A new 2048-style challenge

Yes, the game 2048 is one of those tablet/cellphone games which also run on a desktop browser which have fascinated numerically-inclined people such as myself. It is a wonderful waste of time, something to do while you are doing something else and actually need to pull yourself away for a few moments and clear your mind by diverting your attention to something else for a bit.

As many of you know, it is normally played on a 4×4 grid. When you search for “2048 game” on the Duck Duck Go search engine, at the top of the selection is a game Duck Duck Go has provided:

Duck Duck Go’s attempt to replicate the “classic” game of 2048

My cellphone app offers me a variety of grid dimensions to play on: 4×4, 5×5, 6×6, and 8×8. The bigger the grid size, the easier it gets to obtain the total 2048. Once you obtain that tile, you win, although you can continue playing. You can play until you have no more room to move or merge your tieles. If you run out of room before you reach 2048, you lose.

The 8×8 tileset is so easy, you can have a mess on the grid and still reach 2048 eventually, with plenty of room left on the board to move your tiles around. So if you are just starting out, you might find it easier to play with the larger boards, eventually graduating to the 4×4 board once you have a clear idea what is going on, and have built a sense of intuition. My version of the game allows you to undo your last turn (but no more than that) in case your last move was less than ideal. It is also useful because sometimes I could swipe right and the tiles move in the wrong direction for some reason. They are supposed to move in the direction of the swipe.

Out of curiosity, I went one size down and tried out the 6×6 mode of the app, and so far, I have two 4096 tiles, and a 16384 tile with no end in sight:

A bit blurry, but there are always a ton of empty squares in this game, and the game board, as you can see is less than ideal for tile distribution. But I’m still getting 16384 and beyond. This same game has been played and stored off and on for the past week.

So, since 2048 is fairly easy on grids beyond 6×6, I began to wonder what numbers could pose a challenge on these larger squares. For 8×8, I think I found what might be a suggestion:

Uhhh, woah.

By the title of this game, they appear to be suggesting that I try to reach a tile labelled “2,147,483,648”, a 10-digit number just over 2 billion. I would be curious to see how they achieved fitting such a number on a tile once it is obtained. But how long would that take? On a 4×4 grid, getting that tile could take a couple of hours, if you get there at all.

The original 2048 suggests that 2048 be the winning tile, which is a power of 2, namely 2^{10}. Taking the base-2 log of 2,147,483,648 on my calculator, I find that this new number is equal to 2^{31}. I am beginning to think that I will need to pass this on to my children in my will, assuming I haven’t lost the game yet.

2147483648 is a game offered currently on GitHub, so it is likely the side project of a professional programmer or a pet project for a hobbyist and could still contain bugs (imagine the nightmare in testing and debugging this game into the high numbers). It appears that the only one developing the project is an anonymous, nameless, faceless programmer going by the handle “jiangtyd”. That said, it seems to come with clever innovations, such as allowing the program to auto-move the tiles on its own to create interesting combinations. You can also change base to something like 3, or increase the tiles by way of the Fibonacci sequence. The “3” selection is a bit lame, in that the successive tiles that need to be formed are really 3\times 2^n, so you are really dealing with powers of 2 again, except in a different way.

A Pet Peeve About the Internet

I have said this several times before in many ways, and I will say it again: there is too much importance placed on the internet. It wouldn’t be so bad if the internet was run by the government, since that would make it more accountable. But instead, it is mostly in the hands of large private companies, who are largely unaccountable, and would not be truthful unless regulators, or the threat of regulation, forces their hand.

This was brought out in all its glory yesterday, as the Canadian telecommunications conglomerate, Rogers, experienced a denial of service nationwide, affecting all internet services. It wasn’t just that families were denied Netflix or YouTube, or that you couldn’t receive email or text messages, it was that the entire economy slowed considerably. Interac stopped working, and that meant that people couldn’t make transactions unless they had cash or credit. All major banks and credit unions use Interac, and thus experienced this problem. But in addition, customers were also not able to do e-transfers or pay bills through their bank.

Many vending machines are hooked up to the internet, and some of them were disabled if they had to connect with a Rogers service. Municipal parking, now dependent on the internet as many cities abandon parking meters, were hobbled as municipalities were not able to accept payment for parking. Toronto’s BikeShareTO service, whcih depends on the internet to distribute bikes to users, had to declare their bikes inaccessible for all stations in Toronto. Vancouver had a similar problem with its bike share program. School boards with summer online programs had to go asynchronous and move its deadlines back another day. Public libraries had a stoppage of WiFi service at many locations, as well as self-checkout machines, and book kiosks.

Many retail stores had to close altogether for the day, as was seen in Mississauga’s Square One Mall and Toronto’s Yorkdale Mall. Many condos and apartments experienced a disabling of their buzzer systems due to the outage.

