [subs(f1, [x, y], [1, 1]) == x1,
subs(f1, [x, y], [-1, 1]) == x2,
subs(f1, [x, y], [1, -1]) == x3,
subs(f1, [x, y], [-1, -1]) == x4,
subs(f2, [x, y], [1, 1]) == y1,
subs(f2, [x, y], [-1, 1]) == y2,
subs(f2, [x, y], [1, -1]) == y3,
subs(f2, [x, y], [-1, -1]) == y4];
% Solve for a to i
vars = [a, b, c, d, e, f, g, h, i];
solution = solve(constraint, vars);
% Display the solution
disp('Solution:');
disp(solution);
Which outputs:
a: (x1*x3*y2 - x2*x3*y1 - x1*x4*y2 + x2*x4*y1 - x1*x3*y4 + x1*x4*y3 + x2*x3*y4 - x2*x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
b: -(x1*x2*y3 - x2*x3*y1 - x1*x2*y4 + x1*x4*y2 - x1*x4*y3 + x3*x4*y1 + x2*x3*y4 - x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
c: -(x1*x2*y3 - x1*x3*y2 - x1*x2*y4 + x2*x4*y1 + x1*x3*y4 - x3*x4*y1 - x2*x4*y3 + x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
d: (x1*y2*y3 - x2*y1*y3 - x1*y2*y4 + x2*y1*y4 - x3*y1*y4 + x4*y1*y3 + x3*y2*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
e: -(x1*y2*y3 - x3*y1*y2 - x2*y1*y4 + x4*y1*y2 - x1*y3*y4 + x3*y1*y4 + x2*y3*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
f: -(x2*y1*y3 - x3*y1*y2 - x1*y2*y4 + x4*y1*y2 + x1*y3*y4 - x4*y1*y3 - x2*y3*y4 + x3*y2*y4)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
g: (x1*y3 - x3*y1 - x1*y4 - x2*y3 + x3*y2 + x4*y1 + x2*y4 - x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
h: -(x1*y2 - x2*y1 - x1*y4 + x2*y3 - x3*y2 + x4*y1 + x3*y4 - x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
i: 1
What commands should I run in order to simplify these expressions and find common factors between these formulas?
Thanks!
  댓글 수: 3
Torsten
Torsten 2024년 12월 13일 23:45
편집: Torsten 2024년 12월 13일 23:46
You should normalize variable i to 1 right at the beginning so that you have 8 equations in 8, not 9 unknowns.
% Define symbolic variables
syms a b c d e f g h
syms x1 y1 x2 y2 x3 y3 x4 y4
syms x y
% Define the formula
denom = g*x + h*y + 1;
f1 = (a*x + b*y + c)/denom;
f2 = (d*x + e*y + f)/denom;
% Define the constraint
constraint = [subs(f1, [x, y], [1, 1]) == x1,
subs(f1, [x, y], [-1, 1]) == x2,
subs(f1, [x, y], [1, -1]) == x3,
subs(f1, [x, y], [-1, -1]) == x4,
subs(f2, [x, y], [1, 1]) == y1,
subs(f2, [x, y], [-1, 1]) == y2,
subs(f2, [x, y], [1, -1]) == y3,
subs(f2, [x, y], [-1, -1]) == y4];
% Solve for a to i
vars = [a, b, c, d, e, f, g, h];
solution = solve(constraint, vars);
% Display the solution
disp('Solution:');
Solution:
disp(solution);
a: (x1*x3*y2 - x2*x3*y1 - x1*x4*y2 + x2*x4*y1 - x1*x3*y4 + x1*x4*y3 + x2*x3*y4 - x2*x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) b: -(x1*x2*y3 - x2*x3*y1 - x1*x2*y4 + x1*x4*y2 - x1*x4*y3 + x3*x4*y1 + x2*x3*y4 - x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) c: -(x1*x2*y3 - x1*x3*y2 - x1*x2*y4 + x2*x4*y1 + x1*x3*y4 - x3*x4*y1 - x2*x4*y3 + x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) d: (x1*y2*y3 - x2*y1*y3 - x1*y2*y4 + x2*y1*y4 - x3*y1*y4 + x4*y1*y3 + x3*y2*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) e: -(x1*y2*y3 - x3*y1*y2 - x2*y1*y4 + x4*y1*y2 - x1*y3*y4 + x3*y1*y4 + x2*y3*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) f: -(x2*y1*y3 - x3*y1*y2 - x1*y2*y4 + x4*y1*y2 + x1*y3*y4 - x4*y1*y3 - x2*y3*y4 + x3*y2*y4)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) g: (x1*y3 - x3*y1 - x1*y4 - x2*y3 + x3*y2 + x4*y1 + x2*y4 - x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) h: -(x1*y2 - x2*y1 - x1*y4 + x2*y3 - x3*y2 + x4*y1 + x3*y4 - x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)
Simon
Simon 2024년 12월 14일 0:29
@John D'Errico I know from this thread that it can be simplified https://math.stackexchange.com/questions/186286/get-transformation-matrix-from-points. However as I want to extend the original formulas, having such simplifications automated would be helpful (and useful to know about in general).

