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Connect each red circle with each black circle by drawing a line and the lines should not touch.
From each red circle, 3 lines must be drawn which connect red circles with black circles, but the lines must not touch.

enter image description here

I am 32 years old now. When I was in 4th class, a friend of mine challenged me to solve this problem and I still can't solve it.

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    $\begingroup$ In graph theory speak this is impossible because the graph K 3,3 is non-planar $\endgroup$ – Mitchel Paulin May 27 at 15:13
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    $\begingroup$ See also: Connect 3 houses with 3 wells $\endgroup$ – Rubio May 30 at 0:52
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    $\begingroup$ (Note that the question here isn't "How to solve this" - the title makes it clear the question is, Why is it not possible (otherwise, this would be a dup of the question I just linked, and would be closed as such). Answers that proffer a solution, rather than address why this is not possible, are not answering the question asked and are subject to deletion.) $\endgroup$ – Rubio May 30 at 0:56
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This is a famous problem called the Three utilities problem, which is part of the mathematical field of graph theory. The problem basically asks for a planar embedding of the utility graph, which does not exist.

While it's not exactly know when the puzzle was invented, it was published at least as far back as 1913 and it took mathematicians until 1930 to solve it. That means it's rather difficult to prove why the puzzle is impossible to solve; it requires some advanced mathematics to do so, beyond the scope of almost all readers here. This is again a case of "it's hard to prove a negative".

Simply put, there are just too many lines which need to be drawn and too few points. This is just a 'fact' of two-dimensional mathematics, just like the icosahedron being the largest (three dimensional) polyhedron.

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    $\begingroup$ I would say it is not too difficult to prove once you can assume the Jordan curve theorem. I've essentially outlined the argument in my answer which I think is what they are referring to here $\endgroup$ – hexomino May 24 at 12:46
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    $\begingroup$ Proving that $K_{3,3}$ is nonplanar isn't that hard. Even without the Jordan curve theorem, the Wikipedia page on the three utilities problem includes a short proof based on Euler's formula for planar graphs. $\endgroup$ – user2357112 May 25 at 3:33
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    $\begingroup$ @Gnudiff, not a cylinder. You need a torus like this one. (no financial interest) $\endgroup$ – Euro Micelli May 25 at 23:01
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    $\begingroup$ @Gnudiff this puzzle is entirely solveable when drawn on, say, an orange :) $\endgroup$ – Moo-Juice May 27 at 11:40
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    $\begingroup$ @Moo-Juice, you cannot solve this puzzle on the surface of a sphere. If you were able to do it, you could then select a point on the sphere not covered by your drawing, use the point to make a hole in the sphere and then continuously transform the sphere into a plane. Now you sphere solution would turn into a plane solution. But we already know that plane solution does not exist. Contradiction. Q.E.D. $\endgroup$ – ppgdev May 28 at 4:00
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This is a problem I've thought about quite a bit too and here is my "proof" for why it cannot be done

For convenience, let us label the black circles $B_1, B_2, B_3$ and the red circles $R_1, R_2, R_3$.
Suppose there exists a way to connect all the circles as specified. Then the circles $B_1, R_1, B_2, R_2$ form a closed loop in the plane.
Since the circle $B_3$ and $R_3$ must be connected to each other, they must be both inside the loop or both outside the loop.

Both inside
The interiors of the loops formed by sequences of circles $B_1, R_1, B_3, R_2$ and $B_2, R_2, B_3, R_1$ are disjoint regions whose union is the interior of the original loop. The circle $R_3$ must lie inside one of these loops. If it is inside the first loop, there is no way to connect it to $B_2$ which is outside the first loop. If it is inside the second loop, there is no way to connect it to $B_1$ which is outside the second loop. This is a contradiction.

Both outside
In this case, either the sequence of circles $B_1, R_1, B_3, R_2$ or $B_2, R_2, B_3, R_1$ forms a loop whose finite interior is disjoint from that of the original. Without loss of generality, suppose it is the former. Then $R_3$ must be within this loop or outside of it. If it is within the loop, then there is not way to connect it to $B_2$, which is outside the loop. If it is outside this loop, there is not way to connect it to $B_1$ which is within the larger loop formed by $B_3, R_1, B_2, R_2$. Hence another contradiction.

Further reading

This is a corollary of Kuratowksi's Theorem and is an example of what they call a $K_{3,3}$ graph.

