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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Orange Puzzle Cards
[SOLVED] #054 - Orange - 9 Ball
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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JebJoya
Unfettered


Joined: 13 Apr 2005
Posts: 679
Location: UK

 		Very Happy why thankyou Very Happy

Jeb
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PostPosted: Thu Oct 27, 2005 6:16 am
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Pillzbury
Guest


 		Confirmed

Spoiler (Rollover to View):
2


PostPosted: Sat Dec 24, 2005 11:31 am
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joePM
Boot


Joined: 26 Feb 2006
Posts: 20
Rob_Riv wrote:

1) WEIGH ( 123 ) AGAINST ( 456 )
If equal go to step ( 2 )
If ( 123 ) lighter go to step ( 3 )
If ( 456 ) lighter go to step ( 4 )

2) WEIGH ( 7 ) AGAINST ( 8 )
If equal, BALL 9 is the LIGHT one
If ( 7 ) lighter, BALL 7 is the LIGHT one
If ( 8 ) lighter, BALL 8 is the LIGHT one

3) WEIGH ( 1 ) AGAINST ( 2 )
If equal, BALL 3 is the LIGHT one
If ( 1 ) lighter, BALL 1 is the LIGHT one
If ( 2 ) lighter, BALL 2 is the LIGHT one

4) WEIGH ( 4 ) AGAINST ( 5 )
If equal, BALL 6 is the LIGHT one
If ( 4 ) lighter, BALL 4 is the LIGHT one
If ( 5 ) lighter, BALL 5 is the LIGHT one


Ok, yes, the accepted answer is definitely
Spoiler (Rollover to View):
2


But I just want to log my protest! Mad

I have two problems with the solve as quoted above. First: Step 2 says that if all the other balls are equal, 9 must be the light one. How is this answer any more or less enlightening than if we weighed (1234) against (5678) in the first step?
Spoiler (Rollover to View):
Seems to me you could find a light ball in one step if you got lucky the first time.


Even more simply, though: Pick any two balls at random. If one is heavier than the other, you've confirmed that one of the balls is lighter. Thus the -minimum- number of weighings should be
Spoiler (Rollover to View):
1!


My other issue with this solve is that there are actually 10 regulation balls in a 9-ball game, if you include the one whose weight would be likely to make the most difference: the cue ball!

Someone tell me what I'm overlooking, please.
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PostPosted: Tue Mar 07, 2006 3:09 pm
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donxkey
Kl00


Joined: 09 Apr 2006
Posts: 42

 		I have to agree with joePM. The card only asks for the minimum number, whats to say you wouldnt get lucky first time.

PostPosted: Wed Apr 12, 2006 4:24 am
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Joker
Kilroy

Joined: 19 May 2006
Posts: 1
joePM wrote:
Rob_Riv wrote:

1) WEIGH ( 123 ) AGAINST ( 456 )
If equal go to step ( 2 )
If ( 123 ) lighter go to step ( 3 )
If ( 456 ) lighter go to step ( 4 )

2) WEIGH ( 7 ) AGAINST ( 8 )
If equal, BALL 9 is the LIGHT one
If ( 7 ) lighter, BALL 7 is the LIGHT one
If ( 8 ) lighter, BALL 8 is the LIGHT one

3) WEIGH ( 1 ) AGAINST ( 2 )
If equal, BALL 3 is the LIGHT one
If ( 1 ) lighter, BALL 1 is the LIGHT one
If ( 2 ) lighter, BALL 2 is the LIGHT one

4) WEIGH ( 4 ) AGAINST ( 5 )
If equal, BALL 6 is the LIGHT one
If ( 4 ) lighter, BALL 4 is the LIGHT one
If ( 5 ) lighter, BALL 5 is the LIGHT one


Jeb simplified nmarciano answer correctly. IT ELIMINATES ALL POSSIBILITIES WITH 2 WEIGHINS. Every answer is answered.

Ok, yes, the accepted answer is definitely
Spoiler (Rollover to View):
2


But I just want to log my protest! Mad

I have two problems with the solve as quoted above. First: Step 2 says that if all the other balls are equal, 9 must be the light one. How is this answer any more or less enlightening than if we weighed (1234) against (5678) in the first step?
Spoiler (Rollover to View):
Seems to me you could find a light ball in one step if you got lucky the first time.


True, what if. But what IF you weighed 1234 and 5678 and they were not equal? Then you would have to continue past your "1 weighin" to mulitple weighins. i.e more then 2 hence why your answer is not stable. It leaves room for error

Even more simply, though: Pick any two balls at random. If one is heavier than the other, you've confirmed that one of the balls is lighter. Thus the -minimum- number of weighings should be
Spoiler (Rollover to View):
1!


