【发布时间】：2017-05-31 05:44:03
【问题描述】：
使用OpenFlow 我想计算有多少对流满足以下条件（给定任意两个流，例如f1 和f2）：
-
f1的Src ip必须等于f2的Dst ip。 -
f1的Dst ip必须等于f2的Src ip。 -
f1和f2必须具有相同的协议。
我认为这将是两个元素(NFlows Choose 2) 的组合。感谢this question 我正在使用combinations 的方法，如下所示：
val pairFlows = matchs.combinations(2).filter{ list =>
val f1 = list.head
val f2 = list.tail.head
f1.dl_type == f2.dl_type &&
f1.nw_dst == f2.nw_src &&
f1.nw_src == f2.nw_dst
}
我的问题是，这是执行此计算的最有效方式吗？还是创建一个HashMap 来跟踪每个src、dst 和protocol type 会更好吗？
以下是数据示例：
{
1: [
{
"actions": [
"OUTPUT:2"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 18,
"hard_timeout": 0,
"byte_count": 1764,
"duration_sec": 6884,
"duration_nsec": 5000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:02",
"nw_src": "10.0.0.1",
"in_port": 1,
"nw_dst": "10.0.0.2"
}
},
{
"actions": [
"OUTPUT:2"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 18,
"hard_timeout": 0,
"byte_count": 1764,
"duration_sec": 1123,
"duration_nsec": 181000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:02",
"nw_src": "10.0.0.3",
"in_port": 3,
"nw_dst": "10.0.0.2"
}
},
{
"actions": [
"OUTPUT:1"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 10,
"hard_timeout": 0,
"byte_count": 980,
"duration_sec": 1132,
"duration_nsec": 253000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:01",
"nw_src": "10.0.0.3",
"in_port": 3,
"nw_dst": "10.0.0.1"
}
},
{
"actions": [
"OUTPUT:1"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 16,
"hard_timeout": 0,
"byte_count": 1568,
"duration_sec": 1141,
"duration_nsec": 652000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:01",
"nw_src": "10.0.0.2",
"in_port": 2,
"nw_dst": "10.0.0.1"
}
},
{
"actions": [
"OUTPUT:3"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 20,
"hard_timeout": 0,
"byte_count": 1960,
"duration_sec": 6883,
"duration_nsec": 961000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:03",
"nw_src": "10.0.0.2",
"in_port": 2,
"nw_dst": "10.0.0.3"
}
},
{
"actions": [
"OUTPUT:3"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 12,
"hard_timeout": 0,
"byte_count": 1176,
"duration_sec": 6883,
"duration_nsec": 984000000,
"priority": 1,
"length": 120,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 2048,
"dl_dst": "00:00:00:00:00:03",
"nw_src": "10.0.0.1",
"in_port": 1,
"nw_dst": "10.0.0.3"
}
},
{
"actions": [
"OUTPUT:CONTROLLER"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 0,
"hard_timeout": 0,
"byte_count": 0,
"duration_sec": 9156,
"duration_nsec": 252000000,
"priority": 65535,
"length": 96,
"flags": 0,
"table_id": 0,
"match": {
"dl_type": 35020,
"dl_dst": "01:80:c2:00:00:0e"
}
},
{
"actions": [
"OUTPUT:CONTROLLER"
],
"idle_timeout": 0,
"cookie": 0,
"packet_count": 22,
"hard_timeout": 0,
"byte_count": 1260,
"duration_sec": 9156,
"duration_nsec": 268000000,
"priority": 0,
"length": 80,
"flags": 0,
"table_id": 0,
"match": {}
}
],
}
这里有 8 个流，其中 3 个是对的。
【问题讨论】：
-
从 groupBy(_.dl_type) 开始会给你几个较小的列表，而不是一个巨大的，以检查。根据您的数据的外观，这可能会有所帮助，有一点帮助或根本没有帮助。此外，从 n 个元素的列表中查看 2 个元素的所有组合是 O(n^2)。如果您保留两个列表，一个按 dst 排序，一个按 src 排序（这是在 O(nlogn) 中完成的），那么您可以遍历它们一次并收集所有匹配项。与上面的示例相比，这将为您提供更好的时间复杂度。
-
@Buhb 谢谢，我已经设法在
O(n)中编写了一个解决方案
标签： scala collections combinations openflow