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So I was watching Numberphile with Neil Sloane of OEIS fame on the number of ways to make circles intersect. During the intro, he explicitly forbid kissing (touching, tangent) circles. This made me think about the number of ways to draw circles that must all kiss, where intersections are not allowed. Kind of the dual problem, if you will. So in a nutshell:

  • Circles in the affine plane can be of any finite size
  • Circles must kiss at least one other (no separate circles, all must be connected)
  • Solutions that can be transformed into each other through mirroring, rotation, stretching, moving circles without making new tangent points, are considered the same

For $n=2$ circles there are two ways: touching on the outside, and touching on the inside of the larger circle.

For $n=3$ I believe the answer is 9. Try it, should be easy.

For $n=4$ I started drawing and came up with, so far, and with help from you guys, 38, 39, 42, 43, 44, 45, 47, 44 ways. The colored dots represent the number of kissing circles at the tangent points.

Questions:

  • Did I miss any? I'm sure I did.
  • Do I have inadvertent duplicates?
  • Does an integer sequence exist, or even a formula mapping $n = 2, 3, 4, ...$ to $2, 9, 44, ...$?

enter image description here

Jens
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  • There are some missing figures. For example, modify figure 8 by adding an extra tangent point, say top of the middle circle, and make the small circle tangent at that point instead of at the green point. Also, modify figure 4, to make a smaller circle in the middle, and make it tangent to only one or two circles – Andrei Mar 14 '21 at 19:21
  • I get $9$ ways for $N=3$. – aschepler Mar 14 '21 at 19:24
  • For $n=4$, should $A$ touching $B$ and $C$ touching $D$ and no other tangents count? Or do you require the circles are all connected? – aschepler Mar 14 '21 at 19:31
  • @aschepler No disconnected circles. – Jens Mar 14 '21 at 19:35
  • @Andrei Well spotted, eagle eye! I have updated the image now. – Jens Mar 14 '21 at 20:59
  • Nice. I was thinking also at image 40, with one of the lobes a smaller circle inside the central. – Andrei Mar 14 '21 at 21:05
  • Also, I would group the figures by number of tangent points. Maybe it would be easier to account for all combinations – Andrei Mar 14 '21 at 21:10
  • @Andrei Ok, now at 44. You are right, order is semi-random. I'm drawing this with xfig, and I'm not a wizard. If only there were an algorithm that would iterate over all possible drawings... – Jens Mar 14 '21 at 21:26
  • @aschepler Hmm. Each time a find a 9th for n=3, on closer look it is a duplicate. – Jens Mar 14 '21 at 21:42
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    no one thinks this resembles an operad structure? I mean the following: any configuration has a number of "interior" (= empty) circles; if you take a configuration with three interior circles and three configurations, you can plug them inside and get another such "kissing configuration". This kind of structure is called an operad. Here the space of configurations with, say, three interior circles, is a topological space (you have some real parameters and some constraints that must hold between them) and there is a group acting on it (rotations and bla). – Andrea Marino Mar 14 '21 at 22:12
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    What's the difference between #18 and #20? Also, shouldn't there be a third counterpart to #9 and #10 where the tangency point of the inside circle touches closer to the center of the diagram? – Dan Uznanski Mar 14 '21 at 22:13
  • Why this is convenient? Because as all algebraic structures it should have a number of generators (things that can't be obtained by composing ) satisfying some constraints with respect to composition. When you compose the number of circles increase in an orchestrated way. The question then becomes purely combinatorial. Let's spot generators first!! – Andrea Marino Mar 14 '21 at 22:14
  • For $n=3$, there are $2$ with just a single tangent point, $4$ more where one circle encloses the other two, $1$ more with one internal tangent and one external tangent, and $2$ more where no circle is inside another. – aschepler Mar 14 '21 at 22:15
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    https://math.stackexchange.com/questions/1307018/how-many-planar-arrangements-of-n-circles seems to show nine kissing arranements for three circles. – Gerry Myerson Mar 14 '21 at 22:22
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    What's the difference between #15 and #24? Where is the one that's like #34 but there is no shared tangency point? – Dan Uznanski Mar 14 '21 at 22:26
  • I think that there are small variants to do in order for my reasoning to apply. Firstly, let us suppose that there is a big circle containing all circles, of course tangent to another circle. Kissing configurstions without a "big brother" can be converted to "kissing configurations with big brother" by adding an outer circle. Unfortunately, this can be made in multiple ways, but an easily handable number of ways. The "simple" configurations are now the ones without circles containing another circle (beside the big brother). Another small modification that must be done is that [.. ] – Andrea Marino Mar 14 '21 at 22:33
  • Composition accepts a further parameter: we have to decide "how" to plug the configuration inside a circle. Indeed, you can circular slide the plugged configuration and, depending on how the tangency points interweave with the existing tangency points, you get different configurations, because of the constraint of "moving circles without making new tangency points". You can't pass over multiple tangency points without changing type. That said, I think the problem is very difficult for general n even if circles are not allowed to contain other circles. Think how crazy can circles arrange!!! – Andrea Marino Mar 14 '21 at 22:38
  • Do you allow two circles touching and separately other two circles touching? This seems to miss, unless you want to require that the underlying "hey we are kissing" graph to be connected. I think you are implicitly asking for this :) – Andrea Marino Mar 14 '21 at 23:08
  • to be more precise, the configurations with no circles containing another circles - beside the big outer circle - are determined by the graph where vertices are circles and edges connects two kissing circles. Counting these graphs should be easier - then one has to check individually if such graphs can be realized. The outer circle must be considered "special". Sorry for the many comments: this is intended as a recipe for a brave heart to carry on ccalculations and give an answer, which I unluckily don't have the time to do. Very interesting problem though :D – Andrea Marino Mar 14 '21 at 23:16
  • @DanUznanski Sorry to bother you, I sorted the figures by type and renumbered, which affected your questions referring to numbers. Would you mind checking again? Thanks! – Jens Mar 14 '21 at 23:31
  • Ok: #26 and #28 appear to be the same; #15 and #16 appear to be the same. #21 and #42 appear to have a friend where the child's tangency point is toward the inside. #43 has -- well, now that I've done some more work, five friends with the same circle containment hierarchy. – Dan Uznanski Mar 14 '21 at 23:37
  • @DanUznanski Thanks, I updated the images with black blocks to hide the duplicates. Back to 45. – Jens Mar 14 '21 at 23:47
  • It seems for combinatorial building blocks, we could represent a circle and its contents (or a circle and its exterior) as a total number of circles, a number of tangent points on the outer circle, and a symmetry group. – aschepler Mar 15 '21 at 02:42
  • @Jens I thought the black boxes were there for some other reason ... – Some Guy Mar 17 '21 at 17:56

