let the blue and green chameleon meet first.
Result :
14 - red
14 - green
16 - blue
now 14 times pair of red and green meet to make it all 44 blue
On Sep 5, 11:44 pm, Don dondod...@gmail.com wrote:
Yes, you are right, it is 4 in that case. It seems that f is always
even. It is possible for 44
@SandeepGupta: That ways red will be 15.
On Tue, Sep 6, 2011 at 9:43 PM, sandeep gupta sandeepgupta...@gmail.comwrote:
let the blue and green chameleon meet first.
Result :
14 - red
14 - green
16 - blue
now 14 times pair of red and green meet to make it all 44 blue
On Sep 5, 11:44 pm,
@Sandeep: there are 45 chameleons, not 44.
If blue and green meet first, there will be 15 red, not 14.
You can find all of the possible combinations very quickly using a
46x46 table. Start with all cells set to false except for the first
(13,15). If you know the red and green count, you can
No, f(15,14,16) = 5.
Don
On Sep 5, 8:33 pm, wujin chen wujinchen...@gmail.com wrote:
hi all, i encountered this puzzle (http://www.crackpuzzles.com/?p=236):
At one point, a remote island's population of chameleons was divided as
follows:
- 13 red chameleons
- 15 green chameleons
- 17 blue
hi Don, i think f(15,14,16) =|15-14|+|14-16|+|16-15| = 1+2+1=4, hou do you
get f(15,14,16) = 5?
2011/9/6 Don dondod...@gmail.com
No, f(15,14,16) = 5.
Don
On Sep 5, 8:33 pm, wujin chen wujinchen...@gmail.com wrote:
hi all, i encountered this puzzle (http://www.crackpuzzles.com/?p=236):
Yes, you are right, it is 4 in that case. It seems that f is always
even. It is possible for 44 chameleons to be one color, but then there
is only one left and it cannot change to that color. Any time there
are 43 chameleons of one color, the other two are the same color.
It is true that all the
