This week the #puzzle is still unknown to me, as I write this. Because I am busy. But here’s some other content for you.

Some other very fine fiddlers:

• Benjamin Phillabaum (bphillab.github.io)
• Ed Nigma’s Answer Emporium
• Jamshrivelson
• Rohan Lewis (humanrickshaw)
• Sameer Gauria

If you want to compare 2 large numbers, and remember to carry the one(s), this is how you do it.

First row in the table below: Our units. In the middle, there’s a familiar system: 1, 10, 100. This is how our own numbers are constructed. There’s a position for 1s, a position for 10s, etc. To the left and to the right, the units are not the familiar ones, but hopefully we can understand this anyway.

Second row: A sum of a lot of other numbers. The sum hasn’t been normalized to its shortest from yet. We want to normalize and then compare with the last row, another number from the same source.

Normalization example: In the second row, 15 and 15 are bolded. The jump from the first to the second column is to multiply by 2. (1/120 * 2 = 1/60.) So to normalize, we look at the first 15 and divide by 2.

• a = 1/120, b = 1/60
• 2a = b
• 15/2 = 7.5
• 15 = 1 + 7 * 2
  • 15a + 15b
  • = (1 + 7 * 2)a + 15b
  • = a + 7 * 2a + 15b
  • = a + 7b + 15b
  • = a + 22b

So in the next row, instead of 15 and 15, we have 1 and 22.

Move to the next pair of columns and repeat.

1/120

1/60

1/30

1/10

1/5

1

10

100

300

3000

18000

15	15	24	14	19	26	39	11	7	8	5
1	22	24	14	19	26	39	11	7	8	5
1	0	35	14	19	26	39	11	7	8	5
1	0	2	25	19	26	39	11	7	8	5
1	0	2	1	31	26	39	11	7	8	5
1	0	2	1	1	32	39	11	7	8	5
1	0	2	1	1	2	42	11	7	8	5
1	0	2	1	1	2	2	15	7	8	5
1	0	2	1	1	2	2	0	12	8	5
1	0	2	1	1	2	2	0	2	9	5
1	0	2	1	1	2	2	0	2	3	6
1	0	2	1	1	2	2	1	1	3	6

And yes, the end result is, that the 2 numbers aren’t quite the same.

Background: Recently I read this, in From One to Zero, Georges Ifrah :

This week the #puzzle is: Can You Rile and Grace the Polyhedron? #geometry

An alien—more specifically, an Eridian—named “Rocky” needs to pass a solid xenonite crystal to his human friend in a neighboring spaceship. The crystal is shaped like a regular tetrahedron, as shown below, and all its edges have length 1.

To safely transport the crystal, Rocky needs to create a long cylindrical tunnel between the spaceships, and then orient the tetrahedron so that it fits through the tunnel. It’s okay if the crystal fits snugly inside the tunnel—in this case, it can slide along without any friction.

What is the minimum possible radius for my tunnel so that the crystal will fit through it?

And for extra credit:

Next, Rocky wants to transport a solid crystal shaped like a regular dodecahedron to his human friend. As before, each edge has length 1.

This time, the long tunnel between the space ships can be any right prismatic shape, not necessarily a cylinder. Once again, Rocky needs the crystal to fit through the tunnel, and it’s okay if that fit is snug.

What is the minimum possible cross-sectional area for the tunnel so that the crystal will fit through it?

Can You Rile and Grace the Polyhedron?

Intermission

The results from Q1 are in, and the finest of Fiddlers, who each solved all 24 puzzles, are: 👑 Adnan Haque 👑 from New York 👑 David Kong 👑 from Toronto, Canada 👑 Josh Silverman 👑 from Glen Head, New York 👑 Lise Andreasen 👑 from Valby København Danmark 👑 Peter Ji 👑 from Madison, Wisconsin 👑 Seth Cohen 👑 from Concord, New Hampshire

Woohoo!

Solution, possibly incorrect:

Desmos

My only option this week is to guess. I guess the circle shown on Desmos. Radius 0.5.

Given more time, I would have coded a tetrahedron, that could rotate, so that I could test different options.

And for extra credit:

I don’t even have a guess.

This week the #puzzle is: Can You Pile the Primes? #primes #coding #BruteForce #oeis #subsets

Suppose you want to make two groups with equal sums using the first N2 prime numbers. What is the smallest value of N2 for which you can do this?

The answer is three! (Clearly, that wasn’t actually the puzzle.)

The first three primes are 2, 3, and 5, and you can split them up into two sets: {2, 3} and {5}. Sure enough, 2 + 3 = 5.

Your puzzle involves making three groups with equal sums using the first N3 prime numbers. What is the smallest value of N3 for which you can do this?