3 Answers3

11

Let's be a little systemic about drawing these.

For $n=4$, there are 9 hierarchies of circles (I think this is from OEIS A000081).

  1. (((()))) the circles are all nested inside each other.

There are four of these: the differences come from whether the points of tangency coincide or not. You have them labelled as #26, #28, #29, and #30 in the diagram when I started writing this.

  1. ((()())) two innermost circles surrounded by a nested pair.

There are six of these: the two innermost circles can either 1. each touch only their parent, 2. touch each other but only one touch the parent, or 3. touch each other and both touch the parent, and then the point of tangency with the grandparent can be involved or not. Of these, only one is included in your diagram, as #34.

  1. ((())()) two children, one grandchild.

I'll call the child with its own child A, and the child without its own child B. The children can touch just the parent (and the grandchild can touch or not the outer circle); A can touch only B and not parent (and the grandchild can touch or not B); B can touch only A and not parent (and the grandchild can touch B, the outer circle, or neither); A and B can touch both each other and the parent (and the grandchild can touch B, the outer circle, or neither), for a total of 10. Of these, you have five: #22, #23, #31, #32, and #33.

  1. ((()))() one with child and grandchild, one empty.

There are four of these: the grandchild can touch or not the grandparent, and the child can touch or not its uncle. Of these you have two, #27 and #45.

  1. (()()()) one parent with three children.

