Plus-Plus Hexel Two-Dimensional Tessellations

Plus-Plus, according to the Plus-Plus website, allows you to "Unleash the power of your imagination with a unique design in a single shape.  Plus-Plus offers unlimited opportunities for creativity. From a single repeated element, the possibilities are endless as you bring your ideas to life in both 2D and 3D forms."

Plus-Plus (or sometimes just PlusPlus) is a toy made up of small plastic pieces similar to jigsaw puzzle pieces, which are flexible enough to interlock in either two or three dimensions, and strong enough to stand up and make small towers or other 3D models.

My interest in Plus-Plus is purely geometrical.  I was most interested to discover if it's possible to produce a repeating pattern of PlusPlus pieces (which I think the manufacturers refer to as 'hexels') in a 2D surface with perfect tesselation?  

As an aside, I guess they call them hexels because they approximate a hexagonal shape?  I looked at them and decided they were made up of nine mini-squares all combined together, but a nonel sounds daft.  Either way, they are shaped like two plus symbols that have been overlapped ++

After some logical puzzling with a large pile of hexels, I was able to confirm that it's possible to make a tessellating pattern with them.  I don't know if anybody else has named the pattern, so I'm going to go with HV, for Horizontal Vertical.  Each hexel is connected to another in a straight line, with the orientation for the adjacent hexel being rotated 90 degrees (in the same plane) to the first.  The repeating pattern is Horizontal, Vertical, Horizontal, Vertical, and so on.  The picture shows how it looks in a line, and then connects multiple lines to cover a 2D surface.

Interestingly, HV is not the only way to connect a series of hexels: they can be connected HHH, with a slight diagonal offset (green lines below), or connected VVV (orange lines below).  I want to mention at this point that the two are not the same:  it's possible to have multiple HH lines tessellate by lining them up together (there's no interlocking between adjacent lines, so the structure is physically weak, but it still tessellates). 

However, HH and VV cannot tessellate together.  The two patterns are incompatible.  Rotating a HH line will not make it fit into a VV line. Another way of thinking of it (through rotation) is that one line steps 'upwards' while another steps 'downwards' - see below.

Something I found particularly interesting is that it's possible to combine HH with VH, and also to combine VV with VH.  In the diagram below, the grey hexels are in VH configuration.  The white hexels are in HH, with the lines of HH running from top left to bottom right (it's a short line, and the interface between the patterns runs perpendicular to it). The blue hexels fit into both patterns, and show the interface or the border between the two.

And here is proof that you can combine VH, HH and VV:  the grey is VH as before; the white is HH, while the orange is VV.  The blues show the interface between VH and HH; the yellow shows the interface between VV and VH.  The bands of white, grey and orange can be of any fixed width, but they don't get narrower or wider.

This reminded me of the studying I did at Cambridge University, looking at phases of iron-carbon alloys.  Each 'grain' of a particular alloy would have a certain area (or volume, in 3D) where it was made entirely of a particular phase of iron/carbon, and the composition would be consistent throughout the grain.  There would be a grain boundary where it touched another grain of the same composition, but where the grain had been oriented differently.

Diagram from http://www.gowelding.com/met/carbon.htm 

I have not yet considered 3D shapes - these are clearly possible, and I shall be looking at if it's possible to produce regular polyhedra such as cubes and cuboids, and what limitations there are.

Other Geometry Articles I've Written:

Calculating the tetrahedral bond angle
The angle of elevation of a geostationary satellite

Airband Radio Aerials: Maths in Action

I've been interested in aircraft and airshows for over 40 years - anything military or civil, and I've blogged in the past about how to use a spreadsheet to track down where to watch the Red Arrows fly past on their transit flights.  You didn't think that post was about geometry without some real-life applications?  What is this - "Another day I haven't used algebra"?

Anyway - I've been particularly interested in the Red Arrows and their air-to-air chatter, and the communications between pilots and air traffic control.  Yes, I take my airband radio along to airshows and to airports, and listen to the pilots request and receive clearance to take off or land.  Getting to airports is more of a challenge than it used to be - my children aren't as interested as I am in the whole thing, and standing at the end of a runway in poor weather isn't as much fun as it sounds!

