single line graph puzzle 1

[Olaf Doschke]

How can you connect these nine points with a line drawn in a single stroke with only four segments!

.       .       .



.       .       .



.       .       .

Here's an exemple on how it does not work, because the left middle point is not on the graph.

.-------.-------.
 \_             !
   \_           !
     \_         !
.      \.       .
         \_     !
           \_   !
             \_ !
.-------.------\.

Bye, Olaf.

 

[Skie]

6x6 Solution

Spoiler

[tt].! A B C D E F @
.
. G H I J K L
.
. M N O P Q R
.
. S T U V W X
.
. Y Z 1 2 3 4
.
. 5 6 7 8 9 0 #
.
. $[/tt]
About the same as the 5x5, but I'm going to run through it backwards to make it easier to find where # and ! should be.
X to 8 [X,3,8]
8 to # [9,0]
# to ! [4,U] It's probably easiest to figure out where # and ! are if you draw a line through 4 and U.
! to @ [A,B,C,D,E,F]
@ to $ [R,W,2,7]
$ to G [5,Y,S,M,G]
G to L [H,I,J,K,L]
L to 6 [Q,V,1,6]
6 to N [Z,T,N]
N to P [O,P]
It looks like the solution builds off the previous with the last two lines catching a couple of the extra dots. I have a solution that's about the same, but would add two lines to the begining of this solution. It's just figuring out which two dots the long diagonal (# to !) hits. If we make it a battleship grid it'd run through D3 & E6.

If you read the spoiler, I also solved the 7x7 (I'll post the complete solution if someone asks). I would imagine you can keep building up from this solution to work any n x n grid.

 

[Zathras]

Skie: Congratulations on finding a fourth way to do the 4x4. I had not seen that particular one before. (This is exactly why I posted the puzzle -- I want to find more solutions.)

However, I don't think your 5x5 is correct. Your line from 1 to 4 which passes thru both T and M also passes thru F which makes a second visit to that point.

The 6x6 is beautiful. Nice job!

By all means, please post the 7x7

 

[Skie]

You're right, my 5x5 would hit "F" twice. It doesn't with my not evenly spaced dots with not-quite straights lines.
5x5 Fixed

Spoiler

In order to avoid the hit at "F" twice. Erase the line starting at 1 until it's just over the original starting point. Then move the starting point of L just above it so those 2 lines cross

This should cause it to start at A and end at O

7x7 Solution

Spoiler

[tt]0 1 2 3 4 5 6 7 8
A . . . . . . .
B . . . . . . .
C . . . . . . .
D . . . . . . .
E . . . . . . .
F . . . . . . .
G . . . . . . .
H[/tt]

1: C4 to C2
2: C2 to F2
3: F2 to B6
4: B6 to B1
5: B1 to H1
6: H1 to A8
7: A8 to A0
8: A0 to G8 (Runs through D3 and E6)
9: G8 to G3
10: G3 to D7
11: D7 to F7
12: F7 to F5

 

[Zathras]

Excellent! I think your 5x5 is different from either of the two that I found. That makes at least three distinct ways to do a 5x5, and four distinct ways to do a 4x4.

Nice job on the 7x7. One typo: Line 10 s/b C7, not D7.

I still don't see any general solution, since all of the ones found so far visit virtual dots outside of the current grid, preventing a simple turn and go algorithm.

 

[monksnake]

Skie, you deserve a star.

<.

 

[Skie]

I don't know if there's an algorithm, but the solutions have a common pattern. I quickly worked up an two 8x8 solutions, one that's like 7x7 and one that's like 5x5 (fixed) solution. The only hard part is figuring out where the triangle-joining diagonal is and making sure it doesn't double-dot.

What I wonder is given the multiple solutions for larger girds if it's possible to use less lines.

 

[Skie]

I've come up with two more 6x6 solutions. I haven't been able to find a way to get it down to 9 lines though. It seems I always end up with two dots that require a diagonal.