Common Subexpression Elimination in Go

This challenge involves implementing a compiler optimization technique known as Common Subexpression Elimination (CSE). CSE aims to improve program efficiency by identifying and removing redundant computations of identical subexpressions. By replacing repeated calculations with a single computed value, we can reduce the overall execution time and resource usage of the program.

Problem Description

Your task is to implement a function in Go that performs common subexpression elimination on a simplified representation of intermediate code. You will be given a slice of instructions, where each instruction represents an operation. You need to identify and eliminate duplicate computations of the same subexpression.

Key Requirements:

• Input: A slice of Instruction structs. Each Instruction struct has an Op (operation, e.g., "+", "*"), Dest (destination variable name), Src1 (first source variable/constant name), and Src2 (second source variable/constant name).
• Output: A new slice of Instruction structs representing the optimized intermediate code, with common subexpressions eliminated.
• Identification of Common Subexpressions: A subexpression is considered common if it has the same operation and the same operands (order matters for non-commutative operations, but for simplicity, we'll assume commutative for this challenge, so a + b is the same as b + a).
• Elimination Strategy: When a common subexpression is found, the subsequent occurrences should be replaced with a reference to the variable that holds the result of the first computation of that subexpression.
• Variable Naming: You will need to introduce temporary variables to store the results of common subexpressions that are computed for the first time. These temporary variables should be uniquely named.

Expected Behavior:

The function should iterate through the instructions, maintaining a record of computed subexpressions and their corresponding destination variables. If a subexpression is encountered that has already been computed, the current instruction's destination should be updated to point to the previously computed result. If the subexpression is new, its result should be stored, and a new destination variable (or the original if it's the first occurrence) should be used.

Edge Cases:

• Instructions with only one source operand (e.g., unary operations, though for this challenge, we'll focus on binary).
• Instructions that use constants as operands.
• Cases where the result of a common subexpression is immediately used by another instruction.

Examples

Example 1:

Input:
[
  {Op: "+", Dest: "t1", Src1: "a", Src2: "b"},
  {Op: "*", Dest: "t2", Src1: "c", Src2: "d"},
  {Op: "+", Dest: "t3", Src1: "a", Src2: "b"}, // Common subexpression with the first instruction
  {Op: "-", Dest: "t4", Src1: "t1", Src2: "t2"},
  {Op: "+", Dest: "t5", Src1: "b", Src2: "a"}  // Common subexpression with the first instruction (commutative)
]

Output:
[
  {Op: "+", Dest: "t1", Src1: "a", Src2: "b"},
  {Op: "*", Dest: "t2", Src1: "c", Src2: "d"},
  {Op: "=", Dest: "t3", Src1: "t1", Src2: ""}, // Replaced with assignment from t1
  {Op: "-", Dest: "t4", Src1: "t1", Src2: "t2"},
  {Op: "=", Dest: "t5", Src1: "t1", Src2: ""}  // Replaced with assignment from t1
]

Explanation:
The subexpression "a + b" is computed first and stored in "t1". The third instruction also computes "a + b", so its destination "t3" is updated to be an assignment from "t1". The fifth instruction computes "b + a", which is considered the same as "a + b" due to commutativity, so its destination "t5" is also updated to be an assignment from "t1". The output shows assignments using a generic "=" operation with the source being the original destination of the common subexpression.

Example 2:

Input:
[
  {Op: "*", Dest: "res1", Src1: "x", Src2: "y"},
  {Op: "+", Dest: "res2", Src1: "z", Src2: "w"},
  {Op: "*", Dest: "res3", Src1: "x", Src2: "y"}, // Common subexpression with the first instruction
  {Op: "+", Dest: "res4", Src1: "res1", Src2: "res2"},
  {Op: "*", Dest: "res5", Src1: "y", Src2: "x"}  // Common subexpression with the first instruction (commutative)
]

Output:
[
  {Op: "*", Dest: "res1", Src1: "x", Src2: "y"},
  {Op: "+", Dest: "res2", Src1: "z", Src2: "w"},
  {Op: "=", Dest: "res3", Src1: "res1", Src2: ""},
  {Op: "+", Dest: "res4", Src1: "res1", Src2: "res2"},
  {Op: "=", Dest: "res5", Src1: "res1", Src2: ""}
]

Explanation:
Similar to Example 1, "x * y" is computed first into "res1". Subsequent identical computations for "x * y" or "y * x" have their destinations redirected to "res1".

Example 3: (Handling Constants)

Input:
[
  {Op: "+", Dest: "temp1", Src1: "val1", Src2: "5"},
  {Op: "*", Dest: "temp2", Src1: "val2", Src2: "10"},
  {Op: "+", Dest: "temp3", Src1: "val1", Src2: "5"}, // Common subexpression with the first instruction
  {Op: "*", Dest: "temp4", Src1: "val2", Src2: "10"} // Common subexpression with the second instruction
]

Output:
[
  {Op: "+", Dest: "temp1", Src1: "val1", Src2: "5"},
  {Op: "*", Dest: "temp2", Src1: "val2", Src2: "10"},
  {Op: "=", Dest: "temp3", Src1: "temp1", Src2: ""},
  {Op: "=", Dest: "temp4", Src1: "temp2", Src2: ""}
]

Explanation:
Constants are treated as operands. "val1 + 5" is computed first, and subsequent identical computations are replaced. "val2 * 10" is computed, and subsequent identical computations are replaced.

Constraints

• The input slice of Instruction will contain at least 0 and at most 1000 instructions.
• Operand names (variables and constants) will be strings. Variable names will consist of alphanumeric characters. Constants will be numeric strings.
• Operations (Op) will be limited to "+", "-", "*", "/".
• The goal is to achieve an efficient implementation. The time complexity of your solution should ideally be closer to O(N) or O(N log N), where N is the number of instructions, rather than O(N^2).
• You may assume that source operands are either valid variable names that have been defined in a previous instruction or valid constant strings.
• For simplicity, assume all operations are commutative when identifying common subexpressions (e.g., a + b is the same as b + a, a * b is the same as b * a).

Notes

• You will need a way to keep track of subexpressions that have already been computed and the destination variable holding their result. A map could be very useful here.
• Consider how you will represent a subexpression uniquely as a key in your tracking mechanism. This key should account for the operation and both operands.
• When replacing a common subexpression, the original instruction that computed it should remain. Subsequent identical computations will be replaced by an assignment instruction (Op: "=").
• The order of operations in the input slice is important; it dictates the order of computation. Your CSE implementation should respect this order.
• For simplicity, focus on binary operations. You don't need to handle unary operations or complex control flow.