Do the circles intersect?

Question 1

This question is similar to Do the circles overlap?, except your program should return True if the circumference of one circle intersects the other, and false if they do not. If they touch, this is considered an intersection. Note that concentric circles only intersect when both radii are equal

Input is two x, y coordinates for the circle centres and two radius lengths, as either 6 floats or ints, in any order, and output a boolean value, (you may print True or False if you choose)

test cases are in format (x1, y1 x2, y2, r1, r2):

these inputs should output true:

0 0 2 2 2 2
1 1 1 5 2 2
0.55 0.57 1.97 2.24 1.97 2.12

these should output false:

0 0 0 1 5 3
0 0 5 5 2 2
1.57 3.29 4.45 6.67 1.09 3.25

Question 2

So this is the same as the other one except for if one circle is entirely inside the other?

Question 3

@geokavel essentially, yes

Question 4

The slight variation makes for a very different problem. Good job!

Question 5

Related.

Question 6

You are using an unusual definition of intersection. In Mathematics, one circle being entirely within the other implies there is intersection. So you should perhaps rephrase, replacing "intersection" by "partial intersection"

Question 7

APL (Dyalog), 20 bytes

Uses Anders Kaseorg's formula: 0≤(x1−x2)2−(r1−r2)2+(y1−y2)2≤4r1r2

Takes (x1, r1, y1) as left argument and (x2, r2, y2) as right argument.

(-/2*⍨-)(≤∧0≤⊣)×ばつ

Try it online!

The overall function's structure is a fork (3-train) where the tines are -/2*⍨- and ≤∧0≤⊣ and ×ばつ*. The middle tine takes the results of the side tines as arguments. The side tines use the overall function's arguments.

Right tine:
×ばつ multiply the arguments (x1x2, r1r2, y1y2)
2⊃ pick the second element (r1r2)
×ばつ multiply by four (4r1r2)

Left tine:
- subtract the arguments (x1−x2, r1−r2, y1−y2)
2*⍨ square ((x1−x2)2, (r1−r2)2, (y1−y2)2)
-/ minus reduction i.e. alternate sum* ((x1−x2)2−((r1−r2)2−(y1−y2)2))

Now we use these results as arguments to the middle tine:
0≤⊣ is the left argument greater than or equal to zero
∧ and
≤ the left argument smaller than or equal to the right argument?

* due to APL's right associativity

Question 8

Mathematica 67 bytes

This checks whether the area of the intersection of two disks is positive.

RegionMeasure@RegionIntersection[{#,#2}~Disk~#3,{#4,#5}~Disk~#6]>0&

Question 9

Mathematics. Isn't this Mathematica?

Question 10

APL (Dyalog Classic), 17 bytes

(+/∧.≥+⍨)⎕,|0j1⊥⎕

Try it online!

expects two lines of input: x1 y1 x2 y2 and r1 r2

⎕ read and evaluate a line

0j1 imaginary constant i=sqrt(-1)

0j1⊥ decode from base-i, thus computing: i³x₁ + i²y₁ + i¹x₂ + i⁰y₂ = -i x₁ - y₁ + ix₂ + y₂ = i(x₂-x₁) + (y₂-y₁)

| magnitude of that complex number, i.e. distance between the two points

⎕, read the two radii and prepend them, so we have a list of 3 reals

(+/∧.≥+⍨) flat-tolerant triangle inequality: the sum (+/) must be greater than or equal to (≥) the double of each side (+⍨), and all these results must be true (∧.)

Question 11

Python 3, (削除) 56 (削除ここまで) 55 bytes

lambda X,Y,x,y,R,r:0<=(X-x)**2+(Y-y)**2-(R-r)**2<=4*r*R

Try it online!

Question 12

-1 byte.

Question 13

GolfScript, 20 bytes

~@- 2?@@- 2?+@@+2?>!

Try it online!

