You'll notice that all of these semiregular tilings are based on one of the earlier patterns: regular triangles (RT), regular squares (RS), or regular hexagons (RH).
If more than six shapes meet at p then at least one has an interior angle below 52°, which is impossible.
If six shapes meet at p, the average angle is 60°, which is also the smallest possible angle, hence six triangles meet at p, creating RT. Thus we will assume three four or five shapes meet at p.
Place three shapes around p. We have handled the case where they are all hexagons, so the first shape is something smaller, say a pentagon. That leaves 252° for the remaining two shapes. If the smaller of these is a hexagon, we need a regular n-gon with an interior angle of 132°, and there is no such animal. Move on to the pentagon, and it looks like we've struck paydirt. Two pentagons and a decagon fit perfectly around p. But as we walk around the pentagon we arrive at a contradiction. The shapes adjacent to the pentagon run pentagon decagon pentagon decagon pentagon, and then we're back to start. This puts three pentagons together, which does not match the pattern. If three shapes meet at p, and one of them is odd, the other two better be the same.
Drop down to a triangle and use the walkaround principle, hence the remaining shapes are dodecagons, i.e. 12-gons. This creates our first semiregular tiling. It is based on RH; expand each vertex into a triangle, and the hexagons turn into dodecagons.
Finally let one of the three shapes be a square. If one of the other shapes is odd then the third is also a square. That leaves 180° for the odd shape, which is impossible. Thus the remaining shapes are even, and larger than squares. Two octagons will work. Start with RS and expand each vertex into a small square, thus turning the original squares into octagons.
The last configuration is square hexagon dodecagon, abbreviated 4,6,12. This is the 3,12,12 pattern, pulled apart, with triangles expanded out to hexagons and squares inserted between the dodecagons. Squares and hexagons form the grout between the large dodecagon tiles.
Next assume four shapes meet at p. The smallest angle is at most 90°, and if it is a square we have RS, so assume it is a triangle. The next angle is bounded by 100°, and could be a square. If it is, the next shape is also a square, and the last is a hexagon.
If the squares are adjacent, put three squares together in a row, with a triangle on top of the first and third and a hexagon on top of the second. Place a third triangle atop the hexagon, and another three squares up the left side to the apex, and another three squares up the right side to the apex. The hexagon and three triangles form one larger triangle, which is bordered by nine squares. At each of the three corners, two squares cover a triangle and call for a hexagon. This almost fits our pattern. Bring in the three corner hexagons and the corner vertices are 4,6,4,3, while the six vertices inside the triangle are 4,4,6,3. Rotate and reflect the picture, and the six internal vertices are indistinguishable. At the same time, the threee corner vertices are indistinguishable. This is a quaasiregular tiling. Every vertex joins two squares, a hexagon, and a triangle, but type A vertices run 4,6,4,3 and type B vertices run 4,4,6,3. Type A vertices are indistinguishable and type B vertices are indistinguishable. The pattern repeats forever. To picture this pattern, start with RH and pull the hexagons far apart, then draw connecting spokes in between. Each spoke is three squares in a row. The gaps between the spokes are large triangles, each filled with a hexagon and three triangles. Let's see if this pattern is inevitable.
Assume there is a 4,4,6,3 vertex somewhere in the pattern. There is another one on the other side of the triangle, thus building the corner of our larger, square-enclosed triangle. Place the hexagon in the corner, as we did before, then look below the double square. The only possibility is another hexagon triangle, just as they appear above. This calls for another double square, and so on all the way around the corner hexagon. If one of these duble squares is capped by a triangle, and two more squares on either side, we have a new kind of 4,6,4,3 vertex, where the triangle is surrounded by squares. This is not the same as the 4,6,4,3 vertex we saw before. If we want type A vertices to remain indistinguishable, the double squares extend to triple squares, triangles fill in on either side, and the earlier pattern appears.
To find a new pattern, assume all vertices are 4,6,4,3. Thus each hexagon coats itself in a ring of alternating squares and triangles. Start with RH and pull the tiles apart. Place a square between each pair of hexagons, and a triangle between each trio. Thus squares and triangles form the grout between the hexagon tiles.
Next let two triangles meet two other shapes at p. The remaining shapes could be hexagons. Assume every vertex is 3,6,3,6, thus every hexagon has triangle teeth, like a gear, with more hexagons nestled in these teeth. To picture the entire pattern, start with RH and pull the tiles apart, then rotate each hexagon 30° clockwise, so the gaps become triangles.
If a vertex runs 6,6,3,3, follow the border between the two triangles and find another vertex that must run 6,6,3,3. Follow the border between the hexagons and run into two more triangles, then two more hexagons, and so on forever. This row of double triangles and double hexagons has notches above and below that must be filled with triangles. These form 6,3,6,3 vertices. The result is a smooth rectangular stripe, two hexagons high. Stack these stripes on top of each other to build a quasiregular pattern with alternating columns of hexagons and triangles.
What if a stripe is shifted, relative to the stripe below? Thus the hexagons are above the triangles and the triangles are above the hexagons. This introduces another suite of 6,3,6,3 vertices, where the stripes meet. Each triangle at such a vertex has hexagons on all three sides. Yet there are other 6,3,6,3 vertices inside the stripe, with incident double triangles. So the 6,3,6,3 vertices are not all the same.
"What's wrong with that?" you might ask.
