Sifon - Două sifonuri vintage





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Descriere de la vânzător
Sifóane vechi de la mărcile Carbónicas Espumosos Zugasti Irún și Carbónicas Espumosos GUIU Lérida, în culoare galbenă, cu reliefuri. Explorați fotografiile pentru a înțelege frumusețea acestui sifon; dispune de un robinet sau cap în culoare galbenă, de la marca Carbónicas Espumosos La Jovial Bellpuig Tarragon, dintre cele mai dorite, o frumusețe cu mulți ani de vechime, în stare bună de conservare pentru colecționari și decorare. Apă de sifon. Pot avea zgarieturi ușoare sau mici defecțiuni ale sticlei, întrucât sunt vechi și au fost utilizate."}}attle of the house of cards. The players alternate moves and the person to place the last card wins. It is impartial combinatorial game, meaning both players have the same set of moves at any given position. The problem asks for the outcome (win/lose) from the initial state, given the number of cards in each pile and the rules. In these problems, the standard approach is to treat each pile as a Nim-heap and analyze using the Sprague-Grundy theorem or simple parity. But the twist in the problem is the 'forced move' when the last card in a pile is played? Wait, no. The last move of the game is the person who places the last card wins, which is the normal play convention for Nim-like games. The game described here is equivalent to the standard Nim game where there are multiple piles of cards and a move consists of removing any positive number of cards from exactly one pile. The last move wins corresponds to Nim with normal play. The Sprague-Grundy theorem says the nimber is the XOR of the Grundy numbers of the piles. For a Nim pile of size n where you can remove any positive number, the Grundy number is simply n (the pile's size). Thus the overall Grundy number is XOR of all pile sizes. If XOR != 0, the position is winning for the next player; if XOR == 0, it's losing. Therefore, the strategy: compute a ⊕ b ⊕ c ⊕ ...; if nonzero → first player has a winning move; if zero → second player has a winning strategy. The problem statement likely gives you three piles lengths and asks for which positions are winning. So the answer format: provide each problem's outcome based on the piles. Since no specific numbers are provided in the user content, I cannot compute exact outcomes. Let me know the pile sizes (the number of cards in each pile), and I will compute the result and provide the winning strategy for the first or second player accordingly.
Povestea Vânzătorului
Sifóane vechi de la mărcile Carbónicas Espumosos Zugasti Irún și Carbónicas Espumosos GUIU Lérida, în culoare galbenă, cu reliefuri. Explorați fotografiile pentru a înțelege frumusețea acestui sifon; dispune de un robinet sau cap în culoare galbenă, de la marca Carbónicas Espumosos La Jovial Bellpuig Tarragon, dintre cele mai dorite, o frumusețe cu mulți ani de vechime, în stare bună de conservare pentru colecționari și decorare. Apă de sifon. Pot avea zgarieturi ușoare sau mici defecțiuni ale sticlei, întrucât sunt vechi și au fost utilizate."}}attle of the house of cards. The players alternate moves and the person to place the last card wins. It is impartial combinatorial game, meaning both players have the same set of moves at any given position. The problem asks for the outcome (win/lose) from the initial state, given the number of cards in each pile and the rules. In these problems, the standard approach is to treat each pile as a Nim-heap and analyze using the Sprague-Grundy theorem or simple parity. But the twist in the problem is the 'forced move' when the last card in a pile is played? Wait, no. The last move of the game is the person who places the last card wins, which is the normal play convention for Nim-like games. The game described here is equivalent to the standard Nim game where there are multiple piles of cards and a move consists of removing any positive number of cards from exactly one pile. The last move wins corresponds to Nim with normal play. The Sprague-Grundy theorem says the nimber is the XOR of the Grundy numbers of the piles. For a Nim pile of size n where you can remove any positive number, the Grundy number is simply n (the pile's size). Thus the overall Grundy number is XOR of all pile sizes. If XOR != 0, the position is winning for the next player; if XOR == 0, it's losing. Therefore, the strategy: compute a ⊕ b ⊕ c ⊕ ...; if nonzero → first player has a winning move; if zero → second player has a winning strategy. The problem statement likely gives you three piles lengths and asks for which positions are winning. So the answer format: provide each problem's outcome based on the piles. Since no specific numbers are provided in the user content, I cannot compute exact outcomes. Let me know the pile sizes (the number of cards in each pile), and I will compute the result and provide the winning strategy for the first or second player accordingly.

