Math Puzzle (SOLVED by harold aptroot!)
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Not bad. I added that operator because I wasn't sure it was possible without it. If you have indeed done it without, then great work. :)
it is only / one needs, since that one has a singularity at zero, hence it allows to construct C-like functions (resulting in 0 or 1) like x>0 and x<0, x>=0 and x<=0, and therefore also x==0. :)
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Calm down there cowboy. My "stated usage" was just how I thought up this idea. This is just a fun math puzzle to occupy your time should you choose to let it.
PIEBALDconsult wrote:
it's just integer division
Not sure I agree (mod is the remainder from the division), but no shorthand allowed... show your work please. :)
aspdotnetdev wrote:
PIEBALDconsult wrote: it's just integer division Not sure I agree (mod is the remainder from the division),
Sure it is, since your division rounds: a % b = a - b * (a / b) e.g. 12 % 5 = 12 - 5 * (12 / 5) 2 = 12 - 5 * 2 2 = 2
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aspdotnetdev wrote:
PIEBALDconsult wrote: it's just integer division Not sure I agree (mod is the remainder from the division),
Sure it is, since your division rounds: a % b = a - b * (a / b) e.g. 12 % 5 = 12 - 5 * (12 / 5) 2 = 12 - 5 * 2 2 = 2
absolutely, and furthermore PIEBALD will claim the use of the C/C++ pre-processor to introduce and define new operators. :)
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aspdotnetdev wrote:
PIEBALDconsult wrote: it's just integer division Not sure I agree (mod is the remainder from the division),
Sure it is, since your division rounds: a % b = a - b * (a / b) e.g. 12 % 5 = 12 - 5 * (12 / 5) 2 = 12 - 5 * 2 2 = 2
Thanks for the info. :)
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absolutely, and furthermore PIEBALD will claim the use of the C/C++ pre-processor to introduce and define new operators. :)
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Just so long as he shows his work. :)
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absolutely, and furthermore PIEBALD will claim the use of the C/C++ pre-processor to introduce and define new operators. :)
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C++ Preprocessor? That's so kitsch. Prolog operators all the way!
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it is only / one needs, since that one has a singularity at zero, hence it allows to construct C-like functions (resulting in 0 or 1) like x>0 and x<0, x>=0 and x<=0, and therefore also x==0. :)
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Man, all that money wasted on finding the Higgs Boson, when it was hiding in the divide by zero all along!
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Man, all that money wasted on finding the Higgs Boson, when it was hiding in the divide by zero all along!
it is not only at zero, it is everywhere, we are just too blind to see it. :)
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it is not only at zero, it is everywhere, we are just too blind to see it. :)
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Luc Pattyn wrote:
we are just too blind to see it
That'll teach you to look directly into a singularity. Those accretion discs will get you every time.
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absolutely, and furthermore PIEBALD will claim the use of the C/C++ pre-processor to introduce and define new operators. :)
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:-D I was this close to using C rather than C#... Besides, he invented the
\operator. X| -
:-D I was this close to using C rather than C#... Besides, he invented the
\operator. X|PIEBALDconsult wrote:
Besides, he invented the \ operator
I only know \ from several Basic systems including VB. And I don't really like it, so I did not use it. :)
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aspdotnetdev wrote:
PIEBALDconsult wrote: it's just integer division Not sure I agree (mod is the remainder from the division),
Sure it is, since your division rounds: a % b = a - b * (a / b) e.g. 12 % 5 = 12 - 5 * (12 / 5) 2 = 12 - 5 * 2 2 = 2
Actually, I would likely define it using repeated subtraction; so it should still be allowed. :-D
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PIEBALDconsult wrote:
Besides, he invented the \ operator
I only know \ from several Basic systems including VB. And I don't really like it, so I did not use it. :)
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Huh... I've never heard of it. Turbo BASIC has it. Dartmouth BASIC doesn't seem to. BASIC-Plus doesn't. Seems to be an abomination.
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Luc Pattyn wrote:
we are just too blind to see it
That'll teach you to look directly into a singularity. Those accretion discs will get you every time.
She was only a particle physicist's daughter, but she had big accretion discs. :cool:
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:-D I was this close to using C rather than C#... Besides, he invented the
\operator. X|Different languages have different uses for backslash. I did arbitrarily assign each slash to a particular type of integer division (was not based on any programming language). Gotta keep you on your toes. :)
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Gwenio wrote:
you should really make sure such a problem has an answer before presenting it as a puzzle
You might want to tell these guys that. ;) Also, I intuited that it was possible. If it wasn't, then a proof of why it wasn't would have been equally valid. :)
aspdotnetdev wrote:
You might want to tell these guys that. Wink Also, I intuited that it was possible. If it wasn't, then a proof of why it wasn't would have been equally valid.
Well, you should give a prize then. :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
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aspdotnetdev wrote:
You might want to tell these guys that. Wink Also, I intuited that it was possible. If it wasn't, then a proof of why it wasn't would have been equally valid.
Well, you should give a prize then. :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]I did, I gave him a nice shiny 5 vote. What better prize is there?
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I did, I gave him a nice shiny 5 vote. What better prize is there?
:beer: :pizza: for instance. :-D
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
:beer: :pizza: for instance. :-D
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]*searches freezer for pizza*
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Sorry folks, harold aptroot beat you. Better luck next time. :) So a coworker of mine was just converting between different representations for the day of the week. In one system 0 means Sunday and in the other 7 means Sunday (all other days are equal between both systems). He was able to solve it using a simple if statement, but that got me thinking about a mathematical way of solving that (just for fun). I came up with:
newDay = ((oldDay + 2) / (oldDay + 1) - 1) * 7 + oldDay
That makes new day equal to the old day, except the 0 gets turned into a 7. However, it only works for non-negative integers. Any value less than -1 will be incorrect and -1 will cause a division by 0, which for the sake of the puzzle I am disallowing. Your goal is to create an equation that will handle negatives correctly too. So, the input and output would look like this:
Old Value
New Value
-∞
-∞
...
...
-2
-2
-1
-1
0
7
1
1
2
2
...
...
∞
∞
You are only allowed multiplication, addition, subtraction, negation, integer division, and parentheses for grouping. Assume "/" truncates the result (so, -2.5 becomes -2 and 2.5 becomes 2) and "\" rounds toward negative infinity (so, -2.5 becomes -3 and 2.5 becomes 2). Division by 0 is not allowed. The first one to get a correct solution gets 5 kudos points. Have fun. :) P.S. This is not a homework question. I graduated years ago. Though, teachers, feel free to snag this little puzzle for your math students (or programming students, as this is more a logic puzzle that just happens to employ very basic math). If nobody has solved this by the time I get home in a couple hours, I'll go ahead and give it a go myself then post the answer here (assuming I can figure it out).
modified on Tuesday, April 20, 2010 10:16 PM
Here's a smaller approach:
y = x + 7 \* (1 / (x \* x + 1))Assuming integer division, and assuming numbers don't overflow, this should work nicely. The expression x*x is never negative, so x*x+1 is always positive and never zero. The expression 1/q is one when q=1, and zero if q>1; I don't think things can get any simpler than that.