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# 机器学习代考_Machine Learning代考_ENGG3300 Least General Generalization

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## 机器学习代考_Machine Learning代考_Least General Generalization

ILP takes a bottom-up strategy which directly uses the grounded facts of one or more positive samples as the bottom rules, and then gradually generalizes the rules to improve the coverage. The generalization operation could be deleting literals from the rule body or replacing constants with logic variables.

Taking the watermelon data set $5.0$ as an example. For ease of discussion, suppose $\operatorname{riper}(X, Y)$ only depends on the relations involving $(X, Y)$. Then, the bottom rules of positive samples riper $(1,10)$ and riper $(1,15)$ are, respectively
\begin{aligned} \operatorname{riper}(1,10) & \leftarrow \text { curlier_root }(1,10) \wedge \text { duller_sound }(1,10) \ & \wedge \text { hollower_umbilicus }(1,10) \wedge \text { harder_surface }(1,10) \ \operatorname{riper}(1,15) & \leftarrow \text { curlier_root }(1,15) \wedge \text { hollower_umbilicus }(1,15) \ & \wedge \text { harder_surface }(1,15) \end{aligned}
The above two rules have limited generalization abilities since they only describe two specific relational data samples. Hence, we wish to convert such specific rules to rules that are more general. To achieve this goal, Least General Generalization (LGG) (Plotkin 1970) is the most fundamental technique.
Given first-order formulas $\mathbf{r}_1$ and $\mathbf{r}_2, \mathrm{LGG}$ starts by finding all literals that have the same predicate, and then checks each pair of constants at the same position in the literals: if the two constants are the same, then they remain unchanged, denoted by $\operatorname{LGG}(t, t)=t$; otherwise, the constants are replaced by a new variable that also applies to all other places in the formulas. For example, if the two different constants are $s$ and $t$ and the new variable is $V$, then we have $\mathrm{LGG}(s, t)=V$, which means whenever $s$ or $t$ appears in the formula, we replace it by $V$. Let us take the two rules in the above as an example and LGG would start by comparing riper $(1,10)$ and riper $(1,15)$; since ” $10 ” \neq$ ” $15 “$, both constants are replaced with $Y$, and all pairs of ” $10 “$ and ” 15 ” in $\mathbf{r}_1$ and $\mathbf{r}_2$ are replaced with $Y$. Then, we have
\begin{aligned} \operatorname{riper}(1, Y) & \leftarrow \text { curlier_root }(1, Y) \wedge \text { duller_sound }(1,10) \ & \wedge \text { hollower_umbilicus }(1, Y) \wedge \text { harder_surface }(1, Y) \ \operatorname{riper}(1, Y) & \leftarrow \text { curlier_root }(1, Y) \wedge \text { hollower_umbilicus }(1, Y) \ & \wedge \text { harder_surface }(1, Y) \end{aligned}

## 机器学习代考_Machine Learning代考_Inverse Resolution

In logic, deduction and induction are fundamental approaches that humans use to understand the world. Roughly speaking, deduction infers specific phenomenons from general rules, whereas induction summarizes general rules from specific observations. The poofs of mathematical theorems are representative examples of applying deduction, whereas machine learning falls into induction category. In 1965, logician J. A. Robinson proposed that the deductive reasoning in first-order logic can be described by a simple rule, that is, the famous resolution principle (Robinson 1965). Two decades later, computer scientists S. Muggleton and W. Buntine proposed inverse resolution (Muggleton and Buntine 1988) for inductive reasoning, which played an important role in the development of ILP.

Using the resolution principle, we can link first-order logic rules with background knowledge for simplification; whereas using inverse resolution, we can develop new concepts and relations from background knowledge. Next, we take the simpler propositional reasoning as an example to demonstrate how resolution principle and inverse resolution work.

Suppose logic expressions $C_1$ and $C_2$ hold, containing complementary literals $L_1$ and $L_2$, respectively; without loss of generality, letting $L=L_1=\neg L_2, C_1=A \vee L$, and $C_2=B \vee \neg L$. The resolution principle tells that we can obtain resolvent $C=$ $A \vee B$ by eliminating $L$ using deductive reasoning. If we define the deletion operation on disjunctive normal form as
$$(A \vee B)-{B}=A,$$
then the resolution process can be expressed as
$$C=\left(C_1-{L}\right) \vee\left(C_2-{\neg L}\right),$$
abbreviated by
$$C=C_1 \cdot C_2 .$$

## 机器学习代考_Machine Learning代考_Least General Generalization

$\operatorname{riper}(1,10) \leftarrow$ curlier_root $(1,10) \wedge$ duller_sound $^2(1,10) \wedge$ hollower_umbilicus $(1,10) \wedge$ harder_surface $(1,10)$ riper $(1,15) \leftarrow$ curlier_root

$\operatorname{riper}(1, Y) \leftarrow$ curlier_root $(1, Y) \wedge$ duller_sound $(1,10)^{\operatorname{rin}}(1$, hollower_umbilicus $(1, Y) \wedge$ harder_surface $(1, Y)$ riper $(1, Y) \leftarrow$ curlier_root $(1$.

## 机器学习代考_Machine Learning代考_Inverse Resolution

$C_2=B \vee \neg L$. 解析原理告诉我们可以得到 resolvent $C=A \vee B$ 通过消除 $L$ 使用演绎推理。如果我们将析取范式上的删除操作 定义为
$$(A \vee B)-B=A$$

$$C=\left(C_1-L\right) \vee\left(C_2-\neg L\right),$$

$$C=C_1 \cdot C_2 .$$

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