has been confirmed to be cancerous, while NC means it is non-cancerous. Using this data, a decision tree can be created with information gain used to determine the candidate splits for each node. For the next step, the entropy at parent node t of the above simple decision tree is computed as:H(t) = −[pC,t log2(pC,t) + pNC,t log2(pNC,t)] where, probability of selecting a class ‘C’ sample at node t, pC,t = n(t, C) / n(t), probability of selecting a class ‘NC’ sample at node t, pNC,t = n(t, NC) / n(t), n(t), n(t, C), and n(t, NC) are the number of total samples, ‘C’ samples and ‘NC’ samples at node t respectively.Using this with the example training set, the process for finding information gain beginning with for Mutation 1 is as follows: pC, t = 4/7 pNC, t = 3/7 = −(4/7log2(4/7) + 3/7log2(3/7)) = 0.985 Note: will be the same for all mutations at the root. The relatively high value of entropy (1 is the optimal value) suggests that the root node is highly impure and the constituents of the input at the root node would look like the leftmost figure in the above Entropy Diagram. However, such a set of data is good for learning the attributes of the mutations used to split the node. At a certain node, when the homogeneity of the constituents of the input occurs (as shown in the rightmost figure in the above Entropy Diagram), the dataset would no longer be good for learning. Moving on, the entropy at left and right child nodes of the above decision tree is computed using the formulae:H(tL) = −[pC,L log2(pC,L) + pNC,L log2(pNC,L)]H(tR) = −[pC,R log2(pC,R) + pNC,R log2(pNC,R)]where, probability of selecting a class ‘C’ sample at the left child node, pC,L = n(tL, C) / n(tL), probability of selecting a class ‘NC’ sample at the left child node, pNC,L = n(tL, NC) / n(tL), probability of selecting a class ‘C’ sample at the right child node, pC,R = n(tR, C) / n(tR), probability of selecting a class ‘NC’ sample at the right child node, pNC,R = n(tR, NC) / n(tR), n(tL), n(tL, C), and n(tL, NC) are the total number of samples, ‘C’ samples and ‘NC’ samples at the left child node respectively, n(tR), n(tR, C), and n(tR, NC) are the total number of samples, ‘C’ samples and ‘NC’ samples at the right child node