That wasn’t all. 911 services also stopped working in many areas where the 911 services were managed by Rogers. Chatr Mobile and Fido, offshoot services owned by Rogers, also stopped working. Downstream internet service providers (ISPs) also experienced downtime as a consequence.

It caused the phone lines at the CRTC to go down, since they were using IP telephony provided by Rogers. Of course, this would also be true for any IP telephone, which includes all cell phones served by Rogers or its subsidiaries. These IP telephones also went down in passport offices provided by Service Canada. Rogers also manages the multi-factor authentication systems used by the Canada Revenue Agency, so anyone attempting to log in to the CRA website yesterday could not log in.

One way the internet is oversold is in how we market and use IoT (Interent of Things) devices. Imagine for example, the many forms of digital signage you see around you. A good number of them are connected to the internet, and programmed remotely. Examples are digital highway signs which warn of dangers ahead. So are a good number of household appliances, including televisions and tablets. IoT can also centrally link home security systems, allowing a monotoring service to provide surveillance at low cost as if a security guard was on site. IoT has medical applications, such as providing a way to aid medical professionals to monitor someone who has cardiovascular disease. Some of these things seem pretty essential, but others, such as IoT stoves or refrigerators which are sold to households, not so much. All these would stop working if internet providers had a service stoppage the same way Rogers had done, and these effects would have been already felt with the Rogers denial of service.

Too much is riding on the internet, and too much is riding on only a small handful of service providers. So much so, that we appear to take the internet for granted the way we take water, electricity and sewage for granted. We just assume it works and people are doing their jobs.

But the internet is very different from these other public utilities, in that there are too many variables involved in providing people with decent service. Networks can get hacked, and DDOS attacks are common enough to brandish its own acronym. Weather events happen and can cause regions to have to do without service for some time. As we have seen there are many services, and to have the same provider do them all is like putting all your eggs in the one basket.

It must also be added that the Internet wouldn’t exist without government handouts to the telcoms. It was taxpayer’s money that established the main trunk lines for the internet in the 80s and 90s in Canada and the United States. The infrastructure was practically given away to the major telcoms, and we are now seeing an example of what happens when the internet is controlled by too few companies which are largely unregulated and have little public accountability.

There is more than one major provider in Canada – Bell and Cogeco are other big players that come to mind. But we need more than just a few, so that if a DDOS attack happens, the number of people affected will be limited. Outages such as this can slow down the federal government’s attempt to provide all Canadians with universal high-speed internet by 2030.

With information from The CBC, the Toronto Star, and other websites.

Recreational Math I: Magic Squares: Matters of Primes: Part 7

This article concerns the matter of prime numbers in writing computer programs that generate random magic squares of order n where n is prime. My article, like all the other articles in this series, is aimed at hobbyists.

Suppose you wanted to use the Reichmann (1957) algorithm to generate prime-ordered magic squares of any size n where n ≥ 5. If you asked a user to enter a prime number for the square dimensions, there comes a point where the user won’t exactly know, and shouldn’t be expected to know, if a large number is prime. Could an average person be expected to know, off the top of their head, that 127 is prime, for example? Or what about the number 349,871? Both are prime, but I cheated: I used software. But how does the software do it, since that is what you were trying to write?

We’ll get into the algorithm details in a bit. But the easiest thing to do is to have a detection if a number is not prime. If it isn’t prime, then you can’t use Reichmann’s algorithm, and you then need to ask the user for another (hopefully prime) number that is 5 or more. The user is making at best, quasi-educated guesses. Maybe he knows some powers of 2, and by subtracting 1, he could stumble on a Mersenne prime such as the number 127 mentioned earlier, or 31. Or 7.

Prime numbers are numbers divisible by 1 and itself, and has no other factors. A number like 5 is prime because only 1 and 5 divide it evenly. The same can be said of the number 2, since 1 and 2 are its only factors, making it the only even prime number. Numbers which have more than two factors are called composite numbers. A number like 12 has 1, 2, 3, 4, 6, and 12 as factors, and thus is not prime.

Mersenne primes refer to a special class of primes. They are primes of the general formula 2^p - 1, where the exponent p is also a positive prime number.

A programmer could start with a short list of known primes in an array. Let’s say: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, or the primes from 0 to 100. By the Fundamental Theorem of Arithmetic, a number n is composite if any prime number p \le \lfloor \sqrt{n}\rfloor divides it. The symbols \lfloor,  \rfloor represent the floor function which takes the square root of n and essentially chops off the decimal, returning only the integer part of the number. The array of known primes is thus good up to 100^2 = 10,000.

