Question

I have a C# program that detects incoming TCP/IP packets on any given ethernet device. Every packet is processed in the following struct:

struct Packet{
   String sourceIp;
   DateTime arrivalDate;
}

If I have a List of every incoming Packets (List), how do I get those IPs that have more than X packets in less than Y seconds (say 1 second)?

I have no idea how to approach this problem, any help/tip will be highly appreciated.

Answer 1

Using Linq, it will be something like this:

  List<Packet> allPackets =
     new List<Packet>
        {
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:00:00"), sourceIp = "a"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:00:01"), sourceIp = "a"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:00:01"), sourceIp = "a"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:01:00"), sourceIp = "a"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:00:00"), sourceIp = "b"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:01:00"), sourceIp = "b"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:02:00"), sourceIp = "b"},
           new Packet {arrivalDate = DateTime.Parse("2000-01-01 0:03:00"), sourceIp = "b"},
        };
  var xPackets = 2;
  var interval = TimeSpan.FromSeconds(15);

  // We group all the packets by ip, then within that, order the packets by date.
  var ips =
     allPackets
        .GroupBy(
           p => p.sourceIp,
           (ip, packets) => new
                                {
                                   ip,
                                   packets = packets.OrderBy(p => p.arrivalDate).ToList()
                                })
        .ToList();
  // Build a list of IPs with at least x packets in y interval.
  var rapidIps = new List<string>();
  foreach (var ipPacket in ips)
  {
     for (int i = 0, j = xPackets; j < ipPacket.packets.Count; i++, j++)
     {
        if (ipPacket.packets[i].arrivalDate + interval >= ipPacket.packets[j].arrivalDate)
        {
           rapidIps.Add((ipPacket.ip));
           break;
        }

     }
  }

At the end, rapidIps contains [a].