Mixing
Is this quadrilateral cyclic?
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In mathematics, a cyclic quadrilateral is one whose vertices all lie on the same circle. In other words, every vertex is on the circumcircle of the other three. For more information, see the MathWorld article.
Examples
These quadrilaterals are cyclic:
This trapezoid is not cyclic.
(Images from Wikipedia)
Objective
Given the coordinates of four vertices in counterclockwise order which form a convex quadrilateral, determine if the quadrilateral is cyclic.
Coordinates will be integers (note, however, that the circumcenter coordinates and circumradius are not necessarily integers.) As implied by the previous paragraph, no three points will be co-linear and no two coincident.
I/O
You may take input using any reasonable format. In particular, [[x1,x2,x3,x4],[y1,y2,y3,y4]], [[x1,y1],[x2,y2],[x3,y3],[x4,y4]] and complex numbers are all fine.
Output using any different consistent values for true and false.
Test cases
True:
[0,0], [314,0], [314,1], [0,1]
[-5,5], [5,-5], [1337,42], [42,1337]
[104, -233], [109, -232], [112, -231], [123, -224]
False:
[0,0], [314,0], [314,100], [0,99]
[31,41],[59,26],[53,58],[0,314]
code-golf math decision-problem geometry
edited yesterday
FryAmTheEggman
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
add a comment |
up vote
17
down vote
favorite
In mathematics, a cyclic quadrilateral is one whose vertices all lie on the same circle. In other words, every vertex is on the circumcircle of the other three. For more information, see the MathWorld article.
Examples
These quadrilaterals are cyclic:
This trapezoid is not cyclic.
(Images from Wikipedia)
Objective
Given the coordinates of four vertices in counterclockwise order which form a convex quadrilateral, determine if the quadrilateral is cyclic.
Coordinates will be integers (note, however, that the circumcenter coordinates and circumradius are not necessarily integers.) As implied by the previous paragraph, no three points will be co-linear and no two coincident.
I/O
You may take input using any reasonable format. In particular, [[x1,x2,x3,x4],[y1,y2,y3,y4]], [[x1,y1],[x2,y2],[x3,y3],[x4,y4]] and complex numbers are all fine.
Output using any different consistent values for true and false.
Test cases
True:
[0,0], [314,0], [314,1], [0,1]
[-5,5], [5,-5], [1337,42], [42,1337]
[104, -233], [109, -232], [112, -231], [123, -224]
False:
[0,0], [314,0], [314,100], [0,99]
[31,41],[59,26],[53,58],[0,314]
code-golf math decision-problem geometry
edited yesterday
FryAmTheEggman
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
add a comment |
up vote
17
down vote
favorite
up vote
17
down vote
favorite
In mathematics, a cyclic quadrilateral is one whose vertices all lie on the same circle. In other words, every vertex is on the circumcircle of the other three. For more information, see the MathWorld article.
Examples
These quadrilaterals are cyclic:
This trapezoid is not cyclic.
(Images from Wikipedia)
Objective
Given the coordinates of four vertices in counterclockwise order which form a convex quadrilateral, determine if the quadrilateral is cyclic.
Coordinates will be integers (note, however, that the circumcenter coordinates and circumradius are not necessarily integers.) As implied by the previous paragraph, no three points will be co-linear and no two coincident.
I/O
You may take input using any reasonable format. In particular, [[x1,x2,x3,x4],[y1,y2,y3,y4]], [[x1,y1],[x2,y2],[x3,y3],[x4,y4]] and complex numbers are all fine.
Output using any different consistent values for true and false.
Test cases
True:
[0,0], [314,0], [314,1], [0,1]
[-5,5], [5,-5], [1337,42], [42,1337]
[104, -233], [109, -232], [112, -231], [123, -224]
False:
[0,0], [314,0], [314,100], [0,99]
[31,41],[59,26],[53,58],[0,314]
code-golf math decision-problem geometry
edited yesterday
FryAmTheEggman
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
In mathematics, a cyclic quadrilateral is one whose vertices all lie on the same circle. In other words, every vertex is on the circumcircle of the other three. For more information, see the MathWorld article.
