

Oct 2nd, 2012, 01:42 AM

#16

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Re: Try solving this problem :oh:
2



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Oct 2nd, 2012, 02:43 AM

#17

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Re: Try solving this problem :oh:
haven't read the thread. My answer is 2. Done in 30 sec.
Am I a



Oct 2nd, 2012, 02:44 AM

#18

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Re: Try solving this problem :oh:
Quote:
Originally Posted by .Homme.
0=1
1=0
2=0
3=0
4=
5=
6=1
7=0
8=2
9=1
??

This is how I did it.



Oct 2nd, 2012, 06:18 AM

#19

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Re: Try solving this problem :oh:
Quote:
Originally Posted by PhilePhile

oh, thanks for clarifying
There are two functions. f:Z>Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly nonunique process. Hope that makes sense gurl.
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Oct 2nd, 2012, 10:58 AM

#20

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Join Date: Sep 2009
Posts: 2,645

Re: Try solving this problem :oh:
Quote:
Originally Posted by égalité
oh, thanks for clarifying
There are two functions. f:Z>Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly nonunique process. Hope that makes sense gurl.

Sorry, no . Most of us are preschooler here. Can you elaborate on your solution/concept .



Oct 2nd, 2012, 01:00 PM

#21

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Re: Try solving this problem :oh:
Quote:
Originally Posted by égalité
oh, thanks for clarifying
There are two functions. f:Z>Z assigns, for example, the number 1 to 0, the number 0 to 1, the number 0 to 2, etc. (where Z=the integers). Then there is a quaternary map on Z, written as a word of length 4, that simply produces the sum of the "letters" in the word. The expression "8809=6" means g((f,f,f,f)(8,8,0,9))=6, where (f,f,f,f) is the cartesian product of f with itself 4 times (it is a map from Z^4 to Z^4). The composition g((f,f,f,f)) is certainly not injective, given the brief glimpse into the behavior of the function f that was provided in the statement of the problem, and especially given the fact that decomposing an integer into a sum of 4 other integers is a highly nonunique process. Hope that makes sense gurl.

I'm studying chemical engineering and I don't understand a fuck out of this
I've always hated calculus anyway... Algebra on the other hand
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Oct 2nd, 2012, 01:12 PM

#22

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Re: Try solving this problem :oh:
It's 2.
Spoiler: It's the number of enclosed spaces.
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Oct 2nd, 2012, 04:20 PM

#23

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Re: Try solving this problem :oh:
OK, I'll play.
Basically we are computing the rank of the first homology group of each of the given topological spaces.
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Oct 2nd, 2012, 05:11 PM

#24

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Re: Try solving this problem :oh:
Quote:
Originally Posted by moby
OK, I'll play.
Basically we are computing the rank of the first homology group of each of the given topological spaces.

Oh my I didn't notice that
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Oct 2nd, 2012, 08:52 PM

#25

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Re: Try solving this problem :oh:
Quote:
Originally Posted by PhilePhile
"2" is technically not correct. The "=" should be "≈" for the "2" answer to be correct.
For example,
8096 = 5
and
9881 = 5
is not possible. It's reasonable that "9" = "6" ... but "9" + "6" = "8" ? or "0" = "9" ?

8096
8=2
0=1
9=1
6=1
9881
9=1
8=2
8=2
1=0
???
9=6=1
9+6=2
0123456789=5
0123456789=8096=9881
M=Mental
A=Abuse
T=to
H=Health



Oct 2nd, 2012, 10:13 PM

#26

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Re: Try solving this problem :oh:
Quote:
Originally Posted by moby
OK, I'll play.
Basically we are computing the rank of the first homology group of each of the given topological spaces.

Win.
An extension to homotopy groups would ensure that the problem can be further generalized.
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Oct 2nd, 2012, 10:26 PM

#27

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Re: Try solving this problem :oh:
Quote:
Originally Posted by PhilePhile
Sorry, no . Most of us are preschooler here. Can you elaborate on your solution/concept .

Sorry I was just being unnecessarily pedantic My point is that it's perfectly legitimate for 8096 and 9881 to both "equal" 5. As moby pointed out, they are homotopy equivalent.
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