Winter Shadow: Newsletter Challenge (01/06/09)

Bog me, I can't see the 3x3 one !! I'll have to think.

Meantime, have some revenge ;

If it's too small to read, let me know and I'll post the typed up one. The problem googles easy enough, though it could be disguised (ignore the trivial water question and re-write all the categories). This version I like best, because it has the non-pc smoking fields (1980 something).

Doesn't matter that you can't see the 3x3 dots, it is just that a three by three matrix of nine dots.

And, all except for the Englishman living in the horse with the red door, I can read all of the parchment. No time now though!

It's the green door you need to worry about !

I've had the Norwegian in the left house as I look at them and the left house from his perspective, and, come up with contradictions in both cases so I guess I need to go examine my assumptions.

hee hee heee ! Took me ages when I first ever saw it. When I re-found it in an old file it took an hour or two. No clues yet, but it does work mainly on eliminating contradictions. Only at one point do you have to think ahead a fair way (as you would in chess). There is only one solution. At the risk of stating the obvious, draw up a grid with rows for door colour, nationality, etc etc.

ps - you'd best check this against the picture, here's my typed copy to print out.

Who owns the Zebra ?

This Brain-Teaser can be solved by combining deduction, analysis and

sheer persistence. The Essential facts are as follows;

1. There are five houses, each with a front door of a different colour, and inhabited by men of different nationalities, with different pets and drinks. Each man smokes a different kind of pipe tobacco.
2. The Englishman lives in the house with the red door.
3. The Spaniard owns the dog.
4. Coffee is drunk in the house with the green door.
5. The Ukrainian drinks tea.
6. The house with the green door is immediately to the right (your right) of the house with the ivory door.
7. The medium cut smoker owns snails.
8. Spun cut is smoked in the house with the yellow door.
9. Milk is drunk in the middle house.
10. The Norwegian lives in the first house on the left.
11. The man who smokes mixture lives in the house next to the man with the fox.
12. Spun cut is smoked in the house next to the house where the horse is kept.
13. The flake smoker drinks orange juice.
14. The Japanese smokes Rough cut.
15. The Norwegian lives next to the house with the blue door.

Now, who drinks water and who owns the zebra?

Who owns the Zebra ?

The Norwegian!

The Norwegian lives in the first house on the left.
The Norwegian lives next to the house with the blue door.

The Norwegian's door cannot be red, for that is the color of the Englishman's door (#2). It cannot be green or ivory, for the green door is to the left of the ivory door (#6).

The doors in order from left to right are Yellow - Blue - Red - Green - Ivory.

The Norwegian smokes spun cut (#8). He does not drink orange juice (#13), tea (#5), milk (#2 and #9), or coffee (#4), so he must drink water.

The Norwegian does not own the dog (#3), the snails (#7), the fox (#11), or the horse (#12), so he must own the zebra.

I'll figure out the rest of it later.

Fact 6........

I had seen this some years back and didn't solve it (or didn't really try). Surprisingly, this time around it seemed to work by simply following the clues in the order given (after filling in the items which could be decided more-or-less by inspection), and selecting the first of multiple options in each instance.

If the houses are numbered #1 to #5 from L to R, the following sequence gives the answer without contradictions on the first run itself (clue numbers in brackets):

1. (9) milk #3 -- OK

2. (10) Norway #1 -- OK

1. (15) blue #2 -- OK

4A. (2) England + red #3 -- [alt: #5]

4B. (6) ivory #4 + green #5 -- [alt: #3 + #4]

5. (2+6) yellow #1 -- OK

6. (8) spun #1 -- OK

7. (12) horse #2 -- OK [so far by insp. with first option in step 4A+B]

8. (3) Spain + dog #4 -- [alt: #5]

9. (5) Ukraine + tea #2 -- [alt: subject to (3)]

10. (7) snails + medium #3 -- [alt: #5, or others?]

11. (11) fox #1 + mixture #2 -- [alt: #5 + #4, or others?]

12. (13) juice + flake #4 -- [alt: others?]

13. (14) Japan + rough #4 -- [alt: or others?]

The above results from L to R can be represented as:

Nation: Norway, Ukraine, England, Spain, Japan

Door colour: yellow, blue, red, ivory, green

Pet: fox, horse, snail, dog, ?zebra?

Drink: ?water?, tea, milk, juice, coffee

Tobacco: spun, mixture, medium, flake, rough

There didn't seem to be any contradictions so I didn't even start trying the alternatives. =TeeSquare=

Your brain must be aging nicely !

As I recall, there was one point where I got to an 'if m, then p, the q, then contradiction' type situation. The numbers in the boxes below are my solving sequence, which shows there are various routes. I may have noted some in the wrong order, but it shows the general drift.