This is a shortcut, but if the user enters larger numbers than 10,000, you can try numbers beyond the scope of your array. But since the computer program doesn’t know any primes offhand beyond 97, the program could then attempt to check each whole number after 97, sqeuentially up to \lfloor \sqrt{n}\rfloor to see if anything divides n evenly. If none are found, then n is prime.

That’s a whole lot more efficient than checking each integer one by one up to n (or even \lfloor n/2 \rfloor, or half of n with the decimal chopped off) to see what divides n, if anything.

If no upper limit of n is specified to the user, even your more efficient \lfloor \sqrt{n}\rfloor could take hours or days to determine if n is prime, even on a computer considered to be fast. Also, for an n \times n magic square, it would be nice if there were only enough rows and columns to be reasonably displayed on a spreadsheet or console screen. On a spreadsheet, there is always the scrollbar to help you see the numbers that ran off the display. A 101 \times 101 magic square is possible for example, but requires a lot of scrolling to see the 10,201 cells containing the numbers. Today’s computers have lots of memory, so while it is possible to go far beyond even that, there is a matter of practicality. While it might be possible to check the correctness of a 324,773 \times 324,773 magic square programmatically, the 105.5 billion numbers it contains preclude the idea that it can be done on a normal household desktop computer running with 8 gigs of RAM.

Recreational Math I: Magic Squares: the “really good” kind – Part 6

Welcome to part 6, where the magic squares are 7×7. 7 was the number of known planets in medieval times, as well as the number of known elements, and the number of days in the week. The rest is all alchemy.

The algorithm for a 7×7 matrix is the same as for 5×5. In fact, it should work for any odd prime-ordered matrix. Do the first based on a random sequence of the numbers between 1 and 7:


Now, do the second matrix using 0 and every seventh number after it up to 42:


The resulting matrix is magic, but not “hyper-magic” like the 5x5s were:


The matrix above was done using Excel, and checked using VBA, where I checked it for “magic”. The loss of hypermagic is unfortunate, but I suppose inevitable when you scale up. There are (7!)2 = 25,401,660 magic squares possible by this algorithm. The VBA proved handy later when I used dynamic programming to produce any odd-ordered magic square of any size, and investigated the results.

What if I made a mistake and got the shifting wrong, such as shifting from the second number from the left instead of the right? The result is variable; full magic can only be restored if the rightward diagonal is all 4/s:


Why does this work? This is because the sum of the numbers 1 to 7 in any order is 28. Notice that whatever number is in the opposite diagonal is repeated 7 times. There is only one number, when multiplied by 7, equals 28, and that is 4. All other numbers will result in a loss of magic. This appears to be a way to maintain magic in all odd-ordered magic squares past order 7. This can possibly generalize for squares of prime order n by taking the middle number (n+1)/2 and making it the main diagonal.

When added to the second preliminary square, both diagonals became magic, as well as the columns and rows. But because the opposite diagonal consists of all 4’s, we lose variety with this restriction, but not by much: 7!6! = 3,628,800. Still enough to occupy you on a rainy day.

9×9 magic squares won’t work so well, because the second square is shifted by 3, which is a number that divides 9. This will result in certain rows repeating in one of the pre-squares; and the result generally is that one of the diagonals won’t total the magic number (369 in this case).


This is a consequence of 9 not being a prime number. The algorithm that works so well for 5×5 and 7×7 magic squares breaks down when the odd number is composite. Thus, the algorithm is known to work for 11×11, 13×13, 17×17, and has been tried all the way to 101×101 and beyond on my Excel spreadsheet.

The magic number can be known in advance of any magic square because an order-n magic square obeys the formula (n × (n2 + 1))/2. The best bet is to look for odd-ordered n × n squares where n is prime. In my VBA program, I made double sure by sticking to magic squares where the order is prime. Here is an 11 × 11 square, where the magic number is 671:


Finally, to stretch things out to the boldly absurd, what about an order-23 square whose magic number is 6095?

The 23 columns may not all be visible here.

That probably ran into the neighbouring columns, but it’s just for illustration. The code, which I have used to generate magic squares of as high an order as 113 (magic number 721,505) is fewer than 170 lines in VBA for Excel 2007. And yes, even the above square, and the order 113 square are magic.

Recreational Math I: Magic Squares: the “really good” kind – Part 5

I have met with some disappointment as to how a methodology for creating a 4×4 square should pan out, and instead I have come up with many different algorithms, each resulting in its own small sets of magic squares, but had stumbled upon a set of squares with similar “hyper-magical” properties which I called the Durer Series.

I had met with some disappointment at this, and am still in the middle of writing my own 4×4 square program in Visual Basic .NET (it’s going to use a “brute force” algorithm … sorry!). With that, I had to learn about how .NET does objects. To those of you out of the loop on the recent .NET versions of VB, this language actually allows you to create your own novel objects, thus saving processing time if the right kind of objects are created. It’s a work in progress.

