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Monotonic Sequence Games Bruce Sagan Department of Mathematics - PowerPoint PPT Presentation

Aug 01, 2023 •1.57k likes •2.59k views

Monotonic Sequence Games Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/ sagan and The Otago Theory Group Department of Computer Science University of Otago

  1. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  2. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  3. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  4. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  5. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , 5 2 , Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  6. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , 5 2 , 5 3 , Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  7. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  8. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, j = length of a longest increasing subsequence ending at x i , Similarly build a decreasing list D by reversing the inequalities.

  9. Play the same game with [ mn + 1 ] replaced by Q . As π = x 1 x 2 . . . is built, also build the increasing list I : 1. Initially I = ǫ , the empty sequence. 2. If I = y 1 y 2 . . . when x i is picked, have x i replace the smallest y j > x i or append x i to the right end of I if no such y j exists. Ex. π = 4 2 5 3 1 6 I : 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 ǫ, D : ǫ, 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 Since 6 was placed in the third column of I we have an increasing subsequence of length three ending at 6, e.g., 2 3 6. Theorem (Schensted, 1961) If x i is placed in column j of I, and in column k of D j = length of a longest increasing subsequence ending at x i , k = length of a longest decreasing subsequence ending at x i . Similarly build a decreasing list D by reversing the inequalities.

  10. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows:

  11. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R ,

  12. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B ,

  13. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P .

  14. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ,

  15. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, C : ǫ,

  16. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 C : P , ǫ,

  17. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 C : P , P B , ǫ,

  18. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 C : P , P B , P P , ǫ,

  19. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 C : P , P B , P P , R P B , ǫ,

  20. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 C : P , P B , P P , R P B , P P B , ǫ,

  21. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ,

  22. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ, R and P are called redish and draining red is R ← ǫ , P ← B .

  23. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R .

  24. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ .

  25. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C .

  26. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B , P P , R P B , P P B , P P P ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  27. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, C : ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  28. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 C : P , ǫ, R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  29. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 C : P , ǫ, ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  30. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 C : P , P B , ǫ, ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  31. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 C : P , P B ↑ ǫ, , ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  32. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 C : P , P B ↑ P P , ǫ, , ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  33. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 C : P , P B ↑ P ↑ P , ǫ, , ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  34. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 C : P , P B ↑ P ↑ P , R P B , ǫ, , ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  35. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 C : P , P B ↑ P ↑ P , R P B , ǫ, , ↑ ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  36. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 C : P , P B ↑ P ↑ P , R P B , P P B , ǫ, , ↑ ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  37. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 C : P , P B ↑ P ↑ P , R P B , P P B ↑ ǫ, , , ↑ ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  38. Build a combined list C by assigning colors red ( R ), blue ( B ), and purple ( P ) to the x i as follows: x i ∈ I and x i �∈ D = ⇒ color x i with R , x i �∈ I and x i ∈ D = ⇒ color x i with B , x i ∈ I and x i ∈ I = ⇒ color x i with P . Ex. π = 4 2 5 3 1 6 I : ǫ, 4 , 2 , 2 5 , 2 3 , 1 3 , 1 3 6 D : 4 , 4 2 , 5 2 , 5 3 , 5 3 1 , 6 3 1 ǫ, 4 2 4 2 5 2 3 5 1 3 5 1 3 6 C : P , P B ↑ P ↑ P , R P B , P P B ↑ P P P ǫ, , , ↑ ↑ R and P are called redish and draining red is R ← ǫ , P ← B . B and P are called bluish and draining blue is B ← ǫ , P ← R . Algorithm for C . 1. Initially C = ǫ . 2. Each x i inserts a P into the corresponding space of C . 3. Drain red from the closest redish element to the right of the new P (if any), and drain blue from the closest bluish element to the left of the new P (if any).

  39. Theorem (Otago-S) The winner of the game on Q where m ≤ n is: m \ n 0 1 2 3 4 5 6 7 0 A A A A A A A A 1 B B B B B B B 2 A B A B A B 3 A A A A A 4 A A A A where the patterns continue in each of the first 5 rows.

  40. Outline The Game on an Interval The Game on the Rationals Eine Kleine Game Theory Open Questions

  41. Let G be a 2-person, “last player to move wins” game with no draw positions.

  42. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v .

  43. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . Ex. a T = s � ❅ � ❅ � ❅ c b s s ✁✁ ❆ ✁✁ ❆ ❆ ❆ ✁ ❆ ✁ ❆ e e d f s s s s g g s s

  44. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a T = s � ❅ � ❅ � ❅ c b s s ✁✁ ❆ ✁✁ ❆ ❆ ❆ ✁ ❆ ✁ ❆ e e d f s s s s g g s s

  45. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d f s s s s s s s g g g s s s

  46. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d f s s s s s s s g g g s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v .

