NodePath: Make path computation deterministic

In Python3, set operations seem to result in indeterministic order.

+ Remove verbose logging from topology.py

Change-Id: I47cc90637a0b2969e1686b4216b7d8fe4fe2aaab
Signed-off-by: Vratko Polak <vrpolak@cisco.com>

1 files changed, 6 insertions, 5 deletions

diff --git a/resources/libraries/python/NodePath.py b/resources/libraries/python/NodePath.py
index e97bde87ad..6f08be491b 100644
--- a/resources/libraries/python/NodePath.py
+++ b/resources/libraries/python/NodePath.py
@@ -1,4 +1,4 @@
-# Copyright (c) 2019 Cisco and/or its affiliates.
+# Copyright (c) 2020 Cisco and/or its affiliates.
 # Licensed under the Apache License, Version 2.0 (the "License");
 # you may not use this file except in compliance with the License.
 # You may obtain a copy of the License at:
@@ -120,10 +120,11 @@ class NodePath:
                     f"No link between {node1[u'host']} and {node2[u'host']}"
                 )
 
+            # Not using set operations, as we need deterministic order.
             if always_same_link:
-                l_set = set(links).intersection(self._links)
+                l_set = [link for link in links if link in self._links]
             else:
-                l_set = set(links).difference(self._links)
+                l_set = [link for link in links if link not in self._links]
                 if not l_set:
                     raise RuntimeError(
                         f"No free link between {node1[u'host']} and "
@@ -131,9 +132,9 @@ class NodePath:
                     )
 
             if not l_set:
-                link = links.pop()
+                link = links[0]
             else:
-                link = l_set.pop()
+                link = l_set[0]
 
             self._links.append(link)
             interface1 = topo.get_interface_by_link_name(node1, link)