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Discrete Mathematics
Discrete Mathematics North-Holland
295
118 (1993) 295-297
Corrigendum To: D.A. Holton, D. Lou and M.D. Plummer, “On the 2-extendability of planar graphs”, Discrete Mathematics 96 (1991) 81-99. Received
24 June
1992
There is an error in the proof of Theorem 4 of this paper. On page 91 line 18, the statement “Similarly for Cz” is false. What has been proved to that stage is that if UES-{U~,L’~,U~,U~} then v is joined to either C1 or C, by exactly two edges. The theorem is, however, still true. The following proof of case 1.2 completes the proof. Replace lines 17 and 18 of page 91 by “every vertex v in S- {ui, vi, u2, uz} is joined to either C1 or Cz by exactly two edges”. Then continue as follows. (1.2.1) Suppose ISI = 8. (So IX’1=4.) Then X’ = {x1,x2,x5,xg}. x1 and x2 each send two edges to Ci, x5 and xg each send two edges to C2. If there is a vertex y in C, which is adjacent to both u1 and u1 or both u2 and v2, as C1 accepts six edges from S, {xi, x2, y> is a 3-cut, contradicting the 4-connectedness. So no vertex in C1 is adjacent to both u1 and u1 or both u2 and v2. As each of x1 and x2 is contained in a JT, then suppose there are two triangles Ti (i= 1,2) such that xi is adjacent to two distinct vertices of Ti. So Cl consists of the two triangles and some disjoint JT’s, contradicting the fact that C1 is an odd component. Hence there is an edge in C1 both ends of which are adjacent to both x1 and x2. Then C1 consists of the two ends of the edge and some JT’s, contradicting the fact that C1 is an odd component. (1.2.2) Suppose IS/ = 7 and hence IX’1= 3. Without loss of generality, assume X’ = (x1, x2, x5}, x1 and x2 each send two edges to C1, and x5 sends two edges to C2. Suppose one of x1 and x2 also sends two edges to Cz. Without loss of generality, assume x2 sends two edges to Cz. Then and o(G-S’) = IS -2, where C’= G[ V(C,)u{x2}u I’(C,)] IS an odd component S’ = S - {x2}, contradicting the minimality of S. So neither x1 nor x2 sends two edges to Cz. By the same argument as in Case (1.2.1), we have a contradiction. (1.2.3) Suppose ISI =6 and so IX’1=2. (1.2.3.1) Suppose X’= {xi, x2}, where x1 and x2 each send two edges to Ci. If one of x1 and x2 also sends two edges to Cz, by the same argument as in Case (1.2.2), we have 0012-365X/93/$06.00
c
1993-Elsevier
Science Publishers
B.V. All rights reserved
296
D.A. Hahn,
D. Lou, M.D. Plummer
a contradiction of the minimality of S. So none of xi and x2 sends two edges to Cz. By the same argument as in Case (1.2.1), we have a contradiction. (1.2.3.2) Next, suppose X’= (xi, xs}, where xi sends two edges to Ci and x8 sends two edges to Cz. If x1 also sends two edges to Cz or x8 also sends two edges to Ci, by the same argument as in Case (1.2.2) we have a contradiction to the minimality of S. So assume this does not happen. Since Ci is incident with two digons in G”, a vertex in U sends two edges to C1. Assume ui sends two edges to Ci without loss of generality. e, lies on two triangles in G”. But we cannot have wi # Ci and w2 # C1 in G” as d&u,) =4. And we cannot have wi = C1 and w2= Ci in G” as e, lies on two triangles. Without loss of generality, assume wi = Ci and w2 # Ci in G”. We have two cases. (1.2.3.2.1) w2 # C2 in G”. Then w2 is a singleton component of G-S. By counting the edges from S to Ci, no vertex in Ci is adjacent to both u2 and L’~.Triangle (w,, ui, ~‘i) is contained in a JT of which the other vertex is in Ci. Also x1 is adjacent to two vertices of a triangle contained in Ci. So Ci consists of the triangle, the vertex and some JT’s, contradicting the fact that Ci is an odd component. (1.2.3.2.2) w2 = C2 in G”. As e, lies in two triangles in G”. Both ur and ui are adjacent to both vertices of Ci and vertices of C2. And u1 sends two edges to Ci. As C3 and C4 lie in a JTrespectively, they cannot be adjacent to both u1 and vi as &(ui)=4. So both C3 and C4 are adjacent to both u2 and v2. JT(C3)= JT(C4). Now u2 and u2 cannot send two edges to C2 as &(u,)=d,(v,)=4. But C2 is incident with two digons in G”. So ui sends two edges to C2. Suppose xi sends an edge to C2. Then C2 obtains six edges from S. Let y be the vertex in S adjacent to vertices of Ci which is different from xi, u1 and ui. Then {xi,xs,y} is a 3-cut or 2-cut (when y=xs) separating G[V(C,)u{u,,u,}u V(C,)] from the other part of G, a contradiction. So xi is joined to C3 and C, by 4-regularity. By the same argument as above, x8 is joined to C3 and Cd. Besides xi, ui and ui, Ci is joined to u2 or u2, say u2, as there are four vertices in S adjacent to vertices of Cr. By the same reason, C2 is joined to u2 besides x s, u1 and u, by the 4-regularity. Now G” is homeomorphic to K3,3 with principal vertices ({xi, u2, v2}, {C,, C3, C,}). Hence G is not planar, a contradiction. (1.2.3.3) Suppose X’= {x5,x6}, where xg and x6 each send two edges to C2. This is the same as in Case (1.2.3.1) (1.2.4) Suppose ISI = 5. Thus IX’1= 1. Without loss of generality, assume X’= (xi}, where xi sends two edges to Ci. If xi also sends two edges to C2, by the same argument as in Case (1.2.2), we have a contradiction to the minimality of S. So assume this does not happen. As Ci is incident with two digons in G”, a vertex in U sends two edges to Ci. Without loss of generality, assume that u1 sends two edges to Ci. e, lies on two triangles in G”. So we cannot have that w1 = Ci and w2 = Ci. And we cannot have that w1 # Ci and w2# Ci as dc(u1)=4. Without loss of generality, assume w1 =Ci and w2 # C1 in G”.
Corrigendum
(1.2.4.1) w2 #C2 in G”. Then w2 = C3. By the same argument the fact that Ci is an odd component.
297
as Case (1.2.3.2.1) we have a contradiction
to
(1.2.4.2) w2 = C2 in G”. Then u1 and vi each are adjacent to both vertices of C1 and vertices of Cz. And u1 sends two edges to C1. As C3 lies in a JT, C3 is adjacent to both u1 and v2. If both u1 and v2 send two edges to Ca, then {ur, vr, C,} is a 3-cut, separating G[ V(C2)u{u1, vz}] and the other part of G, contradicting the 4-connectedness of G. As Cz is incident with two digons in G”, u1 sends two edges to C2 and one of u2 and v2 sends two edges to C2. Assume
u2 sends two edges to C2. As triangle
{C3,u2,v2}
lies in a JT of which the
other vertex is in C2, v2 sends an edge to C2. Now ul, vl, u2, v2 send six edges to C2. x1 sends two edges to Cr. By 4-regularity, x1 sends the other two edges to C2 and C3. So x1 sends at least an edge to C2, contradicting the fact that C2 accepts six edges from S. (1.2.5) Suppose lSI=4. So X’=@ and S= U. Now the four end vertices of e, and e, form a cutset separating G into two odd components C1 and C2, contradicting an hypothesis of the Theorem.
295
118 (1993) 295-297
Corrigendum To: D.A. Holton, D. Lou and M.D. Plummer, “On the 2-extendability of planar graphs”, Discrete Mathematics 96 (1991) 81-99. Received
24 June
1992
There is an error in the proof of Theorem 4 of this paper. On page 91 line 18, the statement “Similarly for Cz” is false. What has been proved to that stage is that if UES-{U~,L’~,U~,U~} then v is joined to either C1 or C, by exactly two edges. The theorem is, however, still true. The following proof of case 1.2 completes the proof. Replace lines 17 and 18 of page 91 by “every vertex v in S- {ui, vi, u2, uz} is joined to either C1 or Cz by exactly two edges”. Then continue as follows. (1.2.1) Suppose ISI = 8. (So IX’1=4.) Then X’ = {x1,x2,x5,xg}. x1 and x2 each send two edges to Ci, x5 and xg each send two edges to C2. If there is a vertex y in C, which is adjacent to both u1 and u1 or both u2 and v2, as C1 accepts six edges from S, {xi, x2, y> is a 3-cut, contradicting the 4-connectedness. So no vertex in C1 is adjacent to both u1 and u1 or both u2 and v2. As each of x1 and x2 is contained in a JT, then suppose there are two triangles Ti (i= 1,2) such that xi is adjacent to two distinct vertices of Ti. So Cl consists of the two triangles and some disjoint JT’s, contradicting the fact that C1 is an odd component. Hence there is an edge in C1 both ends of which are adjacent to both x1 and x2. Then C1 consists of the two ends of the edge and some JT’s, contradicting the fact that C1 is an odd component. (1.2.2) Suppose IS/ = 7 and hence IX’1= 3. Without loss of generality, assume X’ = (x1, x2, x5}, x1 and x2 each send two edges to C1, and x5 sends two edges to C2. Suppose one of x1 and x2 also sends two edges to Cz. Without loss of generality, assume x2 sends two edges to Cz. Then and o(G-S’) = IS -2, where C’= G[ V(C,)u{x2}u I’(C,)] IS an odd component S’ = S - {x2}, contradicting the minimality of S. So neither x1 nor x2 sends two edges to Cz. By the same argument as in Case (1.2.1), we have a contradiction. (1.2.3) Suppose ISI =6 and so IX’1=2. (1.2.3.1) Suppose X’= {xi, x2}, where x1 and x2 each send two edges to Ci. If one of x1 and x2 also sends two edges to Cz, by the same argument as in Case (1.2.2), we have 0012-365X/93/$06.00
c
1993-Elsevier
Science Publishers
B.V. All rights reserved
296
D.A. Hahn,
D. Lou, M.D. Plummer