  47. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d f s s s s s s s g g g s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P ,

  48. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d P P f s s s s s s s g g g P s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P ,

  49. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d P P f s s s s s s s g g g P s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P , and work upwards using: (i) if there is a P -child of v then let ℓ ( v ) = N

  50. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b N N s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d P N P f s s s s s s s g g g P s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P , and work upwards using: (i) if there is a P -child of v then let ℓ ( v ) = N

  51. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b N N s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d P N P f s s s s s s s g g g P s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P , and work upwards using: (i) if there is a P -child of v then let ℓ ( v ) = N (ii) if all children of v are N then let ℓ ( v ) = P

  52. Let G be a 2-person, “last player to move wins” game with no draw positions. The game tree, T of G has nodes v = positions of G , children of v = all positions reachable in one move from v . If the same position w is a child of both v and v ′ then identify the copies of w to get a (di)graph T . Ex. a a P T = s T = s � ❅ � ❅ � ❅ � ❅ � ❅ � ❅ c c b b N N s s s s ✁✁ ❆ ✁✁ ❆ ✁✁ ❅ � ❆ ❆ ❆ ❅ � ❆ ✁ ❆ ✁ ❆ ✁ ❅ � ❆ e e e d f d P N P f s s s s s s s g g g P s s s � N if the next player wins from position v , Let ℓ ( v ) = P if the previous player wins from position v . Now label all terminal v ∈ T with P , and work upwards using: (i) if there is a P -child of v then let ℓ ( v ) = N (ii) if all children of v are N then let ℓ ( v ) = P

  53. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N

  54. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A.

  55. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is ǫ t P t � ❅ ❅ � PB t t RP ❆ ❅ � � ✁ P 2 ❅ ✟✟✟ ❆ ❍ ✁ t ❍ ❆ ✁ ❍ ✏ P ✏✏✏✏ P PBP t t P t PRP � ❅ ❅ � ❅ ❅ � ❅ ❅ P � � � P RPB t t t t P 2 B RPB 2 R 2 PB RP 2

  56. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is ǫ t P t � ❅ ❅ � PB t t RP ❆ ❅ � � ✁ P 2 ❅ ✟✟✟ ❆ ❍ ✁ t ❍ ❆ ✁ ❍ ✏ P ✏✏✏✏ P PBP t t P t PRP � ❅ ❅ � ❅ ❅ � ❅ ❅ P � � � P RPB t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P .

  57. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t P t t � ❅ ❅ � ❅ ❅ � � PB t t RP t t ❆ ❅ � � ✁ ❅ � ❆ � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P .

  58. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t P t t � ❅ ❅ � ❅ ❅ � � PB t t RP t t ❆ ❅ � � ✁ ❅ � ❆ � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii).

  59. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � PB t t RP t t ❆ ❅ � � ✁ ❆ ❅ � � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii).

  60. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � PB t t RP t t ❆ ❅ � � ✁ ❆ ❅ � � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i).

  61. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P .

  62. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P .

  63. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii).

  64. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❆ ❅ � � ✁ ❆ ❅ � � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii).

  65. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❆ ❅ � � ✁ ❆ ❅ � � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii). So ℓ ( PBP ) = ℓ ( PRP ) = P by (i) and symmetry.

  66. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❆ ❅ � � ✁ ❅ � ❆ � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P P P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii). So ℓ ( PBP ) = ℓ ( PRP ) = P by (i) and symmetry.

  67. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P P P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii). So ℓ ( PBP ) = ℓ ( PRP ) = P by (i) and symmetry. So ℓ ( P 2 B ) = ℓ ( RPB 2 ) = ℓ ( R 2 PB ) = ℓ ( RP 2 ) = N by (ii).

  68. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P P P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 N N N N Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii). So ℓ ( PBP ) = ℓ ( PRP ) = P by (i) and symmetry. So ℓ ( P 2 B ) = ℓ ( RPB 2 ) = ℓ ( R 2 PB ) = ℓ ( RP 2 ) = N by (ii).

  69. N P ✟✟ ❍ ✟✟ ❍ r r �❅❍ �❅❍ (i) (ii) r r r r r r r r r r P N N N N N Theorem (Otago-S) The winner of the game on Q where m = n ≥ 3 is A. Proof. Part of the top of T is P ǫ t t N P t t � ❅ ❅ � ❅ ❅ � � P P PB t t RP t t ❅ � ❆ � ✁ ❅ � ❆ � ✁ P 2 N ❅ ❅ ✟✟✟ ❆ ❍ ✁ ✟✟✟ ❆ ❍ ✁ t t ❍ ❍ ❆ ✁ ❍ ❆ ✁ ❍ ✏ P ✏ P ✏✏✏✏ P ✏✏✏✏ P P P PBP t t P t PRP t t P t � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ � ❅ ❅ P P � � � P � � � P RPB N t t t t t t t t P 2 B RPB 2 R 2 PB RP 2 N N N N Suppose B wins, so ℓ ( ǫ ) = P . So ℓ ( P ) = N by (ii). So ℓ ( PB ) = P or ℓ ( RP ) = P by (i). But m = n and PB , RP are symmetric so ℓ ( PB ) = ℓ ( RP ) = P . So ℓ ( P 2 ) = ℓ ( RPB ) = N by (ii). So ℓ ( PBP ) = ℓ ( PRP ) = P by (i) and symmetry. So ℓ ( P 2 B ) = ℓ ( RPB 2 ) = ℓ ( R 2 PB ) = ℓ ( RP 2 ) = N by (ii). This contradicts ℓ ( RPB ) = N

  70. Outline The Game on an Interval The Game on the Rationals Eine Kleine Game Theory Open Questions

  71. (1) Who wins the game for general m , n on either [ mn + 1 ] or Q ? In appears as if A wins except when m or n is small.

  72. (1) Who wins the game for general m , n on either [ mn + 1 ] or Q ? In appears as if A wins except when m or n is small. (2) What can be said about playing on other partially ordered sets?

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