For now, I wanted to centre on the pièce de résistance of this series: that of the odd prime-ordered magic squares, those of order 5 and beyond. As I hinted at the beginning of this series, these are special and unique in that an algorithm can be made for an order-n magic square (where n is an odd prime greater than or equal to 5) which can generate (n!)2 squares, all magic (even after weeding out duplicates, the numbers of unique squares will still be in the thousands).

This seems to be a hard-to-find algorithm, except from a book called “The Fascination of Numbers”, written on the year of The Queen’s coronation in 1957 by W. J. Reichmann while he was still a headmaster at a Grammar School located in Spalding, Lincolnshire, England (according to my signed copy of the book, purchased from an English vendor through Amazon used books). I read it for the first time at a university library, and it is likely to be found in a similar univeristy or college collection near you.

What I like about this algorithm is that while I have used it to write computer programs for magic squares, it really requires no more than pencil and paper, and a bit of skill at addition.

First of all, write down the numbers from 1 to 5, in any order at all:

3, 2, 1, 5, 4

A 5×5 square matrix is constructed in a pattern by starting from the fourth number in the sequence, then proceeding with the 5th number, then the first, second and finally the third:

4 2 1 5 3
5 3 4 2 1
2 1 5 3 4
3 4 2 1 5
1 5 3 4 2

Next, scramble the first 5 multiples of 5 (counting 0):

20, 0, 5, 15, 10

Create a 5×5 matrix by shifting the order of these numbers by 2 to the left as follows:

20  0  5 15 10
 5 15 10 20  0
10 20  0  5 15
 0  5 15 10 20
15 10 20  0  5

Now add them together, adding together numbers located in the same columns and rows in both squares, as in matrix addition:


The result is a magic square which has many more ways of obtaining  the magic number 65 than just adding the rows, columns, and diagonals. Taking any 3×3 sub-square on the corners or left/right sides of this square and adding its corner numbers to the middle number will obtain 65. This seems to be true of all magic squares made by this method. Since both squares can be scrambled independently, there will be (5!)2 = 14,400 possible squares before duplicates are weeded out. I have written programs in many languages regarding this 5×5 square, and can attest to the robustness of this algorithm in producing a nearly endless series of 5×5 magic squares, all sharing very similar “magical” properties.

Reichmann also reminds us that the scrambled 5×5 squares we created initially are also magic.

I suspect there may be more than 14,400 magic squares. Are there any squares that cannot be produced by this algorithm? This must be checked against a “brute-force” (read: computational) method for generating all possible magic squares to see if this is really the definitive number. It is certainly a couple of orders of magnitude greater than what Danny Dawson did with his 4×4 squares (around 920 squares or so, with some duplicates). It would also be interesting to know what squares could not have originiated from Reichmann’s algorithm (proven by working backwards to show, supposedly, that duplicate numbers appear in some rows of the first or second matrix, which disappears in the resultant matrix.

It also seems that 4×4 and 5×5 squares can have such “compound” magical properties; it is harder to find for 7×7 matrices, although they are additive to the magic number in just enough ways so as to say they are magic. We’ll leave that for the next journal entry.

Recreational Math I: Magic Squares: the “really good” kind – Part 4

How to Make a Random Square

I have noticed that it has been difficult to elucidate a method for systematically creating even-ordered magic squares of any but the most basic kind. I don’t know why this is, since the art has been alive in Europe for at least 600 years, and probably longer in other cultures. There are a few methodologies out there, but they are perfunctory. Such as, this method offered by Reichmann (1957), which tells us little more than to write the numbers down from 1 to 16 and reverse the order of the diagonals:

 1  2  3  4        16  2  3 13
 5  6  7  8  ====>  5 11 10  8
 9 10 11 12         9  6  7 12
13 14 15 16         4 14 15  1

Well, only a limited number of magic squares are possible that way. You can reverse the order of the numbers, or write the numbers in order down the columns instead of the rows, followed by a similar reversal of the diagonals (these would not be considered to be unique solutions, since these methods amount to mirroring or rotating the square).

After some playing around with known 4×4 magic squares (thanks to the thorough resource offered here), and using some simple rules, I think I may have come upon a method on my own. I would suppose that it could not generate all magic squares possible, but I don’t intend to offer a complete solution. I estimate that I would be able to generate maybe another 10 or so magic squares with this method. The intent here is to allow the reader to create a 4×4 magic square sitting down at a table with little more than pencil and paper. I am sure this methodology is published elsewhere, but I wasn’t able to come up with anything.