Examples
These quadrilaterals are cyclic:
This trapezoid is not cyclic.
(Images from Wikipedia)
Objective
Given the coordinates of four vertices in counterclockwise order which form a convex quadrilateral, determine if the quadrilateral is cyclic.
Coordinates will be integers (note, however, that the circumcenter coordinates and circumradius are not necessarily integers.) As implied by the previous paragraph, no three points will be co-linear and no two coincident.
I/O
You may take input using any reasonable format. In particular, [[x1,x2,x3,x4],[y1,y2,y3,y4]], [[x1,y1],[x2,y2],[x3,y3],[x4,y4]] and complex numbers are all fine.
Output using any different consistent values for true and false.
Test cases
True:
[0,0], [314,0], [314,1], [0,1]
[-5,5], [5,-5], [1337,42], [42,1337]
[104, -233], [109, -232], [112, -231], [123, -224]
False:
[0,0], [314,0], [314,100], [0,99]
[31,41],[59,26],[53,58],[0,314]
code-golf math decision-problem geometry
code-golf math decision-problem geometry
edited yesterday
FryAmTheEggman
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
edited yesterday
FryAmTheEggman
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
edited yesterday
FryAmTheEggman
14.6k32482
edited yesterday
FryAmTheEggman
14.6k32482
edited yesterday
FryAmTheEggman
14.6k32482
14.6k32482
asked 2 days ago
lirtosiast
15.4k436104
asked 2 days ago
lirtosiast
15.4k436104
asked 2 days ago
lirtosiast
15.4k436104
15.4k436104
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5 Answers
5
active
oldest
votes
up vote
9
down vote
Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
Try it online!
Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if $n+1$ points in $mathbb R^n$ are concyclic, provided the last $n$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).
Alternatively, here is a mathematical approach:
Wolfram Language (Mathematica), 29 28 25 24 bytes
Det@{#^2+#2^2,##,1^#}^0&
Try it online!
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.
From the four points $(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)$, this solution constructs the matrix below:
$begin{bmatrix}x_1^2 + y_1^2 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \ x_1 & x_2 & x_3 & x_4 \ y_1 & y_2 & y_3 & y_4 \ 1 & 1 & 1 & 1 end{bmatrix}$
The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.
The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.
answered yesterday
Misha Lavrov
4,000423
add a comment |
up vote
8
down vote
Python 3, 70 bytes
lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8
Try it online!
I use the Ptolemy's theorem.
In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.
b, c, d, e are complex numbers.
answered yesterday
Кирилл Малышев
39115
add a comment |
up vote
6
down vote
Perl 6, 44 bytes
{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}
Try it online!
Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.
answered yesterday
nwellnhof
5,9981122
42? Is it still accurate?
– Jo King
yesterday
1
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
add a comment |
up vote
5
down vote
JavaScript (ES6)
Testing the angles, 114 bytes
Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.
a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI
Try it online!
Computing a determinant, 130 bytes
Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.
This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.
x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))
Try it online!
answered yesterday
Arnauld
69k584291
add a comment |
up vote
2
down vote
TI-Basic (83 series), 21 bytes
e^(ΔList(ln(ΔList(augment(Ans,Ans
not(imag(Ans(1)Ans(3
Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.
This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:
ΔList(augment(Ans,Anscomputes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),
e^(ΔList(ln(of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.- We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.
I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.
answered yesterday
Misha Lavrov
4,000423
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
Try it online!
Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if $n+1$ points in $mathbb R^n$ are concyclic, provided the last $n$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).
Alternatively, here is a mathematical approach:
Wolfram Language (Mathematica), 29 28 25 24 bytes
Det@{#^2+#2^2,##,1^#}^0&
Try it online!
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.