The three children can be in four layouts: they can not touch at all; there can be one pair and one lonely one; there can be three in a line; there can be all three in a triangle. This leads to eleven in the whole class: the three independent siblings can all touch parent #16, the pair can both touch #13 or only one can #35, the line can touch once on the end or in the middle (you have neither) or twice on both ends #12 or end and middle #25 or three times #19, the triangle can touch once (not present) twice (double counted #15 and #24) or three times (#11)

  1. (()())() one with two children, one empty.

This lineup is basically the same as type 2; the only difference is that the "outermost circle" in 2 has become an uncle instead of a grandparent. You have five of these #14 #17 #18 #39 #43, and have double counted two as #20 and #21 - you're missing the uncle-tangent counterpart to #14.

  1. (())(()) two with one child each.

There's only three of these: the question is how many children touch their uncles. It could be zero #38, one #37, or two #36.

  1. (())()() one with one child, two empty.

Geometry finally comes into play! The three outer circles can be in a line or triangle. If they're in a line, the child can be in one of the ends, or in the middle, and it can touch or not an uncle #6, #7, #8, #44; if they're in a triangle, the child can touch one of the uncles #10, or outside the triangle formed by the uncles #9, or inside that triangle (not present)

  1. ()()()() four empties.

I think you actually have all of these: there's line #5, triangle plus tail #1, rhombus #2, square #3, star #40, and then there are three that are a triangle with circle in the middle that touches one #42 two #41 or three #4 of the outer circles.

So in total you appear to have (in the diagram when I started writing this answer double counted three (#20 is the same as #18, #21 is the same as #17, and #24 is the same as #15), and missed sixteen, for a grand total of 58 that I can count.

Dan Uznanski
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    A superb description. Indeed, ordering by the sibling relation is much better to find all solutions than order by tangent point multiplicity. I found all 58 you suggested. – Jens Mar 15 '21 at 17:55
10

Thanks to @DanUznanski here is the complete solution with 58 ways, using his sibling relation to group diagrams.

4 Kissing Circles

Jens
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  • oh good you did it, thank you. I was amidst doing it myself and getting increasingly annoyed at not-terribly-nice numbers! – Dan Uznanski Mar 16 '21 at 11:46
  • @DanUznanski No problem. Now that I stared at 9. four empty for a while, I wonder if the second and the last are the same or not. The adjacency matrix is identical. Move the tiny circle to the 12 o'clock position. Topologically they look different, though. The same holds true for the rhombus and the bottom middle diagram. – Jens Mar 16 '21 at 16:11
  • They are not, because there is no continuous deformation (or reflection) that takes one to the other without going through a known different thing: you can't slide the tiny circle around to the 12 o'clock position without first going through 9-7, a non-kissing layout, 8-6, another non-kissing layout, and 9-3. Similar results make 9-7 distinct from 9-3. Also now that I'm paying attention, you're also still missing the one like 8-5 where the child circle touches inside instead of outside – Dan Uznanski Mar 16 '21 at 16:25
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    @DanUznanski You mean 8-5 with the child falling down and touching at the parent's 6 o'clock position? I'll get right to it! – Jens Mar 16 '21 at 19:12
  • I am staring at 8-3 and imagine iterating this: a total of 7 circles, with 3 empty ones intertwined by 2 with a child. I guess there are 2 (or in fact, 3!) topologically different constellations now: one with both children on the "same side" (say, touching on top, i.e. top-top) and 1 top-bottom - AND the 3rd one, bottom-top, looks like it is distinct again! (Unless reflections of the whole are allowed, not only rotations) Much caution needed... – Wolfgang Mar 18 '21 at 21:59
3

This is not a complete answer, but a look at ways of describing these drawings and forming larger drawings from smaller ones.

Slightly abusing notation, let's say $A \sim B$ if circles $A$ and $B$ touch externally, or $A < B$ if circle $A$ is inside and tangent to circle $B$. Square brackets around a "formula" of three or more circles mean that they all meet at a common tangent point; otherwise, each relation symbol is a tangent point for only two circles.

Then for $n=2$, the two ways are $C_1 \sim C_2$ and $C_1 < C_2$.

