A class of extensions of Restricted (s, t)-Wythoff’s game

Lemma 2.1

For all n ∈ ℤ⁺, An′−An−1′∈{2,4},Bn′−Bn−1′ ∈ {2s + t +δ_t,\ 4s + t + δ_t}. Moreover, Bn′−Bn−1′=2s+t+δt if and only if An′−An−1′=2;Bn′−Bn−1′=4s+t+δt if and only if An′−An−1′=4.

Proof

Clearly 0<An′−An−1′∈Zeven. Now suppose An′−An−1′≥6. Then we have An−1′<An−1′+2<An−1′+4<An′=mex{Ai′,Ai′+1,Bi′,Bi′+1:0≤i<n}. Thus An−1′ +2 and An−1′ + 4 are in the set {Ai′,Ai′+1,Bi′,Bi′+1:0≤i<n}. Now A′ and B′ are disjoint, thus the only possibility is that both An−1′ + 2 and An−1′ + 4 are in B′, i.e., the gap between the adjacent elements in B′ can be 2, which contradicts the fact that Bn′−Bn−1′∈{2s+t+δt,4s+t+δt}. Hence, An′−An−1′∈{2,4}. The remaining part is directly from the definition of Bn′ and the above. □

Lemma 2.2

For every n > m ≥ 0, 2(n − m) ≤ An′−Am′≤4(n−m). In particular, letting m = 0, gives 2n ≤ An′ ≤4n.

Proof

By Lemma 2.1, 2(n − m) ≤ An′−Am′=∑j=mn−1(Aj+1′−Aj′)≤4(n−m).  □

Lemma 2.3

Let ϕ_n = An′ − 2n. For r ∈ ℤ⁰,

In particular, for fixed integer parameters s, t with 2s + t + δ_t > 4, then Aj′=2j if and only if 2 ≤ 2j ≤ 2s + t + δ_t − 2, and Aj′=2j+2 if and only if 2s + t + δ_t ≤ 2 j ≤ 4s + 2t + 2δ_t − 4.

Proof

Let S = {x : 0 < x≤ Aj′ , and x ∈ ℤ^even. By ♯(S) we denote the number of elements of the set S. Put ♯A′=♯(A′∩S)and♯B′=♯(B′∩S). Since A′ ∩ B′ = ∅, A′ ∪ B′ = ℤ^even\ {0}, it is easy to derive that ♯ A′ + ♯ B′ = ♯(S) = Aj′/2,♯A′=j,and so♯B′=Aj′/2−j.

Assume r is exactly the largest index such that Br′<Aj′. Now Bn′ is a strictly increasing sequence. Thus ♯B′ = r. Therefore,

From Eq. (2), A0′=B0′=0,A1′=mex{0,1}=2,andB1′=2s+t+δt. For 2s + t + δ_t > 4, we obtain A2′=mex{0,1,2,3,2s+t+δt,2s+t+δt+1}=4,and thenB2′=4s+2t+2δt. The rest of the proof is obtained by letting r = 0 and r = 1 in Eq. (3), respectively. □

Symmetry of our game rules implies that a game position (a, b) is identical to (b, a) . Clearly, the P-generators added to Restricted (s, t)-Wythoff alter the set of P-positions of Restricted (s, t)-Wythoff. But the set of P-positions of each Γ_i, i ∈ {1, 2, 3, 4 }, is still four sequences of pairs of integers. Denote by {(A_n, B_n), (A_n, B_n + 1), (A_n + 1, B_n), (A_n + 1, B_n + 1)}_{n≥ 0} the set of Γ_i, and for any n ∈ ℤ⁰, A_n, B_n ∈ ℤ^even, A_n ≤ A_n+1 and A_n ≤ B_n. Like in Restricted (s, t)-Wythoff, the pair (A_n, B_n) is a P-generator of P-positions in each Γ_i.

Lemma 2.4

For every n > m ≥ 0, B_n+1 > B_n > A_n > A_m.

