Re: Bullet Weight - Logic Problem for everyone...

29 June 2004, 13:47

Hot Core

Quote:

... I hate statistics so I would use all 12 bullets, and 11 individual weighings (MAX). ...

Hey AC, I read with great interest your thread with lots of fine "statistics" in it.

It seems that I accidentally mis-lead you though. And I imagine this is why it is TOTALLY IMPOSSIBLE for denton to be able to work this problem. We really aren't "weighing" the Bullets. It is a "Comparison Scale" - a pan on each side - like the old scales in the Assayor's Office.

I use "all 12" too, but not quite that many weig...uuuh.. comparisons.

29 June 2004, 13:57

Hot Core

Quote:

If I dropped all 12 bullets onto a Remington 700 safety, would I get an accidental discharge?

Hey 500grains, No, but by dropping those 12 bullets "on the M700 Safety" you will probably create a "Feed Failure" on a "Holy Grail - Controlled Feed - M70" if there is one within a mile of you.

(Reference: Gunsmith Board)

29 June 2004, 14:03

ASS_CLOWN

HotCore,

I am just being an old party pooper. I am thinking about what I would REALLY do, I would simply place one bullet in the center of one pan (eye balling it) and then start swapping out a single bullet at a time in the center of the other pan (again eye balling it) until I found the light one. See I seriously believe I could find the light one quicker this way.

I really do hate statistics. I know how to perform the math, but to me it is a tool, like ratchet, nothing more. I don't love ratchets either (at least most of the time)

Oh yeah, that doesn't mean anyone can take my Snap ons and throw them around the shop!

ASS_CLOWN

29 June 2004, 18:16

Wstrnhuntr

The absoloute minimum is two weighings. Its not too likley, but it is possible.

Heres my hint; 5/5 1/1

30 June 2004, 02:59

squeezer

I came up with two comparisons. Here is my logic.

The problem asks for the minimum number so you need to assume you will get lucky and draw the off-weight bullet the first or second time, but the trick is to insure that you have the off-weight bullet and that you are able to determine if it is heavier of lighter than the norm.

You select two bullets (A and B) and they are equal weight. Now you know what the standard weight (norm)is. You save one of the bullets (A) to use in the next comparison. The next bullet is C which is lighter or heavier than A. Two comparisons and you are done.

Now if the first two bullets are different, A is heavier than B or B is heavier than A, then you need to compare again. Select either A or B and compare with C. Lets say that A is heavier than B, and C and A are the same, then B is the off-weight bullet and it is lighter. If A is heavier than B and C is lighter than A, then C is the off weight bullet and it is lighter. The same logic is used if A is lighter than B. Again you have two comarisons and you are done.

30 June 2004, 13:37

Hot Core

Quote:

There can be no "Luck" in a problem that must be solved with "Absolute Certainty". The minimum number of weighings Must work each and every time with no acceptions to be correct...

Hey Steve, I'll give you one more night to wrestle with it. I know of "one" way to solve it with "Absolute Certainty" in ( ? ) Comparisons.

Actually, there seems to always be more than one way to do things of this sort. I also look forward to seeing how the other folks did it too.

...

Tell you what I'll do, I'll go ahead and post Part 1 and save Part 2 for in the morning. Here is how I get started.

I first break the 12 Bullets into 3 groups:

[1] [2] [3] [4]...{5} {6} {7} {8}...<9> <10> <11> <12>

No comparison was done above.

I initially set <9> <10> <11> <12> off to the side.

For the First Comparison I set it up as follows:

[1] [2] [3] [4] ^{5} {6} {7} {8}

In this instance the Scale is Balanced so therefore the "different" Bullet is in <9> <10> <11> <12>

For the Second Comparison I set it up as follows:

[1] [2] [3] [4] ^{5} <9> <10> <11> and I leave <12> off to the side.

If the Scale remains Balanced, then I know <12> is the "different one.

For the Third Comparison I set it up as follows:

[1] ^ <12>

If the Right side goes UP then <12> is Light, or if the Right side goes DOWN the <12> is Heavy.

...

If however during the Second Comparison the Right side of the Scale went UP then I know either <9> <10> <11> is Light. Or if the Right side went DOWN then either <9> <10> <11> is Heavy.

Lets say it went DOWN meaning either <9> <10> <11> is Heavy.

For the Third Comparison I set it up as follows:

<9> ^ <10>

If it is Balanced, <11> is the Heavy one. If it is not Balanced, which ever side goes down is the Heavy one.

