4 edited body
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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
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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
source | link

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.

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