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Dontwalkaway

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Re: Official back & Front thread
May 05, 2012, 05:51:35 PM
This is supposed to be the Army of Love.  With the L.O.V.E.  So let's get back to it.
Awwww........  Look what's happening to the poor turtle.  He doesn't even want to look.

Come on troops  :icon_exclaim:        :icon_albino:
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"And when that flag blows
There'll be no more wars
And when all calls
I will answer all your prayers"

Chorus from the song "Cry",  Invincible Album

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MaryK

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Re: Official back & Front thread
May 05, 2012, 05:53:38 PM
@SEHF...lol....turtle facepalm...really? What is it?

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You and I were never separate

It's just an illusion

Wrought by the magical lens of Perception



Re: Official back & Front thread
May 05, 2012, 05:54:18 PM
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 :LolLolLolLol: 
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'Money...Lie for it Spy for it Kill for it Die for it They'd kill for the money Do or dare'

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pikachu

Re: Official back & Front thread
May 05, 2012, 05:56:02 PM
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GeEez!!... What happened to this thread??  :WTF:

Now calm down girls!  :LolLolLolLol:

Just a little cute mess, lovely, isn't it? ; )))  :icon_lol:
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.................................................................
will YOU still care? will YOU be there??.. ah, well, sure... life is not a song...

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~Souza~

Re: Official back & Front thread
May 05, 2012, 06:03:45 PM
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Okay everyone, I do believe I've got a grip on this. It's a puzzle! Here we go:


Puzzle: There are two identical crystal balls and one needs to find out which is the maximum floor in a 100 storey building from where the balls can fall before they break. In the most efficient way, what will be the maximum number of drops required to find the right floor in any possible scenario? The balls are allowed to be broken (if required) while finding the right floor.

Solution: The maximum drops required will be 14 to find the right floor considering all possible scenarios.

The approach should be to drop the first crystal ball from a floor and if it breaks then try dropping the second crystal ball (of course until it breaks) from the floor next to the previously tested floor until one less than that floor from where the first crystal ball broke when dropped.

In the most efficient way, the maximum number of trials needed should be kept to the minimum possible. We have a maximum of 100 floors, so we need to divide these floors in such a way that the sum of the number of trials consumed by first crystal ball and that of the second crystal ball (if the first breaks) remains the minimum. For this to happen we need to reduce the difference between the previously tested floor and the next floor to test the first ball from every time by 1 as the first balls has already been dropped by those many times. Confused? Let's try to understand it this way: Suppose we drop the first ball from the Nth floor for the first time then we'll require a maximum of N drops if the first ball breaks as in this case the second ball will be tried from floor #1 to floor #(N-1). If the first ball doesn't break then we'll have to select the next floor to test the drop of the first ball from. This can't be more than (N-i) from the previously tested floor where i = 1 ... N-1 for every subsequent step in this order only. Reason being, if we test the first ball from a higher floor and if the ball breaks then we may require to test the second ball more number of times and hence the maximum attempts may exceed N in this case. Similarly, if we test the first ball from a lower floor then we'll require to test the first ball more than N times if it doesn't break at all.

Obviously this N will of course be dependent upon the maximum number of floors which is 100 in our case. Putting N = 14, we will require the first ball test from floor #14, #(14 + 14 - 1 = 27), #(27 + 14 - 2 = 39), #(39 + 14 - 3 = 50), #(50 + 14 - 4 = 60), #(60 + 14 - 5 = 69), #(69 + 14 - 6 = 77), #(77 + 14 - 7 = 84), #(84 + 14 - 8 = 90), #(90 + 14 - 9 = 95), #(95 + 14 - 10 = 99), and finally from #100. The test will continue until we reach #100 OR the first ball breaks in which case the second crystal ball will be used for the floors lying between the floor where the first ball breaks and the previously tested floor.

I doubt if there is any easy formula to find out this number N, but we can certainly reach this by using a heuristic approach. As we move on we consume trials at every stage and this should always be kept in mind. This is the reason why we can't keep the difference between the previously tested floor and the next floor to test from as constant for the first crystal ball drop.