Proof

First, by the notations above, A_n ≥ A_m and B_n ≥ B_m. Suppose A_n = A_m, this implies B_n ≠ B_m, or else, (A_n, B_n) = (A_m, B_m), impossible. So if B_n > B_m, then there is a move (A_n, B_n)→(A_m, B_m) (Type I, and k = B_n − B_m ∈ ℤ^even), but it contradicts stability property of P-positions. Now suppose B_n < B_m, then (A_m, B_m)→(A_n, B_n), resulting in another contradiction. Thus A_n > A_m. Similarly, B_n > B_m.

We prove below A_n < B_n. Again by the notations, A_n ≤ B_n. If A_n = B_n for some n ∈ ℤ⁺, then we can move (A_n, B_n)→(0,0) (Type II, satisfies condition (1)), another contradiction. Therefore, B_n+1 > B_n > A_n > A_m. □

Lemma 2.5

Put A=⋃i=1∞AiandB=⋃i=1∞Bi. Then A ∩ B = ∅, and A ∪ B = ℤ^even\ {0}.

Proof

We first show A ∩ B = ∅. Suppose A_n = B_m for some m, n ∈ ℤ⁺. By Lemma 2.4, m ≠ n, and B_n > A_n = B_m > A_m. Thus there is a move of Type I that (A_n, B_n)→(A_m, B_m) with B_n → A_m, which is impossible.

Clearly A₀ = B₀ = 0, and A ∪ B ⊆ ℤ^even\ {0}. Suppose there exists u ∈ ℤ^even\ {0} but u ∉ A ∪ B. Thus for every v ∈ ℤ^even, (u,v)∉⋃i=1∞(Ai,Bi), that is, (u, v) is an N-position. We show below that there exists some v₀ ∈ ℤ^even such that (u, v₀) is not an N-position, by proving F(u, v₀)∩ 𝓟 = ∅, i.e., no follower of (u, v₀) is a P-position.

Let n₀ be the largest index such that A_n0 < u. Then from (u, v) we can move to (A_i, B_i) only for some i ≤ n₀. In particular, let m₀ = j be the largest index of Type III move (Aj′,Bj′) from (u, v) . More concretely,

Now Let D = max{B_i: i ≤ n₀} + 2, and v₀ = D + su(t + 1). Note that v₀ ∈ ℤ^even. It is obvious that from (u, v₀) we cannot lead to any position of the form (A_i + 1, B_i), (A_i, B_i + 1) or (A_i + 1, B_i + 1). So it suffices to prove that from (u, v₀) none of Type I, II and III moves leads to (A_i, B_i) with i ≤ n₀.

For every i ≤ n₀, v₀ > max{B_i: i ≤ n₀} ≥ B_i > A_i. Then v₀ ∉ ⋃i=1n0 A_i ∪ ⋃i=1n0 B_i. We know u ∉ ⋃i=1n0 A_i ∪ ⋃i=1n0 B_i. Hence, we cannot move from (u, v₀) to any (A_i, B_i) with i ≤ n₀ by a move of Type I.

For all i ≤ n₀, we have k = u − A_i > 0, ℓ = v₀ − B_i = D − B_i + su(t + 1)> u − A_i = k, and ℓ − k = D − B_i + (s − 1)u + A_i + sut > (s − 1)(u − A_i)+ t = (s − 1)k + t. Therefore, no move of Type II from (u, v₀) leads to (A_i, B_i), i ≤ n₀.

Finally, we consider the move of Type III. By Lemma 2.2, An′≥2n, and so

by virtue of (t+δt)/2≤tandu≥Am0′. By the above, (Am0′,Bm0′) is the maximum Type III move from (u, v₀). However, v0−Bm0′=D+su(t+1)−Bm0′≥D>Bi>Aifor everyi≤n0. Therefore, no move of Type III from (u, v₀) can lead to a P-position.

In conclusion, u ∈ A ∪ B, and then A ∪ B = ℤ^even\ {0}. □

Theorem 3.1

Given s, t ∈ ℤ⁺ with 2s + t + δ_t > 4. For Γ₁, P=⋃n=0∞{(An,Bn),(An,Bn+1),(An+1,Bn),(An+1,Bn+1)}, where for n ≥ 0,

Example 3.2

For s = 2, t ∈ {3, 4}, we display the first few P-generators of Γ₁ in the table below, which show us how to determine 𝓟 by using Eq. (4).

