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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Purple Puzzle Cards
[Ivy Set] #164 Weighs and Means
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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Author Message
Stratman
Veteran

Joined: 03 Feb 2006
Posts: 81
Location: Kettering UK

 		I emailed Erich Friedman (the puzzle designer) and asked his advice...
He kindly replied

My email...
I am trying to solve a series of puzzles in an internet ARG (alternate reality game) called Perplex City.
Is it possible to give me some tips on a strategy to employ in trying to solve this type of puzzle?
Pure trial and error from stratch seems somewhat daunting.
(I think there is a mistake on the card as, although it is not made clear, I suspect each weight has to be unique as in your puzzle page examples).

Reply from Erich...
Spoiler (Rollover to View):

You actually can solve medium size weight puzzles with trial and error. but here's a hint or two.

1) look for balances with few possibilities. if one weight balances three weights, the three weights must be small and the one large, and there might be few possibilities to check.

2) look for divisibility issues. if a weight at distance 1 balances another weight at distance 2, one weight is twice the other. and thus the total of the weights is divisible by 3.

3) invent your own methods...

Quote:

(I think there is a mistake on the card as, although it is not made clear, I suspect each weight has to be unique as in your puzzle page examples).


so true. they weren't clear.

thanks for sending a picture of the card. i hadn't seen it yet.
erich

_________________
There Ain't Half Been Some Clever Bastards...Ian Dury and the Blockheads (1978)

PostPosted: Mon Apr 10, 2006 12:11 pm
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MikeRobinson
Boot


Joined: 10 Apr 2006
Posts: 10
Location: MK

 		Just to say that I've had an email from Perplex City saying that both solutions will now be selected.

I've just this second confirmed that at least one of the solutions with
Spoiler (Rollover to View):
100 on each side
is accepted. Unfortunately 2 others got there before me - but hey, #3's not bad!

PostPosted: Mon Apr 10, 2006 1:09 pm
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Liz
Guest


 		can I be really cheeky and just ask someone to put the answers up for me cos I'm completely confused by this one and can't even get the small set of weights!

cheers! x

PostPosted: Tue Apr 11, 2006 1:18 pm
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Magma
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Joined: 10 Jan 2006
Posts: 119

 		Here's a worked solution (a lot copied from my earlier post):

Spoiler (Rollover to View):

We have this setup:

......5...........2.......4
. _ _ _ _ _ | _ _ _ _ _ _
|.......................|...........|
|.......................|...........|
A......................B..........C

Remember the torque acting on the pivot is equal to the mass x distance, so on the left:

5A acts downwards.

On the right:

2B and 6C acts downwards.

So we have the equation 5A = 2B + 6C, and we need the minimum.

If you work out what is the minimum weight on each of these arms where all of the sub-arms balance, you get:
A=13
B=22
C=9

Considering the right side, leaving B and C as they are, you get 22 x 2 plus 6 x 9, which is 44 plus 54, which is 98. This is the absolute *minimum* torque that the right side can generate. As it is a balance, this is the minimum that the left must balance.

But - remember that you can only use weights in increments of 1kg. This means every weight you add on the left adds 5 torque, as it is 5 units away. So the minimum torque that A can balance is the next number up from 98 that is divisible by 5, which is 100.

So from this you can reason that the lowest possible weight you can put on A is 20kg. What does all this tell us about B and C?

We know that for this lowest state to be possible, 100 must equal 2B + 6C. So if you draw a table starting with the lowest value for B (22) and increase it by 1 every step, then you can work out the corresponding value for C given that 2B + 6C = 100. You can immediately ignore any value that ends in 1/3 or 2/3, because only whole weights are allowed. You can also ignore all values of C below 9, because we already know 9 is the minimum value for C.

It works out there is only one possible set of values for B and C: 23 and 9.

So now we have:
A=20
B=23
C=9

So
5A = 5 x 20 = 100
And
2B + 3C = 2 x 23 + 6 x 9 = 46 + 54 = 100

Now all you have to do is balance the weights on each arm to satisfy those conditions. If you fiddle around with the weights using the spreadheet from this topic's first post, you can find some solutions that work. The one I found is as follows:




SMALL BALANCE
Top row (Left to right) 1, 1, 2.
Bottom row (Left to right) 2, 1.

LARGE BALANCE
Left arm: Top row 2, 2, 3. Middle row 2, 3. Bottom row 5, 1, 2.
Middle arm: Top row 4, 5. Bottom row 10, 1, 1, 1, 1.
Right arm: Top row 3, 1. Bottom row 1, 1, 3.
[spoiler]


PostPosted: Tue Apr 11, 2006 2:35 pm
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padge
Boot

Joined: 05 Jan 2006
Posts: 43

 		So do we know if it is possible to solve this using only the weights 1kg-20kg? (one of each - as in Erich's examples). Could stratman email him again and just confirm that this is possible?