I just have a few simple rules in coming up with the method. First of all, all of the 16 numbers are the result of a sum of two smaller numbers a and b, where a is a number from 1 to 4 such that a = {1, 2, 3, 4}, and b is a multiple of 4: b = {0, 4, 8, 12}. All 16 unique numbers generated will be the result of adding a number from set a to a number in set b.

The numbers 1, 2, 3 and 4 are the result of adding set a to 0 in set b.

The numbers 5, 6, 7, 8 come from adding set a to 4 in set b.

The numbers 9, 10, 11, 12 come from adding set a to 8 in set b.

And the numbers 13, 14, 15, 16 come from adding set a to 12 in set b.

My solution is inspired by Reichmann’s (1957) work on odd-ordered matrices. This solution on even-ordered matrices is quite different than for odd-ordered matrices, but the idea of adding two matrices and getting a magic square appealed to me, so I went with that basic idea.

The first matrix uses numbers from set a, in the following pattern:

4  3  2  1
1  2  3  4
1  2  3  4
4  3  2  1

These numbers may be randomized, but I’ll just stick to numeric order for now. The second matrix is composed of the numbers of set b arranged in a pattern reminiscent of the Durer square:

12  0  0 12
 4  8  8  4
 8  4  4  8
 0 12 12  0

The sum of the two matrices has all the magic of the Durer square, and consist of the unique numbers 1 to 16:

16  3  2 13
 5 10 11  8
 9  6  7 12
 4 15 14  1

In fact, it is the Durer square itself. Now that we know it works, what about mixing things up? Take note of the patterns. Let the first matrix represent numbers {a, b, c, and d} belonging to set a; and {e, f, g, and h} belonging to set b. whatever randomization is chosen, the numbers must fit the following scheme:

a  b  c  d      e  h  h  e
d  c  b  a      f  g  g  f
d  c  b  a   +  g  f  f  g
a  b  c  d      h  e  e  h

a, b, c, and d may be any permutation of the numbers 1 to 4, without apparent limitation. The second and third rows on teh first matrix are the reversals of the first and fourth. The second matrix, however is much more restricted. Considering e and f to be one pair; g and h to be another pair: 12 can only be paired on the same row with 0; and 4 only be paired with 8. Otherwise, no joy.

So, on with a randomized square: let a = {3, 1, 4, 2} and b = {8, 0, 12, 4}

3  1  4  2     8  4  4  8     11  5  8 10
2  4  1  3     0 12 12  0      2 16 13  3
2  4  1  3  + 12  0  0 12  =  14  4  1 15
3  1  4  2     4  8  8  4      7  9 12  6

What about a = {2, 4, 1, 3}, and b = {4, 12, 0, 8}

2  4  1  3      8  4  4  8     10  8  5 11
3  1  4  2     12  0  0 12     15  1  4 14
3  1  4  2  +   0 12 12  0  =   3 13 16  2
2  4  1  3      4  8  8  4      6 12  9  7

You may also treat sets a and b oppositely, while randomizing. This time the restriction is on matrix a — 4 pairs with 1, and 2 pairs with 3:

3  2  2  3     8 12  0  4     11 14  2  7
4  1  1  4     4  0 12  8      8  1 13 12
1  4  4  1  +  4  0 12  8  =   5  4 16  9
2  3  3  2     8 12  0  4     10 15  3  6

The reason I say these squares are limited to about 10 or so, is because as you can see, the individual rows or columns end up being about the same — at best, a shuffling or reversal the same rows found in the Durer square. There are at best 24 such squares, not subtracting tilted (squares the same when laid on its side) or reflected squares (where the squares are mirror images of each other). All that might be said, is that we may have discovered all the ways a magic square may be constructed with the same “hyper-magic” as the Durer square, which may be a pretty cool thing to say actually. It’s just that my goal was more ambitious at the start. I will refer to the set of 4×4 hypermagic squares built on the same rules as “the Durer series.”

Any known magic square, however, is the result of a unique combination of the basic squares built on sets a and b. Respectively, I will call the matrices A and B.  This allows us to work backwards from a 4×4 square we know is magic, and discern any patterns. The above squares were done that way. What patterns can be observed for the following square, which this website offers as #379 out of 900 or so others?

                Matrix "A"      Matrix "B"
13 12  6  3     1  4  2  3     12  8  4  0
10  5 11  8     2  1  3  4      8  4  8  4
 7 16  2  9  =  3  4  2  1  +   4 12  0  8
 4  1 15 14     4  1  3  2      0  0 12 12

One may discern a pattern here, but it is not all that obvious, especially for the square built on set b. It doesn’t seem to resolve itself into simple rules and patterns the way the Durer series of squares did. Also, you might have noticed that this square is not quite as “magic” as the Durer series, though all rows, columns, and both diagonals give you 34.

