I have started to work with this variation on my 30 unique tiles.
The same set of tiles as before, this time in monochrome and with somewhat fatter lines. My first enquiry thread with these will be to think about making patterns by repeating indivdual tiles. I’ll post a series of outcomes soon…
Continuing. I have 30 tiles to work with. (not shown in the original order)
I have a standard grid for my alphabets.
Strictly speaking the 1, 2 and 3 are not part of my standard grid and the # represents the space, which I like to locate in the middle. There are several reasons for laying the alphabet out this way, but I’ll save those for another time.
With the addition of the 1, 2 and 3, I have 30 positions for my 30 tiles.
The tiles in position 1, 2 and 3 have not been given a role in my alphabet as yet, but they could substitute for other punctuation characters (for example). For this alphabet set, I have adjusted the design to suit a set of Moo Lux business cards. A set of cards I had made with the alphabet tiles printed on the back. Here is a snapshot of my journal showing the cards.
They make for a highly portable transient installation kit. Something else I may return to later.
Of course, the tiles can be looked at in many other ways. Here are two adjusted sets.
First, in black and white only, where the symbols and connectors become one.
Second, as black on white symbols without connections. This version looks much more like a font, but lacks the connective nature of the original tile set.
At some point I will post some outcomes from using the business card tile set.
When I started the tile project, the question was: how many unique tessellated tiles can I get by making square tiles with two symmetrical connections on each side? As usual, I was looking for a minimum of 27 to cover the alphabet plus one for a space. So, I set about finding out.
Starting with a bit of maths:
There are eight points to connect one to one other;
After making the first connection, there are six points left;
After making the second connection, there are four points left;
After making the third connection, there are two points left;
Leaving just one possibility for the final connection.
This can be shown to represent this number of possible connections:
7 x 5 x 3 x 1 = 105
That is, after selecting my first point, there are 7 other possible points to connect it to. After making the first connection and selecting another point there are 5 other possible points to connect it to. After making the second connection and selecting another point there are 3 other possible points to connect it to. After making the third connection, there is just one possible connection to be made between the remaining two points.
So, there are 105 different ways to connect all the points one to one other.
Here is a page from my journal showing the first 15 (the other six pages of 15 are hidden behind the first page but you can see the starting pair for each page to the bottom right.
Note: I named the eight points: A, B, C, D, E, F, G, H. I sketched out the 105 tiles and it soon became obvious that there were duplicates. That is, if I rotated, reflected, or rotated and reflected one tile it became identical to another. At this point, I considered writing an algorithm to find the duplicates using some Python code. But, I didn’t want to spend time writing code so, I just did it all by visual comparison. I drew the next tile and then picked it up and turned it every which way while comparing it to previous tiles.
Here is a page from my journal showing some of the duplicates I discovered.
This left me with 30 unique tiles, tiles that remained unique even if I rotated, reflected or rotated and reflected them. 30 tiles is enough to use as a replacement alphabet and I could still recognise a ‘character’ whichever orientation the tile was in, even if I drew it on a transparent sheet and viewed it from the other side.