What is the least total length of the track that can be traversed by the mouse to go from hole to hole?
The pool table is 2 x 1 meter.

Edited September 6, 20169 yr by TimeSpaceLightForce

6 metres 5 metres if the mouse doesnt need to return to original hole

Edited September 6, 20169 yr by Mordred

1 2 3

4 5 6

connecting 1-6,3-4,2-5
is just 1+2*sqt(5)=5.472

but not short enough..

1 2 3

4 5 6

connecting 1-6,3-4,2-5

is just 1+2*sqt(5)=5.472

but not short enough..

But you haven't connected 6 to anything else. After you go 1-6, how do you get to 3? Chief O'Brien or Scotty using the transporter?

via crossroad..

5 meters going round.

via crossroad..

I don't understand what you mean by that.

1 2 3

4 5 6

connecting 1-6,3-4,2-5

is just 1+2*sqt(5)=5.472

but not short enough..

 

Nonsense.

You forgot to mention the one important rule in your game..

The way you described above, one could join 1-4,2-5,3-6 (or 1-4,5-2,3-6). And have 3 total length. But there are missing 4-5 and 2-3!

Therefor the way you explained game in #1 post, the correct answer is 5.

Edited September 7, 20169 yr by Sensei

@sensei
The mouse makes the tracks or pathway that is as short (total length) as possible to access all 6 holes.
It can go to any hole from any hole not necessarily directly or from any hole to the next adjacent hole.
It can go back its track not necessarily increasing the length of track. Dont mind the distance it travels.

Your proposition will be 5 meters too.. because 1-4-1-2-5-2-3-6 is the "E" route to access all pocket holes.
5 meters is wrong because it is not short enough.

@ swansont
The mouse can do the 1-6-middle-3-4-middle-2-5.. eventually making the crossroad where it can go back to
to access the 6 holes.

@ kisai
5 meters is wrong because it is not short enough.

@ Mordred
5 meters is wrong because it is not short enough.

I don't see any solution shorter than 5.

@sensei

The mouse makes the tracks or pathway that is as short (total length) as possible to access all 6 holes.

It can go to any hole from any hole not necessarily directly ....

It can go back its track not necessarily increasing the length of track. Dont mind the distance it travels.

.

Clarify this with Total length travelled

 

in particular it can go back on its track without adding to total length travelled... 😱

Edited September 7, 20169 yr by Mordred

@mordred
-its ok..

-tracks means path,route,course or trail and has length .. not to be mistaken as accumulated length of foot steps.

-no, going back along track will add to total length of travel..not to the track length.

 

HINT : Imagine the pockets are towns.. what is the shortest road system to access all the towns?

What about a path from 1-6, with the other 4 town being reached paths branching and perpendicular to the first path?

What about a path from 1-6, with the other 4 town being reached paths branching and perpendicular to the first path?

No that doesn't work my bad

Least distance is the perimeter of the whole figure.

Least distance is the perimeter of the whole figure.

No it isn't. You can obviously miss out any one link between two holes and still have all of them connected.

 

I'm fairly sure I can do it with 4.828 units

Edited September 8, 20169 yr by John Cuthber

@mullimanz

That is 4.91m..not bad.

 

@John Cuthber

Good sample! But not short enough.

Edited September 9, 20169 yr by TimeSpaceLightForce

 

Angle 120 degrees.

Total length approximately 4.73.

 

I can do it in 4.76 units

Nah - I can do it in 4.66 units. Will post diagram when at full pc

I made SpreadSheet how results changes with various length line from center of table to "middle" of left/right side (A).

 

Here is zoomed in range where it's starting growing:

(25 row is the smallest)

 

A is auto-incremented.

1-A is rest.

B=SQRT((1-A)^2+0.5^2)

 

Total length 4*B+2*A+1

 

 

a = sqrt (.211^2+.789^2) = .8167

b = sqrt (.211^2+.211^2) = .2984

c = 1/[sqrt(3)] =.57753

d = 1 - 2*(1/[2*sqrt(3)] = 1 - 1/sqrt3 = .4226

 

2*a+b+4*c+d = 4.664

I'm a little surprised that the solution (assuming that's an optimum) isn't symmetrical.

@imatfaal

Solved..the 120 degrees network connects n points in space with shortest line segments.

I'm a little surprised that the solution (assuming that's an optimum) isn't symmetrical.

 

To be honest so am I - decided to check non-symmetrical when I split the puzzle into two to simplify. The LHS connects 3 holes the RHS connects 4 holes - one of both in common; rather than each side connecting 4 with two of each in common

And can anyone tell me a nice online app / freeware that would allow me to draw that accurately and easily?

And can anyone tell me a nice online app / freeware that would allow me to draw that accurately and easily?

I was using LightWave 3D. Not free for ordinary user. Although I received it for free.

If we select two points and use Detail > Measure > Measure Points,

it's calculating length of line,

x,y,z abs delta between points.

Edited September 10, 20169 yr by Sensei

@imatfaal

Solved..the 120 degrees network connects n points in space with shortest line segments.

Proof?

 

 

And can anyone tell me a nice online app / freeware that would allow me to draw that accurately and easily?

No, but I used paint which is free and lets you do it really badly.

@John Cuthber
Sorry no proof..and it can not work if there's no need.
If we intersect each 3 points with 3 lines forming "Y" @ 120 deg w/ each other,
that will be the minimum connecting length. Unless the 3 points are making > 120 angle
because the lines connecting them is already the minimum.

 

I tried this with the cubic room where a spider web connects the 8 corners .It works
With 13 segments network.All angles of the adjacent web segment are 120 degrees.
e.i. for the minimum length.

 

AutoCAD is the best for this. No math needed. Just click lines to read measurements.