Proof

Notice first that the condition 2s + t + δ_t > 4 implies that the case s = 1, t ∈ {1, 2 } is not covered. Thus we have either (s = 1, t > 2) or (s > 1, t ≥ 1).

For s = 1 and t > 2, the game was solved in [13], where Theorem 10 is precisely our assertion. But the structure of P-positions of the game is of algebraic form, which provides a poly-time winning strategy for our game. In addition, the methods of proof in [13] are totally different from those for the case s > 1 and t ≥ 1 in this paper.

For s > 1 and t ≥ 1, before we give the proof, some useful properties of the sequences A_n and B_n are required. Analysis similar to that in the proof of Lemmas 2.1 and 2.2 shows that for every n > m ≥ 0, we have (i)2(n−m)≤An−Am≤4(n−m),(ii)An−An−1∈{2,4},(iii)Bn−Bn−1∈{2s+t+δt+2,4s+t+δt}. Particularly, B_n − B_{n− 1} = 2s + t + δ_t + 2 if and only if A_n − A_{n − 1} = 2, B_n − B_{n − 1} = 4s + t + δ_t if and only if A_n − A_{n − 1} = 4.

Let A = ⋃i=1∞ A_i and B = ⋃i=1∞ B_i, then by Lemma 2.5 we have the fact that A ∩ B = ∅ and A ∪ B = ℤ^even\ {0}.

Now it suffices to show two things:

1. Every move from a position u ∈ 𝓟 cannot terminate in 𝓟.

2. From every position v ∉ 𝓟, there exists alegal move v → u ∈ 𝓟.

Proof of I

Let (x, y) with x ≤ y be a position in 𝓟. Assume that a move from (x, y) leads to another position in 𝓟. Since A_n, B_n ∈ ℤ^even and so A_n + 1, B_n + 1 ∈ ℤ^odd, considering our game rules, we have eight possible moves for n > m ≥ 0:

Now none of them could be a Type I move, since A_n and B_n are strictly increasing sequences.

They cannot be moves of Type II either. Indeed, for the first four cases, clearly n > m ≥ 0, and we have

which contradicts Eq. (1). Next consider the last four cases. For every s > 1, t ∈ ℤ⁺ and every n ∈ ℤ⁰, we have B_n = (s − 1)A_n + (t + δ_t + 4)n > A_n. And by above, for every n > m ≥ 0, B_n − B_m ≥ s(A_n − A_m)+2 > A_n − A_m. It follows that B_n − A_m ≥ B_n − B_m>A_n − A_m ≥ A_n − B_m. Thus for these four cases, we have k′=An−Bm≤An−Am=k,ℓ′=Bn−Am≥Bn−Bm=ℓ,andℓ′−k′≥ℓ−k≥(s−1)k+t≥(s−1)k′+t, a contradiction.

It remains to consider the one and only Type III move (A1′,B1′) =(2,2s+ t +δ_t). For the first four cases, note that A_n − A_m = 2 implies n − m = 1. Then we get

a contradiction. For the remaining four cases, if A_n − B_m = A1′ =2 and B_n − A_m = B1′ = 2s+ t +δ_t. Now A_n − A_m ≥ 2(n − m) ≥ 2. Then A1′+B1′ = B_n − B_m + A_n − A_m = s(A_n − A_m) + (t + δ_t + 4)(n − m) ≥ 2s+ t +δ_t + 4, which is impossible. Hence, any one of the hypothetical moves above cannot be a Type III move, ending the proof of I.

Proof of II

Without loss of generality, let (x, y) with x≤y be a position not in 𝓟.

