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gentenaire
Dec 30th, 2004, 03:48 PM
and as usual, I'm stuck.

http://www.albartus.com/motas/

I've put out the fire, but can't find a key to open the gate and I have no idea what to do with the colour thing.

fleemkeģ
Dec 30th, 2004, 03:55 PM
thx Tine :bounce:

:rolleyes: Have to start from the start again :(

gentenaire
Dec 30th, 2004, 04:11 PM
yes, got through!!

And Fleem, I also had to start from the beginning again. They made the green wall easier this time. I had forgotten how I'd done it last time,took me a while back then, but this time there were still a few tiles left black when the wall opened.

Gallofa
Dec 30th, 2004, 08:39 PM
Hmm... I am stuck in level 5... I think I must had forgotten something.

gentenaire
Dec 30th, 2004, 09:31 PM
which level is level 5?

Lynx
Dec 31st, 2004, 05:36 PM
which level is level 5?
I think it's the level with the code-panel with all the logic puzzles (12 smilies → 12 letters → 12 polygons → 12 numbers etc. etc.)

Gallofa: jot down a short description and the number (or position) of the button to push, in case you push a wrong one and are trown back. All in all there are 27 combinations, lol.

Lynx
Dec 31st, 2004, 05:38 PM
BTW, for all those who like puzzles, here is one from before the computer era.

http://www.imagestation.com/picture/sraid152/pbb439c5e8bf9e8d70e3f545db459a095/f5afe2cf.jpg

Given: 12 billiard balls, 1 old-fashioned balance.

The balls are all exactly alike... but one. One of them is a fake, weighting slightly more - or less? - than the others.
You have to find the fake one, using those scales (which will only indicate which side is heavier). Also, you aren't allowed to use the scales more than 3 times.


Enjoy :unsure: :D.

And err... happy New Year. I'm off...

Brian Stewart
Jan 1st, 2005, 09:00 AM
BTW, for all those who like puzzles, here is one from before the computer era.

http://www.imagestation.com/picture/sraid152/pbb439c5e8bf9e8d70e3f545db459a095/f5afe2cf.jpg

Given: 12 billiard balls, 1 old-fashioned balance.

The balls are all exactly alike... but one. One of them is a fake, weighting slightly more - or less? - than the others.
You have to find the fake one, using those scales (which will only indicate which side is heavier). Also, you aren't allowed to use the scales more than 3 times.


Enjoy :unsure: :D.

And err... happy New Year. I'm off...

Are you told in advance which it is, heavier or lighter? If so, it's pretty routine.


***spoiler warning******


You put 6 balls on each side of the scale. One will be heavier than the other. If the bogus ball is heavier than the others, you take the 6 from the lower (heavier) side to the second weighing. Vice-versa if the bogus ball is lighter. Then, you split this group 3 and 3, taking whichever 3 fits the criteria (lower 3 if the bogus ball is supposed to be heavier, higher 3 if it's lighter). Now that you've isolated the bogus ball into one group of 3, you put 1 ball on each side of the scale. If the scales don't balance, you take whichever ball suits the criteria (if the bogus ball is supposed to be heavier, take the one which registers heavier, and vice-versa for if it's supposed to be lighter). If the scales do balance, the bogus ball is the 3rd (unweighed) ball.

Lynx
Jan 1st, 2005, 12:33 PM
Are you told in advance which it is, heavier or lighter?
Nope http://forum.sport.be/ubb/graemlins/nono.gif :lol:. That would be too easy ;).

I forgot to mention: at the end you also have to know if the bogus ball is heavier or lighter. (But if you're doing it right you're going to know that - it almost follows automatically...)

gentenaire
Jan 1st, 2005, 12:42 PM
Lynx, this riddle was posted once in the riddle thread here on WTAworld (long time ago). I remember Crow and I both came up with a different solution and both worked. I've forgotten the solution however, so I'll give it another try later.

Lynx
Jan 1st, 2005, 01:01 PM
Lynx, this riddle was posted once in the riddle thread here on WTAworld (long time ago). I remember Crow and I both came up with a different solution and both worked. I've forgotten the solution however, so I'll give it another try later.
It was? Sorry, didn't know. And I've know this riddle for so long now, I almost consider it mine :p.

I only know one solution (the one I came up with more years ago than I care to remember :o) - I'm thrilled to learn there is another one, although I can't imagine how there could be... :scratch:

(And now you have me puzzling again :( :D)

Lynx
Jan 1st, 2005, 01:11 PM
If I have the time, I'll put my solution in an attachment later tonight.
Have to go visit aunts and the like now... :/

moby
Jan 1st, 2005, 01:25 PM
I only have one solution for it too, although i'm pretty sure there's only one solution.
I worked it out during a particularly boring Maths lecture :o

*****spoiler alert*****

1. Divide the balls into 3 sets of 4
2. Weigh any two sets against each other.
a) The two sets are of the same weight, and so the faulty ball is in the third set.
b) The two sets have different weights.

