Flipped Colour

Currently we have $1101000$ as our starting value with positions $1=1, 2=1, 3=0, 4=1, 5=0, 6=0,$ and $7=0$ .

Analyzing the hints from Alice, we see that

$1+3+5+7 = 1 = odd$

$2+3+6+7= 1 = odd$

$4+5+6+7= 1 = odd$

but $2+3+6+7 = even$ according to Alice. Therefore one of these boxes is the switched box.

Now since the other two statements hold true, we cannot change a box within their sets or we would change the red count to an even number. Therefore we must change a box that is exclusive to $2, 3, 6,$ or $7$ . The only one that applies is box $2$ , therefore we change it to white, giving us $1001000$ as the binary representation.

Conditions 1 and 3 are satisfied, so squares numbered 1, 3, 4, 5, 6, 7 are okay. Condition 2 is false, showing that square 2 is wrong.

This is an example of a 1-bit self-correcting code of the type 7-3(-1). It is designed to work as follows:

• The message is sent as $x_7x_6x_5x_4x_3x_2x_1\ a_2a_1a_0\ e$ , where $a_n$ shows the number of $x_i$ for which bit $n$ is set (1 = odd, 0 = even). Here the original message was $0001001\ 101\ 0$ .

• When the message is received, the receiver calculates $a'_2a'_1a'_0$ in the same manner for the received 7 bits. In this case the message received is $0001011$ making $a' = 111$ .

• Now calculate $c = a\ \text{xor}\ a'$ . In this case, $c = 101\ \text{xor}\ 111 = 010$ . If $c = 0$ the message was received well. Otherwise, $c$ is equal to the bit with the error--here, bit 2. Flip this bit and you have the original message!

• What if the bits of the message are communicated well, but one bit of $a$ is messed up? For that situation, we set along an eleventh bit $e$ , which indicates whether $a$ has an even (0) or odd (1) number of ones. If after reception the values of $a'$ and $e'$ do not match, we know that the error occurred either in $b$ or $e$ ; assuming that no more than one bit was sent wrong, we simply accept the value of $x$ as received.