If x ∈ {0, 1}, then we move y → δ_y, since clearly y − δ_y is even. In what follows, suppose x ≥ 2. Now A ∩ B = ∅ and A ∪ B = ℤ^even\ {0}. So x appears exactly once in exactly one of ⋃n=1∞ {A_n} ∪ ⋃n=1∞ {A_n + 1} and ⋃n=1∞ {B_n} ∪ ⋃n=1∞ {B_n + 1}. Thus we have one of the following two cases:

Case (i) x = B_n or B_n + 1, for some n ∈ ℤ⁺. Then y ≥ x ≥ B_n > A_n + 1 ≥ A_n + δ_y, thus move y → A_n + δ_y since y − A_n − δ_y is even.

Case (ii) x = A_n or A_n + 1, for some n ∈ ℤ⁺. In this case, we have y > B_n + 1 or x ≤ y < B_n. If y > B_n + 1, then move y → B_n + δ_y since 0 < y−B_n−δ_y∈ ℤ^even. If x ≤ y <B_n, we distinguish the following three subcases:

(ii. 1) y = B_n−1 or B_n−2. We know A_n−A_n−1 ∈ {2, 4} for all n ∈ ℤ⁺. If A_n−A_n−1 = 2, then B_n−B_n−1 = 2s + t + δ_t + 2. We move (x, y)→(x−2, y−2s−t−δ_t) by a move of Type III, with k = A1′ = 2, ℓ = B1′ = 2s + t + δ_t. Notice that x−2 = A_n−1 or A_n−1 + 1, and y−2s−t−δ_t = B_n−1 or B_n−1 + 1. Hence, (x − 2, y − 2s−t−δ_t)∈ 𝓟.

If A_n−A_n−1 = 4, then B_n−B_n−1 = 4s + t + δ_t. We move (x, y)→(x−4, y−4s−t−δ_t + 2), Which is a legal move of Type II: First, it is easy to see that x−4 = A_n−1 or A_n−1 + 1, and y−4s−t−δ_t + 2=B_n−1 or B_n−1 + 1. Thus (x−4, y−4s−t−δ_t + 2)∈ 𝓟. Secondly, k = 4, ℓ= 4s + t + δ_t−2, and 0 < k ≤ ℓ = 4s + t + δ_t−2 < sk + t, which satisfies Eq. (1).

(ii.2) x ≤ y < sA_n + t + δ_t. In this subcase, we move (x, y)→(x−A_n, δ_y)∈ 𝓟. This is a move of Type II, with k = A_n and ℓ = y−δ_y. Indeed, both k and ℓ are even, then 0 < k = A_n ≤ y−δ_y = ℓ. And sA_n + t + δ_t is even, which implies y−δ_y ≤ sA_n + t + δ_t−2, and so ℓ−k = y−δ_y−A_n ≤ (s−1)A_n + t + δ_t−2 < (s−1)k + t.

(ii.3) sA_n + t + δ_t ≤ y < B_n−2. Put

where ⌊ x ⌋ denotes the largest integer ≤ x. Then move (x, y)→(x−A_n + A_m, B_m + δ_y)∈ 𝓟 by a move of Type II, which is legal:

First, k = A_n−A_m and ℓ = y−B_m−δ_y are even. Next we have (a) k > 0, (b) ℓ ≥ k > 0, (c) ℓ < sk + t. Indeed,

1. Since A_n ≥ 2n, we have y−(s−1)A_n−2n + 2−δ_y≥sA_n + t + δ_t−(s−1)A_n−2n + 2>0, thereby m ≥ 0. On the other hand, y−(s−1)A_n−2n + 2−δ_y<B_n−(s−1)A_n−2n = (t + δ_t + 2)n, so 0 ≤ m < n. Thus k = A_n−A_m > 0.

2. By the definition of m, we have

   m≤y−(s−1)An−2n+2−δyt+δt+2,

   so y ≥(t + δ_t + 2)m + (s−1)A_n + 2n−2 + δ_y. By (a) n>m and s>1, we get

   ℓ=y−Bm−δy≥(t+δt+2)m+(s−1)An+2n−2−(s−1)Am−(t+δt+4)m=(s−1)(An−Am)+2(n−m)−2≥(s−1)k≥k>0.