Scenario A:
It is pretty obvious what you should do, so I won't insult anyone's intelligence with any detailed description. Weigh two balls from the third set against 2 normal balls. Then narrow it down to one ball.

Scenario B (which is more interesting :) ):
3. Label one set Heavy and one set Light... Give the balls labels like H1, H2, H3 and H4 for the heavy set and L1, L2, L3, L4 for the light set. The H label indicates that if the ball is faulty, it would be because it is heavier, and the L for the ball being lighter.
4. Weigh H1, H2, L1 against H3, H4, L2
5. If H1, H2, L1 is heavier, then, the faulty ball is H1, or H2 or L2.
If the 3 weigh lighter, then the faulty ball is H3, H4 or L1
If they are the same weight, the faulty ball is L3, or L4, in which case it should be obvious how to find out the faulty ball.

6. I'll just describe the scenario for H1, H2, and L1 being heavier... you can infer for the other case. Since the faulty ball is H1, H2, or L2, weigh H1 and L2 against 2 normal balls. If H1 and L2 is heavier, then H1 is the faulty ball. If they are lighter, L2 is the faulty ball. If they are the same weight, H2 is the faulty ball.

moby
Jan 1st, 2005, 04:37 PM
I just finished MOTAS :D
but it's kind of short :(

The green wall was really tough, I had to trial and error my way through it :o
Does anyone have some algorithm to do it systematically?

gentenaire
Jan 1st, 2005, 06:27 PM
The green wall was really tough, I had to trial and error my way through it :o
Does anyone have some algorithm to do it systematically?

It's rather simple really, at least it is now (at first the wall had to be entirely green, took me ages to find the algorithm then, now if there are still 4 black pieces left, it'll move. He probably realised it took people too long otherwise)

anyway, just make them green row by row. First make the bottom row green, then make the second row green only by clicking on the row above, not on the second row itself, so that the bottom row doesn't change, then make third row green by clicking on the fourth row and so on.

Lynx
Jan 1st, 2005, 07:49 PM
Hey moby - yes, I'm inclined to think there is only one solution too... but seeing yours I have to conclude there are at least some minor variations...


scenario A:

Sorry... but your scenario A won't cut it :p. It will give you the bogus ball in two more weighings, true - but there is a chance you won't know if it's heavier or lighter.
(I know: in my first post I forgot to mention you have to find that out too - but I stipulated it in the following post.)

Say in your second weighing you weigh 2 normal balls against two question marks (like you said). If they weigh the same you are left with the 2 last question marks, from which you know nothing at all. Weighing one of them against a normal ball will give you the full answer only if that one IS the bogus ball - but if it isn't you don't have any weighings left to define if the bogus ball is heavier or lighter.

I'm sure you can figure it out for yourself - but here is what I do:
I weigh 3 question marks against 3 normal balls.
> same weight? Then it's the 4th ball - and we can easily determine if it's lighter or heavier.
> different weights? We end up with H1 H2 H3 - or L1 L2 L3, of course. Weighting number 1 against number 2 gives us the solution.

scenario B:

H1, H2, L1 against H3, H4, L2: that's exactly what I do too (although I did concoct a slight variation since I read gentenaire's post, putting L3 L4 and H4 aside and weighing H1, H2, L1 against H3, L2 + 1 normal. It works, but does it count as a "different solution"?)

But as for your point 6)... there is no reason to weigh H1 and L2 against 2 normal balls. Weighing H1 against H2 also leads to a definite conclusion. :)

Lynx
Jan 1st, 2005, 07:58 PM
(...) now if there are still 4 black pieces left, it'll move.
Ah, is that what happened.
I was working from left to right, and I was pretty sure I wasn't done yet when suddenly the wall slid back, lol. I didn't understand why :lol:.

moby
Jan 2nd, 2005, 02:26 PM
yep, sorry my fault :o
in fact, i think that's what i did in my original solution (both weighing 3 balls and point 6) but i couldn't recall it precisely and didn't check my answer cause i remembered that scenario A is the more boring part of the solution :o

i got the question almost a year ago, actually, from a friend, during that very same Maths lecture ;)

moby
Jan 2nd, 2005, 02:29 PM
anyway, just make them green row by row. First make the bottom row green, then make the second row green only by clicking on the row above, not on the second row itself, so that the bottom row doesn't change, then make third row green by clicking on the fourth row and so on.
that's what i did... it isn't very elegant though, you get some gaps in the last row, but thankfully they let it slide. there actually is a solution on the net that will turn the whole wall green in 50 moves or so :worship: