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The original problem was to consider how many ways to make a wiring diagram out of $n$ resistors. When I thought about this I realized that if you can only connect in series and shunt. - Then this is the same as dividing an area with $n-1$ horizontal and vertical lines. When each line only divides one of the current area sections into two smaller ones.

This is also the same as the number of ways to make a set of $n$ (and only $n$) rectangles into a bigger rectangle. If the rectangles can be drawn by dividing the big rectangle, line by line, into the set of rectangles without lose endpoints of the line. - Can someone come to think of "a expression of $n$" which equals this amount, independent of the order of the rectangles or position?

(It is only the relations between the area sections that matters and not left or right, up or down. However dividing an area with a horizontal line is not the same as dividing it with a vertical line.)

Niklas Bäckström
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    Not all subdivisions of a rectangle are of this kind. For instance in a $3\times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos. – Marc van Leeuwen Apr 24 '13 at 09:13
  • Thanks, I notised that recently and will edit the title again... – Niklas Bäckström Apr 24 '13 at 14:48
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    A useful approach to problems like this is to calculate small values by hand and search http://oeis.org/ That often has nice answers and references. The symmetries make this one difficult, even for n=4, though. – Ross Millikan Apr 24 '13 at 15:30
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    I believe this is the serie: A000084 but I'm not sure, can you check it? – Niklas Bäckström Apr 24 '13 at 22:12
  • Here is an illustration of the problem where $f(n)$ is the function we seek: Link – Niklas Bäckström Apr 25 '13 at 19:48
  • I am not sure what symmetries are allowed in A000084. My hand count for $n=2$ gives $6$, not $4$ but I think that is taking two pairs of my two plus 1 to be the same. – Ross Millikan Feb 06 '14 at 15:30
  • @RossMillikan: It seems that in A000084 both operations of serial and parallel composition are fully associative and commutative. The latter is not very intuitively obvious in the serial case, as it makes =--, -=- and --= all three count as the same network (for the parallel case this may seem more natural). In any case I think that for the current question they should be considered associative but not commutative (the three configurations above all count distinct). Then for $n=3$ I find $6$, and for $n=4$ I find $22$. – Marc van Leeuwen Jan 25 '15 at 10:26
  • See https://math.stackexchange.com/questions/1116/number-of-ways-to-partition-a-rectangle-into-n-sub-rectangles – Cheerful Parsnip Jan 27 '18 at 22:06

2 Answers2

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Since you have the big rectangle, and you want to cut it into ‘n’ number of rectangles, the number of cuts will be simply n-1. This is similar to adding a cosecutive series of numbers

Alan duan
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For the original problem, which I understand as dividing a rectangle into $n$ rectangles where each straight cut splits one piece in two, the answer is $n - 1$, as is easy to prove by (strong) induction.

vonbrand
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  • No, he wanted the number of ways to divide a rectangle, not the number of segments used. So with two segments, you can have three neighboring rectangles in two ways (vertical or horizontal) or one by two in four ways, making 6. This is not in the proposed A000084. – Ross Millikan Feb 06 '14 at 15:28