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Bijection Explanation?
Clash Royale CLAN TAG#URR8PPP
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Can anyone help explain what bijection is in the context of this problem, and how exactly it's used to derive the particular solution
From the definition provided it seems as though a bijection is a translation provided to a set of points, but I'm not sure that I understand how that is applied, especially in the context of the problem. Thanks!
probability combinatorics statistics
asked Jul 28 at 1:00
John
476
add a comment |Â
up vote
2
down vote
favorite
Can anyone help explain what bijection is in the context of this problem, and how exactly it's used to derive the particular solution
From the definition provided it seems as though a bijection is a translation provided to a set of points, but I'm not sure that I understand how that is applied, especially in the context of the problem. Thanks!
probability combinatorics statistics
asked Jul 28 at 1:00
John
476
1
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Can anyone help explain what bijection is in the context of this problem, and how exactly it's used to derive the particular solution
From the definition provided it seems as though a bijection is a translation provided to a set of points, but I'm not sure that I understand how that is applied, especially in the context of the problem. Thanks!
probability combinatorics statistics
asked Jul 28 at 1:00
John
476
Can anyone help explain what bijection is in the context of this problem, and how exactly it's used to derive the particular solution
From the definition provided it seems as though a bijection is a translation provided to a set of points, but I'm not sure that I understand how that is applied, especially in the context of the problem. Thanks!
probability combinatorics statistics
asked Jul 28 at 1:00
John
476
asked Jul 28 at 1:00
John
476
asked Jul 28 at 1:00
John
476
asked Jul 28 at 1:00
John
476
476
1
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08
add a comment |Â
1
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08
1
1
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let's look at a smaller example. Suppose there are $7$ spaces in the parking lot, and $4$ cars arrive.
On the left, we list all of the ways that four cars can park so that there are no two adjacent empty spaces. On the right, we list the results of following the instructions in the solution: between each pair of consecutive empty spaces, delete one of the intervening cars.
_ X _ X _ X X _ _ _ X X
_ X _ X X _ X _ _ X _ X
_ X _ X X X _ _ _ X X _
_ X X _ X _ X _ X _ _ X
_ X X _ X X _ _ X _ X _
_ X X X _ X _ _ X X _ _
X _ X _ X _ X X _ _ _ X
X _ X _ X X _ X _ _ X _
X _ X X _ X _ X _ X _ _
X X _ X _ X _ X X _ _ _
On the left, we have a a complicated object, which is unclear how to count. On the right, we see something simpler; every possible way to park $2$ cars in a row of $5$ spots is listed exactly once. The number of such arrangements on the right is clearly $binom52$. Since the left and right columns have the same number of arrangements, this also counts the number of arrangements on the left.
This is the purpose of clever bijections. You make a perfect matching between a mysterious set and a simple one, and use what you know about the simple one to explain the mysterious one.
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let's look at a smaller example. Suppose there are $7$ spaces in the parking lot, and $4$ cars arrive.
On the left, we list all of the ways that four cars can park so that there are no two adjacent empty spaces. On the right, we list the results of following the instructions in the solution: between each pair of consecutive empty spaces, delete one of the intervening cars.
_ X _ X _ X X _ _ _ X X
_ X _ X X _ X _ _ X _ X
_ X _ X X X _ _ _ X X _
_ X X _ X _ X _ X _ _ X
_ X X _ X X _ _ X _ X _
_ X X X _ X _ _ X X _ _
X _ X _ X _ X X _ _ _ X
X _ X _ X X _ X _ _ X _
X _ X X _ X _ X _ X _ _
X X _ X _ X _ X X _ _ _
On the left, we have a a complicated object, which is unclear how to count. On the right, we see something simpler; every possible way to park $2$ cars in a row of $5$ spots is listed exactly once. The number of such arrangements on the right is clearly $binom52$. Since the left and right columns have the same number of arrangements, this also counts the number of arrangements on the left.
This is the purpose of clever bijections. You make a perfect matching between a mysterious set and a simple one, and use what you know about the simple one to explain the mysterious one.
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
add a comment |Â
up vote
1
down vote
accepted
Let's look at a smaller example. Suppose there are $7$ spaces in the parking lot, and $4$ cars arrive.
