URGENT HELP! PARITY BIT TROUBLE! (1 Viewer)

[Citrusacrobat08]

Guys, I know its pretty late in the year for this, but I an getting very confused on how to do any parity bit questions. Cans anyone please lend a helping hand?

 

[jimmysmith560]

Feel free to specify areas of the content that you are not fully confident with and/or post specific questions that you are struggling with.

 

[Citrusacrobat08]

Feel free to specify areas of the content that you are not fully confident with and/or post specific questions that you are struggling with.

I am just wanting explanation on this and how to generally for questions for parity bits, thanks a lot! :

 

[liamkk112]

I am just wanting explanation on this and how to generally for questions for parity bits, thanks a lot! :
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the parity bit is the last bit of every byte, and aims to balance the number of 1s in every byte to be odd/even. this can be used to error check a byte, because if you know the number of 1s should be even for example, yet the byte has an odd number of 1s, then we know some of the bits of the byte have been scrambled and the byte must thus be corrupted.

we can notice that the first byte has odd parity, second has even, third has even. because the second and third bytes have different parity to the first, we can assume that the parity is meant to be odd and that the second and third bytes are corrupt. tbh the question doesn’t consider that corruption of data can still maintain parity, but in the context of hsc i guess you can assume that any corrupted data would have a different parity to what it’s meant to be.

 

[Citrusacrobat08]

the parity bit is the last bit of every byte, and aims to balance the number of 1s in every byte to be odd/even. this can be used to error check a byte, because if you know the number of 1s should be even for example, yet the byte has an odd number of 1s, then we know some of the bits of the byte have been scrambled and the byte must thus be corrupted.

we can notice that the first byte has odd parity, second has even, third has even. because the second and third bytes have different parity to the first, we can assume that the parity is meant to be odd and that the second and third bytes are corrupt. tbh the question doesn’t consider that corruption of data can still maintain parity, but in the context of hsc i guess you can assume that any corrupted data would have a different parity to what it’s meant to be.

Oh My God! That is such a great explanation! Thank you so much, I really appreciate the time you took to explain it to me! Have joyous rest of your week!