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Topic: find the function (Read 4751 times) |
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srn437
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Re: find the function
« Reply #50 on: Sep 23rd, 2007, 9:13pm » |
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copy actually, since cut won't work here. He also gave us an infinite amount of web sites, and out of an infinite amount of web sites, the probability of going to a specific one is zero(positive zero).
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JiNbOtAk
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Re: find the function
« Reply #51 on: Sep 23rd, 2007, 10:09pm » |
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on Sep 23rd, 2007, 9:13pm, srn347 wrote:| copy actually, since cut won't work here. He also gave us an infinite amount of web sites, and out of an infinite amount of web sites, the probability of going to a specific one is zero(positive zero). |
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Wow, an infinite amount of websites, really ? 
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srn437
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Re: find the function
« Reply #52 on: Sep 23rd, 2007, 10:27pm » |
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well, not all at the same time, but more web sites are created each day with no exponential decay, so after an eternity there will be. It is already seemingly infinite and going to be infinite. If you count search result pages, there are infinite since you can search for anything with infinite possibilities.
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towr
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Re: find the function
« Reply #53 on: Sep 23rd, 2007, 11:10pm » |
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on Sep 23rd, 2007, 8:33pm, srn347 wrote:| Do you believe that the same God who gave us reason, purpose, and sense wants us to forgo their use? |
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Then why do you never display any sense or reason and does your only purpose seem to be holding on to your obstinate ignorance?
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amstrad
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Re: find the function
« Reply #54 on: Sep 24th, 2007, 10:02am » |
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How about this for f(n):
f(n) = ( (n%2)==0 ? -n/2 : 2*n )
if n is even f(n) is -n/2 else f(n) is 2n
Still working on g(q)...
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pex
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Re: find the function
« Reply #55 on: Sep 24th, 2007, 10:11am » |
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on Sep 24th, 2007, 10:02am, amstrad wrote:How about this for f(n):
f(n) = ( (n%2)==0 ? -n/2 : 2*n )
if n is even f(n) is -n/2 else f(n) is 2n
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Then f(f(4)) = 1, not -4...
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amstrad
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Re: find the function
« Reply #56 on: Sep 24th, 2007, 12:53pm » |
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on Sep 24th, 2007, 10:11am, pex wrote:
Then f(f(4)) = 1, not -4... |
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Yep doesn't work.
How about this (my coworker's solution, which I think really is correct):
| hidden: |
You need 4 states and a circle process to traverse them: positive even, positive odd, negative even and negative odd. The state transition goes like this:
po -> pe -> no -> ne -> po.......
f(n) =
po: n+1
pe: -n+1
no: n-1
ne: -n-1
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srn437
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Re: find the function
« Reply #57 on: Sep 24th, 2007, 7:13pm » |
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Assuming zero stays zero since it is neither positive or negative(yet). Anyway, inspired by that answer, here is g. g(q)=
integer- q-1+(1/q)
not integer- write it in the representation q-1+1/q where q is an integer and get rid of the q-1.
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| « Last Edit: Sep 30th, 2007, 11:25am by srn437 » |
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Whiskey Tango Foxtrot
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Re: find the function
« Reply #58 on: Sep 24th, 2007, 8:20pm » |
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on Sep 23rd, 2007, 8:33pm, srn347 wrote:| Do you believe that the same God who gave us reason, purpose, and sense wants us to forgo their use? if you ban me, that's what you'll be doing. |
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Another religious concept: I'm done with this sh*t.
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"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forgo their use." - Galileo Galilei
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mikedagr8
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Re: find the function
« Reply #59 on: Sep 24th, 2007, 8:27pm » |
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Good call. 
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amstrad
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Re: find the function
« Reply #60 on: Sep 25th, 2007, 6:46am » |
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Here is my solution for g(q):
For q to be rational it is n/m for some integers n and m.
To mirror my solution for f(n), I make 4 states:
1) n is greater than m and n+m is even (ge)
2) n is greater than m and n+m is odd (go)
3) n is less than m and n+m is even (le)
4) n is less than m and n+m is odd (lo)
again the cycle is ge->go->le->lo->ge.......
so g(n/m) =
if(ge) (n+1)/m
if(go) m/(n-1)
if(le) n/(m+1)
if(lo) (m-1)/n
it is important not to reduce your intermediate results
starting with 8 (or 8/1) you get 8->1/7->1/8->7->8....
starting with 21/5 you get 21/5->5/20->5/21->20/5->21/5...
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Grimbal
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Re: find the function
« Reply #61 on: Sep 25th, 2007, 7:22am » |
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Hm... if you don't simplify your fractions, you are not working with rationals, but with pairs of integers.
You can not define a rational function that does
3/6 -> 5/3
2/4 -> 2/5
and
1/2 -> 1/1
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towr
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Re: find the function
« Reply #62 on: Sep 25th, 2007, 7:22am » |
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on Sep 25th, 2007, 6:46am, amstrad wrote:| it is important not to reduce your intermediate results |
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That's a bit of a flaw then, because fractions keep the same value if you reduce them; they're the same number.
There are alternatives that don't have that problem.
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srn437
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Re: find the function
« Reply #63 on: Sep 25th, 2007, 5:00pm » |
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A fraction can also have both sides be negative, thus reversing the inequality.
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Grimbal
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Re: find the function
« Reply #64 on: Sep 26th, 2007, 12:40am » |
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sure.
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srn437
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the dark lord rises again....
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Re: find the function
« Reply #65 on: Sep 30th, 2007, 9:43pm » |
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At least g was solved for.
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RandomSam
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Re: find the function
« Reply #66 on: Oct 4th, 2007, 5:05pm » |
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Based entirely on amstrad's post, how's this for rationals:| hidden: | For integers a and b
x>1, floor(x) odd : g(x) = x + 1
x>1, floor(x) even: g(x) = (x - 1)-1
x<1, floor(x-1) odd : g(x) = (x-1 + 1)-1
x<1, floor(x-1) even: g(x) = x-1 - 1 |
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| « Last Edit: Oct 4th, 2007, 5:07pm by RandomSam » |
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towr
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Re: find the function
« Reply #67 on: Oct 5th, 2007, 1:33am » |
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on Oct 4th, 2007, 5:05pm, RandomSam wrote:Based entirely on amstrad's post, how's this for rationals:| hidden: | For integers a and b
x>1, floor(x) odd : g(x) = x + 1
x>1, floor(x) even: g(x) = (x - 1)-1
x<1, floor(x-1) odd : g(x) = (x-1 + 1)-1
x<1, floor(x-1) even: g(x) = x-1 - 1 |
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Seems fairly good, although you probably want to check for the absolute value of x, and define what to do with 1.
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RandomSam
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Re: find the function
« Reply #68 on: Oct 5th, 2007, 8:22am » |
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on Oct 5th, 2007, 1:33am, towr wrote:
Seems fairly good, although you probably want to check for the absolute value of x, and define what to do with 1. |
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oops...  I meant to say "x is a positive rational" instead of defining a and b as integers, which aren't used in the rest of the solution!
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