question in computer science...

$2^{nd}$ Normal Form $(2$NF$)$ and $3^{rd}$ Normal Form $(3$NF$)$

In the Designer Task Report Example below, the current Design is violating the $2^{nd}$ Normal Form $(2$NF$)$ and $3^{rd}$ Normal Form $(3$NF$).$

A simpler interpretation of the $2$ NF is that every M:M Relationship Table and every Recursive Relationship Table should only contain Relationship columns$/$attributes$,$ NOT the Entity columns$/$attributes$.$

Recall that a table that violates $2$ NF $,$ by definition, also violates $3$ NF $.$ However, a table that violates $3$ NF does not necessarily violate $2$ NF $.$

A simpler way to ensure $2$ NF and $3$ NF :

$(1)$ Examine each Entity Table: ensure that each non$-$PK column is either an attribute of that Entity or a suitable FK column.

$(2)$ Examine each Relationship Table: ensure that each non$-$PK column is an attribute of that Relationship and not of part of the PK$.$ A non$-$PK column in a Relationship Table should not be an Entity attribute.

According to the client's requirements, there are only $4$ entity classes: Designer, Project, TaskType, and Project. The attributes are as follows.

$(1)$ Designer: Name, Rank, JoinDate, Address, City, State, and Zip.

$(2)$ TaskType: Name, Importance.

$(3)$ Customer: Name, Priority, Address, City, State, and Zip.

$(4)$ Project: Name, Type, Difficulty.

Also, we have designated following Relationships:

$(1)$ Project $(1)$ and Designer $($M$).$

$(2)$ Designer $($M$)$ and TaskType $($M$),$ table TaskProficiency.

$(3)$ Mentor $($Designer$,$ M$)$ and Mentee $($Designer$,$ M$),$ table Mentorship.

$(4)$ Customer $(1)$ and Project $($M$).$

$\backslash$table$[[$Designer$],[$CName$,$PK$/$FK$,$Ref$],[$ID$,$PK$,],[$Name$,,],[$ProjectID$,$FK$,$Project.ID$],[$Rank$,,]]$

$\backslash$table$[[$TaskType$],[$CName$,$PK$/$FK$,$Ref$],[$ID$,$PK$,],[$Name$,,],[$Importance$,,]]$