Circle intersection area

Question

Description :

Given x and y positions of two circles along with their radii, output the area of intersection of the two circle.

Input :

You will be given following input :

array 1 = x and y positions of circle a
array 2 = x and y positions of circle b
radius  = radii of the two congruent circles

Input method :

([12 , 20] , [20 , 18] , 12)     ---> two array and number
([12 , 20 , 20 , 18] , 12)       ---> array and a number
(12 , 20 , 20 , 18 , 12)         ---> all five numbers
('12 20' , '20 18' , 12)         ---> 2 strings and a number
('12 20 20 18' , 12)             ---> string and a number
('12 20 20 18 12')               ---> one string

Output :

A non-negative integer (no decimal) equal to area of intersection of two circles.
A string equal to the above mentioned integer.

Note :

Output must be >= 0, since area can't be negative.
In case of decimal round down to nearest integer

Examples :

([0, 0], [7, 0], 5)                   ---> 14

([0, 0], [0, 10], 10)                 ---> 122

([5, 6], [5, 6], 3)                   ---> 28

([-5, 0], [5, 0], 3)                  ---> 0

([10, 20], [-5, -15], 20)             ---> 15

([-7, 13], [-25, -5], 17)             ---> 132

([-12, 20], [43, -49], 23)            ---> 0

Winning criteria :

This is code-golf so shortest code in bytes for each language wins.

Suggestions :

Provide a TIO link so it can be tested.
Provide an explanation so others can understand your code

These are only suggestions and are not mandatory.

@FrownyFrog: Excuse me ? I am not aware of what you are talking about? nvm check on internet and I am sorry to report that is part of the problem. see the tag that says math and geometry. It is a good excuse to brush up on your math. What do you think. But if you disagree I think I will update the question and add formula. — Muhammad Salman, Commented Apr 15, 2018 at 14:11
@MuhammadSalman Change answer must be positive to answer must be >= 0 - If the circles don't intersect (as in examples 4, 7, 10) then the correct answer is 0, which last I checked is not positive. — manassehkatz-Moving 2 Codidact, Commented Apr 20, 2018 at 20:46

Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (ES6), 72 bytes

Alternate formula suggested by @ceilingcat

Takes input as 5 distinct parameters (x0, y0, x1, y1, r).

with(Math)f=(x,y,X,Y,r)=>-(sin(d=2*acos(hypot(x-X,y-Y)/r/2))-d)*r*r*2>>1

Try it online!

JavaScript (ES7), 81 80 77 bytes

Saved 3 bytes thanks to @Neil

Takes input as 5 distinct parameters (x0, y0, x1, y1, r).

(x,y,X,Y,r,d=Math.hypot(x-X,y-Y))=>(r*=2)*r*Math.acos(d/r)-d*(r*r-d*d)**.5>>1

Try it online!

How?

This is based on a generic formula from MathWorld for non-congruent circles:

A = r².arccos((d² + r² - R²) / 2dr) +
    R².arccos((d² + R² - r²) / 2dR) -
    sqrt((-d + r + R)(d + r - R)(d -r + R)(d + r + R)) / 2

where d is the distance between the two centers and r and R are the radii.

With R = r, this is simplified to:

A = 2r².arccos(d / 2r) + d.sqrt((2r - d) * (2r + d)) / 2

And with r' = 2r:

A = (r'².arccos(d / r') + d.sqrt(r'² - d²)) / 2

Note: If d is greater than 2r, Math.acos() will return NaN, which is coerced to 0 when the right-shift is applied. This is the expected result, because d > 2r means that there's no intersection at all.

@ceilingcat Thanks! Using with(Math) and moving the definition of d saves 2 more bytes. — Arnauld, Commented Apr 17, 2018 at 6:20

Community · Accepted Answer · 2020-06-17 09:04:33Z

Jelly, 27 25 24 22 bytes

×,²I½
÷ÆA×²}_çHḞ
ạ/çḤ}

A full program accepting a list of the two centres as complex co-ordinates and the radius which prints the result (as a dyadic link it returns a list of length 1).

Try it online!

To take the two co-ordinates as pairs add Uḅı to the main link, like this.

How?

×,²I½ - Link 1, get [√(s²d² - s⁴)]: separation of centres, s; diameter, d
 ,    - pair = [s, d]
×     - multiply (vectorises) = [s², sd]
  ²   - square (vectorises) = [s⁴, s²d²]
   I  - incremental differences = [s²d² - s⁴]
    ½ - square root (vectorises) = [√(s²d² - s⁴)]

÷ÆA×²}_çHḞ - Link 2, get intersection area: separation of centres, s; diameter, d
÷          - divide = s/d
 ÆA        - arccos = acos(s/d)
    ²}     - square right = d²
   ×       - multiply = acos(s/d)d²
       ç   - call last Link (1) as a dyad (f(s,d)) = [√(s²d² - s⁴)]
      _    - subtract (vectorises) = [acos(s/d)d² - √(s²d² - s⁴)]
        H  - halve (vectorises) = [(acos(s/d)d² - √(s²d² - s⁴))/2]
         Ḟ - floor = [⌊(acos(s/d)d² - √(s²d² - s⁴))/2⌋]
           -  ...Note: Jelly's Ḟ takes the real part of a complex input so when
           -           the circles are non-overlapping the result is 0 as required

ạ/çḤ} - Main link: centres, a pair of complex numbers, c; radius, r
 /    - reduce c by:
ạ     -   absolute difference = separation of centres, s
      -   ...Note: Jelly's ạ finds the Euclidean distance when inputs are complex
      -            i.e. the norm of the difference
   Ḥ} - double right = 2r = diameter, d
  ç   - call last Link (2) as a dyad (f(s,d))
      - implicit print

numbers only. And what is that [-7+13j,-25+-5j] ? I don't have that example. You might have to explain what you did ? — Muhammad Salman, Commented Apr 15, 2018 at 16:18
I explained it in the answer already... they are co-ordinates on the complex plane... I can do [[x1,y1],[x2,y2]] instead but it costs 3 bytes. (Note also that -7+13j is a number :)) -- the [-7+13j,-25+-5j] corresponds to the example that returns 132, [-7, 13], [-25, -5], 17 — Jonathan Allan, Commented Apr 15, 2018 at 16:21
I don't know Jelly so I am lost on that. Also I sent the message before the explanation. But yeah , sure this works (I guess?) — Muhammad Salman, Commented Apr 15, 2018 at 16:30
It's nothing to do with Jelly per-se, it's just mathematics. A point in 2-space is the same as a complex number. — Jonathan Allan, Commented Apr 15, 2018 at 16:36
Not what I meant. Normal languages I would be able to read and tell what is going on. Jelly and other such languages are a pain to read. — Muhammad Salman, Commented Apr 15, 2018 at 16:39

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Mathematica 66 57 51 bytes

Floor@Area@RegionIntersection[#~Disk~#3,#2~Disk~#3]&

A Disk[{x,y},r]refers to the region circumscribed by the circle centered at {x,y} with a radius of r.

RegionIntersection[a,b] returns the intersection of regions a, b. Area takes the area. IntegerPart rounds down to the nearest integer.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 15, 2018 at 15:18

DavidC

25.4k2 gold badges52 silver badges105 bronze badges

\$\begingroup\$ For the record, I didn't see alephalpha's submission as I was doing my own. His is a shorter (hence a more successful) entry, but I left mine in anyway. \$\endgroup\$
– DavidC
Commented Apr 18, 2018 at 17:21
\$\begingroup\$ You could replace IntegerPart with Floor. \$\endgroup\$
– matrix42
Commented Aug 23, 2019 at 1:05
\$\begingroup\$ @mathe, Thanks. If I use the dedicated Floor brackets, do you know how I am to count the bytes? \$\endgroup\$
– DavidC
Commented Aug 24, 2019 at 13:02
\$\begingroup\$ @DavidC each one is 3 bytes, so the substitution is neutral in this case for byte count. They're useful if the expression would otherwise need parenthesizing, though (-1 byte compared to Floor[ ]). \$\endgroup\$
– att
Commented Aug 24, 2019 at 17:04

Add a comment |

alephalpha · Accepted Answer · 2018-04-15 15:02:42Z

2

Wolfram Language (Mathematica), 50 bytes

Floor@Area@RegionIntersection[#~Disk~#3,Disk@##2]&

Try it online!

answered Apr 15, 2018 at 15:02

alephalpha

50k7 gold badges68 silver badges182 bronze badges

\$\begingroup\$ +1 Floor. Of course! \$\endgroup\$
– DavidC
Commented Apr 16, 2018 at 9:37

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ceilingcat · Accepted Answer · 2018-04-21 08:43:33Z

2

C (gcc), 83 79 71 66 bytes

f(a,b,c,d,e){float g=cacos(hypot(a-c,b-d)/e/2)*2;e*=(g-sin(g))*e;}

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edited Apr 21, 2018 at 8:43

answered Apr 16, 2018 at 0:33

ceilingcat

7,6552 gold badges30 silver badges30 bronze badges

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flaajx · Accepted Answer · 2020-03-13 15:10:43Z

2

Python 3, 106 chars with centers given as lists

from numpy import*
def f(a,b,r):d=min(1,hypot(*subtract(b,a))/2/r);return(arccos(d)-d*(1-d*d)**.5)*r*r//.5

100 chars with center coords separately

from math import*
def f(a,b,A,B,r):d=min(1,hypot(A-a,B-b)/2/r);return(acos(d)-d*(1-d*d)**.5)*r*r//.5

edited Mar 13, 2020 at 15:10

answered Mar 13, 2020 at 14:51

flaajx

212 bronze badges

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Angs · Accepted Answer · 2018-04-15 16:38:31Z

1

Haskell, 83 bytes

(k!l)m n r|d<-sqrt$(k-m)^2+(l-n)^2=floor$2*r^2*acos(d/2/r)-d/2*sqrt(4*r*r-d*d)::Int

Just the formula, really. Type has to be declared as Int for NaN to map to 0 with floor.

