Four Is Not Enough - d2r55xnwy6nx47.cloudfront.net€¦ · Quanta Magazine June 18, 2018 Four Is Not Enough

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https://dev.quantamagazine.org/the-numbers-and-geometry-behind-a-math-coloring-puzzle-20180618/ June 18, 2018

Four Is Not EnoughHow many colors do you need to color an infinite plane so that no points 1 unit apart are the samecolor?

By Patrick Honner

BIG MOUTH for Quanta Magazine

Suppose you want to cover a standard 8-by-8 chessboard entirely with rectangular dominoes thateach cover two squares on the board. It’s easy to imagine how you might do it: You could line thedominoes up horizontally, four in a row, or vertically, four in a column. You could arrange them likestairs, concentric squares or gnashing teeth. There are nearly 13 million ways to do it, and eacharrangement requires exactly 32 dominoes, since the total area of the chessboard is 64 squares andeach domino covers 2 squares.

But what happens if we remove a pair of opposite corners from the chessboard? How many ways canyou cover this new board with 2-by-1 dominoes? Those unfamiliar with this classic problem may be

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surprised to learn that the answer is zero. It is impossible to cover this altered chessboard with 2-by-1 dominoes. How did we go from 13 million to zero so quickly? The answer is remarkably easy tofind, if you know where to look.

The squares on a chessboard alternate in color, between dark and light. This has two importantconsequences for a standard 8-by-8 chessboard: First, there must be an equal number of dark andlight squares, and second, opposing corners must be the same color. This means that when weremove a pair of opposite corners, the number of light squares that remain will be unequal to thenumber of dark squares. But every 2-by-1 domino must cover one dark square and one light square,which means a collection of dominos can only cover a chessboard with an equal number of dark andlight squares. Our altered chessboard cannot be covered.

Of the many wonderful chessboard math problems, this one is my favorite because of its surprisingsolution. Who would expect the coloring of the squares to be the key to unlocking the puzzle? Butmathematicians have been using colorings like this — which indicate structure and implycategorizations — to solve problems for a long time. And a recent advance in plane coloring hasbreathed new life into a problem that has stumped the mathematical community for over 60 years.

Consider the standard geometric plane, an infinite expanse of points in two dimensions. Your task isto color each of the infinitely many points in the plane. You might wish to color the entire plane red,or maybe half red and half blue, or maybe you’d splatter the color like in a Jackson Pollack painting.But there’s one rule in our plane coloring problem: If two points are exactly 1 unit apart, they cannotbe the same color. Can you color every point in the plane without violating this rule?

“Of course!” you might say, “I’ll just use infinitely many colors.” There is a certain elegance to thissneaky approach (setting aside the philosophical question of whether infinitely many colors exist),but can you do it with finitely many colors? And if so, how many different colors would you need?Finding the minimum number of colors needed is known as the Hadwiger-Nelson problem, and it’salso often described as finding the “chromatic number” of the plane. (The chromatic number is aproperty of graphs, which are collections of vertices, or nodes, and the edges, or lines, that connectthem. The chromatic number of a graph is the minimum number of colors needed to color all thevertices so that no two vertices connected by an edge are the same color. This is related to analternate formulation of our plane coloring problem involving so-called unit-distance graphs.) TheHadwiger-Nelson problem was first posed nearly 70 years ago, and we still don’t know the minimumnumber of colors we would need. But the new result has narrowed down the possibilities. So, whatare the possible numbers?

First, let’s see that finitely many colors will do. This red square will help us see why.

https://www.quantamagazine.org/decades-old-graph-problem-yields-to-amateur-mathematician-20180417/

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The red square above has side length $latex \frac {2}{3}$. Suppose you have two points in thissquare: How far apart could they be? The maximum distance between any two points in a square isequal to the length of the square’s diagonal, which we can compute using the Pythagorean theorem.

$latex d^2=\frac {2}{3}^2+\frac {2}{3}^2$

$latex d^2=\frac {4}{9}+\frac {4}{9}$

$latex d^2=\frac {8}{9}$

$latex d=\sqrt{\frac {8}{9}}$

And since

$latex \sqrt{\frac {8}{9}}<\sqrt{\frac {9}{9}}=\sqrt{1}=1$,

we know that the distance between any pair of points in this square is less than 1 unit. This meansthat we can color this whole square red and not violate our plane coloring rule.

Now, let’s make a bigger square out of nine of these little squares of side length $latex \frac{2}{3}$, and we’ll give each one its own color.

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Since each square has a different color, we still have not violated our rule. Of course, since this largesquare (made of nine smaller squares) has a side length of 2, we’ve only covered 4 square units ofthe plane. But we can get the rest by copying and pasting!

