Calcudoku puzzle forumhttps://www.calcudoku.org/forum/ Subtraction and Division with 3 or more squareshttps://www.calcudoku.org/forum/viewtopic.php?f=2&t=4 Page 2 of 3

 Author: mparisi  [ Fri May 20, 2011 2:59 am ] Post subject: 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.

 Author: sneaklyfox  [ Fri May 20, 2011 3:24 am ] Post subject: 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.

 Author: duffifi  [ Sun May 29, 2011 1:05 pm ] Post subject: 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.)

 Author: sneaklyfox  [ Sun May 29, 2011 9:02 pm ] Post subject: 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).

 Author: duffifi  [ Mon May 30, 2011 6:07 am ] Post subject: 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.

 Author: pnm  [ Mon May 30, 2011 12:55 pm ] Post subject: 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 generallyfor 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

 Author: duffifi  [ Mon May 30, 2011 2:07 pm ] Post subject: 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.)

 Author: pnm  [ Mon May 30, 2011 2:11 pm ] Post subject: 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

 Author: maartensmit  [ Thu Jun 02, 2011 2:00 pm ] Post subject: 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?

 Author: pnm  [ Thu Jun 02, 2011 2:09 pm ] Post subject: 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.

 Page 2 of 3 All times are UTC + 1 hour [ DST ] Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Grouphttp://www.phpbb.com/