Connecting Colored Boxes

Logic Level 3

In the above diagram, you must connect every colored box with every other colored box that has the same color. You do this by drawing a (not necessarily straight) line connecting the two boxes.

What is the minimum number of line intersections over all possible ways to connect the boxes?

Details and Assumptions
The connecting lines must not go out of the square board.
There cannot be three or more connecting lines concurrent at one point.
Here is one example of a way to connect the boxes with 12 intersections.

The answer is 0.

1 solution

Daniel Liu
Jul 2, 2014

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A solution with $\boxed{0}$ intersections. (Boy, did that take a long time to figure out and draw!)

Another possible solution (thanks @Jeffery Li):

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Daniel Liu - 6 years, 11 months ago

Interesting Figure :)

Dinesh Chavan - 6 years, 11 months ago

I like Jeffrey's figure. It used a similar approach of starting out with 4 monochromatic triangles, and then (topologically) moving the vertices without having lines intersect. Of course, this assumed that 0 was possible, which needn't always be the case.

Calvin Lin Staff - 6 years, 11 months ago

I solved it by first grouping the colors all together, connecting the boxes, then sliding the boxes to their correct positions, pretending the lines connecting the boxes are strings. If a box slides into a string, it just bends the string around it.

Daniel Liu - 6 years, 11 months ago

Same technique for me. If box A can connect B and B can connect C, then A can also connect to C. Easy pie for that XD

And you don't restrict that it can only be drawn in the unit blocks. So you can make a really thin gap between them. You can just figure the solution that can connect A to B and B to C without A to C. ^_^

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

did in exactly similar manner .

Shriram Lokhande - 6 years, 11 months ago

I got the answer same as Jeffery Li's. I used a clockwise spiral instead of an anticlockwise one. How can I post a picture file to show the solution?

Chew-Seong Cheong - 6 years, 11 months ago

I used the following logic - Connecting any three boxes of same colour would give us a some sort of region or area bounded by the lines joining the boxes , which depends upon our figure , in which way we connect them. Now, two lines will neccesarily interesect if the area bounded by boxes of a certain colour has or includes one or more than one boxes of other colours , in that case the line segments must intersect , so as the boxes in the region are not left isolated. $OR$ Let's take the example of red boxes . Now suppose we somehow connect the boxes , if the region bounded by the lines contains let us say a yellow and blue box, the yellow and blue boxes cannot finish their arrangements unless they intersect the red lines . The region outside the red box region will have two blue and two yellow boxes and to connect the three yellow and three blue boxes the lines must " get into " the red region for which they have to intersect the red lines

Now all I had to do is to find out the minimum number of boxes that could be accumulated in a box region of any colour. As I figured there is no property or rule that neccessiates that there should be atleast a box of defferent colour in a box region of a particular colour , this means it is a possible to have a arraangement in which every box region of all colours have no boxes , so this means they have zero intersections . So , the answer I found , without actually figurinb the arrangement was zero.

And , dont mind my english.

Utkarsh Dwivedi - 6 years, 1 month ago

Connecting Colored Boxes

The answer is 0.

1 solution

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