So, I've started developing my home-based receiver.  In other words, I spent my birthday money on an airband antenna and an extension cable to connect it from outside (cold and sometimes rainy) to my desk (warm and inside) so that I can listen to pilots flying nearby.

Now: nearby is a relative term.

From Stoke on Trent, I've been able to pick up pilot transmissions from about 35 miles away, on the southern edge of Manchester Airport.  That's with a very basic antenna, set on my garden gatepost and about two metres off the ground - not bad for a first attempt.

My dad, on the other hand, has been tracking radio transmissions for decades.  His main areas of interest are long wave (around 200 kHz), medium wave (500-1600 kHz), and TV (UHF, 300 Mhz to 3GHz).  

Airband falls into the Very High Frequency range, around 100-200 MHz.

Here comes the maths:
All radio transmissions travel at the speed of light, c = 2.998 * 10^8 ms-1.
c = f w

Where f (sometimes the Greek letter nu, ν) is the frequency, and w (usually the Greek letter lambda, ƛ) is the wavelength.

So, if we know the frequency range that we want to listen to, we can calculate the wavelength of that transmission.  And this is important, because the length of the antenna (or aerial) that we need will depend on the wavelength.  Ideally, the aerial should be the length of one full wavelength, for maximum reception effectiveness.  Alternatively, a half-wavelength or a quarter-wavelength can be used.

So:  we know the speed of light, c = 2.998 * 10^8 ms-1
And we know the frequency of the transmissions we want to receive, which is around 118 MHz.

c/ν = ƛ

ƛ =   2.5 metres

Which is feasible for an external, wall-mounted aerial.  Can you see where this is going?

Exactly.  And here it is:  

It's just over two metres from end to end, with a feed at the midpoint.  This is the Mark One; the Mark Two will be the same aerial but even higher up, and closer to vertical (with a bracket that will enable it to dodge the eaves of the roof!

Calculator Games: Ulam Sequences: Up, Up and Away!

 Up, Up and Away With Ulam Seqeunces

This article in the ongoing series of ‘mathematical puzzles you could investigate with a calculator’ (that’s why I just call it ‘Calculator Fun and Games’) is the Ulam Sequence.  Ulam sequences, named after mathematician Stanisław Ulam, are fascinating numerical sequences that begin with two specified integers. Each subsequent number in the sequence is defined as the smallest integer that is the sum of two distinct earlier numbers, where such a sum is unique within the sequence. This uniqueness constraint shapes the sequence's progression in an intricate way.

Ulam sequences are studied for their intriguing mathematical properties and their unpredictable, non-linear behavior, which challenges patterns typically found in additive sequences. They have applications in number theory and combinatorics, offering rich grounds for exploration and research.

Let's have a look at them in more detail, and start with the simplest.

How to Generate an Ulam Sequence

Take with two numbers (specifically, positive integers).  A good place to start is with a =1 and b=2.  To find the next number in the sequence, find the smallest integer that can be written as the sum of two distinct earlier numbers in just one way.  Continue with the next number, and the next.

For example, let’s start the Ulam Sequence with the numbers 1 and 2.  These are the first two terms in the sequence.

The next term is 3 (since 1+2=3).
After that comes 4 (since 1+3=4).

The next terms is not 5.  We can write 5 as 1+4 and as 2+3 using the terms that we’ve generated already.

The next term is 6 (since 2+4=6), and we can only write this in one way using our terms.

We can write 7 as 4+3 and as 6+1, so we skip 7.

The next term is 8 (since 2+6=8).

So, the beginning of the sequence is: 1,2,3,4,6,8,… (and it continues
1,2,3,4,6,8,11,13,16,18,26,28,36,38,47,48,53,57,62,69,72,77,82,87,97,99,102)

Note that 5, 7, 9, 10, 12, 14, 15, 17, 19, 20, 21 and 22 can be obtained in multiple ways using the terms before them.

