🚨 Fix DeprecationWarning: invalid escape sequence

Author: nipunsadvilkar

tests/regression/test_issue27-31.py

@@ -14,7 +14,7 @@
         ["Thus, we first compute EMC 3 's response time-i.e., the duration from the initial of a call (from/to a participant in the target region) to the time when the decision of task assignment is made; and then, based on the computed response time, we estimate EMC 3 maximum throughput [28]-i.e., the maximum number of mobile users allowed in the MCS system.",
          "EMC 3 algorithm is implemented with the Java SE platform and is running on a Java HotSpot(TM) 64-Bit Server VM; and the implementation details are given in Appendix, available in the online supplemental material."
         ]),
-    ('#31', "Proof. First let v ∈ V be incident to at least three leaves and suppose there is a minimum power dominating set S of G that does not contain v. If S excludes two or more of the leaves of G incident to v, then those leaves cannot be dominated or forced at any step. Thus, S excludes at most one leaf incident to v, which means S contains at least two leaves ℓ 1 and ℓ 2 incident to v. Then, (S\{ℓ 1 , ℓ 2 }) ∪ {v} is a smaller power dominating set than S, which is a contradiction. Now consider the case in which v ∈ V is incident to exactly two leaves, ℓ 1 and ℓ 2 , and suppose there is a minimum power dominating set S of G such that {v, ℓ 1 , ℓ 2 } ∩ S = ∅. Then neither ℓ 1 nor ℓ 2 can be dominated or forced at any step, contradicting the assumption that S is a power dominating set. If S is a power dominating set that contains ℓ 1 or ℓ 2 , say ℓ 1 , then (S\{ℓ 1 }) ∪ {v} is also a power dominating set and has the same cardinality. Applying this to every vertex incident to exactly two leaves produces the minimum power dominating set required by (3). Definition 3.4. Given a graph G = (V, E) and a set X ⊆ V , define ℓ r (G, X) as the graph obtained by attaching r leaves to each vertex in X. If X = {v 1 , . . . , v k }, we denote the r leaves attached to vertex v i as ℓ",
+    ('#31', r"Proof. First let v ∈ V be incident to at least three leaves and suppose there is a minimum power dominating set S of G that does not contain v. If S excludes two or more of the leaves of G incident to v, then those leaves cannot be dominated or forced at any step. Thus, S excludes at most one leaf incident to v, which means S contains at least two leaves ℓ 1 and ℓ 2 incident to v. Then, (S\{ℓ 1 , ℓ 2 }) ∪ {v} is a smaller power dominating set than S, which is a contradiction. Now consider the case in which v ∈ V is incident to exactly two leaves, ℓ 1 and ℓ 2 , and suppose there is a minimum power dominating set S of G such that {v, ℓ 1 , ℓ 2 } ∩ S = ∅. Then neither ℓ 1 nor ℓ 2 can be dominated or forced at any step, contradicting the assumption that S is a power dominating set. If S is a power dominating set that contains ℓ 1 or ℓ 2 , say ℓ 1 , then (S\{ℓ 1 }) ∪ {v} is also a power dominating set and has the same cardinality. Applying this to every vertex incident to exactly two leaves produces the minimum power dominating set required by (3). Definition 3.4. Given a graph G = (V, E) and a set X ⊆ V , define ℓ r (G, X) as the graph obtained by attaching r leaves to each vertex in X. If X = {v 1 , . . . , v k }, we denote the r leaves attached to vertex v i as ℓ",
         ['Proof.', 'First let v ∈ V be incident to at least three leaves and suppose there is a minimum power dominating set S of G that does not contain v. If S excludes two or more of the leaves of G incident to v, then those leaves cannot be dominated or forced at any step.', 'Thus, S excludes at most one leaf incident to v, which means S contains at least two leaves ℓ 1 and ℓ 2 incident to v. Then, (S\\{ℓ 1 , ℓ 2 }) ∪ {v} is a smaller power dominating set than S, which is a contradiction.', 'Now consider the case in which v ∈ V is incident to exactly two leaves, ℓ 1 and ℓ 2 , and suppose there is a minimum power dominating set S of G such that {v, ℓ 1 , ℓ 2 } ∩ S = ∅.', 'Then neither ℓ 1 nor ℓ 2 can be dominated or forced at any step, contradicting the assumption that S is a power dominating set.', 'If S is a power dominating set that contains ℓ 1 or ℓ 2 , say ℓ 1 , then (S\\{ℓ 1 }) ∪ {v} is also a power dominating set and has the same cardinality.', 'Applying this to every vertex incident to exactly two leaves produces the minimum power dominating set required by (3).', 'Definition 3.4.', 'Given a graph G = (V, E) and a set X ⊆ V , define ℓ r (G, X) as the graph obtained by attaching r leaves to each vertex in X. If X = {v 1 , . . . , v k }, we denote the r leaves attached to vertex v i as ℓ'])
 ]