For $n=3$, there are $9$ ways:

  • $[C_1 < C_2 < C_3]$
  • $C_1 < C_2 < C_3$
  • $C_1 \sim C_2 < C_3$
  • $C_1 < C_3$ and $C_2 < C_3$
  • $C_1 \sim C_2 < C_3$ and $C_1 < C_3$
  • $[C_1 < C_2 \sim C_3]$
  • $C_1 < C_2 \sim C_3$
  • $C_1 \sim C_2 \sim C_3$
  • $C_1 \sim C_2 \sim C_3 \sim C_1$

Note for any connected diagram of this sort with three or more circles, it's always possible to remove some circle with empty interior so that the remaining diagram is still connected. Proof:

Given a circle $C$, call its "inner depth" the largest integer $n$ such that there is a sequence of circles $C_i$ with $C_1 < C_2 < \cdots < C_n < C$. A circle with empty interior has inner depth zero.

If all circles in a diagram of at least three circles have inner depth zero, consider the graph defined by identifying the circles as vertices, with edges between vertices where the circles are tangent. A finite graph with at least two vertices always has at least two vertices which are either a leaf or within a graph cycle, and removing such a vertex from the graph leaves the remainder of the graph connected. So a corresponding circle can be removed and leave the remaining diagram connected, and in this case all the circles have empty interiors.

If not all circles in a diagram of at least three circles have inner depth zero, then there is at least one circle $C_0$ with inner depth $1$. Consider the graph whose vertices are $C_0$ and the circles it contains. There is at least one contained circle, and all these contained circles have inner depth $0$. This graph again has at least two vertices which can be removed without disconnecting the graph. Remove a circle which is not $C_0$ from the diagram. This circle has inner depth zero and empty interior. Since the graph remains connected, all the other remaining circles contained in $C_0$ are connected to $C_0$. All the diagram's circles outside $C_0$ are connected to $C_0$ via paths entirely outside $C_0$ (up to a tangent point), and those paths have not been affected. So the diagram is still connected.

The other way around, then, when looking for diagrams of $n+1$ circles, we can simplify by considering where to add a circle to a diagram of $n$ circles so that the added circle does not enclose any other circle. In the above list of diagrams for $n=3$, the first $5$ can be found from adding the minimal $C_1$ to a diagram $C_2 < C_3$, the next $2$ from adding to a diagram $C_1 < C_2$ OR $C_2 \sim C_3$, and the last $2$ from adding to a diagram $C_2 \sim C_3$.

Without looking back at the OP's drawings, I came up with 52 drawings for $n=4$. They're ordered by the nature of the diagram for $C_2,C_3,C_4$. A parenthesized color code matches the OP's convention of "r" for a point with two circles, "g" a point with three, or "b" a point with 4. A number after that indicates the equivalent drawing in the OP, if any.

