What is the number of ways to divide a rectangle into $n$ smaller rectangles line by line?












9














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.)






combinatorics







share|cite|improve this question




















  • 2




    Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
    – Marc van Leeuwen
    Apr 24 '13 at 9:13










  • Thanks, I notised that recently and will edit the title again...
    – Niklas Bäckström
    Apr 24 '13 at 14:48






  • 1




    A useful approach to problems like this is to calculate small values by hand and search 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






  • 1




    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


















9














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.)






combinatorics







share|cite|improve this question




















  • 2




    Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
    – Marc van Leeuwen
    Apr 24 '13 at 9:13










  • Thanks, I notised that recently and will edit the title again...
    – Niklas Bäckström
    Apr 24 '13 at 14:48






  • 1




    A useful approach to problems like this is to calculate small values by hand and search 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






  • 1




    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
















9












9








9


4





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.)






combinatorics







share|cite|improve this question















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.)






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 24 '13 at 15:11

























asked Apr 24 '13 at 8:39









Niklas Bäckström

12812




12812








  • 2




    Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
    – Marc van Leeuwen
    Apr 24 '13 at 9:13










  • Thanks, I notised that recently and will edit the title again...
    – Niklas Bäckström
    Apr 24 '13 at 14:48






  • 1




    A useful approach to problems like this is to calculate small values by hand and search 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






  • 1




    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
















  • 2




    Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
    – Marc van Leeuwen
    Apr 24 '13 at 9:13










  • Thanks, I notised that recently and will edit the title again...
    – Niklas Bäckström
    Apr 24 '13 at 14:48






  • 1




    A useful approach to problems like this is to calculate small values by hand and search 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






  • 1




    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










2




2




Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
– Marc van Leeuwen
Apr 24 '13 at 9:13




Not all subdivisions of a rectangle are of this kind. For instance in a $3times3$ square you can single out the central square and partition the remaining $8$ squares into $4$ dominos.
– Marc van Leeuwen
Apr 24 '13 at 9:13












Thanks, I notised that recently and will edit the title again...
– Niklas Bäckström
Apr 24 '13 at 14:48




Thanks, I notised that recently and will edit the title again...
– Niklas Bäckström
Apr 24 '13 at 14:48




1




1




A useful approach to problems like this is to calculate small values by hand and search 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




A useful approach to problems like this is to calculate small values by hand and search 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




1




1




I believe this is the serie: A000084 but I'm not sure, can you check it?
– Niklas Bäckström
Apr 24 '13 at 22:12




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






Here is an illustration of the problem where $f(n)$ is the function we seek: Link
– Niklas Bäckström
Apr 25 '13 at 19:48












1 Answer
1






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oldest

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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.






share|cite|improve this answer





















  • 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 6 '14 at 15:28











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active

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active

oldest

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0














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.






share|cite|improve this answer





















  • 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 6 '14 at 15:28
















0














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.






share|cite|improve this answer





















  • 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 6 '14 at 15:28














0












0








0






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 6 '14 at 15:19









vonbrand

19.8k63058




19.8k63058












  • 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 6 '14 at 15:28


















  • 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 6 '14 at 15:28
















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 6 '14 at 15:28




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 6 '14 at 15:28


















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