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 Subtraction and Division with 3 or more squares 
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Posted on: Fri May 20, 2011 2:59 am




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Post Re: Subtraction and Division with 3 or more squares
sneaklyfox wrote:
Can't have two or more zeros in a cage? Ok, maybe I forgot. I just thought maybe that had happened. Maybe it was 0x.


Yes, I remember a 0x with two zeros in it.


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Posted on: Fri May 20, 2011 3:24 am




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Post Re: Subtraction and Division with 3 or more squares
mparisi wrote:
sneaklyfox wrote:
Can't have two or more zeros in a cage? Ok, maybe I forgot. I just thought maybe that had happened. Maybe it was 0x.


Yes, I remember a 0x with two zeros in it.


Yes, of course you can have two or more zeros in a 0x cage. But I think pnm said you can't have two or more zeros in a 0: cage which was different than I remembered. But as he's the one who wrote the program, I have to assume that I remembered incorrectly.


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Posted on: Sun May 29, 2011 1:05 pm




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Post Re: Subtraction and Division with 3 or more squares
If I'm not mistaken, no one has yet given the rule which actually applies to all four arithmetic operations: iterate the binary operation by consistently applying brackets to the left. So it's (x-y)-z, ((x-y)-z)-w, and (x/y)/z. (This is of course correct but superfluous in the case of addition and multiplication.)


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Posted on: Sun May 29, 2011 9:02 pm




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Post Re: Subtraction and Division with 3 or more squares
duffifi wrote:
If I'm not mistaken, no one has yet given the rule which actually applies to all four arithmetic operations: iterate the binary operation by consistently applying brackets to the left. So it's (x-y)-z, ((x-y)-z)-w, and (x/y)/z. (This is of course correct but superfluous in the case of addition and multiplication.)


What about for exponentiation? I think I've come across x^(y^z).


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Posted on: Mon May 30, 2011 6:07 am




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Post Re: Subtraction and Division with 3 or more squares
@sneakly fox: I believe you are correct on both counts, and that rule for bracketing is meant to be implied by the accompanying explanation below puzzles which include exponentiation. In case it wasn't clear, my rule 'bracket to the left' was intended to apply only to +, -, x, and : .

@Patrick: do you think it would be worthwhile to make mention of which way the brackets are associated in the case of non-associative operations? I recall being puzzled myself when I first saw this, and the rule I settled on was, I thought, the clearest way to resolve the ambiguity without the need for case-splitting.


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Posted on: Mon May 30, 2011 12:55 pm




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Post Re: Subtraction and Division with 3 or more squares
duffifi wrote:
@Patrick: do you think it would be worthwhile to make mention of which way the brackets are associated in the case of non-associative operations? I recall being puzzled myself when I first saw this, and the rule I settled on was, I thought, the clearest way to resolve the ambiguity without the need for case-splitting.

Yes, for - and / the rule is: apply left to right (no need to use/mention brackets really).

For exponentiation the rule _was_ the same, until someone pointed out that generally
for three numbers:

Code:
         z
    y^
x^


first y^z is computed, and then x ^ (y^z).
(which can also be inferred from the example at the bottom of the page:
"Or, with three digits: 2^4^2 = 2^16 = 65536")

Patrick


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Posted on: Mon May 30, 2011 2:07 pm




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Post Re: Subtraction and Division with 3 or more squares
Aside: of course, this way of associating iterated exponentials is the standard convention in mathematics, reason being that with the other way (x^y)^z, you could simply use (x^y)^z = x^(yz) to reduce the size of the exponential stack. (It is common to think of exponentiation as a higher-order operation, and the 'true' height of the exponential stack is used to reflect just how high the order is. I guess I could go on about this, at the risk of derailing the discussion.)


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Posted on: Mon May 30, 2011 2:11 pm




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Post Re: Subtraction and Division with 3 or more squares
duffifi wrote:
Aside: of course, this way of associating iterated exponentials is the standard convention in mathematics, reason being that with the other way (x^y)^z, you could simply use (x^y)^z = x^(yz) to reduce the size of the exponential stack. (It is common to think of exponentiation as a higher-order operation, and the 'true' height of the exponential stack is used to reflect just how high the order is. I guess I could go on about this, at the risk of derailing the discussion.)

Maybe start a new topic :-)


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Posted on: Thu Jun 02, 2011 2:00 pm




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Post Re: Subtraction and Division with 3 or more squares
Just to be entirely sure about this, modulo operators don't have any particular order, right?

So 5 and 7 in a cage could be either 2mod or 5mod?

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Posted on: Thu Jun 02, 2011 2:09 pm




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Post Re: Subtraction and Division with 3 or more squares
maartensmit wrote:
Just to be entirely sure about this, modulo operators don't have any particular order, right?
So 5 and 7 in a cage could be either 2mod or 5mod?

Yes, absolutely, the same rule as for the other operators.


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