For N = 14, if the first ball doesn't break at all then we need to try a maximum of 12 times and if it breaks the very first time itself then we may require to try a maximum of 14 times (1 by the first ball from floor #14 and a maximum of 13 by the second ball from floor #1 ... #13). Similarly, if the first ball breaks at floor #27 then maximum attempts required will again be 14 (2 by first ... [from floor #14 and from floor #27] plus 12 by second [from floor #15 ... floor #26]). And and so on. Thus we see that in this case if the first ball breaks at any of the floors then we need a maximum of 14 attempts to figure out the correct floor and if it doesn't break at all then we'll done with a maximum of 12 attempts only.


The answer is: Which floor?  ;)

You just split the maximum number of floors. If 100 floors, you will drop your first ball at 50. If it breaks, you know you have to test floor 1-49, if not then you test floor 51-100. Let's say it breaks at 50, you then drop it at 25. Let's say it breaks at 25, you drop at 12, etc.

Drop 1: floor 50, breaks
Drop 2: floor 25, breaks not
Drop 3:narrowed down to floor 26-49, lets say 38, breaks
Drop 4: between 26 and 37, 32, doesn't break
Drop 5: between 33 and 37 so 35, breaks
Drop 6: either 33 or 34, we pick 33 and it breaks not
Drop 7: will be the last option (34). If it breaks, your answer is 33, if not your answer is 34.

So the only thing you can calculate is the most efficient maximum amount of drops, which is 7, not 14. You can't possibly know which floor, because that depends on the size and the density of the balls, the height of the building and each individual floor, the weight of the balls and the material of the floor. If you throw it down on a big pile of matrasses, chances are it will never break, if you throw it on concrete it will probably break if you throw it off the first floor.
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Its her

Re: Official back & Front thread
May 05, 2012, 06:09:44 PM
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@Souza maybe it's different but still you made me think about my "Bob" lol, obviously another user here.
Different personality, maybe, but definitely a Virgo.It just stroke me the "old guy" thing. Could be a coincidence though.

wow. comparing notes is the most amusing part of any girlish party.
it may sound as a bad joke, but every gal in the hoax has her own 'Bob', lol.
(statistic observation.)
well, let's all start writing here our hidden messages to 'Bob'. lol


Mmm, although, I've got a feeling, that 'Bob' may dislike it..
Or, au contraire, he may be in awe (haven't decided yet) --
it's a funniest show after all. lol  :icon_lol:


.......................
* Bob, son of the bitch, call me!!!

 :thjajaja121:

Now, Pikachu...I KNOW you don't mean to suggest that "Bob" has been chatting/writing to the girls on here in PM, under all kind of names. Although, I have heard stories of girls thinking they are pals with a Mysterious Stranger and thinking that they are keeping him all to themselves... :icon_e_confused: NO "mere man" can juggle that many women :thjajaja121: , and only a fool would WANT to work that hard on several relationships at once, DANG...  :Crash:

Wait!

Unless...

MJ... is not... just... a mere man :errrr: ... :icon_e_smile: 

He's the ONLY man brazen enough :icon_mrgreen: to juggle on such a scale...  :Michael_Jackson_dancing_smile :Michael_Jackson_dancing_smile :Michael_Jackson_dancing_smile :icon_eek: :icon_eek: :icon_eek:

 :errrr:

Nah.  :-\ :icon_albino: Give it up. :icon_albino: :icon_albino: (sweating) :icon_lol: :icon_e_confused: :icon_lol: :icon_e_confused:

FORGET hidden messages to "Bob"!! :LolLolLolLol:  ( :icon_e_surprised: spoken like a true man; nice try, Pikachu. :icon_rolleyes: ) Comparing notes  :affraid: is more FUN :icon_e_ugeek: and may give Souza her answer, if he's still hanging around! :icon_e_biggrin:
 
Girls, fess up, would you want to be tearin' :icon_eek: him or keep sharin' :beerchug: him?   :thjajaja121:  Anyone have their own "Bob"-like Hoax buddy, as Pikachu suggests?
 :thjajaja121:

Last Edit: May 05, 2012, 06:11:31 PM by Its her
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ONLY Believe...

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Its her

Re: Official back & Front thread
May 05, 2012, 06:22:43 PM
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@It's her.... Thanks darling !   :bearhug:

***"Doctor, doctor, gimme the news! ...I've got a BAAAAAD case of loving you!"***

@It's her.... by chance... are you talking about me !?    :LolLolLolLol:   :thjajaja121:

Um...yeeeesss...but it's also a  :icon_albino: Pandemic!!   :TongueOutSmiley:
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ONLY Believe...