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  • $\begingroup$ so its not possible? $\endgroup$ – Navid2132 May 24 at 12:05
  • $\begingroup$ @Navid2132 Yes, unless you do something clever like suggested in Tahel's answer. But just as a planar problem, impossible to solve. $\endgroup$ – hexomino May 24 at 12:08
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    $\begingroup$ @hexomino, nice proof! +1! $\endgroup$ – Domosed May 24 at 17:00
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It has to do with graph theory and non-planar graphs.

If (and only if) a graph is planar, you can draw it on a flat piece of paper without the lines ever crossing.

On the other hand, Kuratowski's theorem states that a finite graph is planar if (and only if) it doesn't contain the K5 graph or the K3,3 graph in any form.

The K5 graph is the completely connected graph with 5 nodes. The K3,3 graph is the complete bipartite graph with 3 nodes on either side, which is also known as "the utility graph" because of this very puzzle. (It's usually given in the form of connecting utilities to houses.)

So this puzzle asks you to draw one of the two "archetypal" non-planar graphs, and therefore it's impossible to solve it on a flat piece of paper. Which is why some silly guys "fixed" the puzzle by printing it on a mug. :-)

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  • $\begingroup$ The "If (and only if)" sentence should be replaced by "a graph is said to be.. " or "a graph is called...". It is good practice to keep words involving causality (in one or both ways) for situations that actually involve theorems and not definitions. $\endgroup$ – Arnaud Mortier May 24 at 22:57
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While hexomino's answer gives a very nice proof from geometric principles, there is a somewhat fast way to see this: simply put, this is impossible because there are too many lines to draw. Especially for mathematicians already familiar with some graph theory, this can serve as a very fast way to solve the problem, but I will expand the theory to, as much as possible, be accessible to all.


This fact follows from something called the "Euler characteristic." Basically, suppose you create a figure obeying the following rules:

  1. You draw some set of points called vertices.

  2. You draw a set of curves called edges, each of which begins and ends at a vertex and does not intersect itself or any other edge.

  3. No pair of edges connect the same pair of vertices.

  4. It is possible to get from any vertex to any other vertex by following some series of edges.

Such a figure is called a connected planar graph. You'll note that such a graph determines a set of faces, which are just the regions cut out by the edges - including one face representing "the outside". If you let $V$ be the number of vertices and $E$ be the number of edges and $F$ be the number of faces, you have the following relation: $$V-E+F=2$$ I won't go into the proof because, while it's straightforward, it's hard to phrase without mathematical jargon - you can try it out, though! Draw any figure according to these rules and this relation will hold. (This is also a standard fact, subject to proofs in standard sources, such as those that appear via Google)

Another property sometimes called the faceshake lemma is that, if you define the degree of a face to be the number of edges bordering the face, then the sum of the degrees of the faces is twice the number of edges, since each edge borders exactly two faces. Note that the edges on the boundary of any face form a cycle - that is, if you traverse them, you end up back where you started. Since the shortest cycle in the desired graph has length $4$, every face would have to have at least four edges around it; this means we have the relation $$4F\leq 2E$$ since the sum of the degrees of the faces is at least $4F$.

Now, we put things together: we are supposed to draw a graph with $6$ vertices and $9$ edges. There must be no more than $4$ faces, for the faceshake lemma to hold. However, $6-9+F$ is supposed to equal $2$, which means that $F$ should equal $5$. This is a problem because $5>4$ - so there is no way to do it!

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Simple proof of impossibility:

Draw the six lines connecting two black circles to all three red circles. This divides the plane into three four-sided "faces", with two black and two red circles on each face. Regardless of where you place the third black circle, it cannot reach the red circle not sharing its face, Q.E.D. Diagram

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  • $\begingroup$ I do not understand this. You could make it a lot clearer by adding some illustrations.  Of course you run the risk that, once your answer becomes comprehensible, it will be revealed to be a duplicate of one of the ones already given. $\endgroup$ – Peregrine Rook Jun 18 at 4:12
  • $\begingroup$ Thanks for the clarification.  Your answer is similar to hexomino’s answer, but it is a slightly different way of looking at it. $\endgroup$ – Peregrine Rook Jun 21 at 1:07
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The problem can be reduced to the following scenario:

utility

where:

both green vertices need to be connected to both red vertices, but the edges cannot enter the grey area. This can be seen be a straight-forward connection in the original problem of A1, A2, A3, B3, and C3, which leaves B1, B2, C1 and C2 left, as in the above graph.

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