And of course, if the Right side had gone up during the Second Comparison that means either <9> <10> <11> is Light. And you follow the same "Logic" as in the Third Comparison to determine which one it is.

...

Now we go back and look at the First Comparison which was set up as follows:

[1] [2] [3] [4] ^{5} {6} {7} {8}

Tomorrow I'll show you how to solve it when the Scale is "NOT" Balanced. That will mean the "different" Bullet is in:

[1] [2] [3] [4] ^{5} {6} {7} {8}

Therefore, we know for sure that <9> <10> <11> <12> are all alike.

Best of luck tonight!

30 June 2004, 17:24

dhs

'course this might help some too... http://www.jdawiseman.com/papers/easymath/weighing.html

(this has gone on too long... )

30 June 2004, 17:24

Wstrnhuntr

The obvious way to solve this problem is going to be a process of elimination. 6.5 Bandit is right. It can be done in two, for certian, no less.

A few things to consider. The question to the problem is "what is the MINIMUM number of weighings required"?

Also, we are already given the FACT that only one of the bullets are odd, so we need not wrestle with that on the scale.

Heres an example. With luck!! 10 bullets are placed on the scale, 5/5 and it ballances. We have just eliminated 10 possibilities of 12. Remove all but one bullet from the scale, place one of the remaining two on the opposing pan. If it happens to be the one then it will show, and you will also know if it is light, or heavy. That is the MINIMUM number of weighings and it can be accomplished several ways. 4/4 1/1, 3/3 1/1, 2/2 1/1. All other answers can be modified within the process according to..... the luck of the draw!

Cool brain wringer Hot core..

30 June 2004, 19:53

Jameister

You guys all blew it. you dont need to weigh any of them on the scale. load all bullets into equal brass and powders. shoot all through a chrony. the one that goes the fastest is the lightest, or the slowest is the heaviest. And I did not touch your darn scale once.

30 June 2004, 05:58

Savage99

I do regret wasting time on this irrelevant topic. What I remember from math probability is that there are 144 combinations.

I promise not to read this topic again so have at it.

01 July 2004, 04:14

Hot Core

OK folks, here is Part 2. There is absolutely no Luck or Guessing involved with this method. By the way, I found "coloring" the Bullets helped me see the "trick".

...

Now we go back and look at the First Comparison which was set up as follows backin Part 1:

[1] [2] [3] [4] ^{5} {6} {7} {8}

But, this time the Scale is "NOT" Balanced. That will mean the "different" Bullet is in:

[1] [2] [3] [4] ^{5} {6} {7} {8}

And, we know for sure that <9> <10> <11> <12> are all alike.

For the Second Comparison set it up as follows:

[1] [2] {5} <9> ^ [3] [4] <10> <11>

And I have {6} {7} (8} and <12> off to the side.

If the Scale is Balanced, I know the different Bullet is either {6} {7} (8} and determine which one is Heavier or Lighter just as we did back up in Part 1.

If the Right side of the Scale goes down, I know that either [1] [2] must be light and the Third Comparison:

[1] ^ [2] would tell me which one is Light.

If however the Right side of the Scale went Up after the Second Comparison, then either {5} is Heavy or [3] or [4] is Light.

For the Third Comparison we do:

[3] ^ [4]

If it is Balanced, then {5} is Heavy.

If either Side goes Up, that is the Light Bullet.

And you are done!

...

The "trick" is grouping things into "3s". No luck and no guessing is involved. This way works irregardless of the position you choose to put the "different" Bullet in the original sequence.

Pick any one as being Heavier or Lighter. Then go back and follow the Logic Steps as outlined above in Part 1 & 2 and you can prove it to yourself.

If there are other ways to do it where no luck or guessing is involved, please include your method and my hat will be off to you.

Glad to see some of you folks enjoyed it!

28 June 2004, 13:13

robe0280

Put the house on 'FOUR' no less no more
Should know which one and if higher or lower everytime with no 'luck'

01 July 2004, 19:33

Jameister

That brick would be two pounds, and a brick and a half would be three pounds, but you would need to load them and shoot them through a chrony to see if the sun was out.

01 July 2004, 03:40

Hot Core

Quote:

You guys all blew it. you dont need to weigh any of them on the scale. load all bullets into equal brass and powders. shoot all through a chrony. the one that goes the fastest is the lightest, or the slowest is the heaviest. And I did not touch your darn scale once.

Hey Jameister, That is indeed an "interesting" idea.

01 July 2004, 06:20

DKing

Hot Core

I'm a little confused.

Please tell me how this scenario would work in either [2] was heavy or [5] was light.

/r