But I do notice that in matrix A the last column is a reversal of pairs of the first column. Also, the third column is built on the middle numbers of the first column, and the second column are built on the middle numbers of the last column.

What of Matrix B? The first and last columns are reversals (of all 4 numbers, not of pairs), the second and third both start with the middle two numbers, then first, then last from their adjacent end columns: “e f g h” becomes: “f g e h”.

When I tried to build on these rules, I got duplicate numbers in my squares (and with that, some missing numbers). It would appear that the observable number patterns in such squares are limited to getting you maybe 4 to 8 unique squares, then you have to make a new set of rules each time. With over 900 squares discovered (give or take a duplicate or two), that’s a lot of rules. The total number of 4×4 squares, both magic and not, are equal to 16! = 21 trillion possible squares. Finding all possible magic squares, weeding out any duplicates, reflections, and sideways duplicates would seem a tad non-trivial. Even at the website I referred to for these squares, it has been shown that the number “924” claimed as the number of possible 4×4 magic squares, might be too high, due to some duplicates found. Their scripts took just under 25 seconds to generate and check these squares — but just to check them for magic, less so for duplication, which sounds like it would take more computer time.

Recreational Math I: Magic Squares: the “really good” kind – Part 3

Notice that to show the rules for making these kind of magic squares, I used only odd-ordered square matrices as examples. What about matrices of even numbers of rows and columns? The rules for these vary.

This is a small part of a 1514 engraving by Albrecht Durer, called Melancholia. The author of the article that houses this graphic asserts that there are 32 possible 4×4 magic squares with the famous pun “1514” in the same position as above. This magic square has a symmetry in the numbers, as explained below.

The famous Durer magic square, with the year of the engraving cleverly made a part of a magic square, has a certain organization in its construction, as well as a certain symmetry. The numbers are constructed, in sequence:

_   3   2   _        _   3   2   _
_   _   _   _        5   _   _   8
_   _   _   _  ====> _   6   7   _
4   _   _   1        4   _   _   1         

_   3   2   _       16   3   2  13
5  10  11   8        5  10  11   8
9   6   7  12  ====> 9   6   7  12
4   _   _   1        4  15  14   1

So, you start from the bottom right and proceed in a horseshoe to the top then the bottom left. The next diagram places the numbers 5-8 in a pattern that is left-to-right u-shape. Then the same u-shape for the numbers 9-12 from left to right, except this time it’s upside-down. Finally ending as we started, the same horseshoe shape (except right side up) from right to left.

Durer’s square has many things about it, apart from its magic number (34) which works on all the attendant diagonals, rows and columns. The middle 4 squares add up to 34 (10 + 11 + 6 + 7 = 34); the four corners add to 34 (16 + 13 + 4 + 1 = 34), and all corner foursomes add to 34: (16 + 3 + 5 + 10); (2 + 13 + 11 + 8); (9 + 6 + 4 + 15); and (7 + 12 + 14 + 1).

The numbers at the ends of the two middle rows add to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

and the numbers at the tops and bottoms of the two middle columns add to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

If we take another symmetrical combination: a rightward-slanting rectangle whose corners are 2, 8, 9, and 15, these also add to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

The leftward-slanting rectangle, whose corners are 5, 3, 12, and 14 also add to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

Starting from 2 and proceeding in an “L”-shape to the left to the number 5, and continuing counter-clockwise in the same manner gets us the corners of a tilted square whose numbers 2, 5, 15, and 12, add to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

Starting from “3” and doing likewise yields the numbers 3, 9, 14, and 8, also adding to 34:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

And what about this talk about “symmetry”? By this, we mean that we may take pairs of numbers at the start and end of any row, and they add up to the same number in a symmetrical place elsewhere. 16 + 3 = 4 + 15, taking the top and bottom of the first and second column. Likewise can be done for the last two columns: 2 + 13 = 14 + 1. The middle two rows have the same property: 5 + 10 = 9 + 6; and 11 + 8 = 7 + 12. On a larger scale, the sums of the middle two rows of columns 1 and 2 are the same as the tops and bottoms of columns 3 and 4: 11 + 8 = 7 + 12 = 16 + 3 = 4 + 15. Likewise, the sums of the middle two rows of columns 3 and 4 are the same as the tops and bottoms of columns 1 and 2: 2 + 13 = 14 + 1 = 5 + 10 = 9 + 6. These two groups of symmetrical numbers are illustrated below in red and green:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

The sums of 15 (green) and 18 (red) across each row form this pattern

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

The downward symmetry is also interesting. Here, the sum of 25 is in gold and the sum of 9 is in blue. In the process, we can discern the patterns that we used to construct the square in the first place:

16   3   2  13
 5  10  11   8
 9   6   7  12
 4  15  14   1

This is an incredible amount of magic, but if you follow the order of filling (horseshoes are right-to-left, u-shapes are left-to-right, along with the peculiar pattern of filling u’s and horseshoes), there really are four possible patterns that have these “hyper-magic” qualities, but you lose the “1514” idea in two of them:

 8  11  10   5      12   7   6   9       4  15  14   1
13   2   3  16       1  14  15   4      12   6   7   9
 1  14  15   4      13   2   3  16       5  11  10   8
12   7   6   9       8  11  10   5      16   3   2  13

Of course, you could reverse all of the numbers in the rows of the first two squares to get your “1514” back.

