4 edited body edited Sep 28 '15 at 7:45 Gamow 43k1010 gold badges128128 silver badges367367 bronze badges This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard. Plain: plain numbering of the squares suffices. This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard. Plain numbering of the squares suffices. This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard: plain numbering of the squares suffices. 3 little improvements edited Apr 18 '15 at 13:31 JLee 12.1k11 gold badge3939 silver badges133133 bronze badges This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard: plain. Plain numbering of the squares suffices. This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard: plain numbering of the squares suffices. This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard. Plain numbering of the squares suffices. 2 little improvements edit approved Apr 18 '15 at 13:31 Marco Bonelli 2,17522 gold badges1313 silver badges2626 bronze badges This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection.). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. (Note Note also that for the trick we do not need to know the geometric structure of the checkerboard. A: plain numbering of the squares suffices.) This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection.) Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. (Note also that for the trick we do not need the geometric structure of the checkerboard. A plain numbering of the squares suffices.) This indeed is an old puzzle. One possible source (but certainly not the first one) is: Andy Liu: Two Applications of a Hamming Code The College Mathematics Journal 40, (Jan 2009), pp. 2-5 The trick on the $$2^k\times2^k$$ board is to associate each of the $$2^{2k}$$ squares with a unique binary number with $$2k$$ bits. (Note that the number of squares equals the number of such binary labels, so that there indeed exists a bijection). Let $$y$$ denote the binary label of the square picked by $$P$$ from the audience. Consider the checkerboard at the moment just after $$P$$ has flipped his coin, and consider the binary labels of all the squares with a white piece. Let $$z$$ denote the XOR of all such white binary labels; hence $$z$$ is another binary number with $$2k$$ bits. Your goal is to flip a piece, so that afterwards the XOR of all white binary labels becomes equal to $$y$$. Mathematically, you would like to find a square whose binary label $$x$$ satisfies $$x \oplus z=y$$. Since the operation $$\oplus$$ is associative, and since $$z\oplus z=0$$ and $$z\oplus 0=z$$ for all $$z$$, this desired equation $$x \oplus z=y$$ is equivalent to $$x\oplus z \oplus z=y\oplus z$$ and furthermore to $$x=y\oplus z$$. Hence if we simply flip the square with binary label $$y\oplus z$$, our partner will compute the XOR of all white binary labels and then announce the corresponding square to the audience. Done. Note also that for the trick we do not need to know the geometric structure of the checkerboard: plain numbering of the squares suffices. 1 answered Mar 17 '15 at 9:54 Gamow 43k1010 gold badges128128 silver badges367367 bronze badges