Same "what if" as above

My other issue with this solve is that there are actually 10 regulation balls in a 9-ball game, if you include the one whose weight would be likely to make the most difference: the cue ball!

TONE DIDN'T TAKE THIS WELL AND CHALLENGED ME TO PROVE IT. "I WILL PROVE IT!" I
ANNOUNCED, TAKING DOWN THE REGULATION WEIGHING SCALES FROM THE WALL (ALL
BARS HAVE THEM, DON'T ASK ME WHY). WHAT WAS THE MINIMUM NUMBER OF WEIGHINGS
OF THE NINE BALL POOLS I NEEDED TO TAKE TO FIND OUT WHICH WAS LIGHTER THAN
THE OTHERS?

Someone tell me what I'm overlooking, please.


Maybe to answer the scale problem if someone has not already asked, right side of the card, top right, you will see a scale in the dark background.

dunno man, puzzles have a way of doing this to people

PostPosted: Fri May 19, 2006 6:08 am
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yodle
Kilroy

Joined: 21 Jul 2006
Posts: 1

 		hmmm
I see your arguement and agree, to be certain, you need 2 weighings - (thats assuming one ball is lighter, which cant be proved in two weighings, and was the intial point of the arguement - but anyway)

The card states what is the minimum number of weighings - it doesn't refer to certainty - so the answer should be 1

Either way, if I was Tone, I wouldn't be happy until at least 3 weighings had taken place

And even then I wouldn't be convinced that 1 lighter ball could ruin a game of 9 ball - the only real way could be if the lighter ball was the 9 ball, so you would really only need 1 weighing i.e. the 9 ball against any other ball...

PostPosted: Fri Jul 21, 2006 8:43 pm
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Platinumflux
Boot


Joined: 03 Aug 2006
Posts: 43
Location: Ireland

 		Hi all! Wink new to this so give me a chance!
If all the balls are weighed together as a group and the result divided by 9, this would give the answer X. Then if a single ball is weighed the amount would not equal X regardless of which ball is weighed which means one has to be lighter.
So the minimum amount of weighings would be
Spoiler (Rollover to View):
2!


I dont think anyone has mentioned this yet but do you suppose that the ball numbers in the picture mean anything?
1,3,9,7,4,8

PostPosted: Thu Aug 03, 2006 1:34 pm
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Corvus
Greenhorn

Joined: 03 Sep 2006
Posts: 8
Location: Vancouver, Canada

 		If I was Tone I wouldn't be satisfied until each ball had been weighed, individually, against another ball that had already be measured as lighter than or the same weight as another ball. This would require 8 weighings.

The posted answer assumes that eight balls have the same weight and the ninth is lighter. This is not at all supported by the preconditions; the balls could all be different weights, or another ball could be as much heavier than the others as the light ball is lighter.

Since the posted answer presupposes what we are trying to prove in the first place, and Tone is prepared to accept that answer, he has already accepted that there is indeed one lighter ball, therefore the correct answer is that it takes zero weighings.

PostPosted: Tue Sep 05, 2006 9:15 am
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smarty33
Guest


 		on the new website it is a multiple choice and i have tried all of them and yet i am wrong....a little help here

the choices are...
1
2
3
4
5
the site needs to make up its mind.

Spoiler (Rollover to View):
my answer is one take the subject ball the put 8 on each side of the scale...but they are even...then the last of the 9 balls that was not weighed is the lightest...we had a problem in math like this last year only with candy bars. Very Happy
so it should be easy but its not...

PostPosted: Mon Feb 19, 2007 1:32 pm
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smartyman
Boot

Joined: 14 Feb 2007
Posts: 22
smarty33 wrote:
on the new website it is a multiple choice and i have tried all of them and yet i am wrong....a little help here

the choices are...
1
2
3
4
5
the site needs to make up its mind.

Spoiler (Rollover to View):
my answer is one take the subject ball the put 8 on each side of the scale...but they are even...then the last of the 9 balls that was not weighed is the lightest...we had a problem in math like this last year only with candy bars. Very Happy
so it should be easy but its not...


I had the same problem with this card and #48 (both utilize drop-down boxes for answers). On a hunch I fired up Firefox and bingo! No problem. Seems the new site has some sort of javascript incompatibilities with Internet Explorer.

PostPosted: Sun Feb 25, 2007 1:24 am
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lonadar
Boot


Joined: 30 Jan 2007
Posts: 50
Location: Indiana, USA

 		I've had issues with several cards now, and I've tried FireFox and Safari.

I've started a topic here about it.

PostPosted: Tue Feb 27, 2007 9:27 pm
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