Well it permits infinitely many tilings. If a stripe is in phase with the base stripe, assign it the digit 0. If it is shifted out of phase, assign it the digit 1. Now the stripes, going up the plane, spell out a binary sequence. Any repeating sequence is fair game. Try 01010101… or 001001001001… or 0001000100010001… etc. So there are infinitely many periodic tilings with the same collection of n-gons at each vertex. I would rather explore regular, semiregular, and quasiregular tilings; there are only a finite number of these.
If two triangles and a pentagon meet, that leaves 132°, and there is no such animal. If three triangles meet they leave 180°, also impossible. The last combination of four shapes is two triangles a square and a dodecagon. Start with a 12,3,4,3 vertex and ask what is on the third side of the triangle. A dodecagon would place two dodecagons at one vertex, and a square would place two squares at one vertex. Thus we have a triangle, with a square between it and the dodecagon, and a dodecagon between it and the square. Unfortunately the same thing happens on the other side, hence the far corner of the square meets two dodecagons, which is impossible. Therefore every vertex runs 12,3,3,4. The dodecagon coats itself in a ring of alternating squares and triple triangles. This places three triangles at a point, which is impossible. That's it for four shapes meeting at p.
When five shapes meet at p, the smallest angle is at most 72°, and that has to be a triangle. The next two smallest shapes are also triangles, and the last two could be either two squares or a triangle and a hexagon. Let's start with the latter.
Let 4 triangles and a hexagon meet at a point. Walk around the perimeter of the hexagon and note that it completely coats itself in triangles. Place another hexagon at the far side of one of these triangles, and it also surrounds itself with triangles. Do this 6 times and the original hexagon is surrounded by six other hexagons, with triangles acting as grout. this is a valid semiregular pattern.
Finally consider two squares and three triangles. The simplest pattern consists of rows of squares and rows of triangles. The triangles, pointed up and pointed down, cause the row of squares above to be shifted by half a square, relative to the row of squares below.
If three squares line up in a row, place five alternating triangles above these, and two more squares above those. A third square is required, to match the first row of squares, and that brings in a sixth triangle. A fourth square would force an infinite row of squares, so let's try to avoid that. We're hoping that the two vertices, where the three squares meet the five triangles above, will correspond under a reflection through the middle square. The third square on the top row of squares swings around and becomes a fourth square, forcing an infinite row of squares. We've already explored alternating rows of squares and triangles, so assume three squares never line up in a row.
If two squares are adjacent, but never three, place three triangles above the double square, and another square above that. Let this be a double square, and let it point up like a chimney on a house. Now the horizontal double square has to look like the chimney and house, laid on its side. That means it has 3 triangles to the right, and a vertical double square to the right of that. Put triangles and a horizontal double square above this, and close the loop with three more triangles. That leaves a square hole inside. This quasiregular pattern repeats forever.
Start with the double square and three triangles as before, but let the chimney point to the side. This brings in a fourth triangle. The upper double square has three more triangles on top of it. Let p be the point where the lower double square meets its triangles, and let q be the point where the upper double square meets its triangles. How can we map p to q? Suppose the picture is reflected through the vertical line at p, then shifted up to q. This means double squares weave in and out, right and left, as we move up and down the column. All the double squares are capped by triangles at the left and right, to avoid three squares in a row. Fill in all the notches at the right of this wavy column. Focus on one of the double squares that is shifted right. It has a cap of three triangles, and another two squares to the right of that. This looks like the house and chimney laying on its side. We've already handled this case, so p does not map to q by reflection.
Instead, the map is a simple translation, and double squares propagate up and to the right forever. Call this a zigzag strip. We could paste a copy of this strip next to itself, it would fit perfectly, but that would place four squares in a row. Instead, take a copy of the strip and reflect it along its length, then place it to the right of the original strip. Do this again and again, building a quasiregular pattern.
Return to the original zigzag strip, and try not to place another one next to it. Add a layer of triangles and squares to the right of this strip, then add another layer to the right of that. This spawns a quasiregular pattern that must repeat forever. Zigzag strips run parallel to each other, with gaps in between. The gap is just wide enough for a layer of squares and triangles. The 4,4,3,3,3 vertices are inside the zigzag strips, and the 4,3,4,3,3 vertices connect the strips to the grout in between. That takes care of the double square on top of the house.
Start with the double square and three triangles as before, and place another square on top. This is not a double square, so it has triangles on all four sides. Two more squares fill the notches at the left and right. These squares are incident to the double square. The same pattern appears below the double square, and that forces three triangles to the right of the double square. This looks like the house and chimney laying on its side, and we've already dealt with that.
Start with the double square and three triangles as before, and place a fourth triangle above the three triangles, making a larger triangle, which must be enclosed in 6 squares. Each corner brings in two more triangles. The same pattern appears below the double square, and that forces three triangles to the right of the double square. This looks like the house and chimney laying on its side, and we've already dealt with that. That takes care of the double square. Hereinafter, squares are not adjacent.
Start with RS and separate the squares just a bit. Now rotate each square clockwise or counterclockwise, in an alternating pattern. Focus on four squares as they rotate. They join corners to create a diamond shaped hole. If this diamond is vertical, the diamonds above, below, left, and right are horizontal. We have a grid of diamond holes that alternate between vertical and horizontal. Fill each diamond with two triangles and find a semiregular pattern.
Is the above inevitable? Remember, we can't put squares together. Let two triangles form a vertical diamond, and place square triangle square on either side. The top and bottom of this diamon have two squares, and must be filled in with two triangles, i.e. horizontal diamonds. These have vertical diamonds at their tips, and so on throughout the entire plane. The pattern is inevitable.
If you've been keeping score, there are 3 regular tilings, 8 semiregular tilings, and 5 quasiregular tilings. That's a total of 16 patterns.