gjm comments on Innate Mathematical Ability - Less Wrong

40 Post author: JonahSinick 18 February 2015 11:11AM

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Comment author: Vaniver 19 February 2015 01:14:17AM *  2 points [-]

I'm trying to figure out if that has to happen -- that is, if the first two rows are xor across, and the first two columns are xor down, and the missing piece fits xor in at least one direction, is it required to also fit xor in the other direction?

That is:

A⊕B=C (1)
D⊕E=F (2)
G⊕H=I (3)

and

A⊕D=G (4)
B⊕E=H (5)

We want to know if it is true that:

C⊕F=I

We begin with our goal, and substitute out C and F using (1) and (2):

(A⊕B)⊕(D⊕E)=I

Now we ask Wikipedia if ⊕ is associative and commutative, and the answer is yes, allowing us to rearrange that as (this is actually multiple steps, condensed):

(A⊕D)⊕(B⊕E)=I

Now we substitute using (4) and (5):

G⊕H=I

This is (3), and thus we have our proof. (Perhaps a more natural way is to start at (3) and work forward to our desired formula, but I like working backwards.)

As a side point, I believe it is the case that most (all?) Raven's patterns are applied both horizontally and vertically.

Comment author: gjm 19 February 2015 02:15:12AM 4 points [-]

I think the proof is simplified by the observation that (+ meaning XOR) a+b=c is the same as a+b+c=0. So if all rows have the XOR property, we find that the XOR of all entries is 0. If two columns have the XOR property, the XOR of their entries is 0, leaving 0 for the XOR of the entries in the last column, and we're done.

Comment author: Vaniver 19 February 2015 03:45:40AM 0 points [-]

Agreed; my proof doesn't make use of the fact that C⊕C=0, and if you use that fact you get there quicker.