However:  23 is not in the sequence because it cannot be obtained using the previous terms at all!    24 can be written as both 16+8 and 18+6, while 25 is not obtainable.

The sequence grows in an irregular, almost random pattern.  Let’s see what happens when we start with 1 and 3 instead of 1 and 2.

4 = 1 + 3 only
5 = 1 + 4 only
6 = 5 + 1 only
7 can be written as 4+3 and 6+1
8 = 5+3
9 is 4+5 and 6+3
10 = 6+4
11 is 6+5 and 8+3

The first 20 terms for the 1,3 Ulam Sequence are:

1,3,4,5,6,8,10, 12,17, 21, 23, 28, 32,34,39,43,48,52,59 and 63.

The Ulam Sequence is an interesting example of how simple rules can lead to complex and intriguing mathematical structures, which makes it ideal for calculator (or spreadsheet) exercises.  For example, here’s a comparison of the Ulam sequences for 1,2 compared with 1,3 (as I’ve calculated above) and then 1,4 and 1,5.  Interestingly, the 1,5 sequence does not race ahead of the 1,2 sequence as I had originally expected.

Term	Ulam (1,2)	Ulam (1,3)	Ulam (1,4)	Ulam (1,5)
1	1	1	1	1
2	2	3	4	5
3	3	4	5	6
4	4	5	6	7
5	6	6	7	8
6	8	8	8	9
7	11	10	10	10
8	13	12	16	12
9	16	17	18	20
10	18	21	19	22
11	26	23	21	23
12	28	28	31	24
13	36	32	32	26
14	38	34	33	38
15	47	39	42	38
16	48	43	46	40
17	53	48	56	41
18	57	52	57	52
19	62	59	66	57
20	69	63	70	69

There’s a balance between the ability to leap to larger numbers (1,5) initially – from 1 to 5 – and the need to fill in more numbers between 5 and 10 (because there are very smaller numbers that can be made in multiple ways).

A quick comparison of the Ulam Sequences for (2,b) is even more interesting.  We have to start with 2,3 since 2,1 is the same as 1,2 above, and 2,2 will only produce the even numbers (which is cute but dull).  In fact, any even number paired with 2 will produce uninteresting results!

Let’s compare 2,3 and 2,5:  These grow at a slower rate compared to the 1,b sequences.  Interestingly, they contain far fewer even numbers than the 1,b sequences; in fact 2,5 only contains the even numbers 2 and 12 in the first 20 terms (with no indication that there are any more even numbers further along the sequence).

Term	2,3	2,5
1	2	2
2	3	5
3	5	7
4	7	9
5	8	11
6	9	12
7	13	13
8	14	15
9	18	19
10	19	23
11	24	27
12	25	29
13	29	35
14	30	37
15	35	41
16	36	43
17	40	45
18	41	49
19	46	51
20	51	55

So there’s plenty of scope for investigation with a spreadsheet for the larger numbers.  For example, I haven’t found anybody else list the Ulam sequence for 10,11… so here it is:  the Ulam Sequence for (10,11)

10, 11, 21, 31, 32, 41, 43, 51, 54, 61, 62, 65…

After huge initial leaps of +10 or +11 between consecutive terms, the growth rate of the sequence starts to slow down.  There is only one term in the 20s, then two in the 30s, 40s and 50s, then three in the 60s.

Further reading:
Wolfram has lists and links for many of the 1,b and 2,b Ulam sequences.

Other articles on this blog on similar themes:
Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams with five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero as rapidly as possible)
Calculator Games: Front to Back
Calculator Games: The Kaprekar Constant

Calculator Games: Front to Back

This puzzle is not from the Calculator Fun and Games book, but another one that would be suitable (if you had time).  Early indications and preliminary reading indicate that this one could take some time to complete, but let's wade in anyway.

The game would be described in this way:

Take a number (three digits to start with)
Reverse the digits to form a new number
Add the two numbers together (this makes a change from subtracting...!)
If the new number is not a palindrome, continue reversing digits and adding.

