Strength Reduction: Optimizing Multiplication with Addition
Strength reduction is a compiler optimization technique that replaces expensive operations with equivalent, less expensive ones. This challenge focuses on a common form of strength reduction: replacing multiplications by constants with equivalent additions or shifts. This optimization can significantly improve performance, especially in loops.
Problem Description
Your task is to implement a function in Go that takes an Abstract Syntax Tree (AST) representing a simple arithmetic expression and performs strength reduction for multiplications by a constant. Specifically, you need to identify expressions of the form x * C where C is a constant integer, and if C is a power of 2, replace it with a left bit shift (x << log2(C)). If C is not a power of 2, you should identify opportunities to replace x * C with a series of additions and shifts if C can be expressed as a sum of powers of 2 (e.g., x * 5 can be (x << 2) + x).
Key Requirements:
- AST Traversal: You will be provided with a simplified AST structure. Your solution should traverse this AST.
- Multiplication by Constant Detection: Identify binary expressions where the operator is multiplication (
*) and the right-hand operand is a constant integer. - Power of 2 Optimization: If the constant multiplier is a power of 2 (e.g., 2, 4, 8, 16), replace the multiplication with an equivalent left bit shift. For example,
x * 8becomesx << 3. - Sum of Powers of 2 Optimization: If the constant multiplier is not a power of 2 but can be represented as a sum of powers of 2 (e.g., 5 = 4 + 1, 6 = 4 + 2), replace the multiplication with a combination of left shifts and additions. For example,
x * 5can become(x << 2) + x. - Preserve Semantics: The transformed AST must produce the same results as the original AST for all valid inputs.
- Handle Edge Cases: Consider cases where the multiplier is 0 or 1.
Expected Behavior:
The function should return a new AST that has undergone the strength reduction. The structure of the AST should be modified in place if feasible, or a new AST can be returned.
Important Considerations:
- This challenge assumes a simplified AST structure for demonstration purposes.
- We are focusing on multiplying a variable or another expression by a constant.
- We are not concerned with floating-point numbers or other complex types.
Examples
Example 1:
Input AST:
Multiply(Variable("x"), Constant(8))
Output AST:
ShiftLeft(Variable("x"), Constant(3))
Explanation: 8 is 2^3, so x * 8 is optimized to x << 3.
Example 2:
Input AST:
Multiply(Variable("y"), Constant(5))
Output AST:
Add(ShiftLeft(Variable("y"), Constant(2)), Variable("y"))
Explanation: 5 = 4 + 1. So y * 5 becomes (y * 4) + (y * 1), which is (y << 2) + y.
Example 3:
Input AST:
Multiply(Constant(0), Variable("z"))
Output AST:
Constant(0)
Explanation: Multiplying by 0 always results in 0, regardless of the other operand.
Example 4:
Input AST:
Multiply(Variable("a"), Constant(1))
Output AST:
Variable("a")
Explanation: Multiplying by 1 does not change the value.
Example 5:
Input AST:
Add(Multiply(Variable("b"), Constant(6)), Constant(3))
Output AST:
Add(Add(ShiftLeft(Variable("b"), Constant(2)), ShiftLeft(Variable("b"), Constant(1))), Constant(3))
Explanation: 6 = 4 + 2. So b * 6 becomes (b * 4) + (b * 2), which is (b << 2) + (b << 1). The constant addition remains.
Constraints
- Operands for multiplication will be variables or constants.
- Constant multipliers will be non-negative integers.
- The AST depth is at most 10.
- The value of constant multipliers will not exceed 255.
- The solution should aim for an efficient traversal, ideally O(N) where N is the number of nodes in the AST.
Notes
- You'll need to define the AST node structures yourself. Think about
Expression,Variable,Constant,BinaryOp(likeAdd,Multiply,ShiftLeft), etc. - For checking if a number is a power of 2, you can use bitwise operations.
- For decomposing a number into powers of 2 for addition, you can iterate through the bits of the number.
- Consider how to represent the operations in your AST nodes.
- The goal is to transform the AST. You do not need to evaluate it.