The Three Ways of XOR
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The Three Ways of XOR
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Re: The Three Ways of XOR
#2> most programming languages don’t have an explicit “logical operator” for it
I guess the author never taught of != . The only thing to be careful with is that it doesn't implicitly convert arguments to boolean, so expressions like " != (flags & Flag)" will go wrong without an explicit conversion "bool(flags & Flag)" or equivalent expression like "((flags & Flag) != 0)".
And let's not forget about the friend, ==. I've more than once seen code like "(a && b) || (!a && !b)".
A similar interesting pattern many don't think of is "bool(a1) + ... + bool(aN) == M" (particularly with M==1) and instead we see unreadable monstrosities :)
Re: The Three Ways of XOR
#3> most programming languages don’t have an explicit “logical operator” for it I guess the author never taught of != . The only thing to be careful with is that it doesn't implicitly convert arguments to boolean, so expressions like " != (flags & Flag)" will go wrong without an explicit conversion "bool(flags & Flag)" or equivalent expression like "((flags & Flag) != 0)". And let's not forget about the friend, ==. I'v…
Is "bool(a1) + ... + bool(aN) == M" not just "a1 || a2...||an"?
Re: The Three Ways of XOR
#4> most programming languages don’t have an explicit “logical operator” for it I guess the author never taught of != . The only thing to be careful with is that it doesn't implicitly convert arguments to boolean, so expressions like " != (flags & Flag)" will go wrong without an explicit conversion "bool(flags & Flag)" or equivalent expression like "((flags & Flag) != 0)". And let's not forget about the friend, ==. I'v…
Is "bool(a1) + ... + bool(aN) == M" not just "a1 || a2...||an"?
Obviously no, the first (for M==1) means exactly one is true, the latter at least one is true. But a1||...||aN is equivalent to bool(a1)+...+bool(aN)>0 (assuming no integer overflow :).
Re: The Three Ways of XOR
#5> most programming languages don’t have an explicit “logical operator” for it I guess the author never taught of != . The only thing to be careful with is that it doesn't implicitly convert arguments to boolean, so expressions like " != (flags & Flag)" will go wrong without an explicit conversion "bool(flags & Flag)" or equivalent expression like "((flags & Flag) != 0)". And let's not forget about the friend, ==. I'v…
Is "bool(a1) + ... + bool(aN) == M" not just "a1 || a2...||an"?
No, it's true when exactly M are true. "a1 || a2 || ... || an" is true when at least one is true.
The latter is the same as "a1 + ... + an == 1 || a1 + ... + an == 2 || ... || a1 + ... + an == n".
Re: The Three Ways of XOR
#6Lovely article, in particular the small but deep excursion into AI history and the AI winter after the publication of Minsky/Papert's "Perceptrons" (though it was 70's, not 80's).
Wonder when the current AI summer will come to an end...
Re: The Three Ways of XOR
#7Another way to look at XOR - it's an adder (the sum, without the carry)
Re: The Three Ways of XOR
#8Earlier quoted context omitted.
Is "bool(a1) + ... + bool(aN) == M" not just "a1 || a2...||an"?
No, it's true when exactly M are true. "a1 || a2 || ... || an" is true when at least one is true. The latter is the same as "a1 + ... + an == 1 || a1 + ... + an == 2 || ... || a1 + ... + an == n".
So your + operator implicitly casts booleans to an integer type, with false=0 and true=1?
Re: The Three Ways of XOR
#9it can also be used to generate simple parity.
Re: The Three Ways of XOR
#10Earlier quoted context omitted.
No, it's true when exactly M are true. "a1 || a2 || ... || an" is true when at least one is true. The latter is the same as "a1 + ... + an == 1 || a1 + ... + an == 2 || ... || a1 + ... + an == n".
So your + operator implicitly casts booleans to an integer type, with false=0 and true=1?
In c++, yes. Not every language will do that though.