1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
10 Can assume there is another leaf outside E6, say n − 1.
2011 ICDMEC
Sketch of the Proof
1 We prove by induction on the number |VG | of vertices in the tree G .
2 If |VG | = 6, then G = E6 is non-degenerate and the assertion holds by Reeder’s result.
3 Suppose |VG | > 6 and u is a movable configuration in G .
4 Then there is a leaf of G not in the subgraph E6, say n.
5 By switching colors, can assume un= 0.
6 Use induction to settle the case when u is movable at some vertex i 6= n.
7 Can assume that n is the unique vertex that u is movable at n.
8 The case that G = E6+ Pn−6+ e and G is non-degenerate can be done by Reeder’s result.
9 The case that G = E6+ Pn−6+ e and G is degenerate can be done by using that 0 is the unique unmovable configuration in G − n.
2011 ICDMEC
n − 1 must be black. Applying the moves by selecting the vertices consecutively along the path from n to n − 1. The colors of n and n − 1 are switched, and a branch vertex z becomes movable, a contradiction.
2011 ICDMEC
Applying the moves by selecting the vertices consecutively along the path from n to n − 1. The colors of n and n − 1 are switched, and a branch vertex z becomes movable, a contradiction.
2011 ICDMEC are switched, and a branch vertex z becomes movable, a contradiction.
2011 ICDMEC
n − 1 must be black. Applying the moves by selecting the vertices
and a branch vertex z becomes movable, a contradiction.
2011 ICDMEC
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2011 ICDMEC
n − 1 must be black. Applying the moves by selecting the vertices
2011 ICDMEC
Figure. The binary star P6,3,1.