HAMILTONICITY OF HANOI GRAPHS Katherine Rock Advisor: Dr. John - PowerPoint PPT Presentation

ON THE PLANARITY AND HAMILTONICITY OF HANOI GRAPHS Katherine Rock Advisor: Dr. John Caughman Fariborz Maseeh Department of Mathematics and Statistics Portland State University MTH 501 Presentation June 2, 2016

The Tower of Hanoi Puzzles The original Tower of Hanoi puzzle Édouard Lucas 1883 1842-1891

The Tower of Hanoi Puzzles • 𝑜 discs arranged on 3 + 𝑛 vertical pegs, with 𝑜, 𝑛 ∈ ℤ ≥0 . • Each disc is a different size. • Regular state: If multiple discs are on the same peg, they are arranged in decreasing size from bottom to top. • Perfect state: A regular state in which all discs are on the same peg.

The Tower of Hanoi Puzzles • Object: To move from one perfect state to another by moving one disc at a time from the topmost position on one peg to the topmost position on another peg. • Divine rule: No larger disc may be placed on top of any smaller disc.

Hanoi Graphs 𝑜 corresponds to the Tower of Hanoi • The Hanoi graph 𝐼 𝑛 puzzle with 3 + 𝑛 pegs and 𝑜 discs. • Label the pegs 0,1, … , 2 + 𝑛 and let 𝑦 𝑗 be the position of the disc with radius 𝑗 , for each 𝑗 = 1,2, … , 𝑜 . • Then each regular state in the puzzle is represented by vertex in the graph, labeled with an 𝑜 -tuple (𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 ) , where each 𝑦 𝑗 ∈ {0,1, … , 2 + 𝑛} . 𝑜 are all the possible legal moves of the • The edges of 𝐼 𝑛 discs. Two vertices are adjacent if and only if their corresponding states can be achieved from one another through a legal move of exactly one disc.

5 Example: 𝐼 𝑛 5 , In the graph 𝐼 𝑛 1,1,1,1,1 ~(0,1,1,1,1) and 0,1,1,1,1 ~(0,2 + 𝑛, 1,1,1) , but 1,1,1,1,1 ≁ 0,2 + 𝑛, 1,1,1 .

Hanoi Graphs Definition Let 𝑜, 𝑛 ∈ ℤ, with 𝑜 > 0 and 𝑛 ≥ 0. 𝑜 is the graph with vertex set 𝑊(𝐼 𝑛 𝑜 ) The Hanoi graph 𝐼 𝑛 given by 𝑜 𝑊 𝐼 𝑛 = 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 0 ≤ 𝑦 𝑗 ≤ 2 + 𝑛, 𝑦 𝑗 ∈ ℤ} and where 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 ~ 𝑧 1 , 𝑧 2 , … , 𝑧 𝑜 if and only if there exists 𝑗 ∈ 1,2, … , 𝑜 such that 𝑦 𝑗 ≠ 𝑧 𝑗 , i. 𝑦 𝑘 = 𝑧 𝑘 for all 𝑗 ≠ 𝑘 , and ii. 𝑦 𝑗 , 𝑧 𝑗 ∩ 𝑦 1 , … , 𝑦 𝑗−1 = ∅ . iii.

3 Example: 𝐼 0

1 & 𝐼 1 2 Example: 𝐼 1

Outline • Introduction (done) • Hamiltonian graphs 𝑜 • Hamiltonicity of 𝐼 𝑛 • Planar graphs • Planarity of Hanoi graphs

Hamiltonian Graphs Definition A graph 𝐻 is called hamiltonian if it contains a cycle that is a spanning subgraph of 𝐻 . 3 Example: 𝐼 0

𝑜 Hamiltonicity of 𝐼 𝑛 Lemma 1 𝑜 , with s 1 ≠ 𝑡 2 Let 𝑡 1 , 𝑡 2 , 𝑡 3 , and 𝑡 4 be perfect states in 𝐼 𝑛 and 𝑡 3 ≠ 𝑡 4 . 𝑜 ) such that Then there exists an automorphism 𝑔 ∈ 𝐵𝑣𝑢(𝐼 𝑛 𝑔 𝑡 1 = 𝑡 3 and 𝑔 𝑡 2 = 𝑡 4 .

𝑜 Hamiltonicity of 𝐼 𝑛 Theorem 1 Every Hanoi graph is hamiltonian. Proof: Fix any 𝑛 ∈ ℤ ≥0 . The proof consists of two parts. • Part I: We will show by induction on 𝑜 that there exists a 𝑜 beginning and ending with vertices hamiltonian path in 𝐼 𝑛 that correspond to distinct perfect states. • Part II: We will use the result of Part I to construct a 𝑜+1 . hamiltonian cycle in 𝐼 𝑛

Theorem 1, Part I Base Case: Let 𝑜 = 1 . 1 is isomorphic to the complete graph The Hanoi graph 𝐼 𝑛 on 3 + 𝑛 vertices, which is hamiltonian, and so contains a hamiltonian path. 1 Example: 𝐼 2

Theorem 1, Part I Induction Hypothesis: 𝑜 has a hamiltonian path Fix any 𝑜 ≥ 1 and suppose 𝐼 𝑛 beginning and ending with vertices that correspond to distinct perfect states. 𝑜+1 corresponds to the puzzle obtained by adding a disc 𝐼 𝑛 with radius 𝑜 + 1 to the Tower of Hanoi puzzle that 𝑜 . correspond to 𝐼 𝑛 𝑜 𝑜+1 𝐼 𝑛 𝐼 𝑛

Theorem 1, Part I Without loss of generality, suppose all discs begin on peg 0. By the induction hypothesis, there is a hamiltonian path 𝑜 . between distinct perfect states in 𝐼 𝑛 By Lemma 1, perfect states are isomorphic, so there is a hamiltonian path between any two distinct perfect states. We can move disc 𝑜 + 1 stepwise through every peg from 0 to 2 + 𝑛 in the following way.

Theorem 1, Part I Before each step moving disc 𝑜 + 1 , we perform a hamiltonian path transferring the 𝑜 -tower of discs to a peg allowing disc 𝑜 + 1 to move. In general, before moving disc 𝑜 + 1 from peg 𝑗 to peg 𝑗 + 1 , we first move the 𝑜 -tower to peg 𝑗 + 2 (mod 3 + 𝑛 ).

Theorem 1, Part I After the last move of disc 𝑜 + 1 to peg 2 + 𝑛 , the 𝑜 - tower can be transferred to peg 2 + 𝑛 as well, again through a hamiltonian path 𝑜 . in 𝐼 𝑛 During this process, every possible state of all 𝑜 + 1 discs is achieved exactly once, completing a 𝑜+1 . hamiltonian path in 𝐼 𝑛

Theorem 1, Part II 𝑜+1 . We now construct a hamiltonian cycle in 𝐼 𝑛 Without loss of generality, let the initial vertex in the cycle 𝑜+1 . be 1,1, … , 1,0 ∈ 𝑊 𝐼 𝑛 By Part I, we can transfer the 𝑜 -tower of smaller discs from peg 1 to peg 2 through a hamiltonian path, followed by moving disc 𝑜 + 1 to peg 1. In this step, we’ve gone through every vertex with a 0 in the last entry, ending on vertex (2,2, … , 2,1 ).

Theorem 1, Part II Continuing in this way, we transfer the 𝑜 -tower through a hamiltonian path from peg 𝑗 + 1 to peg 𝑗 + 2 for each 𝑗 ∈ {0,1, … , 2 + 𝑛} , following each by a single move of disc 𝑜 + 1 from peg 𝑗 to peg 𝑗 + 1 , where each step is modulo 3 + 𝑛 . In each step, we go through every vertex with an 𝑗 in the last entry.

Theorem 1, Part II The process terminates when we transfer the 𝑜 - tower back to peg 1, followed by moving disc 𝑜 + 1 to peg 0. We have completed a path 𝑜+1 that goes through in 𝐼 𝑛 every vertex exactly once and ends on the initial 𝑜+1 contains vertex. Thus 𝐼 𝑛 a hamiltonian cycle. ■

Planar Graphs Definition A graph 𝐻 is called planar if it can be drawn in the plane without any crossings. Example: The complete graph The complete graph 𝐿 5 𝐿 4 is planar. is not planar

𝑜 Planarity of 𝐼 𝑛 Theorem 2 𝑜 , 𝐼 1 1 , and 𝐼 1 2 . The only planar Hanoi graphs are 𝐼 0 Proof: 1 and 𝐼 1 2 are planar by • Part I: We will show that 𝐼 1 constructing planar embeddings of each. 𝑜 is planar for all • Part II: We will show by induction that 𝐼 0 𝑜 ∈ ℕ . 𝑜 is non-planar for all 𝑛 ≥ 2 • Part III: We will show that 𝐼 𝑛 and 𝑜 ≥ 1 . 𝑜 is non-planar for all 𝑜 ≥ 3 . • Part IV: We will show that 𝐼 1

Theorem 2, Part I 1 and 𝐼 1 2 are planar, as demonstrated by planar 𝐼 1 embeddings. 2 is 3-connected (there is no pair of Note that, since 𝐼 1 vertices whose deletion results in a disconnected graph), 2 is essentially unique. this planar embedding of 𝐼 1

Theorem 2, Part I 𝒏, 𝒐 1 2 3 4 5 … 0 1 Y Y 2 3 4 5 ⋮

Theorem 2, Part II 𝑜 allows a planar We will show by induction on 𝑜 that 𝐼 0 embedding, whose infinite face is the complement of an equilateral triangle with side length 2 𝑜 − 1 , and whose corners are the perfect states. Base Case: Let 𝑜 = 1 . 1 corresponds to the Tower of Hanoi puzzle The graph 𝐼 0 with 1 disc on 3 pegs. The disc can move freely between 1 is isomorphic to the complete graph 𝐿 3 . the pegs, so 𝐼 0 1 is planar and it can be drawn as an Thus 𝐼 0 equilateral triangle with side length 1 = 2 1 − 1.

Theorem 2, Part II Induction Hypothesis: 𝑙 can be drawn without Fix any 𝑙 ∈ ℕ and suppose 𝐼 0 crossings such that its infinite face is the complement of an equilateral triangle with side length 2 𝑙 − 1 and the corners are the perfect states. 𝑙 by ( 0 ) , ( 1 ) , and Label the perfect states of 𝐼 0 2 , where ( 𝑗 ) is the 𝑙 -tuple consisting of all 𝑗 ’s.

Theorem 2, Part II 𝑙+1 in the following way. We construct 𝐼 0 𝑙 , one for each possible position of disc • Take 3 copies of 𝐼 0 𝑙 + 1 (peg 0, 1, or 2). • Relabel their vertices with (𝑙 + 1) -tuples ending in 0, 1, and 2, respectively. • Add 3 edges to form the adjacencies 0 , 1 ~( 0 , 2) , 1 , 0 ~( 1 , 2) , and 2 , 0 ~ 2 , 1 . 𝑙 is an equilateral triangle, • Since each of the 3 copies of 𝐼 0 through flips we can arrange them so that each of the three edges added are the middle edges of a new equilateral triangle with side length 2 2 𝑙 − 1 + 1 = 2 𝑙+1 − 1 .

Theorem 2, Part II

Theorem 2, Part II We certainly have the adjacencies 0 , 1 ~( 0 , 2) , 𝑙+1 , since if the 𝑙 - 1 , 0 ~( 1 , 2) , and 2 , 0 ~ 2 , 1 in 𝐼 0 tower of smaller discs are all on one peg, then disc 𝑙 + 1 is free to move between the other two pegs. To verify that exactly 3 edges are added to the 3 copies of 𝑜 𝑙 to form 𝐼 0 𝑙+1 , we can use the edge count formula for 𝐼 𝑛 𝐼 0 𝑜 = (3 + 𝑛)(2 + 𝑛) 3 + 𝑛 𝑜 − 1 + 𝑛 𝑜 𝐹 𝑛 4 to show that 𝑙+1 = 3 𝐹 0 𝑙 + 3. 𝐹 0 𝑜 is planar for all 𝑜 ∈ ℕ . Thus 𝐼 0

Theorem 2, Part II 𝒏, 𝒐 1 2 3 4 5 … 0 … Y Y Y Y Y 1 Y Y 2 3 4 5 ⋮