## Checkerboard Trilogy Question #2

OK, are you ready for Bobo The Bear’s 2nd Checkboard Trilogy problem?

**How many 1-inch diameter checkers can fit on an 8-inch by 8-inch checkerboard (Checkers may be placed anywhere on the board, as long as the entire checker lays flat and does not extend over the edges of the board)?**

*As always, answers can be entered in the comment section below. Submissions to be revealed next week, thanks*

You can check out the 1st Checkboard puzzle here ** **

munna| Profile June 10th, 2010 - 6:52 amAnswer is 68 Checkers (by placing them in Honeycomb patter).

If we place Checkers in Honeycomb patter on the board, then every two row will have 17 Checkers (1st row contain 8 Checkers and 2nd row contain 7 Checkers).

In this patter the breadth covered = [0.5 + {sqrt(3) * (N - 1)}/2 + 0.5] inches. Where N = Whole number representing number of rows.

We have breadth of 8 inches and the number of rows that can be accommodated in 8 inch of breadth can be derived by:

[0.5 + {sqrt(3) * (N - 1)}/2 + 0.5] <= 8 ("Less than or equals" operator is used because the total breadth covered should be less than or equal to 8 inches)

Further solving: {sqrt(3) * (N – 1)}/2 <= 7

(N – 1) <= 14/sqrt(3)

N <= 9.083

Since N is a whole number, we take N = 9. That means we can accommodate 9 rows easily.

9 rows = 5 rows with 8 Checkers + 4 rows with 7 Checkers (provided we start 1st row with 8 Checkers)

Hence the total number of checkers that can be placed on 8-inch by 8-inch board = (5 * 8) + (4 * 7) = 40 + 28 = 68.

suineg| PUZZLE MASTER | Profile June 10th, 2010 - 8:44 amWOW, I think is close to 72, 1 inch diameters checkers, but I solved visually so, it is very inexact, but cool

Shawn| PUZZLE GRANDMASTER | Profile June 10th, 2010 - 8:49 amUsing the given parameters, I suppose that if you stack the checkers in 64 columns you could theoretically place an infinite number of checkers onto a checkerboard.

Of course, at some height we will run out of gravity, and the cumulative weight of the checkers will begin to crush the board and the bottom rows of checkers. I assume that weight would become an issue long before checkers started to float off into space, but I’m not going to do that calculation.

suineg| PUZZLE MASTER | Profile June 10th, 2010 - 8:50 amMy mistake, I count one row twice, so it would be 67 or 68… pretty sure, cool!

suineg| PUZZLE MASTER | Profile June 10th, 2010 - 9:42 amNow mathematically, (dont want to appear lazy all the time man):

Quote: “Gauss proved that the hexagonal lattice is the densest plane lattice packing, and in 1940, L. Fejes Tóth proved that the hexagonal lattice is indeed the densest of all possible plane packings.”

So acording to:

http://www.combinatorics.org/V.....3i1r16.pdf

for k= 8, n= 68, my guess was not bad… coolkit!!!

MarkS| Profile June 10th, 2010 - 3:41 pmAssuming we’re ruling out stacking checkers then without doing the geometry first

I’m going with 68. Start with a row of eight and then alternating

with seven then eight again. The center of the checkers in the rows of eight will

be aligned with the center of the squares (but not centered in the squares) and the

checkers in the rows with seven will be centered on the line between the black and white

squares. This (I’m just guessing here) will allow enough room to add a final ninth row

with eight checkers. So five rows of eight and 4 rows of seven gives 68.

Jimmy Anders| PUZZLE MASTER | Profile June 10th, 2010 - 6:17 pmFill up the first row with checkers. Then for each adjacent pair of checkers, put a checker below that touches both. This pattern repeats as you can put a copy of the first row down so that each member of an adjacent pair touches the checker above that pair, and then copy the second row and so on. Looking at the first two rows, take a pair from the top and their mutually tangent checker below. Draw in radii for each checker that point to the centers of the other two. This makes an equilateral triangle of side length one. Next draw in radii that point straight up for the top two checkers and straight down for the bottom checker. This illustrates that the width of the first two rows is 1 (two radii) plus the height of the equilateral triangle, which is sqrt(3)/2. This means each row you add takes another sqrt(3)/2 inches. After doing the first row you have 7 inches left. That means there is room for 7 / [sqrt(3)/2] > 8 more rows. Since every pair of rows has 15 checkers, after you place 8, there is room for 4 * 15 = 60 checkers, meaning 68 total, which is 4 more than if you just place a checker on every square. This however still leaves more than 13 checkers worth of unfillable space, so maybe there is a more efficient way to fill the board…

infinityaurora314| Profile June 10th, 2010 - 9:10 pm128 checkers, 64 on the front and 64 on the back

infinityaurora314| Profile June 10th, 2010 - 9:26 pmWhat if you stacked up 8 vertical columns of checkers next to each other, stacked another 8 columns parallel to those, and set the checkerboard vertically between them? Then the number would depend on the thickness of the checkers…but it would be a big number!!!

mrman| Profile June 11th, 2010 - 2:42 pm73

I’m assuming this involves the hexagonal packing ratio of circles, but I could be way off.

Bobo The Bear| PUZZLE MASTER | Profile June 17th, 2010 - 3:24 pmThe ansewr I had in mind was 68, using a triangular packing, as most of you realized. I tried to word the problem to exclude the possibility of stacking (I should have stated that each checker does have to come into contact with the board), so I can see the merits of the “stacked checkers” answer as well.

@infinityaurora314: Using the back of the board as well. I like it! Way to think outside (or at least “on the other side” of) the box.

Shawn| PUZZLE GRANDMASTER | Profile June 23rd, 2010 - 1:42 pmI like your avatar, Bobo. Did you make that yourself?

Bobo The Bear| PUZZLE MASTER | Profile June 23rd, 2010 - 5:21 pmYou give me WAY too much credit there, Shawn. It’s the album cover of the Pink Floyd compilation album “ulse”. One of my favorite covers from one of my favorite bands.

Bobo The Bear| PUZZLE MASTER | Profile June 23rd, 2010 - 5:24 pm@Shawn: That’s Pulse, not “ulse”.

@RK: Is there a way that an individual can shut off the “auto-smiley” feature, so as to avoid occurrences such as these?

Shawn| PUZZLE GRANDMASTER | Profile June 24th, 2010 - 3:03 pmAha, don’t have that album. You’re right, it is quite worthy of the band. My personal favorite pf album is The Final Cut; as usual, bucking the trend of normal people everywhere.

RK| Founder | Profile June 24th, 2010 - 5:47 pmBa Ba Ba Ba Should we shoooouuuut….. should we sca-ream… .

What happened… tooo the post war dream?

Love that album!

Wow, those songs bring back vivid memories of my 1st year in college

Shawn| PUZZLE GRANDMASTER | Profile June 25th, 2010 - 3:22 pmAnother lover of perhaps the least popular pf album of all-time? Interesting that our shared trait would be viewed by most people as more of a quirk

They say Gilmour has the better voice, they say the band would be nothing without Gilmour’s silky guitar, said that Waters was obsessed and petty while Gilmour was righteous, but MAN could Waters belt out a tune. He sounded like his vocal cords were about to splinter, left it all on the field (of poppies), and there was never any doubt that he MEANT every word he sang, felt the lyrics down to his core.

Hold oooon to the dreeeeeeeaammm (cue throaty bluesy tenor sax)

I guess I know what I’ll be listening to on the car ride home tonight.