Try solving this problem :oh: - Page 2 - TennisForum.com
TennisForum.com is the premier Women's Tennis forum on the internet. Registered Users do not see the above ads.Please Register - It's Free!

 Oct 2nd, 2012, 01:42 AM #16 ce Senior Member     Join Date: Jan 2007 Location: Belgrade Posts: 17,156 Re: Try solving this problem :oh: 2
 Oct 2nd, 2012, 02:43 AM #17 Moveyourfeet Senior Member   Join Date: Oct 2010 Posts: 1,638 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
Moveyourfeet
Senior Member

Join Date: Oct 2010
Posts: 1,638
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
égalité
Senior Member

Join Date: Jun 2006
Location: Hingistan
Posts: 12,049
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 non-unique process. Hope that makes sense gurl.
__________________
Bitttchhhhhhhhhhhhhhhhhhhhhhhhhhhh You Better Pay It Honey! The Devil Is A Liar!!!!! Bitches Get Interviews And Shit? Where They Do That At Honey? Girls Are Late Honey!!!! The Queen Needs To Get Into It Honey Cause The Girls Is Late Out Here! Yes Honey Im Throwing Epic Shade!

Management!

Oct 2nd, 2012, 10:58 AM   #20
PhilePhile
Senior Member

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 non-unique process. Hope that makes sense gurl.
Sorry, no . Most of us are pre-schooler here. Can you elaborate on your solution/concept .

Oct 2nd, 2012, 01:00 PM   #21
Sammo
Senior Member

Join Date: Feb 2009
Posts: 26,679
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 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
__________________
MIRJANA "MLB Da MVP" LUCIC-BARONI | SAM STOSUR ☯ | MARTINA HINGIS ♚ | DANIELA HANTUCHOVA ✯ | OLGA PUCHKOVA ❤ | MADISON KEYS ☢
| MARTINA NAVRATILOVA (because fuck you that's why)

R.I.P. Bally

 Oct 2nd, 2012, 01:12 PM #22 Sean. Look who's back!     Join Date: Oct 2007 Posts: 32,436 Re: Try solving this problem :oh: It's 2. Spoiler: It's the number of enclosed spaces. __________________ Vera Zvonareva * Ana Ivanović * Li Na * Laura Robson ......
 Oct 2nd, 2012, 04:20 PM #23 moby Senior Member     Join Date: Sep 2001 Posts: 13,608 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. __________________ A single flow'r he sent me, since we met./All tenderly his messenger he chose; Deep-hearted, pure, with scented dew still wet - One perfect rose. I knew the language of the floweret;/'My fragile leaves,' it said, 'his heart enclose.' Love long has taken for his amulet/One perfect rose. Why is it no one ever sent me yet/One perfect limousine, do you suppose? Ah no, it's always just my luck to get/One perfect rose.
Oct 2nd, 2012, 05:11 PM   #24
égalité
Senior Member

Join Date: Jun 2006
Location: Hingistan
Posts: 12,049
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
__________________
Bitttchhhhhhhhhhhhhhhhhhhhhhhhhhhh You Better Pay It Honey! The Devil Is A Liar!!!!! Bitches Get Interviews And Shit? Where They Do That At Honey? Girls Are Late Honey!!!! The Queen Needs To Get Into It Honey Cause The Girls Is Late Out Here! Yes Honey Im Throwing Epic Shade!

Management!

Oct 2nd, 2012, 08:52 PM   #25
Direwolf
Senior Member

Join Date: Dec 2005
Location: Out There
Posts: 13,742
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
Morning Morgan
Senior Member

Join Date: Sep 2001
Location: sddfs
Posts: 2,859
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.
__________________
He looked deep into his eyes and said, "You're perfect."

"No, I'm not. But with you, I don't even care."

Oct 2nd, 2012, 10:26 PM   #27
égalité
Senior Member

Join Date: Jun 2006
Location: Hingistan
Posts: 12,049
Re: Try solving this problem :oh:

Quote:
 Originally Posted by PhilePhile 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.
__________________
Bitttchhhhhhhhhhhhhhhhhhhhhhhhhhhh You Better Pay It Honey! The Devil Is A Liar!!!!! Bitches Get Interviews And Shit? Where They Do That At Honey? Girls Are Late Honey!!!! The Queen Needs To Get Into It Honey Cause The Girls Is Late Out Here! Yes Honey Im Throwing Epic Shade!

Management!