The Framers Forum

Roboframer wrote:I think the clue is in the sealed room.

The room heats up.

Correct! You may pass "GO!" You may collect 200 gold stars!!!

The sealed room and fridge represent a sealed system. Because the fridge is switched on, energy in the form of electricity is entering the closed system. The energy contained within the overall system is therefore increasing. Regardless of any cooling effect from the refriidgeration process, the additional electrical energy is converted ultimately into heat energy, and therefore the temperature in the room will rise over time.

I'll leave John to do the honours on the answer to the lightbulb teaser!

Ok then.

1. Turn on the 1st switch

2. Go and make a cup of tea or something

3. Come back to switches; turn off the 1st switch; turn on the 2nd switch.

4. Enter the room.

One light will be on (2nd switch) so's you can see, one will be hot (1st switch) and one will be cold (3rd switch)

Very inefficient Robo, but a solution.

steps one and two are unnecessary

Instead: -

switch one on for ten seconds

switch two on.

enter room

the warm bulb belongs to switch one, and the cold one to switch three.

Good heavens man, don't you know that the global warming police will be after you? Leaving lights on. Making cups of tea. Little wonder we're all going to hell in a hand-cart!

John wrote:Very inefficient Robo,

I dunno about that John: personally, I think step two has definite merit!

Right - I've got one - years since I did it so's I'll have to work it out again myself.

You have 12 billiard balls, all the same size and colour, but one is a different weight to the rest - you don't know if it is heavier or lighter.

You have a balance and may use it only 3 times to determine the differently weighted ball and state whether it is lighter or heavier.

You have to cover all the bases - no good saying 'Blah Blah ... balance. Blah blah ...... balance ..... blah blah .... tip' Because I'l just ask what happens if it does not balance 1st time etc etc.

Try it with numbered bits of paper or something and your hands as the balance.

Roboframer wrote:
2. Go and make a cup of tea or something

This is no doubt to allow enough warm up time if the bulbs are low energy types. Nice one John!

Roboframer wrote:Right - I've got one - years since I did it so's I'll have to work it out again myself.

You have 12 billiard balls, all the same size and colour, but one is a different weight to the rest - you don't know if it is heavier or lighter.

You have a balance and may use it only 3 times to determine the differently weighted ball and state whether it is lighter or heavier.

You have to cover all the bases - no good saying 'Blah Blah ... balance. Blah blah ...... balance ..... blah blah .... tip' Because I'l just ask what happens if it does not balance 1st time etc etc.

Try it with numbered bits of paper or something and your hands as the balance.

Weighing #1: Place 6 balls on the right scale; the other six on the left. The side that drops contains the heavy ball.

Weighing #2: Take the six from the heavy side. Split the six into three groups of two balls. A2, B2, C2. Place A2 and B2 on opposite sides of the scale. If they balance, the heavier ball is in group C2 – proceed to step #3.

If one side of the A2 and B2 combination drops, that side contains the heavier ball. Take those two ball and divide them and proceed to step #3.

Weighing #3: Place one ball on the right; one on the left. The heavier ball will drop.

… Blah, blah, blah ….

Bill Henry wrote:Weighing #1: Place 6 balls on the right scale; the other six on the left. The side that drops contains the heavy ball.

Could it not be that the side that rises has the light ball?

Clue.
Start with three groups of four.

This is wrecking my head: so far, I've only got it down to using the balance four times to absolutely guarantee identifying the odd ball and the weight difference. I can only get it down to three if I "strike lucky" on one of the first two weighings....

I rememeberd it today and have the solution if y'all give up.

Bill Henry wrote:
Weighing #1: Place 6 balls on the right scale; the other six on the left. The side that drops contains the heavy ball.

That's a waste of a go Bill - as John said, you don't know if there is a lighter ball on one side - so you're still at square one - just knowing that one ball is a different weight.

If you weighed say, oooooooh I don't know ....four on each side and left four off, and the scales balanced, you've got eight good uns there.

Too many clues already ....

John, I can't see how to guarantee selecting the odd ball out every time with only three uses of the balance. Rather than spoil it for others, any chance of a PM to put me out of my misery?

Roboframer wrote:I rememeberd it today and have the solution if y'all give up.
That's a waste of a go Bill - as John said, you don't know if there is a lighter ball on one side - so you're still at square one - just knowing that one ball is a different weight.

If you weighed say, oooooooh I don't know ....four on each side and left four off, and the scales balanced, you've got eight good uns there.

Too many clues already ....

Yeah, sorry, I misread the original post.

I was thinking about this problem as I tried to drift off to sleep last night, and, like Àine, I cannot see how it can be done in less than four. I'm apparently not gettin' sumthin'!

Robo wrote:four on each side and left four off, and the scales balanced, you've got eight good uns there.

I bet Robo has an easier solution, but here's my take on the first part of the answer.

Take one 'good un' from the eight and balance it together with one of the remaining four against two others from the remaining four.

The outcome is one of: -

• a. even balance
  b. 'good un' side heavy
  c. 'good un' side light

If outcome was a. then balance a 'good un' against the un-weighed ball from the remaining four to see if it is the light or heavy oddball.

If the outcome was b. remove a ball from each side, leaving the 'good un' on the scale.

• If they balance then the ball that was with the 'good un' is the heavy oddball
  If the 'good un' side is heavy then the ball on the opposite side is the light oddball

If the outcome was c. remove a ball from each side, leaving the 'good un' on the scale.