Try it online!

answered Apr 15, 2018 at 16:38

Angs

4,9872 gold badges16 silver badges36 bronze badges

\$\begingroup\$ alternate formula \$\endgroup\$
– ceilingcat
Commented Apr 17, 2018 at 7:46

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score 1 · Accepted Answer · 2018-04-20 19:28:23Z

1

JavaScript (Node.js), 69 bytes

with(Math)f=(a,b,c,d,r)=>(-sin(x=2*acos(hypot(a-c,b-d)/2/r))+x)*r*r|0

Try it online!

Short not sure if it can be golfed any further. Any suggestions are welcome

edited Apr 20, 2018 at 19:28

answered Apr 20, 2018 at 17:56

user79855

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nwellnhof · Accepted Answer · 2018-04-16 11:58:30Z

0

Perl 6, 56 bytes

{{1>$_&&{$_-.sin}(2*.acos)}(abs($^p-$^q)/2/$^r)*$r²+|0}

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Takes circle coordinates as complex numbers.

answered Apr 16, 2018 at 11:58

nwellnhof

10.6k1 gold badge19 silver badges38 bronze badges

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Wernisch · Accepted Answer · 2018-04-16 12:09:02Z

0

Excel, 119 bytes

=INT(IFERROR(2*E1^2*ACOS(((C1-A1)^2+(D1-B1)^2)^.5/2/E1)-((4*E1^2-((C1-A1)^2+(D1-B1)^2))*((C1-A1)^2+(D1-B1)^2))^.5/2,0))

Input taken as 5 separate variables:

x-coordinate    y-coordinate    x-coordinate    y-coordinate    radius
     A1              B1             C1                D1          E1

answered Apr 16, 2018 at 12:09

Wernisch

3,03410 silver badges22 bronze badges

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Sunny Patel · Accepted Answer · 2018-04-17 15:42:49Z

0

Python 2, 109 bytes

from math import*
a,b,x,y,r=input()
d,R=hypot(x-a,y-b),2*r
print int(d<R and R*r*acos(d/R)-d*sqrt(R*R-d*d)/2)

Try it online!

Pretty straightforward. Get the distance between circles, and use R=2r as a substituite in the equation. d<R and to short-circuit if circles don't overlap.

answered Apr 17, 2018 at 15:42

Sunny Patel

2942 silver badges8 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Pyth, 63 bytes

J@+^-hhQh@Q1 2^-ehQe@Q1 2 2K*2eQs&<JK-**KeQ.tcJK4c*J@-*KK*JJ2 2

Test suite

Takes input as a triple consisting of two doubles and a number.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 17, 2018 at 17:20

hakr14

5,0141 gold badge11 silver badges18 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

T-SQL, 122 bytes

SELECT FLOOR(Geometry::Parse('POINT'+a).STBuffer(r).STIntersection(
             Geometry::Parse('POINT'+b).STBuffer(r)).STArea())FROM t

(line break for readability only).

Uses MS SQL's support of spatial geometry.

Per our IO standards, SQL can take input from a pre-existing table t with int field r and varchar fields a and b containing coordinates in the format (x y).

My statement parses the coordinates as POINT geometry objects expanded by the radius using the function STBuffer(), then taking the STIntersection() followed by the STArea().

If I am allowed to input the actual geometry objects in the table instead, then my code becomes almost trivial (48 bytes):

SELECT FLOOR(a.STIntersection(b).STArea())FROM t

Stack Exchange Network

Circle intersection area

Description :

Input :

Input method :

Output :

Note :

Examples :

Winning criteria :

Suggestions :

13 Answers 13

JavaScript (ES6), 72 bytes

JavaScript (ES7), 81 80 77 bytes

How?

Jelly, 27 25 24 22 bytes

How?

Mathematica 66 57 51 bytes

Wolfram Language (Mathematica), 50 bytes

C (gcc), 83 79 71 66 bytes

Python 3, 106 chars with centers given as lists

100 chars with center coords separately

Haskell, 83 bytes

JavaScript (Node.js), 69 bytes

Perl 6, 56 bytes

Excel, 119 bytes

Python 2, 109 bytes

Pyth, 63 bytes

T-SQL, 122 bytes

Your Answer

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math
array
geometry
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Hot Network Questions

Circle intersection area

Description :

Input :

Input method :

Output :

Note :

Examples :

Winning criteria :

Suggestions :

13 Answers 13

JavaScript (ES6), 72 bytes

JavaScript (ES7), 81 80 77 bytes

How?

Jelly, 27 25 24 22 bytes

How?

Mathematica 66 57 51 bytes

Wolfram Language (Mathematica), 50 bytes

C (gcc), 83 79 71 66 bytes

Python 3, 106 chars with centers given as lists

100 chars with center coords separately

Haskell, 83 bytes

JavaScript (Node.js), 69 bytes

Perl 6, 56 bytes

Excel, 119 bytes

Python 2, 109 bytes

Pyth, 63 bytes

T-SQL, 122 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged
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