| x | y | x' | y' | x⊕y | x'⊕y' |
----------------------------------
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
60 degrees.
To convert Gray code to binary code you must be familiar with the logical XOR operator. XOR outputs a 1 bit if either of two input bits is 1, but not both. The truth table for XOR, for all possible inputs p and q, is as follows:p q output0 0 00 1 11 0 11 1 0The algorithm to convert from Gray code to binary code is as follows:Step 1: Fix the most-significant bit, the MSB, which is always the same for both codes. If there are no more bits, we're done, otherwise proceed to step 2.Step 2: XOR the most recently fixed binary bit with the next available Gray bit. Fix the result as the next binary bit.Step 3: If there is at least one more Gray bit available, go to step 2. Otherwise we're done.Therefore, to convert 10101111 from Gray to binary, we proceed as follows:Gray = 10101111Fix MSB = 11 XOR 0 = 11 XOR 1 = 00 XOR 0 = 00 XOR 1 = 11 XOR 1 = 00 XOR 1 = 11 XOR 1 = 1Thus: Binary = 11010101Note that we carry the fixed bit (the bold bit) onto the next line as the l-value (left operand) of XOR. The r-value (right operand) of XOR is always the next available Gray bit after the MSB. Reading the fixed bits from top to bottom reveals the binary code.We can also write this as follows:Gray = 10101111Binary = 1 XOR 0 = 1 XOR 1 = 0 XOR 0 = 0 XOR 1 = 1 XOR 1 = 0 XOR 1 = 1 XOR 1 = 1Reading the fixed (bold) bits left to right reveals the binary code.
A pair of complimentary angles has a sum that measures 90 degrees. If your first angle is 40 degrees, the complement must equal 50. 90= 40+X 50=X
+, -, *, /, and, or, xor
objective complement
| x | y | x' | y' | x⊕y | x'⊕y' | ---------------------------------- | 0 | 0 | 1 | 1 | 0 | 0 | | 0 | 1 | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 |
It depends. XNOR is the inverse of XOR. If the N(ot) part is on the inputs, then they are equivalent. If the N(ot) part is on the output, then they are not.
To complement the 6th bit of the BX register in assembly language, you can use the XOR instruction. First, create a mask that has the 6th bit set (binary 00100000, which is 0x20 in hexadecimal). Then, XOR the BX register with this mask to toggle the 6th bit. Here's an example in x86 assembly: mov ax, <value> ; Load AX with some value xor bx, 0x20 ; Complement the 6th bit of BX This will flip the 6th bit of BX without affecting the other bits.
XORing X with 1 gives X', i.e., NOT(X). If we are able to construct a NAND (AND) using XOR, we can also obtain AND (NAND) from it, which makes XOR a universal gate since inverted inputs to a NAND (AND) will give OR (NOR). However XOR is not a universal gate! Therefore we cannot obtain NAND (AND) using XOR. :-) By, Tirtha Sarathi Ghosh Class 10 IIT Kanpur Aspirant
To complement the 6th bit of the BX register in assembly language, you can use the XOR instruction. The 6th bit corresponds to the bit mask 0x20 (binary 0010 0000). The code would look like this: MOV AX, 6 ; Load AX with 6 (not directly relevant to complementing BX) XOR BX, 0x20 ; Complement the 6th bit of BX This will toggle the 6th bit of BX, effectively complementing it.
45o
Yes. I'm assuming this is talking asking about boolean logic (the question makes little sense otherwise). If a and b are equal, then the complement of a and the complement of b are equal.
xor
The bitwise XOR operator is ^, or shift 6. The bitwise XOR assignment operator is ^=.
a XOR b is a^b in C language
1
Half of 90...