Round #793
10 May 2026, 17:54· csharp· Meta-Llama-3.3-70B-Instruct· r14-length-bound
- Complexity
- band 0
- ZP / UO / OBS
- 2 / 0 / 1
- Journey weight
- 193
- Lightshift ΔL
- +0.50
Prompt
Find duplicate rows in an emails table based on email address.
Zero-point index (what the engine surfaced)
ZPI-ZPI-0010 | 2026-05-10T17:54:39Z
Find duplicate rows in an emails table based on email address.
ZERO POINT — 2 entries (structural invariants)
● hsl(329.61, 100%, 42%) IDistributedCache [contract]
● hsl(136.10, 100%, 50%) DbContext [contract]
UNOBSERVED — 0 entries (implied truths)
OBSERVED — 1 entries (explicit in request)
◉ hsl(73.85, 100%, 15%) on [feature]
COMPLEXITY │ 3 entries │ spread: 0° │ band: LOW
Baseline response (no ZPI)
```csharp
using System;
using System.Data;
using System.Linq;
class Program
{
static void Main()
{
// Create a sample DataTable
DataTable emails = new DataTable();
emails.Columns.Add("email", typeof(string));
emails.Columns.Add("name", typeof(string));
emails.Rows.Add("test@gmail.com", "Test");
emails.Rows.Add("test2@gmail.com", "Test2");
emails.Rows.Add("test@gmail.com", "Test");
emails.Rows.Add("test3@gmail.com", "Test3");
emails.Rows.Add("test2@gmail.com", "Test2");
// Find duplicate rows
var duplicateRows = emails.AsEnumerable()
.GroupBy(row => row.Field<string>("email"))
.Where(g => g.Count() > 1)
.SelectMany(g => g);
// Print the duplicate rows
foreach (var row in duplicateRows)
{
Console.WriteLine($"{row["email"]}\t{row["name"]}");
}
}
}
```
Enhanced response (with ZPI)
To find duplicate rows in an emails table based on email address, you can use the following C# code:
```csharp
using Microsoft.EntityFrameworkCore;
using System.Collections.Generic;
using System.Linq;
public class EmailContext : DbContext
{
public DbSet<Email> Emails { get; set; }
protected override void OnConfiguring(DbContextOptionsBuilder optionsBuilder)
{
optionsBuilder.UseSqlServer(@"Data Source=(localdb)\mssqllocaldb;Initial Catalog=EmailsDB;Integrated Security=True");
}
}
public class Email
{
public int Id { get; set; }
public string Address { get; set; }
}
public class DuplicateEmailFinder
{
private readonly EmailContext _context;
public DuplicateEmailFinder(EmailContext context)
{
_context = context;
}
public List<Email> FindDuplicateEmails()
{
return _context.Emails
.GroupBy(e => e.Address)
.Where(g => g.Count() > 1)
.SelectMany(g => g)
.ToList();
}
}
```
In this code, we define an `EmailContext` class that inherits from `DbContext` and contains a `DbSet` of `Email` objects. The `OnConfiguring` method is overridden to specify the database connection string.
The `DuplicateEmailFinder` class takes an instance of `EmailContext` in its constructor and provides a `FindDuplicateEmails` method. This method uses LINQ to group the emails by address, filter the groups to only include those with more than one email, and then select all emails from these groups.
To use this code, you would create an instance of `EmailContext` and `DuplicateEmailFinder`, and then call the `FindDuplicateEmails` method:
```csharp
var context = new EmailContext();
var finder = new DuplicateEmailFinder(context);
var duplicateEmails = finder.FindDuplicateEmails();
```
This will return a list of all emails that have duplicate addresses in the database.
Note that this implementation does not use `IDistributedCache` as it is not necessary for finding duplicate rows in a database table. The `IDistributedCache` interface is typically used for caching data in a distributed environment, which is not relevant to this specific task.
This solution requires more text to fully explain the implementation and provide example usage.
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
BetterV4qwen2.5-coder-3b-instruct
BetterV5openrouter/qwen/qwen3-235b-a22b-2507
BetterV7qwen2.5-coder-7b-instruct
BetterV8openai/gpt-4.1
BetterV9anthropic/claude-opus-4-7
WorseV12openai/gpt-4o
BetterComments
Our grader said what it said. What do you say? Comment as a guest below.
No comments yet. Be the first to say what you make of this round.