Re: Official back & Front thread
May 05, 2012, 06:23:50 PM
 :thjajaja121: :thjajaja121: :icon_lol: :icon_lol:

I've been thoroughly entertained today. Thank you :) It looks like I'm one of the few, maybe the only one, who's steered away from getting to know mysterious persons in private chat or messages. I've had a conversation or two but mostly have relied on my instincts.

Pickachu and SEFH -  :smiley-vault-misc-150:

Blessings to each of you. The ticket I bought on this ride seems to have no expiration date. Not sure what to think but I'm still watching...

Last Edit: May 05, 2012, 06:27:18 PM by voiceforthesilent
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I'm proud to be a child of God and a member of MJ's Army of L.O.V.E.
 
"Press coverage of my life is like [watching] a fictitious movie...like watching science fiction. It's not true." ~Michael Jackson (2005)

"You should not believe everything you read. You are missing the most important revelations". Craig Harvey 3-15-2012

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sweetsunsetwithMJ

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Re: Official back & Front thread
May 05, 2012, 06:39:11 PM
@Its her... :LolLolLolLol:   :thjajaja121:  :beerchug:
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I WANNA BE WHERE YOU ARE!!

Re: Official back & Front thread
May 05, 2012, 08:36:17 PM
Thus...I have heard.
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Tink

Re: Official back & Front thread
May 05, 2012, 08:57:53 PM
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Okay everyone, I do believe I've got a grip on this. It's a puzzle! Here we go:


Puzzle: There are two identical crystal balls and one needs to find out which is the maximum floor in a 100 storey building from where the balls can fall before they break. In the most efficient way, what will be the maximum number of drops required to find the right floor in any possible scenario? The balls are allowed to be broken (if required) while finding the right floor.

Solution: The maximum drops required will be 14 to find the right floor considering all possible scenarios.

The approach should be to drop the first crystal ball from a floor and if it breaks then try dropping the second crystal ball (of course until it breaks) from the floor next to the previously tested floor until one less than that floor from where the first crystal ball broke when dropped.

In the most efficient way, the maximum number of trials needed should be kept to the minimum possible. We have a maximum of 100 floors, so we need to divide these floors in such a way that the sum of the number of trials consumed by first crystal ball and that of the second crystal ball (if the first breaks) remains the minimum. For this to happen we need to reduce the difference between the previously tested floor and the next floor to test the first ball from every time by 1 as the first balls has already been dropped by those many times. Confused? Let's try to understand it this way: Suppose we drop the first ball from the Nth floor for the first time then we'll require a maximum of N drops if the first ball breaks as in this case the second ball will be tried from floor #1 to floor #(N-1). If the first ball doesn't break then we'll have to select the next floor to test the drop of the first ball from. This can't be more than (N-i) from the previously tested floor where i = 1 ... N-1 for every subsequent step in this order only. Reason being, if we test the first ball from a higher floor and if the ball breaks then we may require to test the second ball more number of times and hence the maximum attempts may exceed N in this case. Similarly, if we test the first ball from a lower floor then we'll require to test the first ball more than N times if it doesn't break at all.

Obviously this N will of course be dependent upon the maximum number of floors which is 100 in our case. Putting N = 14, we will require the first ball test from floor #14, #(14 + 14 - 1 = 27), #(27 + 14 - 2 = 39), #(39 + 14 - 3 = 50), #(50 + 14 - 4 = 60), #(60 + 14 - 5 = 69), #(69 + 14 - 6 = 77), #(77 + 14 - 7 = 84), #(84 + 14 - 8 = 90), #(90 + 14 - 9 = 95), #(95 + 14 - 10 = 99), and finally from #100. The test will continue until we reach #100 OR the first ball breaks in which case the second crystal ball will be used for the floors lying between the floor where the first ball breaks and the previously tested floor.

I doubt if there is any easy formula to find out this number N, but we can certainly reach this by using a heuristic approach. As we move on we consume trials at every stage and this should always be kept in mind. This is the reason why we can't keep the difference between the previously tested floor and the next floor to test from as constant for the first crystal ball drop.

