# Communications of the ACM

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## Puzzled: Solutions and Sources

Solution. The object was to cover a gravy stain of area less than one square inch with a plastic sheet containing a grid of side one inch in such a way that no intersection point of the grid fell on the stain. This puzzle (as I was reminded by Andrei Furtuna, a Dartmouth computer science graduate student) may date to the great Lithuanian-born mathematician Hermann Minkowski (18641909).

It suffices to consider only grids oriented North-South-East-West or, equivalently, to assume the plastic sheet is aligned with the table. Now imagine cutting the tablecloth into one-inch squares in an aligned grid pattern, pin one of the stained squares to the table, oriented as it was originally, and stack (without rotating any square) all other stained squares neatly on top of it.

The stain is now within one square, but since the area of the stain is less than one square inch (and can be reduced only through stacking), some squares remain stain-free. Now pick a stain-free point and place the plastic so its intersection points lie directly on the point.

Since all other intersection points are outside the stained square, no intersection point touches the stacked stain. But what if the tablecloth were sewn back together? Each stained square would then be translated by an integral number of tablecloth squares East or West and North or South back to the original position. It would then bear the same relationship to the plastic sheet's grid points it did before; that is, it would miss them.

### 2. Covering Dots on a Table.

Solution. We had to show that any 10 dots on a table can be covered by non-overlapping \$1 coins, in a problem devised by Naoki Inaba and sent to me by his friend, Hirokazu Iwasawa, both puzzle mavens in Japan.

The key is to note that packing disks arranged in a honeycomb pattern cover more than 90% of the plane. But how do we know they do? A disk of radius one fits inside a regular hexagon made up of six equilateral triangles of altitude one. Since each such triangle has area 3/3, the hexagon itself has area 23; since the hexagons tile the plane in a honeycomb pattern, the disks, each with area π, cover π /(23) ~ .9069 of the plane's surface.

It follows that if the disks are placed randomly on the plane, the probability that any particular point is covered is .9069. Therefore, if we randomly place lots of \$1 coins (borrowed) on the table in a hexagonal pattern, on average, 9.069 of our 10 points will be covered, meaning at least some of the time all 10 will be covered. (We need at most only 10 coins so give back the rest.)

What does it mean that the disks cover 90.69% of the infinite plane? The easiest way to answer is to say, perhaps, that the percentage of any large square covered by the disks approaches this value as the square expands. What is "random" about the placement of the disks? One way to think it through is to fix any packing and any disk within it, then pick a point uniformly at random from the honeycomb hexagon containing the disk and move the disk so its center is at the chosen point.

### 3. Placing Coins.

Unsolved. The solution to Puzzle 2 doesn't tell us how to place the coins, only that there is a way to do it. Is there a constructive proof? Yes, and we can use the solution to Puzzle 1 (concerning the stain) to find it. I leave it to your imagination to follow up.

That proof can be used to increase the number of dots to 11 or 12, still using only an aligned hexagonal lattice of coins. However, since we aren't restricted to a lattice, it seems plausible that quite a few dots can be covered, perhaps as many as 25 (see the August column). If you figure out a dot pattern with, say, 30 or fewer points you think can't be covered by unit disks, please send to me, along with your reasoning.

### Author

Peter Winkler (puzzled@cacm.acm.org) is Professor of Mathematics and of Computer Science and Albert Bradley Third Century Professor in the Sciences at Dartmouth College, Hanover, NH.

### Footnotes

All readers are encouraged to submit prospective puzzles for future columns to puzzled@cacm.acm.org.

DOI: http://doi.acm.org/10.1145/1810891.1810917

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##### P Albert

The solution to the first problem was quite elegant, but I got stuck trying to parse "can be reduced only through stacking" (I think "can be only reduced through stacking", i.e., stacking can only reduce the area, not increase the area, was intended). Nonetheless, the proof works.

I am not convinced that that is the case for the second puzzle. You have not actually shown that it is impossible to place 10 points such that one hexagonal pattern cannot simultaneously cover all 10 points.

Because you do not permit overlap, the coins cannot be independently randomly placed. Furthermore, in order to get to the 91% coverage, the first coin can be randomly placed, but then the only remaining degree of freedom - if the hexagonal pattern is to be obtained - is the rotation of the pattern and that is fixed once the second coin is placed.

Because the disks are all not randomly placed, only the hexagonal pattern is randomly placed, the 90.69% value is the probability that one randomly selected point is covered by one randomly selected hexagonal pattern. So, at best, you have proven that when ten hexagonal patterns are selected, one for each point, there must be some instances where every point was covered by its hexagonal pattern, not that there are some instances where all ten points are covered by one pattern.

However, you haven't even proven that. Granted, if a given set of 10-bit words averaged more than 90% 1's, the set must have included at least one 10-bit word that is all 1's. However, it is not a certainty that if bits are randomly selected with p(1)=0.9069 and p(0)=1-p(1) that the the average number of 1's must be greater than 90%. Likely, with a large enough set, but never a certainty.

##### Anonymous

These probabilistic arguments are tricky if you are not used to them.
In the analogy with 10-bit words, if you pick 10 random (independent or dependent) bits with p(1)=0.9069 then the EXPECTED (or "average") number of ones is equal to 10*p(1), which exceeds 90%. (This is a consequence of "Linearity of Expectation.")
-- G. Rote