Brain Teaser No 1

OMER

I think I've made some progress on this one.

The chap at the back counts how many black, and how many white hats there are. As he can see 7 hats, there will be an odd and an even number.

He shouts out the colour that relates to odd eg black.

The second in line can see 6 hats (either two evens, or two odds). Whichever way, using the combination of what he can see, and what the first guy said, he can deduce his colour.

eg, if he can see 3 black, 3 white, then he knows that his must be white, as black is still odd. If he sees 4 white, 2 black, then he knows his is black, as black is now even.

The remaining people in line all have similar information. Therefore 7 are guaranteed to be correct (as long as they keep concentrating!!!). The first guy has a 50:50 chance of being right.

So, it is a kind of binary solution (which someone else suggested earlier) using odd/even to replace black/white.

Very clever Omer!

STEVE D

OMER,

I don't understand what you mean by this. I am
struggling to understand how it can work for more than 2 colours.

STEVE D

I've e-mailed you my solution to my 'odd ball' teaser, just so you can check it and compare it to the solutions you've seen on other web sites.

It's never been checked and it would be awful to publish it here only to find it had a glitch!!

Cheers

Don

Appologies Martin (and others). I have'nt figured out how to use my scanner and Adobe Photodelux programs even after 2 years. Just getting an image less than 30kb and getting it to work on the Forum was a success for me !!

I'll try harder next time!! Meanwhile I could e-mail a copy to anyone who asks and it will probably be more legible (well 2/3 probably !!)

Cheers

Don

Sam Loyd.......was a famous American who devised over 10000 puzzles in his lifetime..... "How far to salisbury" is the old Sam Loyd ferryboat problem, which you have cleverly re-naimed.

One down, only 9,999 to go !!!!!!!

PS I like the 're-naimed' pun at the end of your quote!!!!

Cheers

Don

quote:

---

Originally posted by Don Atkinson:
_I peeked, but couldn't read the graphic._

Appologies Martin (and others). I have'nt figured out how to use my scanner and Adobe Photodelux programs even after 2 years. Just getting an image less than 30kb and getting it to work on the Forum was a success for me !!

---

No, that's fine. I'm just an incurable helpdesk type - have to point out to people how they could have done it better. ("No, Martin, say it isn't true". "We'd not noticed - honest". "And there was me thinking...").

This is one of the areas I found incredibly frustrating when I first started to use graphics programs. They have all these functions you can do, then you come to save your art and you have a list of about 50 file-types that it could save it as, and no help about which one to use.

When would you want to use a bitmap, a TIFF, a JPeG, a GIF or a PNG?

BMP (bitmap) is sort of a lowest common denominator - it stores all of the information(**) in the image, but is very inefficient. No one really uses this for anything of any size, any more.

JPEG is for photographs - millions of colours, natural, smooth gradations between adjacent colour blocks. Many times smaller than BMP. Saving as JPEG, then re-opening, modifiying & re-saving will result in further image degradation every time.

GIF is very poor on photographs, colour quality would be very poor, and big file sizes. Use it for line-art (pen strokes, etc) and simple computer graphics, though, and it's better in every way than JPEG, which doesn't like that sort of thing. GIF likes large swathes of solid colour (all the same tone, like your white paper background), which tends to result in only a couple or maybe a few dozen different colour shades on the page. If used for this type of image, result will be smaller than JPEGging. You can re-open & re-save these image types over & ove without loss of quality.

TIFF is a sort of super-BMP. Tends to be used for storing "original" images without any loss of quality. Images may be opened & resaved without loss of quality (e.g. after red-eye removal or cropping). May be half the size of a BMP if using LZW compression (don't use JPEG-type compression). Once you have got your image safely saved away as a TIFF you can then save a copy of it as JPG. Personally I find it very disapointing that digital cameras don't usually offer a 'save as TIFF' option, as it would retain more image quality now that 64MB cards are available for £25.

I guess that PNG is sort of a modern JPG (mostly), although it's not supported by older web-browsers or some graphics programs.

And you didn't even know you wanted to know that!

cheers, Martin

The three black and white hats question was good

The eight black and white hats in in a line was brilliant

The eight coloured hats in a line is outstanding (i've got the gist of it, but still haven't managed to write it down clearly)

The guy who thought these ones up must be mad as a hatter.

Cheers

Don

Cheers

Don

Matthew

I mean the length of the hole, once cut, is 6"?

Don

Don

The numbers are, of course, right. I suspect Mathew is politely teasing us when he says it is a 'Guess'.

Now if only Bam hadn't thrown away that old A Level maths book on integration, or was it volumes of revolution or.........Any helpers out there?

Difficult to imagine the earth with a 6" long hole drilled through it !!

Cheers

Don

The bit you drill out can be broken down into a cylinder with a spherical segment at each end.

Volume of sphere = 4/3.r3.PI
Volume of cylinder = h2.6.PI
Volume of each spherical segment = PI.1/6.l.(3.h2+l2)
l is depth of spherical cap l=(r-3)
h=sqrt(r2-9)
Volume of ring = (vol sphere)-(vol cylinder)-2.(vol cap)
Dropping the PIs (for the moment)
Volume of ring = 4/3.r3-6.h2-1/3.(r-3)(3.h2+(r-3)2)
re arranging and substituting for h gives volume of ring = 36.(PI) for any r>3.

Yep, It really is THAT boring at work this afternoon.

Simon

[This message was edited by Lo Fi Si on TUESDAY 04 December 2001 at 14:50.]

Neat solution. Your maths book clearly has a more extensive section on volumes of 3-D objects than mine.....or its more topical given the recent newsiness about THE dome.