For N = 14, if the first ball doesn't break at all then we need to try a maximum of 12 times and if it breaks the very first time itself then we may require to try a maximum of 14 times (1 by the first ball from floor #14 and a maximum of 13 by the second ball from floor #1 ... #13). Similarly, if the first ball breaks at floor #27 then maximum attempts required will again be 14 (2 by first ... [from floor #14 and from floor #27] plus 12 by second [from floor #15 ... floor #26]). And and so on. Thus we see that in this case if the first ball breaks at any of the floors then we need a maximum of 14 attempts to figure out the correct floor and if it doesn't break at all then we'll done with a maximum of 12 attempts only.


The answer is: Which floor?  ;)

You just split the maximum number of floors. If 100 floors, you will drop your first ball at 50. If it breaks, you know you have to test floor 1-49, if not then you test floor 51-100. Let's say it breaks at 50, you then drop it at 25. Let's say it breaks at 25, you drop at 12, etc.

Drop 1: floor 50, breaks
Drop 2: floor 25, breaks not
Drop 3:narrowed down to floor 26-49, lets say 38, breaks
Drop 4: between 26 and 37, 32, doesn't break
Drop 5: between 33 and 37 so 35, breaks
Drop 6: either 33 or 34, we pick 33 and it breaks not
Drop 7: will be the last option (34). If it breaks, your answer is 33, if not your answer is 34.

So the only thing you can calculate is the most efficient maximum amount of drops, which is 7, not 14. You can't possibly know which floor, because that depends on the size and the density of the balls, the height of the building and each individual floor, the weight of the balls and the material of the floor. If you throw it down on a big pile of matrasses, chances are it will never break, if you throw it on concrete it will probably break if you throw it off the first floor.
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Black & Proud! I'm like the Oracle/Batgirl, who helps Batman in the comic books. I believe in "Comic Book justice."


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Tink

Re: Official back & Front thread
May 05, 2012, 09:03:01 PM
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Okay everyone, I do believe I've got a grip on this. It's a puzzle! Here we go:


Puzzle: There are two identical crystal balls and one needs to find out which is the maximum floor in a 100 storey building from where the balls can fall before they break. In the most efficient way, what will be the maximum number of drops required to find the right floor in any possible scenario? The balls are allowed to be broken (if required) while finding the right floor.

Solution: The maximum drops required will be 14 to find the right floor considering all possible scenarios.

The approach should be to drop the first crystal ball from a floor and if it breaks then try dropping the second crystal ball (of course until it breaks) from the floor next to the previously tested floor until one less than that floor from where the first crystal ball broke when dropped.

In the most efficient way, the maximum number of trials needed should be kept to the minimum possible. We have a maximum of 100 floors, so we need to divide these floors in such a way that the sum of the number of trials consumed by first crystal ball and that of the second crystal ball (if the first breaks) remains the minimum. For this to happen we need to reduce the difference between the previously tested floor and the next floor to test the first ball from every time by 1 as the first balls has already been dropped by those many times. Confused? Let's try to understand it this way: Suppose we drop the first ball from the Nth floor for the first time then we'll require a maximum of N drops if the first ball breaks as in this case the second ball will be tried from floor #1 to floor #(N-1). If the first ball doesn't break then we'll have to select the next floor to test the drop of the first ball from. This can't be more than (N-i) from the previously tested floor where i = 1 ... N-1 for every subsequent step in this order only. Reason being, if we test the first ball from a higher floor and if the ball breaks then we may require to test the second ball more number of times and hence the maximum attempts may exceed N in this case. Similarly, if we test the first ball from a lower floor then we'll require to test the first ball more than N times if it doesn't break at all.

Obviously this N will of course be dependent upon the maximum number of floors which is 100 in our case. Putting N = 14, we will require the first ball test from floor #14, #(14 + 14 - 1 = 27), #(27 + 14 - 2 = 39), #(39 + 14 - 3 = 50), #(50 + 14 - 4 = 60), #(60 + 14 - 5 = 69), #(69 + 14 - 6 = 77), #(77 + 14 - 7 = 84), #(84 + 14 - 8 = 90), #(90 + 14 - 9 = 95), #(95 + 14 - 10 = 99), and finally from #100. The test will continue until we reach #100 OR the first ball breaks in which case the second crystal ball will be used for the floors lying between the floor where the first ball breaks and the previously tested floor.