PostPosted: Wed Apr 12, 2006 7:27 am
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Reptile
Veteran


Joined: 13 Oct 2005
Posts: 89
Location: Sheffield

 		i got the same as above but with the left arm of the large balance as
Spoiler (Rollover to View):

top row 2 8 1
middle row 3 1
bottom row 3 1 1


its not accepted, am i going wrong somewhere?
_________________
http://perplexcitytrades.com/Reptile

PostPosted: Wed Apr 12, 2006 8:34 am
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Magma
Veteran


Joined: 10 Jan 2006
Posts: 119

 		No, you're not mad. Mind Candy just didn't know how many solutions this puzzle actually has. Send'em an email, they should add it to the growing list of correct answers Smile

PostPosted: Wed Apr 12, 2006 2:29 pm
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simonaubrey
Boot

Joined: 06 Dec 2005
Posts: 37

 		I get the following to balance:

Spoiler (Rollover to View):

1. 2kg
2. 2kg
3. 1kg
4. 2kg
5. 2kg
6. 3kg
7. 1kg
8. 3kg
9. 1kg
10. 5kg
11. 2kg
12. 1kg


PostPosted: Thu Apr 13, 2006 11:54 am
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drizzt
Guest


 		I cannot fathom what you people are talking about.

It appears to me that there are either 11 (in the case of the larger set) or 13 (in the case of both sets combined) weights that need to be solved, but noone has put together an answer in that fashion. Am I missing something here?

Lets say you've got a balance like this:
===Cen|ter===

2---notch---Cen|ter---notch---notch---notch---1

Does it balance? That would seem to be the point of the rod across the string acting as a fulcrum, but I've been known to be wrong

PostPosted: Thu Apr 13, 2006 6:16 pm
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Magma
Veteran


Joined: 10 Jan 2006
Posts: 119
drizzt wrote:
It appears to me that there are either 11 (in the case of the larger set) or 13 (in the case of both sets combined) weights that need to be solved, but noone has put together an answer in that fashion. Am I missing something here?


Because the first balance is simple, most people are focussing on the second, larger one. To solve the first, all you have to do is put the minimum weights to balance the lower section, then the minimum weights to balance the upper one. The force of all the weights balanced below added together pulls on the string attached to the upper balance.


drizzt wrote:
2---notch---Cen|ter---notch---notch---notch---1

Does it balance? That would seem to be the point of the rod across the string acting as a fulcrum, but I've been known to be wrong


The torque on both sides must be the same. Torque is given as force times distance. Let's assume that the notches are 1 metre apart, and the weights exert 1 or 2 Newton of force downwards.

(A common mistake non-physicists make is confusing weight and mass. It doesn't really matter unless you go to places with different gravity though...)

So we have a 2 Newton weight 1 metre away from the fulcrum. This gives a left-hand torque of 2 x 1 = 2 Nm.

There is also a 1 Newton weight 3 metres away from the fulcrum. This gives a right-hand torque of 1 x 3 = 3 Nm.

So the right side gives more torque, and it will topple over that side.
Move the right side weight 1m further in, or make the left weight exert 1 Newton more weight, and it will balance.

PostPosted: Thu Apr 13, 2006 6:33 pm
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marky1124
Boot

Joined: 23 Dec 2005
Posts: 31

 		Was a correct solution rejected?
Hi,

Would someone please be kind enough to check my answer. I can't figure out whether I'm being stupid or whether I need to mail Mind Candy with another solution.

Spoiler (Rollover to View):

Small balance: top row = 1 1 2, bottom row = 2 1
So solution page 1=1, 2=2, 3=1.

Larger balance.

I'm balancing the left branch total weight of 20kg against the middle branch of 23kg & the right branch of 9kg. So (20*5) = (23*2) + (9*6) which is 100 = 100.

Left branch. I have 6 solutions for this branch to total to 20kg.
Here's the one I tried: top = 2 8 1 middle = 3 1 bottom = 3 1 1
So solution page 4=2, 5=1, 9=3, 11=1

Middle branch. top row = 4 5, bottom row = 10 1 1 1 1
So solution page 10=5, 12=1

Right branch. top row = 3 1, bottom row = 1 1 3
So solution page 6=3, 7=1, 8=3


So I entered the solution weights in order as: 1 2 1 2 1 3 1 3 3 5 1 1
and was rejected.

Question Is it me?


Thanks very much,
Cheers,
Mark

PostPosted: Fri Apr 14, 2006 2:05 pm
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Grizy
Veteran

Joined: 19 Nov 2005
Posts: 89
Location: Preston, UK

 		Your solution seems fine Mark.
As others have mentioned earlier, there is more than one solution to this puzzle.
The site already accepts a couple of them. I f it doesn't accept your solution email Mind Candy and they will probably update the site in due course.

PostPosted: Fri Apr 14, 2006 2:33 pm
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marky1124
Boot

Joined: 23 Dec 2005
Posts: 31

 		Thanks Grizy. You gave me the confidence to keep entering answers. The first two I gave it were rejected the third succeeded. Phew. Just avoided a lock out.

I've mailed card support with the 6 solutions that I think should all be accepted.

All the best,
Mark

PostPosted: Fri Apr 14, 2006 3:00 pm
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Reptile
Veteran


Joined: 13 Oct 2005
Posts: 89
Location: Sheffield

 		marky, thats the same solution i have, they're adding it to the database next update
_________________
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PostPosted: Fri Apr 21, 2006 7:17 am
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pheebs
Guest


 		sorry, i'm a little confused lol (its late and i've had a few G&Ts, oops!). which answers have been accepted? the ones i entered werent accepted so i'm guessing MC havent updated yet? which ones do we know are accepted and which ones arent?

i remember vaguely similar problems in maths at GCSE, its a wonder i didnt fail that LOL. Very Happy

PostPosted: Fri Apr 21, 2006 7:26 am
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