Every time I look at that darned Durer square, I keep seeing more patterns. I think there comes a point where one has to leave the remaining observations up to the reader.

There is yet another 4×4 square, and with it we can increase the magic, if that can even be conceivable after all I have said. But there is a square with even more magic than the Durer square. R. J. Reichmann mentioned it in his book “The Fascination of Numbers”, first published in 1957. The square could be constructed like this:

-  -  3  -        -  -  3  6        - 10  3  6        15 10  3  6
4  -  -  - ====>  4  5  -  - ====>  4  5  -  9  ====>  4  5 16  9
-  -  2  -        -  -  2  7        - 11  2  7        14 11  2  7
1  -  -  -        1  8  -  -        1  8  - 12         1  8 13 12

This square has all the magic of the Durer square and then some. One thing this new square has over the Durer square is that any four numbers in square formation will add to 34, from anywhere in the square. These include the foursomes:

10  3     16   9     11   2       4   5
5  16      2   7      8  13      14  11

Recreational Math I: Magic Squares: the “really good” kind – Part 2

Last time I introduced the idea of magic squares. I promised I would show you how to make one. In this post, I will begin by discussing “trivial” squares, or squares made by simple rules of following diagonals and wrapping.

When I say a square is “magic”, I mean that all rows, columns, and diagonals add up to the same number. While other sources, such as Wolfram’s Mathematica, say that only the main diagonal of the matrix need be magic, I will take the more strict requirement that both leftward and rightward diagonals have to be magic.

There are trivial magic squares that begins by following a rule where you start with “1” in the top middle square, then move up and to the right one square, and place a “2” there.

_  1  _
_  _  _
_  _  _

But you may have noticed that if you start at the top, how can you move “up and to the right”? You get around this by “wrapping” to the bottom, treating the bottom of the rightward column as though it is above.

_  1  _
_  _  _
_  _  2

OK, you say, but now there’s no “right” after the last column. Now what? Now you can wrap so that the leftmost column is treated as “right of” the rightmost column:

_  1  _
3  _  _
_  _  2

Now another problem: up and to the right of “3”, there is a “1” in the way. If this happens, you are allowed to place the fourth number below the “3”:

_  1  _
3  _  _
4  _  2

Now, following these rules and exceptions, we can keep going:

8  1  6
3  5  7
4  9  2

The result is a magic square whose rows, columns and diagonals add up to 15.

I found that if I moved the 1 elsewhere and followed these rules in the same manner, some or all of the “magic” is lost. There seems to be only one magic square that can be made using these rules, at least one that adds up to 15 in all of its rows, columns and diagonals. The following 3×3 magic squares were the closest I could come to any credible “magic” by placing the “1” in a different position:

6  8  1                4  9  2
7  3  5 and similarly: 8  1  6
2  4  9                3  5  7

But notice in both cases, neither of the diagonals add up to 15. In the next post, I will discuss a way to break this limitation, making it possible to construct up to 36 3×3 magic squares.

Meanwhile, let’s expand the idea to 5×5, using the same, identical rules. This one seems easier in a way, since there aren’t as many blockages early on:

17  24   1   8  15
23   5   7  14  16
 4   6  13  20  22
10  12  19  21   3
11  18  25   2   9

I particularly like 5×5 squares. But my experience with placing the “1” elsewhere than the exact middle position of the top row has resulted in a loss of “magic”. However, I was lucky on my first attempt with moving the “1” around. The following magic square has a “Mathematica” level of magic:

11  18  25   2   9
17  24   1   8  15
23   5   7  14  16
 4   6  13  20  22
10  12  19  21   3

Later in this series, we can break this limitation, too. But next, we shall discuss some 4×4 magic squares, including one that made history.

Java and Math: Coding for Graphics


In my computer class, a rotation of a graphic of Sonic the Hedgehog was attempted. The graphic was in a jpg format, and the method chosen was to write code in Java which read the graphic into a BufferedImage object. The idea is to copy the graphic into another BufferedImage object, reversing the order of pixel reading to obtain what amounts to a rotation in the new object. The image object is then written to a new JPG file. After the run, the new file is inspected.

This article is a bare-bones know-nothing introduction to how to think of reading and writing graphics that avoid the implementation details as much as possible. Documentation on the details of BufferedImage are plentiful on the internet, as well as information on computer graphics generally.

In a learning activity, students in grades 10 and 11 were asked to modify the statements:

int w2 = x;
int h2 = y;

such that (w2, h2) becomes the new placement for the pixel (x, y) when the graphic is rotated. Students were instructed not to modify any other part of the program, which performed object declarations, file opening and closing, try/catch statements and other concepts considered too advanced for this stage of their course. Students were encouraged to use trial-and error and view the consequences of various general formulas they were to try out. width and height were already declared and given values in the code.

The tutorial below was given to the students.

The graphic in question
In general terms, for any graphic of dimensions a\times b, x can have any value in the domain of 0 to a-1, while y can have values in the range of 0 to b-1.

Graphics are made of little units of colour information which end up on your computer screen as pixels. Think of each unit of information as belonging to a part of that graphic on a Cartesian coordinate plane. Much like the coordinate plane you learn about in grade 9 and 10, x and y represent the position of the pixel horizontally and vertically.

If the graphic is 100 units across by 200 units in the vertical direction, then the graphic is said to have dimensions 100\times 200, in terms of pixels. This means all pixels in a graphic will have some location (x, y). Because x has to start from 0, it can take on values between 0 to 99 for this graphic, while y can take on values between 0 and 199.

In our code, that means that the dimensions of the graphic is heightxwidth, and while x and y are fine (for the original graphic), a formula must be applied so that h2 and w2 (the x and y values for the rotated graphic) become appropriate for a rotation. With the statements written as they are, all you will get are two copies of the same graphic.

It is expected that the formula should be simple, but beware of values going beyond the range of the graphic dimensions. That will result in a crash. The dimensions of the graphic are given as heightxwidth, and it shouldn’t matter all that much what their real values are, you might want to insert printf statements where you can trace h2 and w2 against the values of height and width to see what is going on.

It is also noteworthy that you can only carry the Cartesian plane analogy so far. (0, 0) represents the pixel in the top left corner of the graphic, while for a graphic of dimensions hxw, the coordinate (w-1, h-1) will represent the pixel on the bottom right. The first difference is that no negative values are possible. The second difference is that the y-axis is upside-down, because it increases downward. The x-axis is still fine, however, increasing from left to right as usual.

Recreational Math I: Magic Squares: the “really good” kind


ONE OF THE few things you see on the web these days is how to do a really good magic square. There are many websites that tell you about how spiralling arrangements of sequential numbers on a square matrix is magic, but for me, that’s dull. You are limited to doing seemingly less than a dozen such magic squares, so I don’t find them too interesting.

Recall that magic squares are numbers arranged in a square matrix such that each of its rows and columns, and normally both diagonals add up to the same number. Usually, a square of n numbers to a side which has numbers total, will be populated with the entire set of numbers from 1 to n inclusive, in some quasi-random order. These numbers would be arranged in such a manner that the total of each of its rows, columns, and both diagonals equal the same “magic number”, which is different depending on the dimensions of the square. By using random methods suggested in this article, the number of magic squares possible, when n is odd, is equal to (n!)2.

For the 5×5 square, you apparently have to start by moving from the current position to the “top right” square (wrapping to the opposite edge if necessary), and if that square is occupied, move down by 1 square. This non-random, deterministic method apparently works for all squares greater than 5×5 (with odd dimensions).

I read from an old book on recreational math (The Fascination of Numbers, by W. J. Reichmann (1958)), that

  1. Squares of even dimensions (4×4, 6×6) have to be arranged by a different algorithm than squares of odd prime dimension (5×5, 7×7, 11×11, …).
  2. A randomly-generated 5×5 magic square can be made which uses the sum of two matrices.

The number of possible permutations of 5×5 matrices is equal to (5!)2.

Reichmann’s book was the only place where I could find such an algorithm. This seems to be a rare algorithm, even on an internet search. But it is the only method that leads to “magic” results in a variety of ways. These squares seem to be the most robust in terms of the number of ways their “magic” qualities can be determined. They have inspired my writing computer programs that generate such squares as a way of practicing programming several years ago. I have written magic square programs following Reichmann’s algorithm (not sure if he originated it) in VB5, Visual Basic .NET, VB for applications (in Excel), and in Microsoft Quick Basic 4.5. The 16-bit QB 4.5 version does not run on my 64-bit machine, and for similar reasons, neither does the VB5 version, whose runtime DLL is no longer supported by later versions of MS Windows.

In the next instalments, starting this coming Saturday, I will begin to discuss the making of 3×3 and 5×5 squares, and discuss their magic properties.