And for extra credit:

Now you want to make six groups with equal sums using the first N6 prime numbers. What is the smallest value of N6 for which you can do this?

Can You Pile the Primes?

Intermission

I am very excited, because this is the last fiddler of Q1, and I still have maximum points. Wish me luck.

Solution, possibly incorrect:

Program

Method 1: Write a brute force program. This actually finds a solution rather quickly. N₂ is confirmed, among other things.

Testing N1 =   1. Testing these primes: 2
................. Group 1: 2
................. Sum of groups: 2
................. N1 = 1

Testing N2 =   2. Sum of primes    5 isn't a multiple of 2, rejected.
Testing N2 =   3. Testing these primes: 5,3,2
................. Group 1: 5
................. Group 2: 3,2
................. Sum of groups: 5
................. N2 = 3

Testing N3 =   3. Sum of primes   10 isn't a multiple of 3, rejected.
Testing N3 =   4. Sum of primes   17 isn't a multiple of 3, rejected.
Testing N3 =   5. Sum of primes   28 isn't a multiple of 3, rejected.
Testing N3 =   6. Sum of primes   41 isn't a multiple of 3, rejected.
Testing N3 =   7. Sum of primes   58 isn't a multiple of 3, rejected.
Testing N3 =   8. Sum of primes   77 isn't a multiple of 3, rejected.
Testing N3 =   9. Sum of primes  100 isn't a multiple of 3, rejected.
Testing N3 =  10. Testing these primes: 29,23,19,17,13,11,7,5,3,2
................. Group 1: 29,7,5,2
................. Group 2: 23,17,3
................. Group 3: 19,13,11
................. Sum of groups: 43
................. N3 = 10

Result: N₃ = 10.

And for extra credit:

OEIS

Method 1: The same program.

N₆ has to be at least 57. But then my program gives up. Heat death of the universe kind of thing.

Testing N6 =  57.
Testing these primes:
269,263,257,251,241,239,233,229,227,223,
211,199,197,193,191,181,179,173,167,163,
157,151,149,139,137,131,127,113,109,107,
103,101,97,89,83,79,73,71,67,61,59,53,
47,43,41,37,31,29,23,19,17,13,11,7,5,3,2
Or rather, test left as an exercise for the reader.

Method 2: Try to find a solution with the first 57 primes. And I do!

• 269,199,197,139,137,89,59,37,19
• 263,211,181,163,109,103,53,47,13,2
• 257,223,173,167,107,101,71,41,5
• 251,227,179,157,127,83,61,31,29
• 241,229,193,151,113,97,67,43,11
• 239,233,191,149,131,79,73,23,17,7,3

N₆ = 57.

Method 3: Look for an OEIS sequence. Found. Results confirmed.

This week the #puzzle is: Can You or “Cantor” You Find the Distance? #probabilities #infinity #coding #montecarlo #permutations #distance

I start with a number line from 0 to 1, and then I remove the middle third. Then I take each of my remaining pieces (the segment from 0 to 1/3 and the segment from 2/3 to 1) and remove their middle thirds. Now I have four segments (from 0 to 1/9, 2/9 to 1/3, 2/3 to 7/9, and 8/9 to 1) and remove their middle thirds. I do this over and over again, infinitely many times.

The points I’m left with are collectively known as the Cantor set. The Cantor set is not empty; in fact, it contains infinitely many points on the number line, such as 0, 1, 1/3, and even 1/4.

It’s possible to pick a random point in the Cantor set in the following way: Start with the entire number line from 0 to 1. Then, every time you remove a middle third, you give yourself a 50 percent chance of being on the left remaining third and a 50 percent chance of being on the right remaining third. Then, when you remove the middle third of that segment, you again give yourself a 50 percent chance of being on the left vs. the right, and so on.

Suppose I independently pick two random points in the Cantor set. On average, how far apart can I expect them to be?

And for extra credit:

Suppose I independently pick three random points in the Cantor set. Each point has some value between 0 and 1.

What is the probability that these three values can be the side lengths of a triangle?

Can You or “Cantor” You Find the Distance?

Solution, possibly incorrect:

Program

Method 1:

• Monte Carlo. I try different “sensitivities”. This corresponds with how many decimals (ternimals?), I have in my ternary number. Using 30 digits and 1000000 tries, I get an average of 0.40. Really? I am baffled.

Method 2:

• Brute force. For a couple of different numbers of digits (1-12), I try all the numbers with that many digits, and try these in all combinations. My average converges nicely to 0.40 again.

Method 3:

• Let’s try and look at this using probabilities. Let’s look at all the 2 digit ternary numbers.