You can find more on the problem if you google "The Zebra problem".

Kudos to you, TeeSquare !

C'mon, we all know it was the Frenchman that kept the snails

Thanks Kris,

But I must admit it took me a couple of hours to work out a strategy for recording the trials sequence compactly. Can't retain more than two or three steps in my head, and writing things out in words is painful and also confusing. (I could solve Rubik's cube only after I spent several days literally inventing a code language for representing moves and colour patterns, but that was long, long ago).

The difficult clues seem to be listed in the 'right' sequence. Had they been rearranged I would probably have struggled for hours or just given up!

Couple of minor typos in my post : Step 3 was listed as 1, and in step 13 it should have been #5 and not #4, both obvious I guess. =TeeSquare=

Checked google as suggested, and was surprised to find this is a classic 'celebrated' problem of sorts. I had approached it in a dull, plodding manner and it just fell into place. If the number of options is limited it amounts to manual number crunching, and is perhaps in the same class as the boxes-and-coins or twelve coins type of problem. I would have thought there is an elegant solution to this one. The computer people make even simple solutions look complicated and uninteresting (e.g. the pdf file from google!).

For what it's worth, my steps after filling in the first few easy items are tabulated below:

The straightforward strategy is to run through the clues in sequence until one reaches a contradiction, in which case one should score out the final row(s) and repeat with a different alternative (here in step 5 or 4 or 2 or back to 1).

Finally all the columns should be full, except for one pet and one drink. For the present problem, just following the clues in the given sequence and taking the first option from the left yielded the solution without any back-tracking.

Such a procedure seems fairly logical, but formulating a puzzle like this must take some real ingenuity! =TeeSquare=

Yeah, I was staggered at how complex some of the solutions are. Like yourself, all I did was start with a blank table then begin filling in. The complex solutions, I'm guessing, are the folk who like to look at it from a computer point of view. Somewhere I came across a program that allows you to vary the number of fields. Wish I'd kept note of where ! The ones that attribute it to Einstein are a bit far -fetched. Nobody seems to know where it originated, which is curious since it doesn't seem to resemble any of the classical problems. Still, it's fun, and great to stick on a mate one evening - it's irresistible not to try and wastes a good couple of hours for most folk.

Just as an aside, my preference is for cryptic crosswords. They defy all logic that a computer is capable of, and solving them is often a matter of assorted bits of knowledge coming together spontaneously. It might take me hours with some clues, then all of a sudden the 'eureka' moment hits with no apparent explanation. Sudoku is fun for a bit of brain training, but a bit boring in that it boils down to timing personal best. They're also very amenable to computer solving <yawn>.

I wonder if they'll ever post the Rubber bullet solution. I asked for the solution to this one by PM to Chris (+ more than a little play ), so somebody else can have a shot now. I'm getting the impression that it's only us who follow these things to the end !

It's astonishing how well they can climb, and they can run almost faster then I can!

Yeah, but, have you ever seen one try to jump about six feet from one branch to another about 15 feet off the ground!?

Maybe this is one-----------------V

LOl - I shall file that under "Erotica"

Since this thread is getting populated by non-official challenges, entirely off-topic, here's another for those who seek distraction while contemplating the colour of their winter shadows:

What is the shortest (great circle) distance between any two places A and B on the earth (assumed to be a perfect sphere of radius R), given only their latitude and longitude (Θ _A, Φ _A) and (Θ _B, Φ _B), expressed as a single formula, valid for locations in any hemisphere.

That means Θ ranges from -90 to +90 deg below/above the equator, and Φ from -180 to +180 deg about Greenwich.

The final result should be a fairly compact trigonometric expression. The proof need not be posted, as it may be cumbersome.

The answer will be posted right here on CR4 by next doomsday. (Conditions apply)** =TeeSquare=

**

1. the pc survives, and emerges long enough from its normally hung status

2. the teesquare survives, is in town, and feels so inclined

3. connection to internet is successful

4. CR4 web site is accessible and acknowledges my existence

5. the posting, formatting, reviewing sequence is successful

6. the usual force majeure items (act of God, war, pestilence, plague famine, gill bates, etc.)

7. etc.

I'd express the locations in terms of direction cosines with repect to x,y,z axis, such that ;

A = l,m,n

B= l',m',n'

The angle between them, ψ, is found from..

cos ψ = l.l' + m.m' + n.n'

Then you can use the radius of the Earth, R, to get distance d = 2ΠR x ψ/360

Gets a bit mangled to rearange the start terms, and duty calls elsewhere.....