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Old Oct 2nd, 2012, 12:42 AM   #16
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Re: Try solving this problem :oh:

2
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Old Oct 2nd, 2012, 01:43 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
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Old Oct 2nd, 2012, 01:44 AM   #18
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Re: Try solving this problem :oh:

Quote:
Originally Posted by .Homme. View Post
0=1
1=0
2=0
3=0
4=-
5=-
6=1
7=0
8=2
9=1

??
This is how I did it.
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Old Oct 2nd, 2012, 05:18 AM   #19
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Re: Try solving this problem :oh:

Quote:
Originally Posted by PhilePhile View Post
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 non-unique process. Hope that makes sense gurl.
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Old Oct 2nd, 2012, 09:58 AM   #20
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Re: Try solving this problem :oh:

Quote:
Originally Posted by égalité View Post
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 non-unique process. Hope that makes sense gurl.
Sorry, no . Most of us are pre-schooler here. Can you elaborate on your solution/concept .

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Old Oct 2nd, 2012, 12:00 PM   #21
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Re: Try solving this problem :oh:

Quote:
Originally Posted by égalité View Post
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 non-unique 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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Old Oct 2nd, 2012, 12:12 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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Old Oct 2nd, 2012, 03:20 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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Old Oct 2nd, 2012, 04:11 PM   #24
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Re: Try solving this problem :oh:

Quote:
Originally Posted by moby View Post
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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Old Oct 2nd, 2012, 07:52 PM   #25
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Re: Try solving this problem :oh:

Quote:
Originally Posted by PhilePhile View Post
"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
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Old Oct 2nd, 2012, 09:13 PM   #26
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Re: Try solving this problem :oh:

Quote:
Originally Posted by moby View Post
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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Old Oct 2nd, 2012, 09:26 PM   #27
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Re: Try solving this problem :oh:

Quote:
Originally Posted by PhilePhile View Post
Sorry, no . Most of us are pre-schooler 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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