MHB Solving Bitwise Arithmetic: What Trick Makes it Easy?

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The discussion focuses on solving bitwise arithmetic expressions, specifically 1234567 ^ 7 ^ ~~1234567 and 5 & (12345678 ^ ~~12345678). The key trick mentioned involves understanding basic properties of bitwise operations: negating a bit twice (~~x) returns the original value, and the exclusive-or operation (^) has properties such as x ^ x = 0 and being both commutative and associative. These properties simplify the expressions and make solving them easier.
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1234567 ^ 7 ^ ~~1234567

5 & (12345678 ^ ~~ 12345678)

My prof said there is an easy way to solve these as there is a trick to it. Does anyone know what trick is being referred to?
 
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tmt said:
1234567 ^ 7 ^ ~~1234567

5 & (12345678 ^ ~~ 12345678)

My prof said there is an easy way to solve these as there is a trick to it. Does anyone know what trick is being referred to?

The "tricks" here are just the basic properties of bitwise arithmetic, namely that ~~x is equivalent to x (negating a bit, and then negating it again, gives you back the original bit) and x ^ x is equivalent to 0 (look at the truth table for exclusive-or). Finally, ^ is commutative. Does that make it easier?
 
Bacterius said:
Finally, ^ is commutative.

Nitpicking. We also need that ^ is associative, which it is. ;)
 
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