Pre-Grundy Games Games And Graphs Workshop 2017 In collaboration - - PowerPoint PPT Presentation

Octal Games Pre-Grundy Games thks

Pre-Grundy Games

Games And Graphs Workshop 2017 In collaboration with : ´ Eric Duchˆ ene, Antoine Dailly and Urban Larsson Gabrielle Paris

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Octal Games Pre-Grundy Games thks

Are Pre-Grundy games boring ?

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Sommaire

1

Octal Games

2

Pre-Grundy Games

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Definition

Octal games: [Winning Ways] A game played on heaps where each player: 1. 2. 3.

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Definition

Octal games: [Winning Ways] A game played on heaps where each player:

• 1. removes all the tokens of a heap, or

1. 2. 3.

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Definition

Octal games: [Winning Ways] A game played on heaps where each player:

• 1. removes all the tokens of a heap, or
• 2. removes only some, at the end, or

1. 2. 3.

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Definition

Octal games: [Winning Ways] A game played on heaps where each player:

• 1. removes all the tokens of a heap, or
• 2. removes only some, at the end, or
• 3. removes only some in the middle

1. 2. 3.

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Definition

Octal games: [Winning Ways] A game played on heaps where each player:

• 1. removes all the tokens of a heap, or
• 2. removes only some, at the end, or
• 3. removes only some in the middle

Coded by an octal number 0.d1 . . . dt 1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...
• Kayles: 0.77.

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...
• Kayles: 0.77.

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...
• Kayles: 0.77.
• Dawson Chess: 0.137.

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...
• Kayles: 0.77.
• Dawson Chess: 0.137.

1. 2. 3.

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Example of octal games

• Nim: 0.33333333333333333...
• Kayles: 0.77.
• Dawson Chess: 0.137.

1. 2. 3. Conjecture (Guy) Octal games with finite code have a Grundy sequence ultimately periodic.

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Hexadecimal games

Why stop at two heaps ? 4.

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Hexadecimal games

Why stop at two heaps ? 4. 5. ⇒ we can simply generalize to leaving any number of heaps...

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Grundy sequences

5 10 15 20 5 10 15

Game 0.F2 : periodic

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Grundy sequences

5 10 15 20 5 10 15

Game 0.17FF : arithmetic-periodic

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Periodicity and arithmetic-periodicity results

Theorem (WW) Let H be the hexadecimal game 0.d1 . . . dt. If there exist e and p such that: ∀e < n ≤ 3(e + p) + t, G(n + p) = G(n) then the Grundy sequence of H is periodic of period p and with pre-period e.

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Periodicity and arithmetic-periodicity results

Theorem (WW) Let H be the hexadecimal game 0.d1 . . . dt. If there exist e and p such that: ∀e < n ≤ 3(e + p) + t, G(n + p) = G(n) then the Grundy sequence of H is periodic of period p and with pre-period e. Theorem (R. Nowakowski) Let H be the hexadecimal game 0.d1 . . . dt. If there exist e, 3p ≥ t + 2, s = 2γ−1 + j, j < 2γ−1 such that:

• 1. ∀e < n < t+αe,γ,jp,

G(n + p) = G(n) + s,

• 2. G(0, e) ⊂ 0, s − 1 and G(0, e + p) ⊂ 0, 2s − 1,
• 3. ∃d2v+1, dv ≥ 8, ∀g ∈ 0, 2s − 1, ∃n > 0,G(n) = g or

∃du ≥ 8, ∀g ∈ 0, 2s − 1, ∃2v + 1, 2w ≥ 0,G(2v + 1) = G(2w) = g, then the Grundy sequence of H is arithmetic-periodic of period p, pre-period e and saltus s.

