Math solution for number of possible window monitor positions using 1, 2, or 3 divisions

Posted on 2017-01-29
Last Modified: 2017-01-30

What math equation describes the number of possible window positions when dividing a computer screen (rectangle) using one, two, & three sections in one or both (horizontal & vertical) axes?

For example, one division obviously has 1 possibility:
Fig. 1And it appears that with one or two divisions in each direction there are 4 + 2 + 2 = 8 possibilities:
Fig. 2Fig. 3But what about with one, two, or three divisions? I initially counted 9 + 3 +3 + 4 + 12 = 31
Fig. 4Fig. 5Fig. 6Fig. 7but then I realized I had forgotten the 2-3 combinations so that's 12 more (ie 31 + 12 = 43).
Fig. 8I suspect there may be others because 43 seems like a bit of an odd total. That's why it needs math!

And what about having up to four divisions? I didn't even try to start doing that graphically.

Question by:WeThotUWasAToad
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LVL 34

Accepted Solution

sarabande earned 350 total points
ID: 41985381
the problem is not well defined yet. why should rectangles (0, 0, 8, 12) and (4, 0, 8, 12) be valid solutions (3rd and 5th yellow) but (0, 0, 6, 12) is not?

if you say all windows must have same size and a solution covers the whole screen, you nicely have

-           0 split lines 1 window
-           1 split line   3 windows
- up to 2 split lines 5 windows
- up to 3 split lines 7 windows

LVL 33

Assisted Solution

by:Rob Henson
Rob Henson earned 150 total points
ID: 41985573
Is it not a simple mathematical function:

(Vertical Dividers +1) x (Horizontal Dividers +1)

Vert      Horiz
0+1  x   0+1 = 1
1+1  x   0+1 = 2
2+1  x   1+1 = 6

Rob H

Author Comment

ID: 41986195
Thanks for the responses.

the problem is not well defined yet. why should rectangles (0, 0, 8, 12) and (4, 0, 8, 12) be valid solutions (3rd and 5th yellow) but (0, 0, 6, 12) is not?
Apologies for the ambiguity in my OP. I did state that the possibilities for a given number of divisions includes those for a lower number but I neglected to account for the lower numbers and create a total . In other words:

1 division = 1 possibility
2 divisions = 1 (from 1 division) + 8 (as stated in OP) = 9 total possibilities
3 divisions = 1 (from 1 division) + 8 (from 2 divisions) + 43* (as stated in OP) = 52 total possibilities  

To avoid the same head scratching among other experts, I think the best step is to award points and close this thread and then re-post the question in a new thread.

Thanks again for the input.

Author Closing Comment

ID: 41986199
See new thread.

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