Let pi be the probability that an arbitrary tuple in D belongs to class Ci, estimated by |C(i,D)|/|D|.
Expected information need to classify a tuple in D:
$$Info(D) = -\sum\limits_{i=1}^n{P_i}log_2P_i$$
Information needed (after using A to split D into v partitions)to classify D:
$$Info_A(D) = \sum\limits_{j=1}^v\frac{|D_j|}{|D|}Info(D_j)$$
So, information gained by branching on attribute A:
$$Gain(A) = Info(D)-Info_A(D)$$

Class P:buys_computer = "yes"
Class N:buys_computer = "no"
the number of classs P is 9,while the number of class N is 5.

So,$$Info(D) = -\frac{9}{14}log_2\frac{9}{14} - \frac{5}{14}log_2\frac{5}{14} = 0.940$$

In Attribute age, the number of class P is 2,while the number of class N is 3.
So,$$Info(D_{
Similarly,
$$Info(D_{31...40}) = 0$$,$$Info(D_{>40}) = 0.971$$
Then,$$Info_{age}(D) = \frac{5}{14}Info(D_{40}) = 0.694$$
Therefore,$$Gain(age) = Info(D) - Info_age(D) = 0.246$$
Similarly,
$$Gain(income) = 0.029$$
$$Gain(Student) = 0.151$$
$$Gain(credit_rating) = 0.048$$

What if the attribute's value is continuous? How can we handle it?
1.The best split point for A
+Sort the value A in increasing order
+Typically, the midpoint between each pair of adjacent values is considered as a possible split point
-(a i +a i+1 )/2 is the midpoint between the values of a i and a i+1
+The point with the minimum expected information requirement for A is selected as the split-point for A
2.Split:
+D1 is the set of tuples in D satisfying A ≤ split-point, and D2 is the set of tuples in D satisfying A > split-point.

$$SpiltInfo_A(D) = -\sum\limits_{j=1}^v\frac{|D_j|}{|D|}*log_2\frac{|D_j|}{|D|}$$
Then the GainRatio equals to,
$$GainRatio(A=Gain(A)/SplitInfo(A)$$
The attribute with the maximun gain ratio is selected as the splitting attribute.

If a data set D contains examples from n classes, gini index gini(D) is defined as
$$gini(D) = 1 - \sum\limits_{j=1}^nP_j^2$$
where p_j is the relative frequency of class j in D.
If Data set D is split on A which have n classes.Then
$$gini_A(D) = \sum\limits_{i=1}^n\frac{D_i}{D}gini(D_i)$$
Reduction in Impurity
$$\Delta gini(A) = gini(D)-gini_A(D)$$