On the left, we list all of the ways that four cars can park so that there are no two adjacent empty spaces. On the right, we list the results of following the instructions in the solution: between each pair of consecutive empty spaces, delete one of the intervening cars.
_ X _ X _ X X _ _ _ X X
_ X _ X X _ X _ _ X _ X
_ X _ X X X _ _ _ X X _
_ X X _ X _ X _ X _ _ X
_ X X _ X X _ _ X _ X _
_ X X X _ X _ _ X X _ _
X _ X _ X _ X X _ _ _ X
X _ X _ X X _ X _ _ X _
X _ X X _ X _ X _ X _ _
X X _ X _ X _ X X _ _ _
On the left, we have a a complicated object, which is unclear how to count. On the right, we see something simpler; every possible way to park $2$ cars in a row of $5$ spots is listed exactly once. The number of such arrangements on the right is clearly $binom52$. Since the left and right columns have the same number of arrangements, this also counts the number of arrangements on the left.
This is the purpose of clever bijections. You make a perfect matching between a mysterious set and a simple one, and use what you know about the simple one to explain the mysterious one.
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let's look at a smaller example. Suppose there are $7$ spaces in the parking lot, and $4$ cars arrive.
On the left, we list all of the ways that four cars can park so that there are no two adjacent empty spaces. On the right, we list the results of following the instructions in the solution: between each pair of consecutive empty spaces, delete one of the intervening cars.
_ X _ X _ X X _ _ _ X X
_ X _ X X _ X _ _ X _ X
_ X _ X X X _ _ _ X X _
_ X X _ X _ X _ X _ _ X
_ X X _ X X _ _ X _ X _
_ X X X _ X _ _ X X _ _
X _ X _ X _ X X _ _ _ X
X _ X _ X X _ X _ _ X _
X _ X X _ X _ X _ X _ _
X X _ X _ X _ X X _ _ _
On the left, we have a a complicated object, which is unclear how to count. On the right, we see something simpler; every possible way to park $2$ cars in a row of $5$ spots is listed exactly once. The number of such arrangements on the right is clearly $binom52$. Since the left and right columns have the same number of arrangements, this also counts the number of arrangements on the left.
This is the purpose of clever bijections. You make a perfect matching between a mysterious set and a simple one, and use what you know about the simple one to explain the mysterious one.
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
Let's look at a smaller example. Suppose there are $7$ spaces in the parking lot, and $4$ cars arrive.
On the left, we list all of the ways that four cars can park so that there are no two adjacent empty spaces. On the right, we list the results of following the instructions in the solution: between each pair of consecutive empty spaces, delete one of the intervening cars.
_ X _ X _ X X _ _ _ X X
_ X _ X X _ X _ _ X _ X
_ X _ X X X _ _ _ X X _
_ X X _ X _ X _ X _ _ X
_ X X _ X X _ _ X _ X _
_ X X X _ X _ _ X X _ _
X _ X _ X _ X X _ _ _ X
X _ X _ X X _ X _ _ X _
X _ X X _ X _ X _ X _ _
X X _ X _ X _ X X _ _ _
On the left, we have a a complicated object, which is unclear how to count. On the right, we see something simpler; every possible way to park $2$ cars in a row of $5$ spots is listed exactly once. The number of such arrangements on the right is clearly $binom52$. Since the left and right columns have the same number of arrangements, this also counts the number of arrangements on the left.
This is the purpose of clever bijections. You make a perfect matching between a mysterious set and a simple one, and use what you know about the simple one to explain the mysterious one.
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
answered Jul 28 at 2:10
Mike Earnest
14.9k11644
14.9k11644
add a comment |Â
add a comment |Â
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1
A bijection is a function from one set to another such that every element from the domain is mapped to exactly one element in the codomain and further every element in the codomain has exactly one element mapping to it from the domain. In the context of this problem and problems like it, finding such a convenient bijection is in essence "rewording the problem" in such a way that the answer to the reworded problem is more apparent and in such a way that the answers to both the original and the reworded problems are clearly the same.
– JMoravitz
Jul 28 at 1:05
"We can form a bijection between..." It is just a fancy way of saying that the unable cases can be counted by counting how many ways there are to pick four spots out of $13$ such that none of them touch.
– Cristhian Grundmann
Jul 28 at 1:08