Is this possible?

[vicdan]

As I said, if you figure out how I know that having a branchless 4-ball weighing guarantees branchless weighing of any number of balls*, you will know how to compose such a weighing.

I'm still missing your point. Your branchless 4-ball weighing includes ballast balls, so the guarantee for any number of balls presumably likewise requires ballast balls.

No. In any larger solution, previous steps provide ballast balls.

But we are considering complete branchless solutions (in particular, for 12 balls) where there are no ballast balls to start with. How can your guarantee then apply?

it can. Trust me. :)

[Laird]

vicdan: As I said, if you figure out how I know that having a branchless 4-ball weighing guarantees branchless weighing of any number of balls*, you will know how to compose such a weighing.

Laird: I'm still missing your point. Your branchless 4-ball weighing includes ballast balls, so the guarantee for any number of balls presumably likewise requires ballast balls.

vicdan: No. In any larger solution, previous steps provide ballast balls.

I'm talking about backing up to the point where there are no previous steps. We seem to be talking at cross-purposes. I'm trying to find a branchless solution for the 12-ball problem, where we start at the first weighing - i.e. no ballast balls. Then you come along and start talking about beginning halfway through where there are ballast balls because there have been previous weighings... no, Vic, no: no previous weighings dude!

[vicdan]

<shrug> if you don't want to solve the problem, that's your business; but trust me, I am talking about a real universal solution for branchless weighing.

Just try to figure out how it could be done, instead of trying to argue about how i am doing it wrong. i am using standard proof-theoretic methods to arrive at a solution, and solving it for 4 balls with ballast balls gives you a universal solution for any higher number of balls (at least the number maximal for given # of steps, e.g. 12, 36, 108, etc.) without ballast.

See, I solved a reductive case (which I had shown), and i can prove that any larger case (e.g. 12, 36, 108, etc. balls) can be reduced to my reductive solution as stated. That is, i can prove how you can solve the branchless 12-ball problem without ballast by reducing it to the 4-ball problem with ballast.

if you want to solve this problem, what you should be trying to figure out is how the 12-ball no-ballast problem can be turned into 4-ball with-ballast problem*.

* Actually it's '4-ball with-ballast problem or an even simpler one'. The 4-ball problem is the keystone.

[Laird]

if you want to solve this problem, what you should be trying to figure out is how the 12-ball no-ballast problem can be turned into 4-ball with-ballast problem*.

That's the sentence that I needed to read. I'll get thinking.

[Laird]

Nah, it's just not happening for me Vic. For a start, your supposed solution for four balls:
1) AB <=> C#
2) AC <=> ##
can't say anything about D. Yes, if both weighings are equal we know that D is the culprit, but we don't know whether it was heavier or lighter.

Next, I just can't make any relationship between this and the 9-ball branchless solution that I provided back on page 1, let alone derive a general principle from the two.

I'm going to call you on this: are you willing to provide a 12-ball branchless solution and explain the general principles behind branchless solutions?

[vicdan]

But we don't care if it's heavier or lighter -- we are just looking for the odd ball. however, if you do care, toss D on instead of one of the ballast balls in step #2.

And yes, i can explain the principle behind the branchless solution. :) Do you want me to do it, or would you rather give it another shot?

[Laird]

But we don't care if it's heavier or lighter -- we are just looking for the odd ball.

Hey, what can I say - I'm a caring kinda guy.

however, if you do care, toss D on instead of one of the ballast balls in step #2.

Yeah, that works.

And yes, i can explain the principle behind the branchless solution. :) Do you want me to do it, or would you rather give it another shot?

I'm outta patience and ideas. Go right ahead.

[vicdan]

I'm outta patience and ideas. Go right ahead.

OK, I will explain the principle first, and give you a chance to work out the actual solution.

See, after the first weighing, all the balls are partitioned into two sets -- the balls which are ballast, and the ones which could contain the culprit. You can put ballast balls on the scales without disturbing the measurement.

So what you do it, you compute the branched solution for each branch separately, and merge the branches, letting the balls from the other side play the role of ballast; i.e. you place both the balls from the set ABCDEFGH on the scale (as if scale unbalanced on the first weighing) and balls from the set IJKL on the scale (as if the scale balanced) -- and at the end, if the scale had indeed balanced, you can simply ignore the first 8 balls except inasmuch as they provide ballast.