I will award you 7/10, whilst giving Bam 5/10 for his actual volume. For Mathew T, who I think is holding back (and Bam and others) I am holding back a possible 10/10 for a more fundamental or more elegant solution.

Now, i've never been sure about the history of classical maths. So I have no idea whether Eucalyptus (or was it the builder of large constuction trucks-Euclid?) knew that the volume of a sphere was (4/3)Pi*R^3 let alone that the volume of the top of your egg when chopped off for breakfast was (Pi/6)*l(3h^2+l^2)where l=depth of dome and h=radius of base of dome, or top of egg, as the case might be. I am aware that Sir Issac probably managed the necessary manipulation of a sphere's volume and probably a bead as well, with his new-fangled calculus. Whether he got there before Euclid or others, I just don't know. BUT, to my humble mind, a solution based on the old calculus of Sir I looks a lot more elegant than Simon's proposal.

So, for that ULTIMATE accolade of 10/10, can any body do better than Simon?

Cheers

Don

Well, I never imagined how thin the 6" 'ring' of earth would be. I recon 1.2*10^-8 (to 2 s.f). On average, of course.

Probably more like 2*10^-8 max. Any one care to help out here?

Cheers

Don

Simon

P.S. I do mean weight not mass

[This message was edited by Lo Fi Si on WEDNESDAY 05 December 2001 at 07:36.]

I don't get the 10/10. I knew that the volume of the bead/ring was always the same. But I can't remember the elegant proof. Will continue to think about it and be completely astounded if I figure it out.

Matthew

Well, at least I hope the solution is attached and (for the benefit of Martin P) readable.

Cheers

Don

PS Looks like the file size was too big (32kb). I will try to compress it somehow?

Don

PS ok, not the best of pictures, having scanned and reduced the original to get below 32kb. If any one would like a better copy, please e-mail me

[This message was edited by Don Atkinson on FRIDAY 07 December 2001 at 20:25.]

You pull out an integer I.
The number of integers >I is F-I.
The number of integers <I is F+I.

After some strong narcotics you may therefore convince yourself that F+I>F-I and so you have a greater chance if you guess the second number is less than the first when I is >0 and the opposite if I<0. If I=0 then you are hosed.

BAM

Number the balls 1 to 12 (simply for reference). Divide them into 3 Groups of 4

1st Weighing.
Compare Group 1,2,3,4 with Group 5,6,7,8

If Group 1,2,3,4 is heavier than Group 5,6,7,8 then IF the odd ball is in Group 1,2,3,4 it is heavy and IF the odd ball is in Group 5,6,7,8 it is light. Go to 2A
If Groups 1,2,3,4 and 5,6,7,8 do balance, the odd ball is in Group 9,10,11,12. Go to 2P

2nd Weighing
2A Compare Group 1,2,7,8 with Group 5,9,10,11
If Group 1,2,7,8 is heavy then either 1 or 2 is heavy or 5 is light. Go to 3A
If Group 1,2,7,8 is light then either 7 or 8 is light. Go to 3B
If Group 1,2,7,8 and Group 5,9,10,11 balance then 3,4 or 6 is the odd ball. Go to 3C

2P Compare Group 1,2,3 (in fact any 3 balls from 1 to 8) with Group 9,10,11.
If Group 9,10,11 is heavy then the odd ball is 9,10 or 11 and is heavy. Go to 3P
If Group 9,10,11 is light then the odd ball is 9,10 or 11 and is light. Go to 3Q
If Group 1,2,3 and Group 9,10,11 balance then the odd ball is 12. Go to 3R

3rd Weighing
3A Compare 1 and 2 and recall 5
If 1 is heavy then 1 = odd ball
If 2 is heavy then 2 = odd ball
If 1 and 2 balance then 5 is the odd ball and is light

3B Compare 7 and 8
If 7 is light then 7 = odd ball
If 8 is light then 8 = odd ball

3C Compare 2 and 4 and recall 6
If 3 is heavy then 3 = odd ball
If 4 is heavy then 4 = odd ball
If 3 and 4 balance then 6 is the odd ball and is light

3P Compare 9 and 10 and recall 11
If 9 is heavy then 9 = odd ball
If 10 is heavy then 10 = odd ball
If 9 and 10 balance then 11 is the odd ball and is heavy

3Q Compare 9 and 10 and recall 11
If 9 is light then 9 = odd ball
If 10 is light then 10 = odd ball
If 9 and 10 balance then 11 is the odd ball and is light

3R Compare 12 with any ball
If 12 is heavy it is the odd ball
If 12 is light it is the odd ball

Hope this makes sense

Cheers

Don

I hope the attachment is readable. However, the guts of it are:-

For a thin horizontal slice R0^2-Ri^2 = l^2-h^2 where h=height of slice above x-x axis and l=length of hole and where Ro and Ri are outer and inner radii.

Area of thin horizontal slice A = pi(Ro^2-Ri^2)

Volume of bead = 2*pi*[integral of](l^2-h^2) (from h=0 to h=l) ie INDEPENDENT of R!!

Volume = 2pi*(l^3 - L^3/3) = 4/3*(pi*l^3)

An e-mail will elicit a readable copy of the attachment

Now back to probabilities and statistics!!!!!

Cheers

Don

Matthew

If they do have the same mass then the rotational inertia of the cylinder and ring will be the same. Thus I predict the ball will reach the bottom first and the ring and cylinder equal second place.

If the objects do not have the same mass but let's assume they are made of the same material then the ring will reach the bottom first since it has the smallest rotational inertia (least mass too). The second place would depend upon the dimensions of the cylinder - it's length and thickness, ie its rotational inertia compared with the ball.