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We can cover the entire plane like this, but have we satisfied our rule about using different colorsfor points 1 unit apart? Just think about the red squares. We’ve already shown that no two pointsinside any one red square are 1 unit apart. Now, since each little square has side length 2/3, theminimum distance between any two distinct red squares is $latex \frac {2}{3}$ + $latex \frac{2}{3}$ = $latex \frac {4}{3}$, which means that any two points inside two different red squareswill be more than 1 unit apart. Thus, no two red points in the plane are exactly 1 unit apart. Andsince the same argument applies to all nine colors, this is indeed a valid coloring of the plane inwhich no two points of the same color are exactly 1 unit apart. This shows not only that thechromatic number of the plane is finite, but it is at most 9.

A slightly more complicated argument using regular hexagons shows that seven colors are actuallyenough. You can surround a regular hexagon with six others and give them all different colors.

As with the squares above, you can extend this coloring in every direction.

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By making the hexagon the right size — a diameter of slightly less than 1 will do — and extendingthe pattern correctly, we can be sure that no two points exactly 1 unit apart will be the same color.

So we’ve established an upper bound of 7 on the chromatic number of the plane. Can we easilyestablish a lower bound?

Well, it’s easy to see that we need more than one color. Take a point, P, in the plane: P needs a color,so let’s color it blue.

Since the plane extends infinitely in every direction, we know we can find a point, Q, that is exactly 1unit away from P.

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Point Q needs a color too, but our rule says Q can’t be blue. So we need a second color. Let’s color Qred.

Thus, the chromatic number of the plane is at least 2. Are two colors enough? Since the planeextends infinitely in every direction, there are actually infinitely many points 1 unit away from P. Infact, this is the definition of the circle of radius one centered at P.

Since P is blue and every point on the circle is 1 unit away from P, no point on the circle can be blue.With only two colors, that would mean that every point on the circle would have to be red.

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But now think about things from Q’s perspective: Nothing 1 unit from Q can be red. But, just as withP, there is an entire circle of points exactly 1 unit away from Q.

This new circle around Q intersects the red circle centered at P at two points: Let’s call them R andS. Points R and S can’t be blue, because they are 1 unit from P, and they can’t be red, because theyare 1 unit from Q. So we need a third color. Let’s color R and S green.

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Notice the familiar geometric shapes emerging here: PQR and PQS are both equilateral triangles,and PSQR is a rhombus. Some elementary geometry shows us that $latex SR = \sqrt{3}>1$ , so weknow it’s OK to color both S and R green.

This demonstrates that the chromatic number of the plane is at least 3. A clever trick will get us to 4.Consider that rhombus we just made.

Suppose we have two copies of this rhombus sitting on top of each other. Imagine that we drive a

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nail through R, so that the two rhombuses hang like pictures on a wall. Now we grab the bottomvertices of the two rhombuses and slowly pull them apart.

The rhombuses stay rigid as we pull them apart, and all the colorings remain valid. That is, until wepull them far enough apart so that the two bottom vertices are exactly 1 unit apart.

Now we have a problem. The two bottom vertices can’t both be green, because they are exactly 1unit apart. Nor can they be red or blue. We need a fourth color! This diagram is called the Moserspindle, named after Leo and William Moser, and it proves that the chromatic number of the plane isat least 4.

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In summary, we now know that the chromatic number of the plane is at least 4 but at most 7. Andfor nearly 60 years, this was everything we knew about the problem. Until Aubrey de Greyannounced the discovery of this mathematical object in April 2018.

Source: https://arxiv.org/pdf/1804.02385.pdf

Just as the Moser spindle creates intricate interdependencies among its seven points that cannot besatisfied using only three colors, the 1,581-point construction by de Grey shown above cannot becolored using only four colors without a pair of points 1 unit apart being colored the same. Butunlike with the Moser spindle, it’s not easy to see this just by looking — in fact, it takes a computersearch to verify that four colors really aren’t enough. Thanks to de Grey’s discovery, we now knowthat the chromatic number of the plane is at least 5. Combined with our previous knowledge, thatmeans we know that the chromatic number of the plane is either 5, 6 or 7. We just don’t knowwhich!

This surprising result from de Grey, a biologist and recreational mathematician, has re-energized themathematical community around the problem of coloring the plane. A collaborative online effortquickly sprang up to verify de Grey’s graph and find simpler versions (there are now examples usingonly around 800 points), and mathematicians are exploring whether de Grey’s techniques might beextended to get the lower bound to six colors or even seven. But some mathematicians suspect thatthe true answer to the Hadwiger-Nelson problem is unknowable, and that the chromatic number ofthe plane depends on independent issues like how we allow ourselves to think about and choose

https://arxiv.org/abs/1804.02385

https://arxiv.org/pdf/1804.02385.pdf

https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/

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from infinite sets of points.

Still, there’s no question that de Grey’s result has gotten mathematicians excited about thiselementary open problem, and many believe that an answer may be near at hand. Will we find it?Color me optimistic.

Download the “Four Is Not Enough” PDF worksheet to share with students.

https://d2r55xnwy6nx47.cloudfront.net/uploads/2018/06/MathematicalColoringWorksheet.pdf

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Four Is Not Enough - d2r55xnwy6nx47.cloudfront.net€¦ · Quanta Magazine June 18, 2018 Four Is Not Enough