Let's start with 456

456+654 = 1110

1110+0111 = 1221 so it's a palindrome.

Let's try 782

782+287 = 1069

1069 + 6901 = 10670

10670 + 7601 = 18271

18271 + 17281 = 35552

35552 + 25553 = 61105

50116 + 61105 = 111221

111221 + 122111 = 233332  which is a palindrome.

And let's try a smaller three-digit number, 165

165 + 651 = 816
816 + 618 = 1434
1434 + 4341 = 5775 which is a palindrome.

This is called the Lychrel process, and it's not named after a famous mathematician.  It makes a change!  It was named by a man called Wade van Landingham in 2002, who created the name as a rough anagram of his girlfriend's name, Cheryl.  Unlike most mathematical concepts, this one is not hundreds of years old - this also makes a pleasant change!

There is one number which has not yet been found to form a palindrom after many, many iterations of the Lychrel process, and that's 196.  Innocuous, isn't it?

196 + 691 = 887

887 + 788 = 1675

1675 + 5761 = 7436

7436 + 6347 = 13783

13783 + 38731 = 52514

52514 + 41525 = 94039

94039 + 93049 = 187088

187088 + 880781 = 1067869

1067869 + 9687601 = 10755470

10755470 + 7455701 = 18211171

18211171 + 17111281 = 35322452

35322452 + 25422353 = 60744805

60744805 + 50844706 = 

And at this point, my calculator says "Enough!"  I can't get all the digits any more, and this number still isn't reaching a palindrome.

My spreadsheet goes a little further:

60744805 + 50844706 = 111589511

111589511 + 115985111 = 227574622

227574622 + 226475722 = 454050344

454050344 + 443050454 = 897100798

897100798 + 897001798 = 1794102596

And then starts throwing "#VALUE!" messages at me, without reaching a palindrome.

The general definition for a Lychrel number is one that does not reach a palindrome in fewer than 500 iterations. This is easier to measure compared to 'never reaches a palindrome', and that means that the Lychrel numbers (more than 500 iterations) include 295, 394, 493, 592 and 689.

Some numbers immediately become palindromes after one iteration - these are trivial, commonplace and not very interesting!  For example, 110 + 11 = 121, and any other number where the units value is zero, and the hundreds and tens are both less than five.  The longer ones are definitely more interesting, because there's no obvious pattern (and it reminds me of the Collatz conjecture, which I'll be revisiting soon).  Larger numbers which need more than 500 iterations include 10538, 10553 and 10585.

So: can you find numbers which reach a palindrome before they make your calculator (or your spreadsheet) explode?

Further reading:

Lychrel Number Tester - dcode.fr
Lychrel Numbers - Geeks for Geeks

Other Calculator Fun and Games articles:

Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams with five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero as rapidly as possible)
Calculator Games: Front to Back
Calculator Games: Up, up and away with Ulam sequences
Calcualtor Games: The Kaprekar Constant

Calculator Fun and Games: Premium Prime Numbers

 PRIME NUMBERS

 Calculator Fun and Games is a maths puzzle book that’s worth its salt: it has a section on prime numbers, as any good maths book does.

Prime numbers are only divisible by themselves and 1, they have no other factors.  1 is not a prime number (by definition), but 2 is, making 2 the only even prime number.  The study of prime numbers dates back thousands of years. Some of the earliest records come from the ancient Greeks, particularly the mathematicians of Pythagoras' school (around 500 BC to 300 BC), who explored numbers for their mystical properties.  I recently looked at Pythagoras's Triples - there's always an overlap between different parts of maths.

One of the most significant contributions came from Euclid (circa 300 BC), who proved that there are infinitely many prime numbers. His work in Elements laid the foundation for number theory, including the Fundamental Theorem of Arithmetic, which states that every integer can be uniquely factored into prime numbers.  That's worth its own investigation, and is a puzzle in itself.

Calculator Fun and Games lists the prime numbers up to 1000, just for good measure.  But is there any relationship between them?  Are they connected?  Are there ways of finding (or generating) them?  And can all this be done with a humble calculator?  Maybe one day.

