rumus Apel Vs Jeruk


T= Type / ClassTe
\mathcal{T} = \{ T | \exists(s,w3type T) \in D\} Te

I = instance Te
I(T,D)= set of instance with type T in dataset DTe
I(T,D) = \{ s | \exists(s,w3type T) \in D\}Te

P(T) = set of distinct property in Type $T$Te
P(T) = \{ p | s \in I(T,D) and \exists(s,p,o) \in D\}Te

OC(p,I(T,D) = occurrence property $p$ in $I(T,D$Te
OC(p,I(T,D) = | \{ s | s \in I(T,D) and \exists(s,p,o) \in D\} |Te

Coverage
CV(T,D) = \frac{\sum_{p \in P(T) OC(p,I(T,D)}}{|P(T)| \times |I(T,D)| }Te

Weight
WT(CV(T,D)) = \frac{|P(T)| \times |I(T,D)|}{\sum_{T' \in \mathcal{T}} |P(T')|+|I(T',D)}Te

Coherence
CH(\mathcal{T},D) = \sum_{T \in \mathcal{T}} WT(CV(T,D)) \times CV(T,D)Te

D = real datasetTe
D' = new dataset after removing coinTe
|D| < |D'| and $D \subset D’$Te

what is coin ?
removing a set of triples with the same subject and propertyTe

\mathcal{T}(s)= \{Ts^1,...,Ts^n\}Te

A1 \Longrightarrow We do not completely remove property $p$ from any of the types {$Ts^1$,…,$Ts^n$ }. That is, after the removal, for each type there will exist instances that have property p.Te

A2 \Longrightarrow We do not completely remove instance $s$ from the dataset. This can be very easily enforced by keeping the triples {$s$, rdf:type, $Ts^i$ } in the dataset.Te

Weight is the same after removing the triples
but the coverage is changed : Te
CV(T,D)' = \frac{\sum_{q \in P(T)-p} OC(q,I(T,D)) + OC(p,I(T,D) -1) }{|P(T)| \times |I(T,D)|}Te

CH(\mathcal{T}, D')= \mathcal{T}Te
|D'| = \sigmaTe

coin(\mathcal{T}(s),p)=CH(\mathcal{T},D) - CH(\mathcal{T},D)'Te

|coin(S,p)|= number of subjects that are instance of all the types in $S$ and have at least one triple with property $p$Te
|coin(S,p)|=|\{s \in \bigcap_{T \in S} I(T,D)| \exists(s,p,v) \in D \}|Te

C1 \Longrightarrow the amount by which we decrease coherence (by removing coins) should be less than or equal than the amount we need to remove to get from $CH(\tau, D)$ (the coherence of the original dataset) to \lambda (the desired coherence).Te

X(S,p) the integer programming variable representing the number of coins to remove for each type of coin.Te
\tau = number of types
\pi = the number of properties in the dataset
worst case the number of variables for D can be 2^\tau \piTe

sets x and y, x \subseteq y if all elements of x are also elements of y
Te
C1 \Longrightarrow \sum_{S \subseteq \mathcal{T},p} coin (S,p) \times X(S,p) \leq CH(\mathcal{T},D) - \lambda Te

M \Longrightarrow the amount by which we decrease coherence should be maximized.

M \Longrightarrow MAXIMIZE \sum_{S \subseteq D,p} coin (S,p) \times X(S,p) Te

C2 \Longrightarrow \forall S \subseteq ,p 0 \leq X(S,p) \leq |coin(S,p)|-1 Te

ct(S,p) =average number of triples per coin typeTe
C3 \Longrightarrow (1-\rho) \times (|D| - \sigma ) \leq \sum_{S \subseteq \mathcal{T},p } X(S,p) \times ct(S,p) Te

$latexC4 \Longrightarrow \sum_{S \subseteq \mathcal{T},p } X(S,p) \times ct(S,p) \leq (1+ \rho ) \times (|D| – \sigma )$

sumber : https://researcher.ibm.com/researcher/files/us-sduan/sigmod2011_RDF_benchmark_duan.pdf

  1. wah, ilmu baru neh bagi saya
    bener-bener baru…

  2. Waw..hanya orang berilmu yang bisa membaca kode-kode itu.
    Saya tidak bisa menulis rumus apel vs jeruk, tapi kalo mangga dan lengkeng dikawin sama durian, bisa….

  3. pusing dah …kalu dah ngomongin ttg rumus ..
    hehe

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