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A really simple connect-the-dots problem


bonanova
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A classic puzzle notes that 3 miles of road is sufficient to connect four cities situated on the corners of a square one mile on a side. Yes, the cities are really small; point-like, for our purposes. The question is then asked: what length of road(s) is necessary to connect them? I.e. the shortest distance of road(s) required.

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Essentially, the roads would look like this: )—(

The ends of each curve meets with the cities and the horizontal line connects each 'pair' of cities

The curved roads are from unit circles.

To get the curved roads I simply did: (1/3 * pi) = 1.0472 The 1/3 comes from the two curves combined

For the horizontal line, I needed to know the distance between the furthest point of each curve to the side of the imaginary box

This required pythagoras theorem and the previous unit circle. I did: sqrt of (1^2 - 0.5^2) which resulted in sqrt of 0.75 = 0.8660. However, this only took into account one curve. So for two curves the horizontal line is: 1 - 0.8660 = 0.1340 * 2 = 0.2680, 1 - 0.2680 = 0.7320

0.7320 + 1.0472 = 1.7792 miles

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A classic puzzle notes that 3 miles of road is sufficient to connect four cities situated on the corners of a square one mile on a side. Yes, the cities are really small; point-like, for our purposes. The question is then asked: what length of road(s) is necessary to connect them? I.e. the shortest distance of road(s) required.

Here's an attempt,

Let the cities be arranged on a grid with coordinates (0,0), (1,0), (0,1), and (1,1). We draw the lines with the following shape: >-<. Essentially, we make two way points, each joining two cities. We then draw a straight line between the two waypoint.

If we let the coordinates of the waypoints be ( sqrt( 1/12), 1/2 ) and ( 1 - sqrt( 1/12), 1/2 ), then the total distance used is about 2.732.

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Here's an attempt,

Let the cities be arranged on a grid with coordinates (0,0), (1,0), (0,1), and (1,1). We draw the lines with the following shape: >-<. Essentially, we make two way points, each joining two cities. We then draw a straight line between the two waypoint.

If we let the coordinates of the waypoints be ( sqrt( 1/12), 1/2 ) and ( 1 - sqrt( 1/12), 1/2 ), then the total distance used is about 2.732.

I think you're right

Calling the straight line distance x, the total distance is (x+2sqrt(2-2x+x^2). If I minimize that function (taking the derivative and setting to 0), I get x=(1-1/sqrt(3)), which is equivalent.

The case of drawing two diagonals is a special case of this, where x=0.

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I can connect them with a road that is slightly over 1 mile long.

How?

The road is exactly one mile wide.

Very clever. I like this solution so much that I submitted it to the combined Town Boards of the ABCD Municipality for approval. They texted me back and said they were willing to proceed, even to bear the cost of all that additional asphalt, provided that

the location of the necessary six-inch-wide yellow divider line. The Board is extremely safety conscious, it seems. By the way, the Town Cartographer chimed in that the length of the road will be recorded as the total length of the yellow line.

The Board awaits our specification.

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Very clever. I like this solution so much that I submitted it to the combined Town Boards of the ABCD Municipality for approval. They texted me back and said they were willing to proceed, even to bear the cost of all that additional asphalt, provided that

the location of the necessary six-inch-wide yellow divider line. The Board is extremely safety conscious, it seems. By the way, the Town Cartographer chimed in that the length of the road will be recorded as the total length of the yellow line.

The Board awaits our specification.

...is a giant roundabout.

:P So what if Americans have some weird difficulty with them? They're safe and efficient and if you're going to pave the entire square, then stick a sign in the middle indicating it's a roundabout and paint white dividers for the lanes. Length of road->0
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Illustration:

road.jpg

Yoruichi-san's solution is elegant, but I want to sell concessions at that median divider, maybe set up a traffic island, possibly a theme park.

Lol...what they didn't tell you was that these four towns are all towns of super-genius scientists (like Eureka ;P), and they've all established teleporters to each other's towns already, and this whole exercise is a diversion to keep the board occupied so they don't have to try to explain the safety procedures of teleportation to a bunch of non-scientists.

Tony Stark to Bruce Banner in The Avengers, discussing gamma radiation: "Finally, someone who speaks English!"

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Very clever. I like this solution so much that I submitted it to the combined Town Boards of the ABCD Municipality for approval. They texted me back and said they were willing to proceed, even to bear the cost of all that additional asphalt, provided that

the location of the necessary six-inch-wide yellow divider line. The Board is extremely safety conscious, it seems. By the way, the Town Cartographer chimed in that the length of the road will be recorded as the total length of the yellow line.

The Board awaits our specification.

Just make the line a color that isn't yellow. (Or make an infinitely short yellow line somewhere along the road).

:)
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Just make the line a color that isn't yellow. (Or make an infinitely short yellow line somewhere along the road).

:)

Traffic to and from each city is separated by a yellow line.

The Board was willing to concede that a rose by any other name smelled just as sweet.

But when I forwarded your proposal to paint a yellow line of a non-yellow color,

they just kind of got glassy eyed and looked confused.

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Traffic to and from each city is separated by a yellow line.

The Board was willing to concede that a rose by any other name smelled just as sweet.

But when I forwarded your proposal to paint a yellow line of a non-yellow color,

they just kind of got glassy eyed and looked confused.

If we cover the yellow line, we can put the line in a quantum superposition of states. Then, it's both yellow and not yellow at the same time! Now, I'm not sure how that would help. (Maybe we can confuse the board enough that they'll agree with our proposals.) But anyway, why can't we make a line that's shorter than the road. There's nothing that says that the line has to go all the way along the road, merely that the traffic has to be separated. Or, we could put a sequence of infinitely short yellow lines. The cars will always be separated by at least one, but the length is still zero.

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If we cover the yellow line, we can put the line in a quantum superposition of states. Then, it's both yellow and not yellow at the same time! Now, I'm not sure how that would help. (Maybe we can confuse the board enough that they'll agree with our proposals.) But anyway, why can't we make a line that's shorter than the road. There's nothing that says that the line has to go all the way along the road, merely that the traffic has to be separated. Or, we could put a sequence of infinitely short yellow lines. The cars will always be separated by at least one, but the length is still zero.

The board has this weird notion that every vehicle must have a yellow line on its left every inch of the way between every pair of cities, fearing legal head-on collisions otherwise. So, like it or not, we've got to minimize the total length of that yellow line.

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Anyway ... Overdue kudos to the solvers bushindo and Yoruichi-san.

Here's an attempt,

Let the cities be arranged on a grid with coordinates (0,0), (1,0), (0,1), and (1,1). We draw the lines with the following shape: >-<. Essentially, we make two way points, each joining two cities. We then draw a straight line between the two waypoint.

If we let the coordinates of the waypoints be ( sqrt( 1/12), 1/2 ) and ( 1 - sqrt( 1/12), 1/2 ), then the total distance used is about 2.732.

That's it.

I think you're right

Calling the straight line distance x, the total distance is (x+2sqrt(2-2x+x^2). If I minimize that function (taking the derivative and setting to 0), I get x=(1-1/sqrt(3)), which is equivalent.

The case of drawing two diagonals is a special case of this, where x=0.

120o angles at the waypoints.

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