Takes arguments in the following form r1 r2 x1 y1 x2 y2

Implements the formula below

(x2 - x1)^2 + (y2 - y1)^2 <= (r1 + r2)^2

GolfScript Does not inherently support floating point types. See this post for more details about passing floats https://codegolf.stackexchange.com/a/26553

My original intention was to use GolfScript's zip and fold operations.

GolfScript, 26 bytes

1:a;~zip{{a*+2?-1:a;}*}/+>

Try it online!

Question 14

@JamesHolderness I forgot to mention that little quirk in GolfScript. See the edit concerning this.

Question 15

Just updated the input to be an implementation of the truthy float test. The values aren't exact, but it still works.

Question 16

JavaScript, 52 bytes

(X,Y,x,y,R,r)=>4*r*R/((X-x)**2+(Y-y)**2-(R-r)**2)>=1

Some how Based on Anders Kaseorg Python solution

Question 17

Scala, 59 bytes

This might be golfable.

val u=(a-x)*(a-x)+(b-y)*(b-y)
(q-r)*(q-r)<=u&u<=(q+r)*(q+r)

Try it online!

Scala, 66 bytes

I do like this one, but sadly the previous was shorter. I hate Scala for forcing defining parameter type...

var p=math.pow(_:Double,2)
val u=p(a-x)+p(b-y)
p(q-r)<=u&u<=p(q+r)

Try it online!

Adám 31.8k4 gold badges131 silver badges292 bronze badges · Answer 1 · 2017-07-27 08:40:03Z

APL (Dyalog), 20 bytes

Uses Anders Kaseorg's formula: 0≤(x1−x2)2−(r1−r2)2+(y1−y2)2≤4r1r2

Takes (x1, r1, y1) as left argument and (x2, r2, y2) as right argument.

(-/2*⍨-)(≤∧0≤⊣)×ばつ

Try it online!

The overall function's structure is a fork (3-train) where the tines are -/2*⍨- and ≤∧0≤⊣ and ×ばつ*. The middle tine takes the results of the side tines as arguments. The side tines use the overall function's arguments.

Right tine:
×ばつ multiply the arguments (x1x2, r1r2, y1y2)
2⊃ pick the second element (r1r2)
×ばつ multiply by four (4r1r2)

Left tine:
- subtract the arguments (x1−x2, r1−r2, y1−y2)
2*⍨ square ((x1−x2)2, (r1−r2)2, (y1−y2)2)
-/ minus reduction i.e. alternate sum* ((x1−x2)2−((r1−r2)2−(y1−y2)2))

Now we use these results as arguments to the middle tine:
0≤⊣ is the left argument greater than or equal to zero
∧ and
≤ the left argument smaller than or equal to the right argument?

* due to APL's right associativity

DavidC 25.5k2 gold badges53 silver badges106 bronze badges · Answer 2 · 2017-07-27 08:30:16Z

3

\$\begingroup\$

Mathematica 67 bytes

This checks whether the area of the intersection of two disks is positive.

RegionMeasure@RegionIntersection[{#,#2}~Disk~#3,{#4,#5}~Disk~#6]>0&

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edited Jun 17, 2020 at 9:04

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1

answered Jul 27, 2017 at 8:30

DavidC's user avatar

DavidC

25.5k2 gold badges53 silver badges106 bronze badges

\$\endgroup\$

1

1

\$\begingroup\$ Mathematics. Isn't this Mathematica? \$\endgroup\$

Mr. Xcoder
– Mr. Xcoder

2017年07月27日 08:49:42 +00:00
Commented Jul 27, 2017 at 8:49

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ngn 15.6k2 gold badges45 silver badges90 bronze badges · Answer 3 · 2018-02-04 14:52:25Z

APL (Dyalog Classic), 17 bytes

(+/∧.≥+⍨)⎕,|0j1⊥⎕

Try it online!

