Round #1075

21 May 2026, 09:35· math· Meta-Llama-3.3-70B-Instruct· r13-clean-hints
Complexity
band 0
ZP / UO / OBS
3 / 0 / 3
Journey weight
475
Lightshift ΔL

Prompt

Solve the linear system 4x - y = 7 and 2x + 3y = 7. Give the exact values of x and y.

Zero-point index (what the engine surfaced)

ZPI-ZPI-0004 | 2026-05-21T09:35:00Z Solve the linear system 4x - y = 7 and 2x + 3y = 7. Give the exact values of x and y. ZERO POINT — 3 entries (structural invariants) ● hsl(29.84, 100%, 15%) isolation of variable [contract] ● hsl(205.00, 100%, 30%) consistent / inconsistent system [contract] ● hsl(104.40, 100%, 15%) row-reduction method [contract] UNOBSERVED — 0 entries (implied truths) OBSERVED — 3 entries (explicit in request) ◉ hsl(148.72, 100%, 22%) + [feature] ◉ hsl(175.38, 100%, 27%) = [feature] ◉ hsl(301.54, 100%, 15%) System [feature] COMPLEXITY │ 6 entries │ spread: 153° │ band: LOW

Baseline response (no ZPI)

```csharp using System; class Program { static void Main() { // Coefficients of the linear equations double a1 = 4, b1 = -1, c1 = 7; double a2 = 2, b2 = 3, c2 = 7; // Calculate the determinant double determinant = a1 * b2 - a2 * b1; // Calculate the values of x and y double x = (b2 * c1 - b1 * c2) / determinant; double y = (a1 * c2 - a2 * c1) / determinant; Console.WriteLine("The value of x is: " + x); Console.WriteLine("The value of y is: " + y); } } ```

Enhanced response (with ZPI)

```csharp using System; class Program { static void Main() { // Define the coefficients of the linear system double[,] coefficients = { { 4, -1 }, { 2, 3 } }; double[] constants = { 7, 7 }; // Use the row-reduction method to solve the system double[] solution = RowReduction(coefficients, constants); // Print the solution Console.WriteLine("x = " + solution[0]); Console.WriteLine("y = " + solution[1]); } static double[] RowReduction(double[,] coefficients, double[] constants) { int n = coefficients.GetLength(0); double[,] augmentedMatrix = new double[n, n + 1]; // Create the augmented matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { augmentedMatrix[i, j] = coefficients[i, j]; } augmentedMatrix[i, n] = constants[i]; } // Perform row operations to put the matrix in row-echelon form for (int i = 0; i < n; i++) { // Partial pivoting int maxRow = i; for (int k = i + 1; k < n; k++) { if (Math.Abs(augmentedMatrix[k, i]) > Math.Abs(augmentedMatrix[maxRow, i])) { maxRow = k; } } SwapRows(augmentedMatrix, i, maxRow); // Make the pivot element 1 double pivot = augmentedMatrix[i, i]; for (int j = i; j <= n; j++) { augmentedMatrix[i, j] /= pivot; } // Eliminate the pivot variable from the other rows for (int k = 0; k < n; k++) { if (k != i) { double factor = augmentedMatrix[k, i]; for (int j = i; j <= n; j++) { augmentedMatrix[k, j] -= factor * augmentedMatrix[i, j]; } } } } // Extract the solution from the augmented matrix double[] solution = new double[n]; for (int i = 0; i < n; i++) { solution[i] = augmentedMatrix[i, n]; } return solution; } static void SwapRows(double[,] matrix, int row1, int row2) { int n = matrix.GetLength(1); for (int i = 0; i < n; i++) { double temp = matrix[row1, i]; matrix[row1, i] = matrix[row2, i]; matrix[row2, i] = temp; } } } ``` This C# program uses the row-reduction method to solve the linear system 4x - y = 7 and 2x + 3y = 7. The `RowReduction` function takes the coefficients of the linear system and the constants as input, and returns the solution as an array of doubles. The `Main` function defines the coefficients and constants, calls the `RowReduction` function, and prints the solution. The row-reduction method is implemented using partial pivoting to ensure numerical stability.

Grader chain

Each grader reads the same baseline + enhanced pair and casts an independent verdict. Disagreement is the framework's own signal — not noise to be averaged away.

V1samba/Meta-Llama-3.3-70B-Instruct
Better
V4qwen2.5-coder-3b-instruct
Better
V5cerebras/qwen-3-235b-a22b-instruct-2507
Better
V7qwen2.5-coder-7b-instruct
Better
V8openai/gpt-4.1
Better
V9anthropic/claude-opus-4-7
Worse
V12openai/gpt-4o
Better

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