Arrange n tiles in a rectangle pattern
Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from 1 to 9.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at n smaller than 100 but would rather not have to hardcode everything.
geometry square-numbers rectangles
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from 1 to 9.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at n smaller than 100 but would rather not have to hardcode everything.
geometry square-numbers rectangles
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from 1 to 9.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at n smaller than 100 but would rather not have to hardcode everything.
geometry square-numbers rectangles
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.
For example, here are the results I expect from 1 to 9.
- a 1 by 1 square, fairly simple.
- a 1 by 2 rectangle.
- a 1 by 3 rectangle.
- a 2 by 2 square.
- this is impossible because a 1 by 5 rectangle is thinner than the previous shape.
- a 2 by 3 rectangle.
- this is impossible.
- a 2 by 4 rectangle.
- a 3 by 3 square.
I am only looking at n smaller than 100 but would rather not have to hardcode everything.
geometry square-numbers rectangles
geometry square-numbers rectangles
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Alex Laquerre
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Alex Laquerre
61
asked 3 hours ago
Alex Laquerre
61
61
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 hours ago
John Omielan
72418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
|
show 7 more comments
Your Answer
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Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 hours ago
John Omielan
72418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
|
show 7 more comments
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 hours ago
John Omielan
72418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
|
show 7 more comments
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 hours ago
John Omielan
72418
Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.
If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.
To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.
Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.
answered 2 hours ago
John Omielan
72418
edited 41 mins ago
answered 2 hours ago
John Omielan
72418
answered 2 hours ago
John Omielan
72418
answered 2 hours ago
John Omielan
72418
72418
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
|
show 7 more comments
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
– Alex Laquerre
2 hours ago
2
2
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
– John Omielan
2 hours ago
1
1
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
– Gerry Myerson
2 hours ago
1
1
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
– Ross Millikan
1 hour ago
1
1
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
You are welcome to edit it into your answer, which will make it more accessible to readers.
– Ross Millikan
1 hour ago
|
show 7 more comments
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
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