这种自然连接操作是否正确使用？ （关系代数）
Is this natural join operation used correctly? (Relational Algebra)
教授给我的任务如下：R-E Modell
- Assume the companies may be located in several cities. Find all companies located in every city in which “Small Bank Corporation” is
located.
现在教授的解决方案如下：
s ← Π city (σ company_name=’Small Bank Corporation’ (company))
temp1 ← Π comp_id, company_name (company)
temp2 ← Π comp_id, company_name ((temp1 × s) − company)
result ← Π company_name (temp1 − temp2)
我自己找到了一个完全不同的解决方案，它有一个自然的连接操作，看起来简单得多：
我尝试做的是使用自然联合操作，它定义为如下关系 r 和 s 在它们的共同属性上连接。因此，我尝试通过使用 company_name“Small Bank Cooperation”对所有公司的选择进行投影来获取所有城市名称。之后我加入了 table 和公司 table 的城市名称，这样我就得到了所有包含城市名称的公司条目。
company ⋈ Π city (σ company_name=”Small Bank Cooperation” (company)))
我现在的问题是我的解决方案是否也有效，因为它看起来有点微不足道？
你的不一样。
My answer here says how to query relationally. It uses a version of the relational algebra where headings are sets of attribute names. My answer here summarizes it:
Every query expression has an associated (characteristic)
predicate--statement template parameterized by attributes. The tuples
that make the predicate into a true proposition--statement--are in
the relation.
We are given the predicates for expressions that are relation names.
Let query expression E have predicate e. Then:
R ⨝ S has predicate r and sR ∪ S has predicate r or sR - S has predicate r and not sσ p (R) has predicate r and pπ A (R) has predicate exists non-A attributes of R [r]
When we want the tuples satisfying a certain predicate we find a way
to express that predicate in terms of relation operator
transformations of given relation predicates. The corresponding query
returns/calculates the tuples.
您的解决方案
company ⋈ Π city (σ company_name=”Small Bank Corporation” (company)))
是
所在的行
company company_id named company_name is in city
AND FOR SOME company_id & company_name [
company company_id named company_name is in city
AND company_name=”Small Bank Corporation”]
即
company company_id named company_name is in city
AND FOR SOME company_id [
company company_id named ”Small Bank Corporation” is in city]
即
company company_id named company_name is in city
AND some company named ”Small Bank Corporation” is in city
您正在 returning 行的列数多于 company_name。但是你的公司不是要求的公司。
将您的行投影到 company_name 得到行
some company named company_name is in some city
AND some company named ”Small Bank Corporation” is in that city
After that I joined the table with the city names with the company
table, so that I get all company entrys which have the city names in
it.
不清楚你得到了什么。但是，您所在行中的公司至少位于 SBC 城市之一。该请求适用于所有 SBC 城市：
companies located in every city in which “Small Bank Corporation” is located
我提供的链接告诉您如何编写查询，以及如何在查询结果规范和关系代数表达式之间转换return结果。
当您看到查询匹配某些其他行的“每个”或“所有”行时，您可以预期查询的那部分涉及 relational-division 或一些相关的习语。确切的代数取决于 - 经常 poorly/ambiguously 表达的 - 要求的意图。例如，“companys located in every city in which”是否应该是没有公司（部门）或所有公司（相关成语），当没有这样的城市时。 （你的作业的正常数学解释是后者。）例如，他们是否希望在所有这些城市或至少在所有这些城市都有公司。
（这有助于避免在“find”和“return”之后使用“all”和“every”，无论如何都是多余的。）
How to find all pizzerias that serve every pizza eaten by people over 30?
教授给我的任务如下：R-E Modell
- Assume the companies may be located in several cities. Find all companies located in every city in which “Small Bank Corporation” is located.
现在教授的解决方案如下：
s ← Π city (σ company_name=’Small Bank Corporation’ (company)) temp1 ← Π comp_id, company_name (company) temp2 ← Π comp_id, company_name ((temp1 × s) − company) result ← Π company_name (temp1 − temp2)
我自己找到了一个完全不同的解决方案，它有一个自然的连接操作，看起来简单得多：
我尝试做的是使用自然联合操作，它定义为如下关系 r 和 s 在它们的共同属性上连接。因此，我尝试通过使用 company_name“Small Bank Cooperation”对所有公司的选择进行投影来获取所有城市名称。之后我加入了 table 和公司 table 的城市名称，这样我就得到了所有包含城市名称的公司条目。
company ⋈ Π city (σ company_name=”Small Bank Cooperation” (company)))
我现在的问题是我的解决方案是否也有效，因为它看起来有点微不足道？
你的不一样。
My answer here says how to query relationally. It uses a version of the relational algebra where headings are sets of attribute names. My answer here summarizes it:
Every query expression has an associated (characteristic) predicate--statement template parameterized by attributes. The tuples that make the predicate into a true proposition--statement--are in the relation.
We are given the predicates for expressions that are relation names.
Let query expression E have predicate e. Then:
R ⨝ Shas predicater and sR ∪ Shas predicater or sR - Shas predicater and not sσ p (R)has predicater and pπ A (R)has predicateexistsnon-A attributes of R[r]When we want the tuples satisfying a certain predicate we find a way to express that predicate in terms of relation operator transformations of given relation predicates. The corresponding query returns/calculates the tuples.
您的解决方案
company ⋈ Π city (σ company_name=”Small Bank Corporation” (company)))
是
所在的行 company company_id named company_name is in city
AND FOR SOME company_id & company_name [
company company_id named company_name is in city
AND company_name=”Small Bank Corporation”]
即
company company_id named company_name is in city
AND FOR SOME company_id [
company company_id named ”Small Bank Corporation” is in city]
即
company company_id named company_name is in city
AND some company named ”Small Bank Corporation” is in city
您正在 returning 行的列数多于 company_name。但是你的公司不是要求的公司。
将您的行投影到 company_name 得到行
some company named company_name is in some city
AND some company named ”Small Bank Corporation” is in that city
After that I joined the table with the city names with the company table, so that I get all company entrys which have the city names in it.
不清楚你得到了什么。但是，您所在行中的公司至少位于 SBC 城市之一。该请求适用于所有 SBC 城市：
companies located in every city in which “Small Bank Corporation” is located
我提供的链接告诉您如何编写查询，以及如何在查询结果规范和关系代数表达式之间转换return结果。
当您看到查询匹配某些其他行的“每个”或“所有”行时，您可以预期查询的那部分涉及 relational-division 或一些相关的习语。确切的代数取决于 - 经常 poorly/ambiguously 表达的 - 要求的意图。例如，“companys located in every city in which”是否应该是没有公司（部门）或所有公司（相关成语），当没有这样的城市时。 （你的作业的正常数学解释是后者。）例如，他们是否希望在所有这些城市或至少在所有这些城市都有公司。
（这有助于避免在“find”和“return”之后使用“all”和“every”，无论如何都是多余的。）
How to find all pizzerias that serve every pizza eaten by people over 30?