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Every single time if I set y = 0...
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That looks suspiciously like Griffs answer, and it's still just as true (and incomplete).
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Well I'm an IT guy after all 
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That means you copy from someone else? 
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Story of a programmer 
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If the programmer and/or the pc are drunk enough everything related to work is wrong
Microsoft ... the only place where VARIANT_TRUE != true
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Maths
---------------------------------
Obscurum per obscurius.
Ad astra per alas porci.
Quidquid latine dictum sit, altum videtur .
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Ah Dave is starting the Great Britain versus the world on the usage of English again.
Every day, thousands of innocent plants are killed by vegetarians.
Help end the violence EAT BACON
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I thought it was "English-speaking North America versus the World"?
It is often shortened to maths or, in English-speaking North America, math.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Richard Deeming wrote: I thought it was "Not-Quite-English-speaking North America versus the World"?
ftfy
---------------------------------
Obscurum per obscurius.
Ad astra per alas porci.
Quidquid latine dictum sit, altum videtur .
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Congratulations, you win.
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Ok, but why? (Please  )
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Each position has 4 options 0,0 0,1 1,0 or 1,1 3 of which will give 0 and there are 32 positions.
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Thanks, as usual, it makes sense when looked at from the right angle
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That's hard! The number is huge, and thinking about it has lead to delirium. Please help. For example, I can see the answer for all pairs where the first of the numbers is of the form 10000000000000000000000000000000, 01000000000000000000000000000000, etc. (i.e. a single bit is flipped. - each has 1^31 paired values.) But my head swims when I try to extend this. Usually, there's some other clever way of simplifying the problem, but I sure don't see it.
Is there a simple way to think about this, or is it a lot of calculation?
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There is a simple way of course, I wouldn't ask you guys to try 264 combinations Ok I might, but I didn't.
Here's a start: consider the 1-bit case. 3 out of 4 inputs give 0.
There are several different ways to go from there.
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You could also "cheat" and use algorithms and data structures instead of just math, for example: made an ROBDD representing the function "x & y == 0" (that BDD is pretty small) and count its solutions (in time depending mainly on the size of the BDD, not the magnitude of the answer). (that's what my site[^] does)
That way you might not see that it's 332 though, just the raw number, and it's a little hard to take 1853020188851841 and figure out where it came from, so to speak.
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Time yourself.
I reckon under 30 seconds for most people here:
Solve for x:
(x+1)(x+2)(x+6) = x^3+9x^2+4(7x-1)
---------------------------------
Obscurum per obscurius.
Ad astra per alas porci.
Quidquid latine dictum sit, altum videtur .
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x=4
speramus in juniperus
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x=42
There are only 10 types of people in the world, those who understand binary and those who don't.
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x = beer
Microsoft ... the only place where VARIANT_TRUE != true
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You are only half right:
x = beer + bacon
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(x+1)(x+2)(x+6) =
x^3+9x^2+20x+12 = x^3+9x^2+4(7x-1)
20x+12=4(7x-1)
5x+3=7x-1
4=2x
x = 2
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Congratulations! I'm sure you get an A in GSCE maths for that, or possibly even an A at A-level these days.
Not that they've dumbed down or anything...
Regards,
Rob Philpott.
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