expects two lines of input: x1 y1 x2 y2 and r1 r2

⎕ read and evaluate a line

0j1 imaginary constant i=sqrt(-1)

0j1⊥ decode from base-i, thus computing: i³x₁ + i²y₁ + i¹x₂ + i⁰y₂ = -i x₁ - y₁ + ix₂ + y₂ = i(x₂-x₁) + (y₂-y₁)

| magnitude of that complex number, i.e. distance between the two points

⎕, read the two radii and prepend them, so we have a list of 3 reals

(+/∧.≥+⍨) flat-tolerant triangle inequality: the sum (+/) must be greater than or equal to (≥) the double of each side (+⍨), and all these results must be true (∧.)

Anders Kaseorg 40.7k3 gold badges76 silver badges149 bronze badges · Answer 4 · 2017-07-27 02:48:23Z

2

\$\begingroup\$

Python 3, (削除) 56 (削除ここまで) 55 bytes

lambda X,Y,x,y,R,r:0<=(X-x)**2+(Y-y)**2-(R-r)**2<=4*r*R

Try it online!

Share

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edited Jul 27, 2017 at 3:28

answered Jul 27, 2017 at 2:48

Anders Kaseorg's user avatar

Anders Kaseorg

40.7k3 gold badges76 silver badges149 bronze badges

\$\endgroup\$

1

2

\$\begingroup\$ -1 byte. \$\endgroup\$

notjagan
– notjagan

2017年07月27日 03:14:09 +00:00
Commented Jul 27, 2017 at 3:14

Add a comment |

Marcos 1814 bronze badges · Answer 5 · 2017-12-03 09:35:25Z

GolfScript, 20 bytes

~@- 2?@@- 2?+@@+2?>!

Try it online!

Takes arguments in the following form r1 r2 x1 y1 x2 y2

Implements the formula below

(x2 - x1)^2 + (y2 - y1)^2 <= (r1 + r2)^2

GolfScript Does not inherently support floating point types. See this post for more details about passing floats https://codegolf.stackexchange.com/a/26553

My original intention was to use GolfScript's zip and fold operations.

GolfScript, 26 bytes

1:a;~zip{{a*+2?-1:a;}*}/+>

Try it online!

@JamesHolderness I forgot to mention that little quirk in GolfScript. See the edit concerning this.
Just updated the input to be an implementation of the truthy float test. The values aren't exact, but it still works.

tsh 36.2k2 gold badges36 silver badges133 bronze badges · Answer 6 · 2017-07-27 10:00:04Z

1

\$\begingroup\$

JavaScript, 52 bytes

(X,Y,x,y,R,r)=>4*r*R/((X-x)**2+(Y-y)**2-(R-r)**2)>=1

Some how Based on Anders Kaseorg Python solution

Share

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answered Jul 27, 2017 at 10:00

tsh's user avatar

tsh

36.2k2 gold badges36 silver badges133 bronze badges

\$\endgroup\$

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V. Courtois 9565 silver badges15 bronze badges · Answer 7 · 2017-07-27 09:18:13Z

Scala, 59 bytes

This might be golfable.

val u=(a-x)*(a-x)+(b-y)*(b-y)
(q-r)*(q-r)<=u&u<=(q+r)*(q+r)

Try it online!

Scala, 66 bytes

I do like this one, but sadly the previous was shorter. I hate Scala for forcing defining parameter type...

var p=math.pow(_:Double,2)
val u=p(a-x)+p(b-y)
p(q-r)<=u&u<=p(q+r)

Try it online!

Stack Exchange Network

Do the circles intersect?

7 Answers 7

APL (Dyalog), 20 bytes

Mathematica 67 bytes

APL (Dyalog Classic), 17 bytes

Python 3, (削除) 56 (削除ここまで) 55 bytes

GolfScript, 20 bytes

GolfScript, 26 bytes

JavaScript, 52 bytes

Scala, 59 bytes

Scala, 66 bytes

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