• If they balance then the ball that was with the 'good un' is the light oddball
  If the 'good un' side is light then the ball on the opposite side is the heavy oddball

Does this make sense? I think I've started to confuse myself!

Anyone else want to work out the other half of the solution where the first weighing does not balance?

Nurse!

OK I PM'd Áine with the solution - but I don't think I've remembered it right.

I do remember it took me a month on and off to solve it 1st time!

The basic principles are correct I know - maybe I confused myself numbering the balls - have a look anyway.

The underlined numbers (edit - forgot to underline them ) are balls on scales - in brackets off the scales.

Let's say they balance 1st go and call all those balls 'Y's and the ones off the scale 'X's


1 2 3 4 ............... 5 6 7 8

(9 10 11 12)


Take one Y off one side of the scale (in this case the left) and replace it with an X. Put the other three Xs on the other side with one Y .....

Obviously they won't balance this time - let's say they do this .....


1 2 3 9
......................... 10 11 12 8


(4 5 6 7)

So - either No 9 is lighter, or one of 10, 11 & 12 is heavier . .... so swap (say) No 10 with N0 9, remove No 11 and replace it with a Y ball and keep No 12 where it is - re-weigh.

If the scale does this....



1 2 3 10
............................. 9 4 5 12



(11 6 7 8 )

You have not moved the odd ball - 12 is heavier (you know that from the 2nd weigh)

If it balances you HAVE moved the odd ball - 11 is heavier

If the scales tip the other way - again you have moved the odd ball, 10 is lighter.

Hope I've explained it OK!

Regardless of what the scale does 1st time the same sort of procedure should get you the correct answer.

Phew!


ERROR ALERT - ERROR ALERT - No 10 cannot be lighter - it went down on the second go.

Oh dear, oh VERY dear!!

Why not redirect this thread to the Mythbusters then we can just watch instead of rack the grey stuff? Some of us are still on holiday and the brain has not checked in yet.

OK, hang on to your booleans.


Ist Go.....

Divide the balls into groups of 4

Put one group on each side of the scales and the remaining one leave to one side.

3 things can happen:

Balance: All the balls on the scale are equal and the odd one must be in the
unweighed group -> go to 2nd Go/condition 1


Left side up: One of the 4 balls on the left is light or one of the 4 balls on the
right is heavy-> go to 2nd Go/condition 2.


Left side down: One of the 4 balls on the right is light or one of the 4 balls on the
left is heavy-> go to 2nd Go/condition 3

###############################################
2nd Go......

condition 1

You know that all the balls on the scales are equal and the odd one is in the unweighed group. I'll call
the unweighed ones unknowns and the weighed ones standard.

Put 3 unknowns on the left side of the scales and 3 standards on the right.

3 things can happen:


Balance: You know that the odd ball is the remaining unknown. You have cracked the case in 2 goes.

Left side down: You know one of the 3 balls on the left is a possible heavy. -> go to 3rd Go/condition 2

Left side up: You know one of the 3 balls on the left is a possible light. -> go to 3rd Go/condition 2


condition 2

Put two possible heavy ones and one possible light one on the left side and a possible heavy, a possible
light and a known standard one on the right. This leaves two posible lights and a possible heavy on the side

3 things can happen:

Balance: The odd ball is one of the balls on the side. -> go to 3rd Go/condition 4

Left side down: The odd ball is one of the left two possible heavys or the right possible light. -> go to 3rd Go/condition 5

Left side up: The odd ball is either the right possible heavy or the left possible light. -> go to 3rd Go/condition 6



###############################################

3rd Go........

condition 1 Already solved it

condition 2 You know one of the left balls is heavy. Compare any 2.

3 things can happen:

Balance: > The other one is the odd ball

Left side down > The left one is the odd ball

Left side up > The right one is the odd ball


condition 3 You know one of the left balls is light. Compare any 2.

3 things can happen:

Balance: > The other one is the odd ball

Left side down > The right one is the odd ball

Left side up > The left one is the odd ball



condition 4 Compare two of the previously unweighed possible lights.

3 things can happen:

Balance: >The odd ball is the other unweighed one (heavier)

Left side down > The right one is the odd ball(lighter)

Left side up > The left one is the odd ball(lighter)


condition 5 Compare two of the previously unweighed possible heavys.

3 things can happen:

Balance: >The odd ball is the other unweighed one (lighter)

Left side down > The left one is the odd ball(heavier)

Left side up > The right one is the odd ball(heavier)


condition 6 Compare one of the possible heavys (on left) with a known standard ball(on right).

3 things can happen:

Balance: >The odd ball is the other unweighed possible light one (lighter)

Left side down > The odd ball is the possible heavy on the scale(heavier)

Left side up > can't happen




My brain hurts.

A simpler one.....

You have four wine glasses in a row.

first 2 are full, second two are empty.

F F E E

Rearrange them so they go F E F E

only moving one glass.

Numbering glasses 1-4 from left to right, pour all of the the contents of glass 2 into glass 3 (very carefully, so as not to spill any), then return glass 2 in its original position?

Yes, easy init? I takes genius to spot the bleeding obvious.

Heres a nice one.

Part1. The Easy Bit

How many shapes can you arrange 5 equal squares into? (miiror images not included)

I assume that the squares need to be touching, Prospero. Are there any other constraints, e.g:

1. Exclude additional shapes that can be mapped rotationally?

2. Squares may be connected via edges (I would assume yes).

3. Can shapes be made with just the vertices touching?