3. By the definition of m, we have

   y−(s−1)An−2n+2−δyt+δt+2−1<m,

   i.e., y−δ_y < (t + δ_t + 2)(m + 1) + (s−1)A_n + 2n−2. Further, note that both sides of this inequality are even, then y−δ_y ≤ (t + δ_t + 2)(m + 1) + (s−1)A_n + 2n−4. It follows from 2(n−m) ≤ (A_n−A_m) that

   ℓ=y−Bm−δy≤(t+δt+2)(m+1)+(s−1)An+2n−4−(s−1)Am−(t+δt+4)m=(s−1)(An−Am)+t+δt+2(n−m)−2≤s(An−Am)+t+δt−2<sk+t.

The proof is completed. □

Theorem 3.1 provides a recursive winning strategy in terms of the mex function, which is exponential in the input size log xy of any game position (x, y)∈ ℤ⁰× ℤ⁰. For more theory of computing complexity of heap games, see [6]. Next, the central question we address here is whether our game has a better strategy, such as a polynomial time winning strategy.

We introduce a numeration system that turns out to be relevant to our game Γ₁. For fixed integers s > 1 and t ≥1, let u−1 = 2/(s−1), u₀ = 2, and put u_n = (s + ⌈ t/2⌉)u_n−1 + (s−1)u_n−2(n ≥ 1). Here and subsequently, ⌈ x ⌉ stands for the smallest integer ≥ x. Denote by 𝒰 the numeration system with bases u₀, u₁,… and digits d_i∈{0, 1,…, s + ⌈ t/2 ⌉}. Note that an integer such as u_n has two representations: u_n itself and (s + ⌈ t/2⌉)u_n−1 + (s−1)u_n−2. Since we would like to have uniqueness of representation, it is natural to stipulate that d_i = s + ⌈ t/2⌉ ⟹ d_i−1 < s−1 (i ≥ 1). Then we claim two facts:

1. For every n ≥1, u_n∈ ℤ^even, clearly by induction on n.

2. Every decimal number N ∈ ℤ^even has a unique representation R(N) over 𝒰. This is a special case of Theorem 3 in [14]. Note that the greedy algorithm of repeatedly dividing N or its remainder by the largest u_i not exceeding this remainder gives the unique representation.

Those whose representations R(N) end in an even number of 0s are called vile numbers, and those whose representations R(N) end in an odd number of 0s are called dopey numbers (for an etymology of the terms vile, dopey, see [15]). In addition, by LR(N) we denote the “left shift” of R(N), i.e., LR(N) is obtained from R(N) by adjoining 0 to the right end of R(N).

Example 3.3

We consider Γ₁ of Example 3.2, where s = 2, t∈ {3, 4}, and so ⌈ t/2⌉ = 2. Thus, u₋₁ = 2, u₀ = 2, u₁ = 10, u₂ = 42, u₃ = 178, …. The representations R(N) over 𝒰 of the first few numbers N∈ ℤ^even appear in Table 2.

A question we just might ask at this point is what the connection is between Tables 1 and 2. By scanning the first few entries of both tables, we may be tempted to conclude that all A_ns in Table 1 are vile, also it seems that all B_ns are dopey. Moreover, we consider a P-generator say (12, 52) in Γ₁ with representations R(12, 52) = (11, 110). It is obvious that LR(12) = L(11) = 110 = R(52). Now consider an N-position (20, 84) in Γ₁, whose representations are R(20, 84) = (20, 200), we also have LR(20) = R(84). Therefore, it appears that there is no causal relation between the facts that (x, y)∈ 𝓟 in Γ₁ and the condition R(y) = LR(x). We next show how to determine P- positions of Γ₁ via 𝒰.