a contradiction of the minimality of S. So none of xi and x2 sends two edges to Cz. By the same argument as in Case (1.2.1), we have a contradiction. (1.2.3.2) Next, suppose X’= (xi, xs}, where xi sends two edges to Ci and x8 sends two edges to Cz. If x1 also sends two edges to Cz or x8 also sends two edges to Ci, by the same argument as in Case (1.2.2) we have a contradiction to the minimality of S. So assume this does not happen. Since Ci is incident with two digons in G”, a vertex in U sends two edges to C1. Assume ui sends two edges to Ci without loss of generality. e, lies on two triangles in G”. But we cannot have wi # Ci and w2 # C1 in G” as d&u,) =4. And we cannot have wi = C1 and w2= Ci in G” as e, lies on two triangles. Without loss of generality, assume wi = Ci and w2 # Ci in G”. We have two cases. (1.2.3.2.1) w2 # C2 in G”. Then w2 is a singleton component of G-S. By counting the edges from S to Ci, no vertex in Ci is adjacent to both u2 and L’~.Triangle (w,, ui, ~‘i) is contained in a JT of which the other vertex is in Ci. Also x1 is adjacent to two vertices of a triangle contained in Ci. So Ci consists of the triangle, the vertex and some JT’s, contradicting the fact that Ci is an odd component. (1.2.3.2.2) w2 = C2 in G”. As e, lies in two triangles in G”. Both ur and ui are adjacent to both vertices of Ci and vertices of C2. And u1 sends two edges to Ci. As C3 and C4 lie in a JTrespectively, they cannot be adjacent to both u1 and vi as &(ui)=4. So both C3 and C4 are adjacent to both u2 and v2. JT(C3)= JT(C4). Now u2 and u2 cannot send two edges to C2 as &(u,)=d,(v,)=4. But C2 is incident with two digons in G”. So ui sends two edges to C2. Suppose xi sends an edge to C2. Then C2 obtains six edges from S. Let y be the vertex in S adjacent to vertices of Ci which is different from xi, u1 and ui. Then {xi,xs,y} is a 3-cut or 2-cut (when y=xs) separating G[V(C,)u{u,,u,}u V(C,)] from the other part of G, a contradiction. So xi is joined to C3 and C, by 4-regularity. By the same argument as above, x8 is joined to C3 and Cd. Besides xi, ui and ui, Ci is joined to u2 or u2, say u2, as there are four vertices in S adjacent to vertices of Cr. By the same reason, C2 is joined to u2 besides x s, u1 and u, by the 4-regularity. Now G” is homeomorphic to K3,3 with principal vertices ({xi, u2, v2}, {C,, C3, C,}). Hence G is not planar, a contradiction. (1.2.3.3) Suppose X’= {x5,x6}, where xg and x6 each send two edges to C2. This is the same as in Case (1.2.3.1) (1.2.4) Suppose ISI = 5. Thus IX’1= 1. Without loss of generality, assume X’= (xi}, where xi sends two edges to Ci. If xi also sends two edges to C2, by the same argument as in Case (1.2.2), we have a contradiction to the minimality of S. So assume this does not happen. As Ci is incident with two digons in G”, a vertex in U sends two edges to Ci. Without loss of generality, assume that u1 sends two edges to Ci. e, lies on two triangles in G”. So we cannot have that w1 = Ci and w2 = Ci. And we cannot have that w1 # Ci and w2# Ci as dc(u1)=4. Without loss of generality, assume w1 =Ci and w2 # C1 in G”.
Corrigendum
(1.2.4.1) w2 #C2 in G”. Then w2 = C3. By the same argument the fact that Ci is an odd component.
297
as Case (1.2.3.2.1) we have a contradiction
to
(1.2.4.2) w2 = C2 in G”. Then u1 and vi each are adjacent to both vertices of C1 and vertices of Cz. And u1 sends two edges to C1. As C3 lies in a JT, C3 is adjacent to both u1 and v2. If both u1 and v2 send two edges to Ca, then {ur, vr, C,} is a 3-cut, separating G[ V(C2)u{u1, vz}] and the other part of G, contradicting the 4-connectedness of G. As Cz is incident with two digons in G”, u1 sends two edges to C2 and one of u2 and v2 sends two edges to C2. Assume
u2 sends two edges to C2. As triangle
{C3,u2,v2}
lies in a JT of which the
other vertex is in C2, v2 sends an edge to C2. Now ul, vl, u2, v2 send six edges to C2. x1 sends two edges to Cr. By 4-regularity, x1 sends the other two edges to C2 and C3. So x1 sends at least an edge to C2, contradicting the fact that C2 accepts six edges from S. (1.2.5) Suppose lSI=4. So X’=@ and S= U. Now the four end vertices of e, and e, form a cutset separating G into two odd components C1 and C2, contradicting an hypothesis of the Theorem.