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Periodicity and arithmetic-periodicity results

Theorem (WW) Let H be the hexadecimal game 0.d1 . . . dt. If there exist e and p such that: ∀e < n ≤ 3(e + p) + t, G(n + p) = G(n) then the Grundy sequence of H is periodic of period p and with pre-period e. Theorem (R. Nowakowski) Let H be the hexadecimal game 0.d1 . . . dt. If there exist e, 3p ≥ t + 2, s = 2γ−1 + j, j < 2γ−1 such that:

• 1. ∀e < n < t+αe,γ,jp,

G(n + p) = G(n) + s,

• 2. G(0, e) ⊂ 0, s − 1 and G(0, e + p) ⊂ 0, 2s − 1,
• 3. ∃d2v+1, dv ≥ 8, ∀g ∈ 0, 2s − 1, ∃n > 0,G(n) = g or

∃du ≥ 8, ∀g ∈ 0, 2s − 1, ∃2v + 1, 2w ≥ 0,G(2v + 1) = G(2w) = g, then the Grundy sequence of H is arithmetic-periodic of period p, pre-period e and saltus s.

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Sommaire

1

Octal Games

2

Pre-Grundy Games

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Grundy’s Game

A move consists in taking a heap and splitting it into two non-empty heaps of different sizes. No removing allowed.

1 1

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Grundy’s Game

A move consists in taking a heap and splitting it into two non-empty heaps of different sizes. No removing allowed.

1 1

Conjecture (Belekamp, Conway, Guy) The Grundy sequence of Grundy’s game is ultimately periodic. Remark: 235 first values computed (Flammenkamp) without any further clue...

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Pre-Grundy Games

Definition Let L = {ℓ1, . . . , ℓk} be a list of positive integers. A move on the game PreG(L) consist on splitting a heap of n ≥ ℓj tokens into ℓj + 1 non-empty heaps. L = {1, 3}

1 2 3 1

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Behavior

5 10 15 20 3

L = {1, 3} : P p = 2

5 10 15 20 5 9

L = {1, 3, 4} : AP p = 4; s = 2

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. First: G(n) ≥ a.

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. First: G(n) ≥ a.

• ℓ1 even:

Oℓ1 = (iℓ1 + b + 1, a − i, . . . , a − i) G(Oℓ1) = G(iℓ1 + b + 1) = i

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. First: G(n) ≥ a.

• ℓ1 odd:

− if a − i odd: Oℓ1 =

• iℓ1 + b + 1, 1, a − i − 1

2 ℓ1 + 1, a − i − 1 2 ℓ1 + 1, 1, . . . , 1

• G(Oℓ1) = G(iℓ1 + b + 1) ⊕ G(1) = i

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. First: G(n) ≥ a.

• ℓ1 odd:

− if a − i even: Oℓ1 =

• iℓ1 + b + 1, 2, a − i − 1

2 ℓ1 + 1 2, a − i − 1 2 ℓ1 + 1 2, 1, . . . , 1

• G(Oℓ1) = G(iℓ1 + b + 1) ⊕ G(2) = i

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. First: G(n) ≥ a. In all cases G(Oℓ1) = i for i < a ⇒ G(n) ≥ a.

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. Second: G(n) ≤ a. Assume G(Oℓ) = a for some option of n.

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. Second: G(n) ≤ a. Assume G(Oℓ) = a for some option of n. Oℓ = (a0ℓ1 + b0 + 1, . . . , aℓℓ1 + bℓ + 1)

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. Second: G(n) ≤ a. Assume G(Oℓ) = a for some option of n. Oℓ = (a0ℓ1 + b0 + 1, . . . , aℓℓ1 + bℓ + 1)

• G(Oℓ) = a

⇒ ai = a

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. Second: G(n) ≤ a. Assume G(Oℓ) = a for some option of n. Oℓ = (a0ℓ1 + b0 + 1, . . . , aℓℓ1 + bℓ + 1)

• G(Oℓ) = a

⇒ ai = a

• aℓ1 + b + 1 = aiℓ1 + bi + 1

⇒ ai ≤ a

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) Proofs: For L = {ℓ1, . . . , ℓk}, ℓi > 1. For n = aℓ1 + b + 1, b < ℓ1, let us prove that G(n) = a. Second: G(n) ≤ a. Assume G(Oℓ) = a for some option of n. Oℓ = (a0ℓ1 + b0 + 1, . . . , aℓℓ1 + bℓ + 1)

• G(Oℓ) = a

⇒ ai = a

• aℓ1 + b + 1 = aiℓ1 + bi + 1

⇒ ai ≤ a ⇒ ai = a and b ≥ ℓ ≥ ℓ1... absurd.