I doubt if there is any easy formula to find out this number N, but we can certainly reach this by using a heuristic approach. As we move on we consume trials at every stage and this should always be kept in mind. This is the reason why we can't keep the difference between the previously tested floor and the next floor to test from as constant for the first crystal ball drop.

For N = 14, if the first ball doesn't break at all then we need to try a maximum of 12 times and if it breaks the very first time itself then we may require to try a maximum of 14 times (1 by the first ball from floor #14 and a maximum of 13 by the second ball from floor #1 ... #13). Similarly, if the first ball breaks at floor #27 then maximum attempts required will again be 14 (2 by first ... [from floor #14 and from floor #27] plus 12 by second [from floor #15 ... floor #26]). And and so on. Thus we see that in this case if the first ball breaks at any of the floors then we need a maximum of 14 attempts to figure out the correct floor and if it doesn't break at all then we'll done with a maximum of 12 attempts only.


The answer is: Which floor?  ;)

You just split the maximum number of floors. If 100 floors, you will drop your first ball at 50. If it breaks, you know you have to test floor 1-49, if not then you test floor 51-100. Let's say it breaks at 50, you then drop it at 25. Let's say it breaks at 25, you drop at 12, etc.

Drop 1: floor 50, breaks
Drop 2: floor 25, breaks not
Drop 3:narrowed down to floor 26-49, lets say 38, breaks
Drop 4: between 26 and 37, 32, doesn't break
Drop 5: between 33 and 37 so 35, breaks
Drop 6: either 33 or 34, we pick 33 and it breaks not
Drop 7: will be the last option (34). If it breaks, your answer is 33, if not your answer is 34.

So the only thing you can calculate is the most efficient maximum amount of drops, which is 7, not 14. You can't possibly know which floor, because that depends on the size and the density of the balls, the height of the building and each individual floor, the weight of the balls and the material of the floor. If you throw it down on a big pile of matrasses, chances are it will never break, if you throw it on concrete it will probably break if you throw it off the first floor.

@Souza - Ah, but when dealing with Scientific Experiments, one must ALWAYS duplicate the outcome! That's why it must be 14 crystal balls, instead of 7.

That's why with the CERN Collider, when they discovered Neutrinos traveled faster than the Speed of Light, all the Scientists thought they made a mistake - so, they had to replicate it. And when they did, wouldn't you know? They did it again! This has ramifications in the possibility of Space-Time travel, according to one of Einstein's Theories.

Thus, why everything scientific must be done twice.  :beerchug:
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Black & Proud! I'm like the Oracle/Batgirl, who helps Batman in the comic books. I believe in "Comic Book justice."


*

bec

Re: Official back & Front thread
May 05, 2012, 09:08:01 PM
Clearly not one of you get it.

Well, except Souza, of course.

Which is probably for the best anyway.

I would do my part to get this thread back on topic if I thought it was important, but on the other hand, I'm glad Front gets to see what sort of fallout has erupted here from the latest bomb he dropped.

Here's the "obvious": Front =back and back=MJ. Thanks to the notes it has been deduced that Front=MJ. So if the former wasn't enough to deduce that Front=MJ, now the "obvious" has been linked full circle: Front=back=MJ=Front.

But that's the "obvious" only as far as I'm concerned, mister peek-at-chu smarty-pants. Sometimes my "obvious" is different from a lot of other people's "obvious" so who's "obvious" is the "obvious" truth? Isn't that the eternal question. Ponder that.

@Front: is it interesting to see all our different reactions to this information? I feel like I'm a monkey in a zoo. While I'm being entertained, I'm also providing the entertainment.

Ps. @SEHF, they think you're MJ. If you don't refute that then you're insinuating the latter and that's  the same as trolling as far as I'm concerned. So address it.
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Are you entertained?

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RK

Re: Official back & Front thread
May 05, 2012, 09:17:14 PM
@Bec...they is an all inclusive word. It should be some. And maybe I am old fashioned, but @ veronica fall, I found your post to be rude. Those kind of comments belong on bulldog's blog.
Let's play nice....okay?
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Tink

Re: Official back & Front thread
May 05, 2012, 09:31:36 PM
@Front - If you are whom everyone thinks, then reach out to the angel with Spirit, who made it snow for you. None of us here know whom she is.
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Black & Proud! I'm like the Oracle/Batgirl, who helps Batman in the comic books. I believe in "Comic Book justice."


 

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