Answer:
5,10; 6,12; 7,14
Explanation:
We will demonstrate the iteration of the loop:
First iteration: Number = 7, Count = 5 at the beginning. We will check if Count <= Number? Since it is correct, prints 5,10. Increment the Count by 1.
Second iteration: Number = 7, Count = 6. We will check if Count <= Number? Since it is correct, prints 6,12. Increment the Count by 1.
Third iteration: Number = 7, Count = 7. We will check if Count <= Number? Since it is correct, prints 7,14. Increment the Count by 1.
Forth iteration: Number = 7, Count = 8. We will check if Count <= Number? Since it is not correct, the loop stops.
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
Reproduction. One layed some eggs then another and another and then BOOM here they are.
