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[Solved]: Shouldn't the "even parity" function map 1101 to 0?

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From the book Computer organization and design by Patterson&Hennessy:

Parity is a function in which the output depends on the number of 1s in in the input. For an even parity function, the output is 1 if the input has an even number of ones. Suppose a ROM is used to implement an even parity function with a 4-bit input. Then the contents of the ROM is $$\text{Address} \ 0 : \ 0 \\ \text{Address} \ 1: \ 1 \\ \text{Address} \ 2 : \ 0 \\ \text{Address} \ 3 : \ 1 \\ \vdots \\ \text{Address} \ 13 : \ 1 \\ \text{Address} \ 14 : \ 0 \\ \text{Address} \ 15 : \ 1$$

As per my understanding, ROM which implements the even parity function should store 0 at both the Address 1 and the Address 2, 1 at the Address 3, ... 0 at both the Address 13 and 14, then 1 at the Address 15, for the Address $k$ to represent the map-value of $(k)_{\text{base}2}$.

According to this the concept defined above is not clear enough, Can someone clarify the doubt?

Asked By : giuscri

Answered By : David Richerby

In this context, implementing something as a ROM just means a look-up table. If you want to know the parity of $x$, you put the binary coding of $x$ on the ROM's address wires and the value you read out is the value stored at that memory location within the ROM, which will be either 0 or 1.

And, yes, the contents of the ROM that you've quoted are wrong: they seem to be implementing parity in the sense that the output is 1 if, and only if, the input is an odd number, instead of implementing the even parity function.

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