This is why i said that it can all be reduced to a 4-ball case -- because you can pursue multiple branches simultaneously, so solving the 12-ball problem is really like solving simultaneously the 4-ball problem and the 4-pseudo-ball problem (because if the scales unbalance, you will end up having 4 balls possibly light and 4 balls possibly heavy, which is actually simpler than 4 balls either light or heavy). And solving 36-ball problem is like solving simultaneously the 12-ball problem and 12-pseudo-ball problem.

You just have to make sure that the one ball you leave off to solve the 4-ball problem is mirrored by the one ball you leave off to solve the 4 pseudo-ball problems. In case you are wondering, yes, I have in front of me a 2-step branchless solution for 4 pseudoballs, i.e. 8 probabilistic half-balls abcdEFGH, which involves an odd number of ballast balls in each step.

At the end of the three weighings, you will end with a sequence of three trits forming a number between 1 and 27, which will uniquely identify the culprit (though there might be a couple of numbers identifying the same culprit, because we are dealing with real balls and not IT pseudo-balls); e.g. something like the result '\-/' (left side went down, it balanced, left side went up -- that's how you meant the notation, right?) will point to the solution d (ball D is light). Each of the 27 possible sequence of weighing outcomes will, collectively, identify one culprit ball and its weight.

[Laird]

Victor, being that I have now transcended the mundane, I now know how petty is this piddling around with mathematical symbols. It is abominable to contemplate how many hours are wasted in similar endeavours by minds that could instead be Immersing themselves in The Ultimate. There is, however, hope that - despite your intransigent contempt for the Logic of Enlightenment - you will at last turn to it when you are so caused. I am confident that such causes will only arise when you tire of the barren sterility of academic logic and, in the empty void that is left in your life when this pacifier is yours no longer, suffer like you have never suffered before. For truly suffering is the launching pad from which the rocketship of Enlightenment departs. To this end, I will continue to engage with you in this thread, pushing you further and further, closer and closer to the point of exasperation where finally you throw up your hands and say: "Enough! I give up these God-forsaken practices!"

OK, I will explain the principle first, and give you a chance to work out the actual solution.

And a cunning principle it turns out to be, upon careful contemplation of your description.

In case you are wondering, yes, I have in front of me a 2-step branchless solution for 4 pseudoballs, i.e. 8 probabilistic half-balls abcdEFGH, which involves an odd number of ballast balls in each step.

And I cannot see how that is possible. I have a solution, however it contains an even number of ballast balls in the first weighing:
abEF vs dG##
acEG vs F###

Now we need to combine this with the 4-ball solution:
HI vs J#
IJ vs K#
[oops, this should have used IJKL rather than HIJK]

However, I cannot see how to combine these solutions. I come up with this:
ABEFJ vs DGHI?
ACEGK vs FIJ??

e.g. something like the result '\-/' (left side went down, it balanced, left side went up -- that's how you meant the notation, right?)

Actually I meant '\' as "left side went up" and '/' as "left side went down".

[vicdan]

Victor, being that I have now transcended the mundane, I now know how petty is this piddling around with mathematical symbols. It is abominable to contemplate how many hours are wasted in similar endeavours by minds that could instead be Immersing themselves in The Ultimate. There is, however, hope that - despite your intransigent contempt for the Logic of Enlightenment - you will at last turn to it when you are so caused. I am confident that such causes will only arise when you tire of the barren sterility of academic logic and, in the empty void that is left in your life when this pacifier is yours no longer, suffer like you have never suffered before. For truly suffering is the launching pad from which the rocketship of Enlightenment departs. To this end, I will continue to engage with you in this thread, pushing you further and further, closer and closer to the point of exasperation where finally you throw up your hands and say: "Enough! I give up these God-forsaken practices!"

And I cannot see how that is possible. I have a solution, however it contains an even number of ballast balls in the first weighing:
abEF vs dG##
acEG vs F###

Now we need to combine this with the 4-ball solution:
HI vs J#
IJ vs K#

However, I cannot see how to combine these solutions.

You can't. An insuperable condition is that the number of ballast balls in each weighing of each branch (i.e. the 4-ball branch and the 8-half-ball branch) must have the same parity; i.e. either both must be odd or both must be even. Here is a sample 8-half-ball solution for you:

abcdEFGH

1) abG# <=> E##d
2) aFG# <=> bH##

Reasoning as follows: if at first weighing the left side went up, the culprit is among abE, so we weigh a<=>b. if the left side went down, the culprit is among dG, so we weigh G. if the scale balanced, the culprit is among cFH, so we weigh FH against each other. See the partitioning pattern? note how the 2nd step incorporates all three tests into it -- a<=>b, F<=>H, and G.