From the four points $(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)$, this solution constructs the matrix below:
$begin{bmatrix}x_1^2 + y_1^2 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \ x_1 & x_2 & x_3 & x_4 \ y_1 & y_2 & y_3 & y_4 \ 1 & 1 & 1 & 1 end{bmatrix}$
The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.
The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.
answered yesterday
Misha Lavrov
4,000423
add a comment |
up vote
9
down vote
Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
Try it online!
Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if $n+1$ points in $mathbb R^n$ are concyclic, provided the last $n$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).
Alternatively, here is a mathematical approach:
Wolfram Language (Mathematica), 29 28 25 24 bytes
Det@{#^2+#2^2,##,1^#}^0&
Try it online!
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.
From the four points $(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)$, this solution constructs the matrix below:
$begin{bmatrix}x_1^2 + y_1^2 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \ x_1 & x_2 & x_3 & x_4 \ y_1 & y_2 & y_3 & y_4 \ 1 & 1 & 1 & 1 end{bmatrix}$
The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.
The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.
answered yesterday
Misha Lavrov
4,000423
add a comment |
up vote
9
down vote
up vote
9
down vote
Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
Try it online!
Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if $n+1$ points in $mathbb R^n$ are concyclic, provided the last $n$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).
Alternatively, here is a mathematical approach:
Wolfram Language (Mathematica), 29 28 25 24 bytes
Det@{#^2+#2^2,##,1^#}^0&
Try it online!
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.
From the four points $(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)$, this solution constructs the matrix below:
$begin{bmatrix}x_1^2 + y_1^2 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \ x_1 & x_2 & x_3 & x_4 \ y_1 & y_2 & y_3 & y_4 \ 1 & 1 & 1 & 1 end{bmatrix}$
The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.
The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.
answered yesterday
Misha Lavrov
4,000423
Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
Try it online!
Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if $n+1$ points in $mathbb R^n$ are concyclic, provided the last $n$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).
Alternatively, here is a mathematical approach:
Wolfram Language (Mathematica), 29 28 25 24 bytes
Det@{#^2+#2^2,##,1^#}^0&
Try it online!
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.
From the four points $(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)$, this solution constructs the matrix below:
$begin{bmatrix}x_1^2 + y_1^2 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \ x_1 & x_2 & x_3 & x_4 \ y_1 & y_2 & y_3 & y_4 \ 1 & 1 & 1 & 1 end{bmatrix}$
The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.
The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.
answered yesterday
Misha Lavrov
4,000423
edited yesterday
answered yesterday
Misha Lavrov
4,000423
answered yesterday
Misha Lavrov
4,000423
answered yesterday
Misha Lavrov
4,000423
4,000423
add a comment |
add a comment |
up vote
8
down vote
Python 3, 70 bytes
lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8
Try it online!
I use the Ptolemy's theorem.
In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.
b, c, d, e are complex numbers.
answered yesterday
Кирилл Малышев
39115
add a comment |
up vote
8
down vote
Python 3, 70 bytes
lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8
Try it online!
I use the Ptolemy's theorem.
In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.
b, c, d, e are complex numbers.
answered yesterday
Кирилл Малышев
39115
add a comment |
up vote
8
down vote
up vote
8
down vote
Python 3, 70 bytes
lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8
Try it online!
I use the Ptolemy's theorem.
In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.
b, c, d, e are complex numbers.
answered yesterday
Кирилл Малышев
39115
Python 3, 70 bytes
lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8
Try it online!
I use the Ptolemy's theorem.
In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.
b, c, d, e are complex numbers.
answered yesterday
Кирилл Малышев
39115
edited yesterday
answered yesterday
Кирилл Малышев
39115
answered yesterday
Кирилл Малышев
39115
answered yesterday
Кирилл Малышев
39115
39115
add a comment |
add a comment |
up vote
6
down vote
Perl 6, 44 bytes
{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}
Try it online!
Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.
answered yesterday
nwellnhof
5,9981122
42? Is it still accurate?
– Jo King
yesterday
1
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
add a comment |
up vote
6
down vote
Perl 6, 44 bytes
{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}
Try it online!
Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.
answered yesterday
nwellnhof
5,9981122
42? Is it still accurate?
– Jo King
yesterday
1
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
add a comment |
up vote
6
down vote
up vote
6
down vote
Perl 6, 44 bytes
{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}
Try it online!
Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.
answered yesterday
nwellnhof
5,9981122
Perl 6, 44 bytes
{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}
Try it online!
Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.
answered yesterday
nwellnhof
5,9981122
edited yesterday
answered yesterday
nwellnhof
5,9981122
answered yesterday
nwellnhof
5,9981122
answered yesterday
nwellnhof
5,9981122
5,9981122
42? Is it still accurate?
– Jo King
yesterday
1
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
add a comment |
42? Is it still accurate?
– Jo King
yesterday
1
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
42? Is it still accurate?
– Jo King
yesterday
42? Is it still accurate?
– Jo King
yesterday
1
1
@JoKing No, it's not.
– nwellnhof
yesterday
@JoKing No, it's not.
– nwellnhof
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
What does the colon do in this case? It's definitely not a label, and also not a method call.
– user202729
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
@user202729 It is a method call with indirect invocant syntax.
– nwellnhof
yesterday
add a comment |
up vote
5
down vote
JavaScript (ES6)
Testing the angles, 114 bytes
Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.
a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI
Try it online!
Computing a determinant, 130 bytes
Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.
This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.
x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))
Try it online!
answered yesterday
Arnauld
69k584291
add a comment |
up vote
5
down vote
JavaScript (ES6)
Testing the angles, 114 bytes
Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.
a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI
Try it online!
Computing a determinant, 130 bytes
Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.
This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.
x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))
Try it online!
answered yesterday
Arnauld
69k584291
add a comment |
up vote
5
down vote
up vote
5
down vote
JavaScript (ES6)
Testing the angles, 114 bytes
Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.
a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI
Try it online!
Computing a determinant, 130 bytes
Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.
This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.
x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))
Try it online!
answered yesterday
Arnauld
69k584291
JavaScript (ES6)
Testing the angles, 114 bytes
Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.
a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI
Try it online!
Computing a determinant, 130 bytes
Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.
This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.
x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))
Try it online!
answered yesterday
Arnauld
69k584291
edited yesterday
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
69k584291
add a comment |
add a comment |
up vote
2
down vote
TI-Basic (83 series), 21 bytes
e^(ΔList(ln(ΔList(augment(Ans,Ans
not(imag(Ans(1)Ans(3
Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.
This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:
ΔList(augment(Ans,Anscomputes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),
e^(ΔList(ln(of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.- We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.
I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.
answered yesterday
Misha Lavrov
4,000423
add a comment |
up vote
2
down vote
TI-Basic (83 series), 21 bytes
e^(ΔList(ln(ΔList(augment(Ans,Ans
not(imag(Ans(1)Ans(3
Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.
This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:
ΔList(augment(Ans,Anscomputes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),
e^(ΔList(ln(of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.- We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.
I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.
answered yesterday
Misha Lavrov
4,000423
add a comment |
up vote
2
down vote
up vote
2
down vote
TI-Basic (83 series), 21 bytes
e^(ΔList(ln(ΔList(augment(Ans,Ans
not(imag(Ans(1)Ans(3
Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.
This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:
ΔList(augment(Ans,Anscomputes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),
e^(ΔList(ln(of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.- We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.
I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.
answered yesterday
Misha Lavrov
4,000423
TI-Basic (83 series), 21 bytes
e^(ΔList(ln(ΔList(augment(Ans,Ans
not(imag(Ans(1)Ans(3
Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.
This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:
ΔList(augment(Ans,Anscomputes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),
e^(ΔList(ln(of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.- We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.
I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.
answered yesterday
Misha Lavrov
4,000423
answered yesterday
Misha Lavrov
4,000423
answered yesterday
Misha Lavrov
4,000423
answered yesterday
Misha Lavrov
4,000423
4,000423
add a comment |
add a comment |
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