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1)

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) {1, ℓ2, . . . , ℓk}(ℓi odd ) P (01)∗

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Firsts results

L = {ℓ1, . . . , ℓk} type Sequence 1 / ∈ L AP (0)ℓ1 (+1) {1, ℓ2, . . . , ℓk}(ℓi odd ) P (01)∗ {1, 2, 3, ℓ4, . . . , ℓk} AP (0)1 (+1) {1, 3, 2ℓ}(ℓ ≥ 2) AP (01)ℓ (+2)

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Main result: arithmetic-periodicity test

Definition The game PreG({ℓ1, . . . , ℓk}), ℓk ≥ 2, satisfies the test AP if there exists p > 0 and s = 2j such that:

• AP1. For n ≤ 3p, G(n + p) = G(n) + s
• AP2. G(1, p) = 0, s − 1
• AP3. For n ∈ 3p + 1, 4p, g ∈ 0, s − 1, there is an option Oℓ,n of n

such that G(Oℓ,n) = g and ℓ ≥ 2.

0p + 1 0s 1p + 1 1s 2p + 1 2s 3p + 1 3s 4p + 1 4s n ℓ ≥ 2

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Main result: arithmetic-periodicity test

Definition The game PreG({ℓ1, . . . , ℓk}), ℓk ≥ 2, satisfies the test AP if there exists p > 0 and s = 2j such that:

• AP1. For n ≤ 3p, G(n + p) = G(n) + s
• AP2. G(1, p) = 0, s − 1
• AP3. For n ∈ 3p + 1, 4p, g ∈ 0, s − 1, there is an option Oℓ,n of n

such that G(Oℓ,n) = g and ℓ ≥ 2.

0p + 1 0s 1p + 1 1s 2p + 1 2s 3p + 1 3s 4p + 1 4s n ℓ ≥ 2

Remarks:

• Sometimes (AP1 and AP2) ⇒ AP3.

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AP-Theorem

Theorem Let L = {ℓ1, . . . , ℓk}, ℓk ≥ 2, such that PreG(L) verifies the AP-test. Then for all n ≥ 1, G(n + p) = G(n) + s.

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AP-Theorem

Theorem Let L = {ℓ1, . . . , ℓk}, ℓk ≥ 2, such that PreG(L) verifies the AP-test. Then for all n ≥ 1, G(n + p) = G(n) + s. Let us show that for n = ap + 1 + b: (A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g.

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AP-Theorem

Theorem Let L = {ℓ1, . . . , ℓk}, ℓk ≥ 2, such that PreG(L) verifies the AP-test. Then for all n ≥ 1, G(n + p) = G(n) + s. Let us show that for n = ap + 1 + b: (A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g. (Almost) clear for n ≤ 4: Lemma If n ≤ 4p, then n verifies (B). Let n ≥ 4p.

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Proof: G(n) ≤ as + G(1 + b)

(A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g.

0p + 1 0s 1p + 1 1s 2p + 1 2s (a − 2)p + 1 (a − 2)s (a − 1)p + 1 (a − 1)s ap + 1 as n as + G(1 + b) n − 2p (a − 2)s + G(1 + b)

X G(Oℓ,n) = as + G(1 + b) ⇒ ∃Oℓ,n−2p, G(Oℓ,n−2p) = G(n − 2p)...

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Proof: G(n) ≥ (a − 1)s

(A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g.

0p + 1 0s 1p + 1 1s 2p + 1 2s (a − 3)p + 1 (a − 3)s (a − 2)p + 1 (a − 2)s (a − 1)p + 1 (a − 1)s ap + 1 as n n − 2p n − 2p n ℓ ≥ 2

G(Oℓ,n−2p) = g, g < (a − 3)s, ℓ ≥ 2 ⇒ ∃Oℓ,n, G(Oℓ,n) = g

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Proof: G(n) ≥ (a − 1)s

(A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g.