Distance	003	023	203	223
003 = 0	0	2/9	6/9	8/9
023 = 2/9	2/9	0	4/9	6/9
203 = 6/9	6/9	4/9	0	2/9
223 = 8/9	8/9	6/9	2/9	0

• The average in this tiny example is 7/18 or about 0.3888. Very close to 0.40 already.
• In fact I should be looking at intervals. Not 00₃, but 00₃-01₃.

Distance	003-013	023-103	203-213	223-1003
003-013 or 0/9-1/9	0/9-1/9	1/9-3/9	5/9-7/9	7/9-9/9
023-103 or 2/9-3/9	1/9-3/9	0/9-1/9	3/9-5/9	5/9-7/9
203-213 or 6/9-7/9	5/9-7/9	3/9-5/9	0/9-1/9	1/9-3/9
223-1003 or 8/9-9/9	7/9-9/9	5/9-7/9	1/9-3/9	0/9-1/9

Distance interval	Distance max	Frequency	Acc. freq.
0/9-1/9	1/9	4/16	4/16
1/9-3/9	3/9	4/16	8/16
3/9-5/9	5/9	2/16	10/16
5/9-7/9	7/9	4/16	14/16
7/9-9/9	9/9	2/16	16/16

• Half of the distances are 3/9 or below.
• The other half of the distances aren’t distributed the same way (that would be half of them 6/9 or above). So the average should be dragged down a little from simply being 0.5.

I trust my results. The average is 0.40.

And for extra credit:

Same program, same logic etc. Some of the same bafflement.

The ratio is 0.40.

I actually got curious and conducted the same experiments without the cantor set limitation. Then my average would have been 1/3 and my ratio would have been 1/5. So it does make a difference.

This week the #puzzle is: Can You Shovel the Snow? #geometry #area #equidistant #probabilities #coding #montecarlo

Two teams of shovelers plan to remove all the snow from a parking lot that’s shaped like a regular hexagon. Team Vertex initially places each of its six shovelers at the six corners of the lot. Meanwhile, Team Centroid initially places all its shovelers at the very center of the lot.

Each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team.

What fraction of the lot’s snow is Team Centroid responsible for shoveling?

And for extra credit:

Having completed the shoveling of the hexagonal lot, the two teams move on to a circular lot. This time, Team Vertex initially places all their shovelers at one random point inside the circle, while Team Centroid initially places all their shovelers at one other, independently chosen random point inside the circle. (To be clear, by “random point” in the circle, I mean that the probability of being in any given region of the circle is proportional to that region’s area.)

As before, each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team.

Inevitably, one team is responsible for shoveling a greater fraction of the lot than the other. On average, what would you expect this greater fraction to be?

Can You Shovel the Snow?

Solution, possibly incorrect:

Desmos

This image nicely captures my thoughts (click to enlarge):

• For each point within the hexagon, it will be closest to one particular point on the hexagon.
• The hexagon slices up into 6 parts, defined by being between 2 neighbor points on the hexagon and the center.
• On the image, a slice is shown with red borders, on the right part of the hexagon. The points on the hexagon are at (c,d) and (c,-d). c and d are calculated from the angle at the center.
• The slices are defined by lines passing through the center and a point on the hexagon.
• Within such a slice, there are further 2 halfs. In this case let’s concentrate on the upper half. In this half, points within the slice are either closer to the center or to (c,d). Find the midpoint between these 2 points and draw the perpendicular. (All the relevant perpendiculars, the parts we need at least, are drawn with blue lines.)
• On one side of the perpendicular, points are closest to the center.
• Calculate the area of the whole slice and of the purple triangle, divide, and we have our ratio.

The purple triangle is between these 3 points:

The area of the slice is:

The area of the triangle is:

The fraction of the points closest to the center:

The result is 1/3.

And for extra credit:

Program Desmos 1 Desmos 2

This could of course be solved using integrals. But I don’t know how… So I monte carlo.

In part my program is based on what this image shows:

Given 2 random points (red and blue — the red one was trying to leave the circle there, sigh), find the midpoint between them (green). Translate the midpoint so that it’s on a line going through the center. Then rotate that new midpoint around the center, so that it lands on the x axis. Then draw the perpendicular through that point (black). Calculate the area of the circular segment on the left and calculate the fraction. (I actually calculate the smaller fraction, so it’s then subtracted from 1.)

Do this a lot of times and calculate the average.

Some of the calculations. Find the midpoint between the 2 random points:

Translate that midpoint onto the line going through the center. First find the slant of the line through the 2 random points:

Then find the translated midpoint:

Distance from translated midpoint to center:

Find angle:

Find area:

The area of the whole circle:

The fraction, remember we need the largest one:

The result, from the program, about 0.64.

I plotted the distribution in Desmos too: