Steiner
Tap or drag across open land to lay one connected network of road.
Build. Tap or drag over open land. Drag back along your road to pull it up again.
Read the map. Water blocks you. Open land costs 1 a square, + squares are road already laid and cost nothing, the brown squares are highland and cost 3, and two squares marked with the same letter are the two ends of a portal.
Share the route. Branch off one short trunk rather than running a separate road out to each town.
Wrap. Dashed sides join left and right in the same row. Dashed top and bottom edges also join the same column. Water still blocks the crossing.
Link every town first. Check to reveal the target and, if you want it, the best route.
Colouring
Colour every dot so neighbours joined by a line use different colours.
Pick a colour, then tap a dot. Tap it again with the same colour to erase.
Watch the lines. Red means two neighbours clash.
Use fewer colours. A triangle needs three; start there and press Check.
Bonus: count them. Under the buttons you can guess how many different colourings exist at the palette size you are on. Only the grouping matters; swapping two colours around gives the same answer, and every colour must be used. Be warned that this is markedly harder than the colouring itself; wrong guesses are told higher or lower, there is a full method behind “Show me how to count these”, and it never blocks the main win.
Bonus: how many different ways can this graph be coloured with 3 colours?
Only the grouping counts: two colourings that put the same dots together are one answer, however the colours are swapped. Every colour has to be used, so a grouping that leaves one spare does not count.
Fair warning: this is the hard half. Finding one colouring is a puzzle; counting all of them is a different order of difficulty, and there is no known shortcut for it in general. Wrong guesses are told higher or lower, so treat it as a long game. The main puzzle is won without it.
1. Anchor on a clique, and the colour-swapping disappears. Find a set of dots that are all joined to each other. If there are as many of them as you have colours, they must take one colour each. Because swaps do not count, you may simply fix them: this dot is colour 1, that one colour 2, and so on. That single move turns a symmetric mess into an ordinary count. If the biggest clique is smaller than your palette, anchor on it anyway and handle the leftover colours separately.
2. Take the forced dots next. A dot already touching every colour but one has no choice at all. Fill those in and look again; each one you place often forces another. Forced dots never multiply your total, so clearing them all before you start branching keeps the arithmetic small.
3. Multiply independent choices. Once the forced dots are placed, look for dots whose remaining options do not affect each other, typically dots hanging off the anchored part with no path between them except through fixed dots. If one has 2 options and another has 2, that region contributes 2 × 2. Worked example: a triangle with one spare dot hanging off each corner. The triangle is anchored, so it counts once; each spare dot is barred from its own corner's colour and can take either of the other two. The answer is 2 × 2 × 2 = 8.
4. Branch on the most constrained dot left. When nothing is forced and nothing separates, pick the dot with the fewest legal colours, count each case as its own smaller problem, and add the results. Fewest options first keeps the number of cases down.
5. Split on a pair instead of a dot. Take two dots that are not joined. Every colouring either puts them together or keeps them apart, so the total is (colourings with the pair merged into one dot) + (colourings with a new line drawn between them). Both are smaller graphs. Choosing a pair with many shared neighbours shrinks the problem fastest.
6. Use the symmetry of the picture. If the graph looks the same after a rotation or a mirror, its colourings pair up under that same move. Count the ones that are unchanged by it, then work out the rest in matching pairs. This often halves the effort on the ring-shaped and star-shaped days.
7. Sanity-check with the no-lines ceiling. With no lines at all, n dots split into k groups in a fixed number of ways: 5 dots into 2 groups is 15, into 3 groups is 25. Every line only removes possibilities, so your answer must come in under that ceiling. If it does not, recount.
8. Do not count the groupings that use fewer colours. A grouping that leaves a colour spare is not an answer here. The 6-ring is the clean example: with 3 colours it has 10 answers, plus a single 2-colour grouping that does not count. Miss that and you will be one over.
Two to test yourself on. The 5-ring with 3 colours has 5 answers. The 6-ring with 3 has 10. Both are worth doing by hand before trusting your method on a full board.
When to give up on exactness. Below the optimum the answer is always 0; a palette smaller than the graph needs has nothing to count. And if you are hunting a five-figure number, bisect: the higher/lower reply is all you need, and it costs nothing but guesses.
Graphle
Rebuild the hidden 7–8 vertex graph. Tap two dots to toggle a link. The second dot stays selected, so tapping 1 2 3 draws a chain. Tap it again, or tap the empty board, to lift the pen. Then Guess: you have six tries.
Each guess is scored on five numbers. An arrow means the hidden graph's value is higher ↑ or lower ↓ than yours.
- E
- Edges. How many links you have drawn. Seven dots allow 21 possible links; eight allow 28.
- χ
- Chromatic number. The fewest colours needed to paint the dots so that no link joins two of the same colour. 1 means no links at all, 2 means the dots split into two sides with every link crossing between them, and 3 or more means there is an odd-length ring somewhere.
- △
- Triangles. How many sets of three dots are all linked to each other. Adding one link to a dot with k shared neighbours creates k triangles at once, so this number climbs quickly in a busy graph.
- Cyc
- Cycles. How many distinct closed loops of three or more dots exist, counting a loop once however you walk it. Every triangle is also a cycle, so Cyc is never below △; a big gap between them means long loops rather than tight clumps.
- Diam
- Diameter. Take the two dots that are furthest apart and count the links on the shortest route between them. Small means tightly knit, large means strung out, and ∞ means some dot cannot be reached at all.
Facility
Site the depots. A fixed number of depots go anywhere on the walkable ground — an island world, a desert, a snowfield, or a plain of lava. Every town then walks to its nearest one, and your score is the total of those walks — the lower the better.
Place and move. Tap land to drop a depot, tap it again to lift it. Drag a depot to slide it somewhere else — useful once they are all out.
Some ground cannot be crossed. Sea, canyon, lava — whichever world the day drew, nobody gets over it. A depot on the far side of one is no use however near it looks, so any patch of ground with people on it always needs its own depot.
The rest costs. A step over plain ground counts 1, over the middle grade 2, over the roughest 3, and a step between two grades costs the rougher of the two. A ridge is not a wall, but going round one is often cheaper than going over it. The key under the map names all three for the world you are in.
Read the map. Each town shows how far it currently walks — a town with no route at all shows —. The amber shading is the ground those walks actually cross, so the amber on the board is your score.
Try before you place. Hover any land cell and the map shows which towns would switch to a depot there, the routes they would take, and what the total would become.
Greed is not enough. Placing each depot where it helps most right now is almost never the best answer — every board here beats that. Try pulling one depot off the crowd to cover the far towns and see what the total does.
Serve every town, then Check to reveal the target and, if you want it, the best placement.