Note that the number of ballast balls is odd in each weighing, and that the nine possible outcomes of the two-step weighing will definitively identify one of the 8 balls as the culprit. it should now be obvious how you can merge this with the 4-ball 2-step solution to arrive at a branchless 12-ball discernment algorithm -- and how to extend it to a larger number of balls.

[Laird]

it should now be obvious how you can merge this with the 4-ball 2-step solution to arrive at a branchless 12-ball discernment algorithm

Indeed, and for completeness's sake I include the solution:
ABCD vs EFGH
ABGK vs EDIJ
AFGL vs BHJK

...and its validation (that none of the trinary numbers below is repeated validates the solution):
A heavy ///
A light \\\
B heavy //\
B light \\/
C heavy /--
C light \--
D heavy /\-
D light \/-
E heavy \\-
E light //-
F heavy \-/
F light /-\
G heavy \//
G light /\\
H heavy \-\
H light /-/
I heavy -\-
I light -/-
J heavy -\\
J light -//
K heavy -/\
K light -\/
L heavy --/
L light --\

[Shahrazad]

Laird, I can't believe you're wasting your time with this, now that you've been enlightened for a whole day.

[Philosophaster]

He explained why he was doing it.

Thou shalt not question the reasoning of the Enlightened!

[The Dude]

Heh.

[Shahrazad]

Thou shalt not question the reasoning of the Enlightened!

Oh, I question them all the time. They can handle it.

[The Dude]

Laird, it's just a stupid riddle. There's nobody in this forum that cares about the solution. Why the fuck are you wasting your time with it?

[vicdan]

TheDude,

It's not the solution that matters, it's understanding the problem. Once you understand what is really going on here, the actual solution is child's play, mere pencil pushing.

To be honest, once i figured out how this problem worked, I didn't even bother coming up with a solution until Laird needed a specific example. :)

[The Dude]

Thinking about you, just now, I solved the riddle from Hitchhiker's Guide to the Galaxy.

Marvin, a depressed android "...with a brain the size of a planet..." is once asked about life.
"Life?" he sighs, his logic unable to grasp this simple notion, "don't speak to me of life."

The solution? GET A LIFE!

[Laird]

Laird, I can't believe you're wasting your time with this, now that you've been enlightened for a whole day.

Earlier and in a separate thread I commented on the fine potential that you exhibit for The Attainment of Wisdom. Now it is by logical necessity that I inform you of that which is detaining you from your objective: namely, Deluded Criticism. This term I will explain for your benefit, for you are yet to realise your innate Potential for Genius and thus are not yet capable of Logically Deriving Genius Definitions. "Deluded criticism" refers to the tendency of one whose Mind is not yet Attuned with Ultimate Reality to misapprehend the ways of Genius, to the extent that he (you will note that I recognise your transcendence of biological femininity and that therefore you are deserving of the masculine pronoun) feels obliged to "point out errors" - the errors of which are naturally in his own Deluded Mind. Indeed, your failure in this regard is all the more pertinent given that, as Philosophaster notes, I had already logically proven to you the necessity of my actions.

[vicdan]

Thinking about you, just now, I solved the riddle from Hitchhiker's Guide to the Galaxy.

Marvin, a depressed android "...with a brain the size of a planet..." is once asked about life.
"Life?" he sighs, his logic unable to grasp this simple notion, "don't speak to me of life."

The solution? GET A LIFE!

Oy, Nat, are you running low on Cheetos again? Getting jumpy from withdrawal?

[Dan Rowden]

It's not Nat. One look at the profile should show that.

[vicdan]

I don't see how, but if you compared IP ranges, i will take your word for it.

[The Dude]

Vic: Naw, I'm just an old fan. And I loved your argument with Nat. Thought I'd get a word in, since you two were bickering like an old married couple.... :)

[Trevor Salyzyn]

Victor: you moderated The Ponderers' Guild. Taught me everything I know about philosophy. I was "mookestink". I didn't get ternary logic then, and I'm sure as hell not getting it now. My brain isn't built for that many variables... :)