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답변 (1개)

Star Strider
Star Strider 2024년 12월 13일 22:22
I can’t be certain that this is a significant improvement, however it is the best I can do with your data —
% Define symbolic variables
syms a b c d e f g h i
syms x1 y1 x2 y2 x3 y3 x4 y4
syms x y
% Define the formula
denom = g*x + h*y + i;
f1 = (a*x + b*y + c)/denom;
f2 = (d*x + e*y + f)/denom;
% Define the constraint
constraint = [subs(f1, [x, y], [1, 1]) == x1,
subs(f1, [x, y], [-1, 1]) == x2,
subs(f1, [x, y], [1, -1]) == x3,
subs(f1, [x, y], [-1, -1]) == x4,
subs(f2, [x, y], [1, 1]) == y1,
subs(f2, [x, y], [-1, 1]) == y2,
subs(f2, [x, y], [1, -1]) == y3,
subs(f2, [x, y], [-1, -1]) == y4];
% Solve for a to i
vars = [a, b, c, d, e, f, g, h, i];
solution = solve(constraint, vars);
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
sc = struct2cell(solution);
solutions = cell(size(sc));
for k = 1:numel(sc)
solutions{k,:} = [vars(k) simplify(sc{k}, 500)];
end
% Display the solution
disp('Solution:');
Solution:
disp(solutions);
{[a (x1*x3*y2 - x2*x3*y1 - x1*x4*y2 + x2*x4*y1 - x1*x3*y4 + x1*x4*y3 + x2*x3*y4 - x2*x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) ]} {[b -(x1*x2*y3 - x2*x3*y1 - x1*x2*y4 + x1*x4*y2 - x1*x4*y3 + x3*x4*y1 + x2*x3*y4 - x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)]} {[c -(x1*x2*y3 - x1*x3*y2 - x1*x2*y4 + x2*x4*y1 + x1*x3*y4 - x3*x4*y1 - x2*x4*y3 + x3*x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)]} {[d (x1*y2*y3 - x2*y1*y3 - x1*y2*y4 + x2*y1*y4 - x3*y1*y4 + x4*y1*y3 + x3*y2*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) ]} {[e -(x1*y2*y3 - x3*y1*y2 - x2*y1*y4 + x4*y1*y2 - x1*y3*y4 + x3*y1*y4 + x2*y3*y4 - x4*y2*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)]} {[f -(x2*y1*y3 - x3*y1*y2 - x1*y2*y4 + x4*y1*y2 + x1*y3*y4 - x4*y1*y3 - x2*y3*y4 + x3*y2*y4)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3)]} {[g (x1*y3 - x3*y1 - x1*y4 - x2*y3 + x3*y2 + x4*y1 + x2*y4 - x4*y2)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) ]} {[h -(x1*y2 - x2*y1 - x1*y4 + x2*y3 - x3*y2 + x4*y1 + x3*y4 - x4*y3)/(x1*y2 - x2*y1 - x1*y3 + x3*y1 + x2*y4 - x4*y2 - x3*y4 + x4*y3) ]} {[i 1 ]}
.
  댓글 수: 2
Simon
Simon 2024년 12월 14일 0:33
Thanks! I did try similar things already without seeing too much difference. I was hoping there was a way for matlab to identify common factors between the outputted formulas, like here https://math.stackexchange.com/questions/186286/get-transformation-matrix-from-points but maybe it's not capable of doing that (as I was going to extend the original equations).
Star Strider
Star Strider 2024년 12월 14일 2:00
My pleasure!
In my experience, simplify uses every method (such as collect) at its disposal to provide the best result.
You could use coeffs to see if you can isolate common factors, however I doubt that the result would be in any way illuminating.

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