XOR and XNOR 3Port Boolean for Hamming Code

XOR 3 port

3Port Boolean

Q = (A B` C) + (A` B C) + ( A B C`)

A`means inverted. AB`C means B is inverted and A and C is ordinary, not inverted. How to read : 0 means the switch is OFF, light is off. 1 means light is ON, switch is turned on. The AND gate is an electrical Series copling, and using a * multiply operand between the bits. And the OR gate is an electrical Parallell copling, using a + Sign between the bits.

Q = (A*B`*C) + (A`*B*C) + ( A*B*C`)

Truth Table

                               AND           AND         AND             OR

A B C   A`B`C`   AB`C Q1   A`BC Q2    ABC` Q3    Q1+Q2+Q3 = Q4

0 0 0     1 1 1    0*1*0 = 0  1*0*0 = 0  0*0*1 = 0         0+0+0 = 0

0 0 1     1 1 0    0*1*1 = 0  1*0*1 = 0  0*0*0 = 0         0+0+0 = 0

0 1 0     1 0 1    0*0*0 = 0  1*1*0 = 0  0*1*1 = 0         0+0+0 = 0

0 1 1     1 0 0    0*0*1 = 0  1*1*1 = 1  0*1*0 = 0         0+1+0 = 1

1 0 0     0 1 1    1*1*0 = 0  0*0*0 = 0  1*0*1 = 0         0+0+0 = 0

1 0 1     0 1 0    1*1*1 = 1  0*0*1 = 0  1*0*0 = 0         1+0+0 = 1

1 1 0     0 0 1    1*0*0 = 0  0*1*0 = 0  1*1*1 = 1         0+0+1 = 1

1 1 1     0 0 0    1*0*1 = 0  0*1*1 = 0  1*1*0 = 0         0+0+0 = 0

I am kinda stuck at
P1 = m3 XNOR m5 XNOR m7
P2 = m3 XNOR m6 XNOR m7
P4 = m5 XNOR m6 XNOR m7

for the Hamming Error detection code
p1, p2, m3, p4, m5, m6, m7

for a 4 Bit message code with 3 Parity bit correctors in a 7Bit word

And had to figure the XNOR  for a 3 port.

Ask Table. ( compressed truth table )

ABC    Q

0 0 0    0    odd of 1

0 0 1    0    odd

0 1 0    0    odd

0 1 1    1    pair of 1`s

1 0 0    0    odd

1 0 1    1    pair

1 1 0    1    pair

1 1 1    0    odd

This is used in the Hamming Code. However the Boolean is often overlooked, and a " trick is used" that is to look for 2 of the same bits then write manually  odd or pair. I want to do it machine wise .

The point of this is that it ouputs a Binary 1 where there is 2 of the BIN1 in the XNOR Truth Tabler .

XNOR

3Port Boolean

(A B`C`) + (A`B C`) + ( A`B`C)

This produce the opposite. where there is an pair of 0`s one output a Q=1.

Truth Table

A B C  A`B`C`  AB`C` Q1   A`BC` Q2    A`B`C Q3    Q1+Q2+Q3 = Q4

0 0 0    1 1 1    0*1*1 = 0    1*0*1 = 0    1*1*0 = 0          0+0+0 = 0

0 0 1    1 1 0    0*1*0 = 0    1*0*0 = 0    1*1*1 = 1          0+0+1 = 1

0 1 0    1 0 1    0*0*1 = 0    1*1*1 = 1    1*0*0 = 0          0+1+0 = 1

0 1 1    1 0 0    0*0*0 = 0    1*1*0 = 0    1*0*1 = 0          0+0+0 = 0

1 0 0    0 1 1    1*1*1 = 1    0*0*1 = 0    0*1*0 = 0          1+0+0 = 1

1 0 1    0 1 0    1*1*0 = 0    0*0*0 = 0    0*1*1 = 0          0+0+0 = 0

1 1 0    0 0 1    1*0*1 = 0    0*1*1 = 0    0*0*0 = 0          0+0+0 = 0

1 1 1    0 0 0    1*0*0 = 0    0*1*0 = 0    0*0*1 = 0          0+0+0 = 0

Generating a Syndrom word

I see there is some errors in the Syndrome word here. I presume it will be used a XNOR 4 port for the last conversion . not a XNOR 3 port ....

A sample preview of a 4Bit XOR. will output 1 on ODDS and 0 at PAIRS .