The whole key s,q is not needed to determine t or u, so there are partial key dependences. Here, s->t, and q->u, so we can take t and u out of R. We must create R2 and R3 to match these two determinants, making the LHS of them keys and the RHS non-keys. However, in R2 and R3 both LHS and RHS attributes have been made keys, and this is incorrect.
Looking at R1, it states that s->q. There is no determinate which states this, so this option MUST be wrong.
The whole key s,q is not needed to determine t or u, so there are partial key dependences. Here, s->t, and q->u, so we can take t and u out of R.
This is actually true, as s,q->t and s,q->u also means s,q->t,u, which is as stated in R. However, it still is not normalised. You need to take into account the s->t and q->u determinants.
This is actually a valid relation, as it is exactly what is stated in R. However, it still is not normalised. You need to take into account the s->t and q->u determinants.
Consider the following functional dependencies
| a,b | => | c,d | e,g,h | => | f,j | |
| a,c | => | b,d | p,q | => | r,s | |
| e,f,g | => | h,i | s | => | t | |
| f,g | => | j | q | => | u | |
| g,h | => | i |
The schema R1(s,q) R2(s,t) R3(q,u)The schema R1(s,q) R2(q,t) R3(t,u)The schema R1(s,q) R2(s,t) R3(q,u)The schema R1(s,q,t) R2(s,q,u)The schema R(s,q,t,u)
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