Random Forest (ID3 algorithm)

Created

Aug 7, 2025

Last Modified

2 weeks ago

Random Forest (ID3 algorithm)

Random Forest is a machine-learning algorithm that builds many decision trees and then combines their results to make a better final prediction.
It reduces overfitting and increases accuracy by taking the “majority vote” (classification) or “average” (regression) from all trees

Numerical For Constructing Random Forest

Day	Outlook	Temp	Humidity	Wind	Can Play
D1	Sunny	Hot	High	Weak	No
D2	Sunny	Hot	High	Strong	No
D3	Overcast	Mild	High	Weak	Yes
D4	Rain	Cool	High	Weak	Yes
D5	Rain	Cool	Normal	Weak	Yes
D6	Rain	Cool	Normal	Strong	No
D7	Overcast	Cool	Normal	Strong	Yes
D8	Sunny	Mild	High	Weak	No
D9	Sunny	Cool	Normal	Weak	Yes
D10	Rain	Mild	Normal	Weak	Yes
D11	Sunny	Mild	Normal	Strong	Yes
D12	Overcast	Mild	High	Strong	Yes
D13	Overcast	Hot	Normal	Weak	Yes
D14	Rain	Mild	High	Strong	No

Find the Class for the unseen data point

Outlook	Temp	Humidity	Wind
Overcast	Mild	Normal	Weak

Model 1

Day	Outlook	Temp	Humidity	Wind	Can Play
D1	Sunny	Hot	High	Weak	No
D2	Sunny	Hot	High	Strong	No
D3	Overcast	Mild	High	Weak	Yes
D4	Rain	Cool	High	Weak	Yes
D5	Rain	Cool	Normal	Weak	No
D6	Rain	Cool	Normal	Strong	Yes
D7	Overcast	Cool	Normal	Strong	Yes
D8	Sunny	Mild	High	Weak	No
D9	Sunny	Cool	Normal	Weak	Yes
D10	Rain	Mild	Normal	Weak	Yes

Set (S)

E n t r o p y (S) = - i = 1 \sum c p_{i} lo g_{2} p_{i}

- (\frac{6}{10} lo g_{2} \frac{6}{10}) - (\frac{4}{10} lo g_{2} \frac{4}{10})

0.442 + 0.529

0.971

Calculate information gain for all attributes.

Attribute 1 OUTLOOK

Sunny [No, No, No, Yes]
E = $- (\frac{3}{4} lo g_{2} \frac{3}{4}) - (\frac{1}{4} lo g_{2} \frac{1}{4})$
E = 0.811
Overcast [yes, yes]
E = 0 pure
Rain [Yes, Yes, Yes, No]
E = 0.811

Gain(S, outlook)

0.9711 - (\frac{4}{10} \times 0.811 + \frac{2}{10} \times 0 + \frac{4}{10} \times 0.811)

gain(S, outlook) = 0.322

Attribute 2 Temp

Hot	[No, NO]	E = 0
Mild	[Yes, Yes, No]	E = 0.918
Cool	[Yes, Yes, Yes, Yes, Yes, No]	E = 0.722

Gain(S, Temp)

0.9711 - (\frac{2}{10} \times 0 + \frac{3}{10} \times 0.918 + \frac{5}{10} \times 0.722)

gain(S, Temp) = 0.3347

Attribute 3 Humidity

High	[No, No, No, Yes, Yes]	E = 0.9711
Normal	[No, Yes, Yes, Yes, Yes]	E = 0.722

Gain(S, Humidity)

0.9711 - (\frac{5}{10} \times 0.971 + \frac{5}{10} \times 0.722)

gain(S, Humidity) = 0.125

Attribute 4 Wind

Weak	[No, No, Yes, Yes, Yes, Yes, Yes]	E = 0.9711
Strong	[No, No, Yes]	E = 0.722

Gain(S, Wind)

0.9711 - (\frac{4}{10} \times 0.86 + \frac{3}{10} \times 0.92)

gain(S, wind) = 0.093

Temp has Max gain

Random forest decision tree initial split based on temperature feature showing branches hot, mild, and cool

Branch Temp - mild ->

Set (S1)

Outlook	Temp	Humidity	Wind	Can Play
Overcast	Mild	High	Weak	Yes
Sunny	Mild	High	Weak	No
Rain	Mild	Normal	Weak	Yes

Attribute 1 OUTLOOK

Overcast E = 0 (pure)
sunny E = 0 (pure)
Rain E = 0 (pure)

Gain = 0.918

Attribute 2 Humidity

High E = 1
Normal E = 0

Gain = $0.918 \times (\frac{2}{3} \times 1 + \frac{1}{3} \times 1)$

Gain = 0.251

Attribute 3 Wind

Gain = 0

Wind has only one value, which is "weak"; thus, gain will be 0

Outlook has max gain

Random forest decision tree with multiple splits using temperature and outlook features leading to final classification outcomes

Branch Temp - Cool ->

Set (S2)

Outlook	Humidity	Wind	Can Play
Rain	High	Weak	Yes
Rain	Normal	Weak	Yes
Rain	Normal	Strong	No
Overcast	Normal	Strong	Yes
Sunny	Normal	Weak	Yes

$\sum = 0.722$

Attribute 1 OUTLOOK

Overcast E = 0 (pure)
sunny E = 0 (pure)
Rain E = 0.918

Gain = $0.722 - (\frac{3}{5} \times 0.918)$

Gain = 0.1712

Attribute 2 Humidity

High E = 0
Normal E = 0

Gain = $0.722 - (\frac{2}{3} \times 1 + \frac{1}{3} \times 1)$

Gain = 0.251