Lemma 3.4

Let {V_m}_m≥0 denote the set of all vile numbers in 𝒰 with 0 = V₀ < V₁ < V₂< …, and put R(D_m) = LR(V_m) for all m. Then

Proof

Induction on m. Clearly for m = 0. Suppose D_m−(s−1)V_m = (2⌈ t/2⌉ + 4)m for arbitrary but fixed m. It suffices to prove that the assertion holds for m + 1. Let V_m = ∑i=0ndiui. Since R(D_m) = LR(V_m), D_m = ∑i=0ndjui+1. So we have

Let q = s−2 and r = s + ⌈ t/2 ⌉. Then the linear recurrence of 𝒰 has the form u_n = ru_n−1 + (q + 1)u_n−2 (n≥1), with the digits d_i∈{0, 1, …, r} such that d_{i + 1} = r ⇒ d_i ⟹ q(i≥0). We proceed by distinguish three cases, because the tail of R(V_m) must be one of the following three forms:

1. The tail of R(V_m) has digits

   d2kd2k−1d2k−2…d2d1d0=d2krq…rqrq,

   for some k ∈ ℤ⁰, where d_2k ∈ {0, 1,…, q} and d_2k = q ⇒ d_2k+1 < r. Then it is not hard to see that V_m + 2 = (d_2k+1)u_2k + ∑i=2k+1ndiui, and so V_m + 2 is vile. Thus V_m+1 = V_m + 2, since V_i is even for every i∈ ℤ⁰.

   un	…	u2k	…	u3	u2	u1	u0	N
   dn	…	d2k	…	r	q	r	q	Vm
   dn	…	d2k+1	…	0	0	0	0	Vm+2

   Thus

   Dm+1−(s−1)Vm+1=(d2k+1)(u2k+1−(s−1)u2k)+∑i=2k+1ndi(ui+1−(s−1)ui)=u2k+1−(s−1)u2k+∑i=2kndi(ui+1−(s−1)ui). (7)

   On the other hand, by Eq. (6) and the recurrence ru_n−1 + (q + 1)u_n−2 = u_n, we have

   (2⌈t/2⌉+4)m=q(u1−(s−1)u0)+r(u2−(s−1)u1)+q(u3−(s−1)u2)+⋯+q(u2k−1−(s−1)u2k−2)+r(u2k−(s−1)u2k−1)+∑i=2kndi(ui+1−(s−1)ui)=u2k+1−u1−(s−1)u2k+(s−1)u0+∑i=2kndi(ui+1−(s−1)ui)=u2k+1−(s−1)u2k−(2⌈t/2⌉+4)+∑i=2kndi(ui+1−(s−1)ui).

   Therefore, it follows from Eq. (7) that

   (2⌈t/2⌉+4)(m+1)=u2k+1−(s−1)u2k+∑i=2kndi(ui+1−(s−1)ui)=Dm+1−(s−1)Vm+1.

2. The tail of R(V_m) has digits

   d2k+1d2kd2k−1…d2d1d0=d2k+1rq…rqrqr,

   for some k ∈ ℤ⁰, where d_2k+1 ∈ {0, 1,…, q} and d_{2k + 1} = q ⇒ d_2k+2 < r. Then V_m + 2 = (d_2k+1 + 1)u_2k+1 + ∑i=2k+2ndiui, and so V_m + 2 is dopey. But V_m + 4 is vile, since R(V_m + 4) ends in 1. Thus V_{m + 1} = V_m + 4 = u₀ + (d_{2k + 1} + 1)u_2k+1 + ∑i=2k+2ndiui.

   un	…	u2k+1	…	u2	u1	u0	N
   dn	…	d2k+1	…	r	q	r	Vm
   dn	…	d2k+1+1	…	0	0	0	Vm+2
   dn	…	d2k+1+1	…	0	0	1	Vm+4

   Hence,

   Dm+1−(s−1)Vm+1=(u1−(s−1)u0)+(d2k+1+1)(u2k+2−(s−1)u2k+1)+∑i=2k+2ndi(ui+1−(s−1)ui)=(2⌈t/2⌉+4)+u2k+2−(s−1)u2k+1+∑i=2k+1ndi(ui+1−(s−1)ui). (8)

   On the other hand, by Eq. (6),

   (2⌈t/2⌉+4)m=r(u1−(s−1)u0)+q(u2−(s−1)u1)+⋯+q(u2k−(s−1)u2k−1)+r(u2k+1−(s−1)u2k)+∑i=2k+1ndi(ui+1−(s−1)ui)=u2k+2−(s−1)u2k+1+∑i=2k+1ndi(ui+1−(s−1)ui).