  • (b 45): $[C_1 < C_2 < C_3 < C_4]$
  • (gr 37): $[C_2 < C_3 < C_4]$ and $C_1 < C_2$
  • (gr): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_2$
  • (grr 43): $[C_2 < C_3 < C_4]$ and $C_2 \sim C_1 < C_3$
  • (gr): $[C_2 < C_3 < C_4]$ and $C_1 < C_3$
  • (gr 36): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_3$
  • (grr 41): $[C_2 < C_3 < C_4]$ and $C_3 \sim C_1 < C_4$
  • (gr 38): $[C_2 < C_3 < C_4]$ and $C_1 < C_4$
  • (b 46): $[C_2 < C_3 < C_4 \sim C_1]$
  • (gr): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_4$
  • (gr 35): $[C_1 < C_2 < C_3]$ and $C_3 < C_4$
  • (rrr 5): $C_1 < C_2 < C_3 < C_4$
  • (rrr 9): $C_1 \sim C_2 < C_3 < C_4$
  • (rrrr): $C_2 < C_3 < C_4$ and $C_2 \sim C_1 < C_3$
  • (rrr): $C_2 < C_3 < C_4$ and $C_1 < C_3$
  • (gr): $[C_2 < C_3 \sim C_1]$ and $C_3 < C_4$
  • (grr 44): $[C_2 < C_3 \sim C_1]$ and $C_1 < C_4$ and $C_3 < C_4$
  • (rrr): $C_2 < C_3 < C_4$ and $C_1 \sim C_3$
  • (rrrr 19): $C_2 < C_3 < C_4$ and $C_3 \sim C_1 < C_4$
  • (rrr 4): $C_2 < C_3 < C_4$ and $C_1 < C_4$
  • $C_2 < C_3$ and $[C_3 < C_4 \sim C_1]$
  • (rrr 12): $C_2 < C_3 < C_4 \sim C_1$
  • (rrrr 15): $C_4 > C_2 \sim C_3 < C_4 \sim C_1$
  • (grr 39=40): $[C_3 < C_4 \sim C_1]$ and $C_3 \sim C_2 < C_4$
  • (rrrrrr 30): $C_1 \sim C_2 \sim C_3 \sim C_1$ and $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$
  • (rrrrr 26): $C_1 \sim C_2 \sim C_3 \sim C_1$ and $C_2 < C_4$ and $C_3 < C_4$
  • (rrrrr 25): $C_1 \sim C_2 \sim C_3$ and $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$
  • (rrrr 22): $C_1 \sim C_2 \sim C_3$ and $C_2 < C_4$ and $C_3 < C_4$
  • (rrrr 20): $C_4 > C_2 \sim C_3 < C_4$ and $C_1 < C_4$
  • (gr): $[C_1 < C_2 \sim C_3]$ and $C_3 < C_4$
  • (rrr): $C_1 < C_2 \sim C_3 < C_4$
  • (rrr): $C_1 \sim C_2 \sim C_3 < C_4$
  • (rrrr 17): $C_4 > C_1 \sim C_2 \sim C_3 < C_4$
  • (rrrr): $C_3 \sim C_1 \sim C_2 \sim C_3 < C_4$
  • (rrr 7): $C_1 < C_4 > C_3 \sim C_2$
  • (gr): $[C_3 < C_4 \sim C_1]$ and $C_2 \sim C_3$
  • (rrr 11): $C_2 \sim C_3 < C_4 \sim C_1$
  • (rrr 6): $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$
  • (gr 34): $[C_3 < C_4 \sim C_1]$ and $C_2 < C_4$
  • (rrr 13): $C_2 < C_4$ and $C_3 < C_4$ and $C_1 \sim C_4$
  • (grr 42): $[C_2 < C_3 \sim C_4]$ and $C_3 \sim C_1 \sim C_4$
  • (gr 32): $[C_2 < C_3 \sim C_4]$ and $C_1 \sim C_3$
  • (gr 33): $[C_2 < C_3 \sim C_4]$ and $C_1 \sim C_4$
  • (rrrr 21): $C_1 \sim C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$
  • (rrr 8): $C_1 \sim C_3 \sim C_4$ and $C_2 < C_3$
  • (rrr 2): $C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$
  • (rrr 1): $C_1 \sim C_2 \sim C_3 \sim C_4$
  • (rrr 10): $C_2 \sim C_3 \sim C_4$ and $C_1 \sim C_3$
  • (rrrr 14): $C_1 \sim C_2 \sim C_3 \sim C_4 \sim C_1$
  • (rrrr 18 (and 23)): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_4$
  • (rrrrr 24 (and 27)): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_3$ and $C_1 \sim C_4$
  • (rrrrrr 29): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_2$ and $C_1 \sim C_3$ and $C_1 \sim C_4$

Comparing to the OP's drawings, I see that my descriptions for numbers 18 and 24 also describe 23 and 27, respectively. Among the OP's drawings, I also missed

  • (rrr 3): $C_2 < C_3 \sim C_4 > C_1$
  • (gr 31): $[C_2 < C_3 \sim C_4]$ and $C_1 < C_4$
  • (b 47): $[C_1 < C_2 \sim C_3 > C_4]$

So this is at least 57 drawings for $n=4$. I think the 58th mentioned by @DanUznanski's answer is the variation of (rrrr 21: $C_1 \sim C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$) where $C_2$ touches $C_3$ inside the triangle formed by the other three tangent points.

I don't think this notation is sufficient alone for higher numbers. With a loss of symmetry, it can matter which order the tangent points come in for a circle with three or more tangent points. We already see this sort of trouble for $n=4$ with pairs $18, 23$ and $24, 27$.

aschepler
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