# Access Road System (Formerly Path of the Rat)

There are six towns (see map below) that needs a road system to have access with each other. What is the minimum total length of 10ft. wide road pavement that can be constructed?

o---1mile---o---1mile---o

1 mile

o---1mile---o---1mile---o


Formerly (What is the shortest total length of path that the Rat can make to access all the pocket holes of a 1 x 2 snooker table?)

• Yes @ArbitraryKangaroo, it is either trivial, or TSLF considers the holes to be circles (which the rat must touch), instead of points. – Puzzle Prime Oct 8 '16 at 5:08
• @TSLF then update the question to correct it please. – Jonathan Allan Oct 8 '16 at 7:53
• @YowE3K apparently it is the latter, and OP has explicitly declined to update the question to state that. I agree the question as worded strongly suggests it seeks a minimal trip length, not a minimal total aggregate length of all path segments.... and clearly, I'm not alone in thinking that. – Rubio Oct 8 '16 at 10:33
• I am VTC because puzzlers should not need to read through comments to dissect a question that needs to be reworded. – Jonathan Allan Oct 8 '16 at 13:46
• @TSLF, you problems are very nice, but if you don't start improving the presentation, all of them will be heavily voted down, which is a shame. At least let the more experienced users make changes, otherwise very few people will even understand what you are asking for. – Puzzle Prime Oct 8 '16 at 14:30

Not a complete solution but a lead based on the interpretation that the rat will wear a network of paths into the felt that allows the rat to scurry from any hole to any other hole while staying on that network.

The trail map of such a networked path with the shortest total length of segments is called a...

...Steiner tree, as seen on YouTube: Steiner tree approximation 6 points For anyone who wants to figure this out exactly...
All 3-way intersections consist solely of 120° angles.

That image can be flipped and rotated so that the 120° angles have more familiar alignments. (Rotational symmetry is assumed. Lengths are even multiples for easier calculation. The lengths of the leftmost $\small 4a$ and $\small 2b$ segments are deduced from parallelism helped by the parallelogram with sides $\small 2a$ and $\small 2b$.) The diagram, with $\small\triangle \rm ABC \cong \small\triangle \rm CDE$, compels the following equalities.

\small\begin{align} 4a+b & ~ = ~ \surd3c + \surd3b \\[1ex] 2a + c & ~ = ~ b + \surd3 b \\[1ex] (\surd 2)^2 & ~ = ~ (\surd 3c)^2 + \, (4a+b+\surd3b)^2 \end{align}

These shake out to:

\small\begin{align} a & ~ = ~ \sqrt{ \scriptsize \frac{4-\surd3}{78} } \\[1ex] b & ~ = ~ \tfrac12 (1+\surd3) \, a \\[1ex] c & ~ = ~ \surd3 \, a \end{align}

So the total length is:

\small\begin{align} \textsf{length} & ~ = ~ 4a + 2b + 2c + 2a + 2b +2a + 2c + 4a + 2b \\[1ex] & ~ = ~ (15 + 7\surd3) \sqrt{ \scriptsize \frac{4-\surd3}{78} } \\ & ~ = ~ \sqrt{ 11+6\surd3} \\[1ex] & ~ = ~ 4.6251816... \end{align}

This matches what 2012rcampion reported (and probably derived with less behind-the-scenes algebra) in a comment, along with interior coordinates. • Using CAD the length can be determined easily..minimum=4.62 & jamal's 2nd solution is only 4.66 assuming Y angles are 120degrees – TSLF Oct 8 '16 at 17:42
• Okay I've formulated this on paper to get 4.6251816..., barring mistakes. I'd better post it while this answer is checked. Cleaning it up and preparing a diagram. – humn Oct 9 '16 at 1:04
• With this configuration I get an exact answer of $\sqrt{11+6\sqrt{3}}$, with the intermediate points at $(9+\sqrt{3},6+5\sqrt{3})/39$ and $(69-\sqrt{3},33-5\sqrt{3})/78$ (with the other two symmetrically positioned). And if you're curious, the paths are offset at an an angle of $\cos^{-1}\left(\sqrt{(41+6\sqrt{3})/52}\right)$. – 2012rcampion Oct 9 '16 at 1:49
• I have the feeling that there are two questions here which are mixed up. There is a question of "what is the shortest path a rat should make while visiting each and every holes on the snooker table (start and end points are chosen arbitrarily)?"; and between "what is the shortest network joining all holes together?". I believe that the intentions of the OP was to ask the question answered here, nevertheless I am not at all convinced that this is the optimal solution once a rat is actually travelling, and visiting each of the holes; thus turning around at several points. – Matsmath Oct 9 '16 at 6:20
• Also, there is a difference between demonstrating that such-and-such configuration has such-and-such length, whereas to argue that in fact no shorter path is possible whatsoever. – Matsmath Oct 9 '16 at 6:22

Go to the nearest pocket then around the perimeter for the other 5 holes (1 unit between each hole) for a total of $5+x$ where $x<1$ is the distance from the mouse to the bottom right corner

• Okay.. with that path he can go from hole to hole by traveling back and forth.. 5 is not short enough. – TSLF Oct 8 '16 at 6:48
• Should 5x be 5 + x? – YowE3K Oct 8 '16 at 9:02
• @YowE3K it was 5.x (equivelent to 5+x) but it got edited on me. – gtwebb Oct 8 '16 at 15:15
• Thinking about it in the morning it was a bad way of writing it just basically wanted to say it was between 5 and 6. – gtwebb Oct 8 '16 at 15:30

I think this is the shortest path

• Nice improve. But not short enough – TSLF Oct 8 '16 at 7:37
• Jamal's route is shorter than 5 – TSLF Oct 8 '16 at 7:48
• this might be the most efficient way to to join all the dots but not the shortest path by a long shot. Almost every lines will have to be crossed 2 times and the result will be way over 6. Or is that the catch? rather than total distance to be traversed by the rat, we are looking for an actual path length? – stack reader Oct 8 '16 at 9:10
• @stack reader- Yes. Shortest path length.(see Jamal's Answer) – TSLF Oct 8 '16 at 9:30
• @Rubio- Distance traveled by the Rat do not add to the length of path when going back along the path to access holes – TSLF Oct 8 '16 at 9:39

I couldn't think of anything better than to just follow the perimeter so i tried some outside the box thinking.

I assume it's a pay to play table which locks the balls after one time.
I would say the best way is to jump in hole number one and then automatically slide through all the holes.
By stretching it a lot, one could say that the "path" that had to be taken to "access" all the holes was until the first hole.
And if that is not acceptable then it just comes back to the whole perimeter thing except that the rat can enjoy the ride.

• The Snooker table dont have automatic slide down system. – TSLF Oct 8 '16 at 6:44
• more to the point - snooker tables have ball pockets, so there is no other sane path to follow from one pocket to another than traversing the tabletop. – Rubio Oct 8 '16 at 7:42

I'm not sure if this is allowed but...

a human could suspend the rat slightly off the pool table, then the pool table could be maneuvered beneath the rat. this makes the path 0 (not including any vertical movement)