   The last equality follows from that summing the positive terms, adding and subtracting (s−1)u₀, leads to u_2k+2− 2(s−1), and that summing the negative terms, subtracting and adding (s−1)²u−1 = 2(s−1), leads to −(s − 1)u_2k+1 + 2(s−1). Thus, by Eq. (8), D_1m+1−(s−1)V_m+1 = (2⌈ t/2⌉ + 4)(m + 1), as required.

3. The digit d₀ satisfies q < d₀ < r. By the definition of 𝒰, d₁ < r, and so V_m+1 = V_m + 2. Thus it follows from Eq. (6) that

   Dm+1−(s−1)Vm+1=(d0+1)(u1−(s−1)u0)+∑i=1ndi(ui+1−(s−1)ui)=(2⌈t/2⌉+4)+∑i=0ndi(ui+1−(s−1)ui)=(2⌈t/2⌉+4)(m+1).

Finally, note that if V_m ends in an even number of 0s, it is of the form (i). The proof is completed. □

Theorem 3.5

For all n ∈ ℤ⁰, (V_n, D_n) = (A_n, B_n).

Proof

Obviously, (D_n, D₀) = (A₀, B₀) = (0, 0). For n ≥ 1, we have two facts:

1. A_n, B_n, V_n, D_n∈ ℤ^even;

2. For all t∈ ℤ⁺, t + δ_t = 2⌈ t/2⌉.

By Eqs. (4) and (5), together with Fact (ii), B_n and D_n have the same formation law. It remains only to show that A_n and V_n have the same inductive formation rule, i.e., V_n = mex{V_i, V_i + 1, D_i, D_i + 1 : 0 ≤ i < n}.

Now for every n ≥ 1, D_n is dopey, since V_n is vile and R(D_n) = LR(V_n). By Fact (i), D_n>V_n + 1 > V_n. It follows that ⋃i=1∞Viand⋃i=1∞Di are complementary with respect to ℤ^even(>0), since every positive integer has precisely one representation in 𝒰₁, either vile as V_ns do or dopey as the D_ns do. Let S = {V_i, V_i + 1, D_i, D_i + 1 : 0 ≤ i < n}. Suppose mex S = D_j for some j ≥ n. This implies that V_j ∈ S on account of V_j < D_j, and then j < n, a contradiction. Therefore, mex S = V_n, the assertion follows. □

Corollary 3.6

Given a position (x, y)∈ Γ₁ with s > 1. (x, y) is a P-position if and only if there exists some n ∈ ℤ⁰ such that (⌊ x/2⌋, ⌊ y/2⌋) = (V_n, D_n).

Proof

If (x, y) is a P-position of Γ₁ with its P-generator (⌊ x/2 ⌋, ⌊ y/2 ⌋) = (A_n, B_n) for some n ∈ ℤ ⁰, then by Theorem 3.5, (A_n, B_n) = (V_n, D_n), and vice versa. □

Given any game position (x, y) with 0 < x ≤ y, compute R(x) using the greedy algorithm mentioned above. If R(x) ends in an odd number of 0s, then x = B_n for some n ∈ ℤ⁺. Then we move y → A_n, where R(B_n) = LR(A_n). If R(x) ends in an even number of 0s, then x = A_n for some n ∈ ℤ⁺. The relative size of y and B_n can also be tested, since R(B_n) = LR(A_n). Thus the complexity of this computation, up to a multiplication constant, is that of computing R(x), which is linear in log x. Hence, this winning strategy bases on 𝒰 is polynomial in the input size log x.

Lemma 4.1

Let A_n and B_n be determined by Eq. (4). Then for every n > m ≥ 0, let A_n−A_m = 2(n−m) + 2r. If r ≥ 1, then A_n−A_m≥(r−1)(2s + t + δ_t + 2) + 4.

Proof

Let S = {x : A_m < x ≤ A_n, x ∈ ℤ^even}. As in the proof of Lemma 2.3, ♯(S) = (A_n−A_m)/2, ♯_A = ♯(A ∩ S) = n−m, and ♯ B = ♯(B ∩ S) = ♯(S)− ♯ A = r.