0p + 1 0s 1p + 1 1s 2p + 1 2s (a − 3)p + 1 (a − 3)s (a − 2)p + 1 (a − 2)s (a − 1)p + 1 (a − 1)s ap + 1 as n n − 2p n − 2p n ℓ ≥ 2

G(Oℓ,n−2p) = g, g < (a − 3)s, ℓ ≥ 2 ⇒ ∃Oℓ,n, G(Oℓ,n) = g + 2s Moreover, n verifies (B).

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Proof: G(n) ≥ as + G(1 + b)

(A) G(n) = as + G(1 + b) (B) for g ∈ 0, (a − 1)s − 1, ∃Oℓ,n of n, with ℓ ≥ 2 and G(Oℓ,n) = g.

0p + 1 0s 1p + 1 1s 2p + 1 2s (a − 3)p + 1 (a − 3)s (a − 2)p + 1 (a − 2)s (a − 1)p + 1 (a − 1)s ap + 1 as n n − (a − 1)p n − (a − 1)p n

G(Oℓ,n−(a−1)p) = g, g < s + G(1 + b) ⇒ ∃Oℓ,n, G(Oℓ,n) = g + (a − 1)s ⇒ n verifies also (A)

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Conjectures and remarks

In summary: L = {ℓ1, . . . , ℓk} type Sequence S 1 / ∈ L AP (0)ℓ1 (+1)

• {1, ℓ2, . . . , ℓk}(ℓi odd )

P (01)∗

• {1, 2, 3, ℓ4, . . . , ℓk}

AP (0)1 (+1)

• {1, 3, 2ℓ}(ℓ ≥ 2)

AP (01)ℓ (+2)

• {1, 2ℓ}(ℓ ≥ 2)

AP (01)ℓ(23)ℓ14(54)ℓ−1(32)ℓ(45)ℓ(67)ℓ (+8) C {1, 2ℓ1, 2ℓ2 + 1}, ℓi > 1 AP (01)ℓ1 (+2) C Conjectures:

• 1. {1, 2ℓ}, ℓ ≥ 2 : OK for ℓ ∈ {2, 3, 4} by AP-test
• 2. ℓ1, ℓ2 > 1, {1, 2ℓ1, 2ℓ2 + 1} : OK for {1, 4, 5}, {1, 4, 7}, {1, 5, 6} by AP-test

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Conjectures and remarks

In summary: L = {ℓ1, . . . , ℓk} type Sequence S 1 / ∈ L AP (0)ℓ1 (+1)

• {1, ℓ2, . . . , ℓk}(ℓi odd )

P (01)∗

• {1, 2, 3, ℓ4, . . . , ℓk}

AP (0)1 (+1)

• {1, 3, 2ℓ}(ℓ ≥ 2)

AP (01)ℓ (+2)

• {1, 2ℓ}(ℓ ≥ 2)

AP (01)ℓ(23)ℓ14(54)ℓ−1(32)ℓ(45)ℓ(67)ℓ (+8) C {1, 2ℓ1, 2ℓ2 + 1}, ℓi > 1 AP (01)ℓ1 (+2) C {1, 2} ??? ??? Conjectures:

• 1. {1, 2ℓ}, ℓ ≥ 2 : OK for ℓ ∈ {2, 3, 4} by AP-test
• 2. ℓ1, ℓ2 > 1, {1, 2ℓ1, 2ℓ2 + 1} : OK for {1, 4, 5}, {1, 4, 7}, {1, 5, 6} by AP-test

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L = {1, 2}

10 20 30 40 50 60 70 80 90 100 5 10 15 20 25 30 35 40 45 50

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L = {1, 2}

500 1000 1500 2000 2500 3000 3500 4000 500 1000 1500 2000

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...maybe not!

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Thank you!

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