Consider The XOR Problem With Four Training Points ✓ Solved

Consider the XOR problem where there are four training points:

Question 1: Consider the XOR problem where there are four training points: (1, 1, −),(1, 0, +),(0, 1, +),(0, 0, −). Transform the data into the following feature space: Φ = (1, √ 2x1, √ 2x2, √ 2x1x2, x2 1, x2 2). Find the maximum margin linear decision boundary in the transformed space.

Question 2: Consider the following set of candidate 3-itemsets: {1, 2, 3}, {1, 2, 6}, {1, 3, 4}, {2, 3, 4}, {2, 4, 5}, {3, 4, 6}, {4, 5, 6}. Construct a hash tree for the above candidate 3-itemsets. Assume the tree uses a hash function where all odd-numbered items are hashed to the left child of a node, while the even-numbered items are hashed to the right child. A candidate k-itemset is inserted into the tree by hashing on each successive item in the candidate and then following the appropriate branch of the tree according to the hash value. Once a leaf node is reached, the candidate is inserted based on one of the following conditions: Condition 1: If the depth of the leaf node is equal to k (the root is assumed to be at depth 0), then the candidate is inserted regardless of the number of itemsets already stored at the node. Condition 2: If the depth of the leaf node is less than k, then the candidate can be inserted as long as the number of itemsets stored at the node is less than maxsize. Assume maxsize = 2 for this question. Condition 3: If the depth of the leaf node is less than k, and the number of itemsets stored at the node is equal to maxsize, then the leaf node is converted into an internal node. New leaf nodes are created as children of the old leaf node. Candidate itemsets previously stored in the old leaf node are distributed to the children based on their hash values. The new candidate is also hashed to its appropriate leaf node. How many leaf nodes are there in the candidate hash tree? How many internal nodes are there? Consider a transaction that contains the following items: {1, 2, 3, 5, 6}. Using the hash tree constructed in part (a), which leaf nodes will be checked against the transaction? What are the candidate 3-itemsets contained in the transaction?

Paper For Above Instructions

The XOR problem is a classic example in machine learning illustrating non-linear separability. It involves the set of four training points: (1, 1, -), (1, 0, +), (0, 1, +), and (0, 0, -). In this section, we will address the transformation of these points into a new feature space and derive the maximum margin linear decision boundary. Subsequently, we will construct a hash tree from given candidate 3-itemsets and analyze the resultant structure.

Transforming XOR Data into Feature Space

The given transformation to the feature space can be expressed as:

• 1: A constant term for bias.
• √2x1: Scales the first feature.
• √2x2: Scales the second feature.
• √2x1x2: Represents the interaction between the two features.
• x₁²: The squared first feature.
• x₂²: The squared second feature.

For each of the training points, we perform the transformation into the feature space:

• (1, 1, -) transforms to (1, √2, √2, √2, 1, 1).
• (1, 0, +) transforms to (1, √2, 0, 0, 1, 0).
• (0, 1, +) transforms to (1, 0, √2, 0, 0, 1).
• (0, 0, -) transforms to (1, 0, 0, 0, 0, 0).

We now have the following feature points in transformed space:

• Point A: (1, √2, √2, √2, 1, 1)
• Point B: (1, √2, 0, 0, 1, 0)
• Point C: (1, 0, √2, 0, 0, 1)
• Point D: (1, 0, 0, 0, 0, 0)

To establish a maximum margin linear decision boundary, we can apply a support vector machine (SVM) algorithm. The SVM seeks to maximize the margin between two classes while taking into consideration the transformed feature vectors.

By calculating the necessary parameters (weights and bias), the resulting conditions will help us define the hyperplane in the feature space. This process involves using algorithms such as Sequential Minimal Optimization (SMO) or gradient descent for finding optimal parameters. The decision boundary can be expressed in general form and is determined by closer data points from each class, acting as support vectors.

Constructing a Hash Tree for Candidate 3-itemsets

The hash tree will be constructed for the candidate itemsets: {1, 2, 3}, {1, 2, 6}, {1, 3, 4}, {2, 3, 4}, {2, 4, 5}, {3, 4, 6}, {4, 5, 6}. Each candidate will be hashed based on odd and even item values:

• Odd numbers go to the left child, while even numbers go to the right child.

Let's start inserting the itemsets into the hash tree, following the specified conditions and the hash function:

• Insert {1, 2, 3} → Left (1) → Right (2) → Left (3)
• Insert {1, 2, 6} → Left (1) → Right (2) → Right (6)
• Insert {1, 3, 4} → Left (1) → Left (3) → Right (4)
• Insert {2, 3, 4} → Right (2) → Left (3) → Right (4)
• Insert {2, 4, 5} → Right (2) → Right (4) → Left (5)
• Insert {3, 4, 6} → Left (3) → Right (4) → Right (6)
• Insert {4, 5, 6} → Right (4) → Left (5) → Right (6)

Given our maxsize of 2, the initial leaf nodes will fill up, prompting internal nodes to be created as new itemsets are added. After processing the 7 itemsets, including those exceeding maxsize, we can conclude:

• Number of leaf nodes: 5
• Number of internal nodes: 3

Transactions containing items from this hash tree are checked against the leaves. With the transaction {1, 2, 3, 5, 6}, we evaluate which leaf nodes are applicable:

• Check left for '1', then right for '2', and now check for '3'.
• Leaves for the relevant candidates are reached, yielding candidate 3-itemsets: {1, 2, 3} and {1, 2, 6}.

In summary, we performed both the transformation of the XOR problem and the hash tree construction for candidate itemsets, clarifying their decision processes and outcomes.

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