However, the book points out an interesting fact:  if you

Take a prime number (greater than 3)
Square it
Add 14
Divide by 12

Then the remainder is always 3.

For example:
5² = 25
25 + 14 = 39
39/12 = 3, with a remainder of 3

Let’s take a larger example: 577
577² = 332929
332929 + 14 = 332943
332943 / 12 = 27745 remainder 3

Alternatively, x² + 11 always divides exactly by 12.  I’m not sure why Ben Hamilton decided on adding 14… maybe he likes the remainder 3?

Does this apply in reverse?  If I take a number, and subtract 11 and take the square root, do we always get a prime? 

No.  Even if the square root is an integer, it doesn’t mean that the starting number was a prime.  For example, 92-11 = 81, and the square root of 81 is 9.  Only a subset of a = (SQRT(b-11)) will give a as a prime.

TWIN PRIMES

Twin primes are primes with a difference of two; for example, 11 and 13, or 17 and 19.  These are rarer than prime numbers, but it still seems that there will be an infinite number of twin primes.  An example of a larger twin prime is 971 and 973; while the largest known twin primes have 388,342 digits:  they are:  2996863034895×2^1290000±1.

Not a number you’d fit on your calculator.  The largest prime number which will fit on an eight-digit calculator (with a good old-fashioned LCD display) is 99,999,989.

Looking at all those nines, I’d like to play Over and Out with it, and see how long it would take to get it down to zero! 😊

Other recent Calculator Fun and Games articles:

Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams, five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero in as few steps as possible)

Calculator Games: The Kaprekar Constant

The Kaprekar Constant: 6174

This is one of those problems that doesn't come from Calculator Fun and Games, but would be an interesting offshoot, and can certainly be done with a calculator, pen and pencil.

Take any four digit number.  It must contain at least two different digits (i.e. 9933 is permitted, but 8888 is not).  Sequence the digits in descending order (e.g. 2536 becomes 6532), and ascending order (2536 becomes 2356).  Subtract the smaller number from the larger, then repeat the process of finding the descending and ascending numbers, and subtracting.  To quote Calculator Fun and Games (and countless vague Maths GCSE questions):  What do you notice?

First example:

1874
Rearrange:  8741-1478 = 7263
Rearrange: 7632 - 2367 = 5265
6652 - 2566 = 4086
8640 - 0468 = 8172
8721 - 1278 = 7443
7443 - 3447 = 3996
9963 - 3699 = 6264
6642 - 2466 = 4176
(7641 - 1467 = 6174, which goes round in a loop)

Some wider reading shows that 6174 is the Kaprekar constant, and it is the end result for all four-digit numbers, and that all numbers terminate there after no more than seven steps.  Never one to shirk a challenge (especially if it involves a calculator), I tried out some more numbers:

9954 --> 5553 --> 9981 --> 8870 --> 8532 --> 6174

9732 --> 7533 --> 6174

1110 --> 9900 --> 9810 --> 9621 --> 6174

I repeated this several times, with upwards of 30 different numbers, and in order to summarise my results, I have drawn out a network of key paths to 6174.  I've consistently used a colour scheme - dark green for 6714, light green for numbers which lead directly to 6174; yellow for numbers which need two steps, light blue need three steps, and so on.

Firstly:  pathways for numbers which require all seven steps to reach 6174.  I've found four numbers that require seven steps, and they follow the paths below; three of them go through 9954, and they all go through 8532.  My work on the Kaprekar constant tried to focus on numbers which either need all seven steps, or resolve to 6174 in one step.

And interestingly, five six-step numbers that all follow the same pathway:

And finally, a collection of one-step and two-step numbers which have no parents:

One step:  6642, 8532, 9621, 7533, 7531, 6200, 8754 and 8752.  Bold numbers have no parents - no other numbers (that I have found) go through them to 6174.  7533 has six parents, but none of these parents have any parents of their own.

These findings are all based on my own work, and I fully acknowledge that they could be incorrect due to my incomplete research.


Extension:  a five-digit Kaprekar process... another time!