# A Real Bear of a Problem

Michael Morris
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Posts: 3451
This one has nothing to do with programming aside from reasoning. It's one of my favorites of all time. Here's the problem:
A hunter rises early one morning and walks one mile due south from his camp and spots a bear. He shoots the bear, wounding it and the bear goes due east one mile where the hunter finally catches up with him and finishes him off. The hunter drags his quarry exactly one mile due north back to his camp. What color was the bear?

Michael Ernest
High Plains Drifter
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Does the blood count?

John Lee
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Posts: 2545
depends on which color is the light.

Michael Morris
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Posts: 3451
Does the blood count?
He shot him with a cauterizing laser. No blood, no mark.
depends on which color is the light.
Well that depends on the season.

Jessica Sant
Sheriff
Posts: 4313
White -- its a polar bear
You go South 1 mile, East 1 mile, and due North 1 mile (even if you'd gone West 1 mile instead of east) it will put you back at your camp - cause your camp is right smack dab on the North Pole.

Jim Yingst
Wanderer
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Posts: 18671
In a related question: if the hunter goes f-=39 and t+=2899, where does he end up?
[ June 02, 2003: Message edited by: Jim Yingst ]

Jason Menard
Sheriff
Posts: 6450
I haven't read the other answers so if this has been said already, forgive me. It was a white bear, a polar bear. The hunter was at the North Pole. His walk one mile due east, after going one mile south, was basically a circle that brought him to his starting point. He then only needed to walk the one mile north to return to camp.
[Looks like Jessica got it before me.]
[ June 02, 2003: Message edited by: Jason Menard ]

Bear Bibeault
Author and ninkuma
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Posts: 65216
95
A Real Bear of a Problem

That's what they call UI issues here at work that I'm expected to solve!
Devo 1: Hey, I'm having a wierd problem with this servlet doing something-really-bizarre-or-other.
Devo 2: Sounds like that's a real Bear of a problem.
Bear: Sigh! (writes it down as an action item)
bear

Michael Morris
Ranch Hand
Posts: 3451
Yep Jess and Jason, good old circular trig. From the North Pole where is South? I'm always amazed at how many people have trouble with understanding that one let alone trying to figure out why the Pythagorean theorem doesn't apply in this situation.
That's what they call UI issues here at work that I'm expected to solve!
We all have our crosses to Bear

In a related question: if the hunter goes f-=39 and t+=2899, where does he end up?

Hate to show my ignorance, but I'm not following your nomenclature. Remember the the last circular triangle I solved was probably circa 1970.

Jim Yingst
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Posts: 18671
That's OK, I just made it up. But there is a meaning, which will be reasonably self-evident if you look at it the right way. Well, maybe. Let's just say they're not the sort of spatial coordinates you're likely to be thinking of. Gotta think outside the box. Or sphere in this case.

Jim Yingst
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Posts: 18671
Also, for anyone who hasn't seen this problem before: take the bear out of the picture. From the rest of Michael's problem statement, where else could the hunter be?

Jason Menard
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Posts: 6450
Originally posted by Jim Yingst:
Also, for anyone who hasn't seen this problem before: take the bear out of the picture. From the rest of Michael's problem statement, where else could the hunter be?

If polar bears were not involved, his camp could be one mile north of the South Pole. Of course his walk one mile east would be a bit monotonous, and he would have to make sure he went one mile north on the correct heading.

Jim Yingst
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If you're right at the south pole, I don't think it's possible to say you're going east, or west. Though I suppose this is arguable. I was thinking of a different solution.
[ June 03, 2003: Message edited by: Jim Yingst ]

David O'Meara
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Posts: 13459
Originally posted by Jim Yingst:
[QB]f-=39 and t+=2899

I though I was getting to the center of the problem, but my maths didn't work out

Roy Tock
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Posts: 83
The other solution set is infinite: it is the set of all points exactly (1 + (1/PI))miles north of the south pole.
At those points, the hunter walks south a mile, then east in a circle having a one mile circumference, then retraces his steps north to his starting point.
If you're having trouble picturing it, think of those wire easter egg dippers you find in easter egg coloring kits.

[ June 04, 2003: Message edited by: Roy Tock ]

Bert Bates
author
Sheriff
Posts: 8900
5
Even more solutions...
1 + n / pi
The hunter could walk around the circle 'n' times before heading north again

Jim Yingst
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Posts: 18671
Well, almost. Where exactly do you end up for n = 1000, for example?
For those who were perplexed by the mysterious coordinates in my previous post - take a look at the URL of this page.

Jim Yingst
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Posts: 18671
And Roy - so if you walk south one mile, how far are you from the pole? And what exactly is the circumference of a circle at this radius?

Michael Morris
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Posts: 3451
Originally posted by Jim Yingst:
Well, almost. Where exactly do you end up for n = 1000, for example?
For those who were perplexed by the mysterious coordinates in my previous post - take a look at the URL of this page.

Here?

Jim Yingst
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Yup M^2, that's what I meant.

Bert Bates
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1 / (n * pi) , yup that's what I meant! :roll:

Jim Yingst
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Posts: 18671
Almost...

Bert Bates
author
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Posts: 8900
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Would you believe...
1 + 1 / (n * pi)

Jim Yingst
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Posts: 18671
Well, no.

Bert Bates
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Having thoroughly proven that I can't do it in my head, I've applied pencil to paper...
(Let's see if I can do it now!)
1 + 1 / (2n * pi )
:roll:

Michael Morris
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Posts: 3451
Ok, I'm getting lost here (pun intended). I don't think Roy's solution is correct and I'm not even sure what Bert is trying to do. At first thought it would seem that you would have to be 1 mile north of the circle that is cut by the plane parallel to and the same distance from the equator as the one in the northern hemisphere that was the path of our hunter going east. That circle should have a radius of R*sin(1/R). The circumference of that circle is 2*PI*R*sin(1/R) which is very close to 2*PI not 1 as Roy stated. But would that circle do the trick? As you wander from the South Pole then north has a specific direction, from the South Pole all directions are north. What am I missing?

Bert Bates
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Michael -
What I'm 'trying' , to do is map all the circles whose circumference is 1/n miles. Then, once you've walked south a mile, to a point on one of these 1/n circles, you walk around the circle n times and then head north again.

Jim Yingst
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Posts: 18671
(Let's see if I can do it now!)
1 + 1 / (2n * pi )

That's what I was hoping to see. We don't want that hunter ending up on the other side of the world, do we? (Even if that's just a few miles away.)
[M^2]: Ok, I'm getting lost here (pun intended). I don't think Roy's solution is correct and I'm not even sure what Bert is trying to do.
Roy was doing the same sort of thing Bert has just explained - but also overlooked a factor of 2.
At first thought it would seem that you would have to be 1 mile north of the circle that is cut by the plane parallel to and the same distance from the equator as the one in the northern hemisphere that was the path of our hunter going east. That circle should have a radius of R*sin(1/R). The circumference of that circle is 2*PI*R*sin(1/R) which is very close to 2*PI not 1
as Roy stated. But would that circle do the trick? As you wander from the South Pole then north has a specific direction, from the South Pole all directions are north. What am I missing?

Ummm... I'm not sure I followed that. As Bert said, we're looking for circles whose exact radius is 1/n, which means their exact (flat) radius is 1/(2n*PI). Due to the curvature of the earth there's technically an extra x/(sin x) factor here, where x is the angle from the south pole (in radians) but 1/(2n*PI) is close enough, all things considered. Let's just flatten the pole out and not worry about it.
[ June 29, 2003: Message edited by: Jim Yingst ]

Michael Morris
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Ummm... I'm not sure I followed that. As Bert said, we're looking for circles whose exact radius is 1/n, which means their exact (flat) radius is 1/(2n*PI). Due to the curvature of the earth there's technically an extra x/(sin x) factor here, where x is the angle from the south pole (in radians) but 1/(2n*PI) is close enough, all things considered. Let's just flatten the pole out and not worry about it.
Maybe I don't understand what problem we are trying to solve. Are we trying to find any alternate location on the earth where a person can walk south one mile, east or west one mile and return to the exact place of beginning by walking one mile north? If that is the problem, there is no alternate spot. To show this we need to define some criteria. First can we agree that in order to move in a true south or true north direction we will remain on the same longitude, in other words if I walk due north from Greenwich 100 miles, my lattitude will change but I will still be on the prime meridian? The same is true for east and west except that the lattitude will remain constant while the longitude changes. Next, there is an infinite set of longitudes on the face of the earth and they intersect at only two points: the North and South poles. With that in mind, let's take a trip to Antarctica. Even though we can find a circle near the South Pole where we can almost pull this off it becomes obvious that we cannot do the two things required: return to the exact beginning and remain on the same longitude. The last leg of the journey would be slightly longer and ever so slightly west of north. This becomes much more apparent when the arc length of each leg of the spherical triangle is sufficiently large.
[ June 29, 2003: Message edited by: Michael Morris ]

Bert Bates
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Michael -
Huh?
Are you saying that we can't find a latitude whose circumference is EXACTLY 1/n miles? Help!

prashant komaragiri
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Its a Black bear.. cause there is a zoo bang at the north pole and the Black Bear is in that zoo

Jim Yingst
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Maybe I don't understand what problem we are trying to solve. Are we trying to find any alternate location on the earth where a person can walk south one mile, east or west one mile and return to the exact place of beginning by walking one mile north? If that is the problem, there is no alternate spot. To show this we need to define some criteria. First can we agree that in order to move in a true south or true north direction we will remain on the same longitude, in other words if I walk due north from Greenwich 100 miles, my lattitude will change but I will still be on the prime meridian? The same is true for east and west except that the lattitude will remain constant while the longitude changes. Next, there is an infinite set of longitudes on the face of the earth and they intersect at only two points: the North and South poles.
I agree with everything so far.
With that in mind, let's take a trip to Antarctica.
Well OK, if you're paying for it.
Even though we can find a circle near the South Pole where we can almost pull this off it becomes obvious that we cannot do the two things required: return to the exact beginning and remain on the same longitude. The last leg of the journey would be slightly longer and ever so slightly west of north. This becomes much more apparent when the arc length of each leg of the spherical triangle is sufficiently large.
Eh, I just don't see where this is coming from. Are you asserting that we can't find a circle such that after traveling 1 mile east (let's stick with east for the sake of argument) we end up exactly where we were before traveling 1 mile east? Or are you saying that even if our circle takes us right to the start of the circle again, the subsequent 1 mile north will diverge from the earlier 1 mile south?
Let's standardize some terminology here. The position the hunter starts at is point A. Traveling 1 mile south takes him to B. Traveling 1 mile east takes him to C. Traveling 1 mile north again takes him to D. OK - we're looking for solutions where A and D overlap exactly. For the solutions Bert, Roy and I are talking about, this is achieved by having B and C overlap exactly, which means all of AB and CD overlap exactly. In between is BC - this is a circle along a "parallel" (meaning a path of constant latitude, as it doesn't look very parallel when it's a circle). The circumference of this circle is exactly 1 - or more generally, 1/n. So that if you travel 1 mile "east" along it, you go all the way around the circle exactly n times, and arrive where you started, B = C.
So, where do these statements differ from your own mental picture?

Michael Morris
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Posts: 3451
Eh, I just don't see where this is coming from. Are you asserting that we can't find a circle such that after traveling 1 mile east (let's stick with east for the sake of argument) we end up exactly where we were before traveling 1 mile east? Or are you saying that even if our circle takes us right to the start of the circle again, the subsequent 1 mile north will diverge from the earlier 1 mile south?
OK, I'm slow somtimes. Now, I see where you're going. Our hunter is taking a trip around the world. Let's hope it doesn't take him 80 days.

Bert Bates
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Yeah!!!
I think we've got concensus!!!

Jim Yingst
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Woo Hoo! I was afraid we were going to have to send a polar bear to Texas to take care of M^2 for us, but I guess not.

Michael Morris
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Originally posted by Jim Yingst:
Woo Hoo! I was afraid we were going to have to send a polar bear to Texas to take care of M^2 for us, but I guess not.

A polar bear wouldn't stand a chance against the fire ants and mosquitos in Texas.

Jim Yingst
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Well that's why I was concerned. It seemed dreadfully unfair to the bear. But I just didn't see another option.

Dave Jochim
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Posts: 31
Sorry to open this up after you achieved 'consensus'.
The solution accepted was
1 + 1 / (2n * pi ) correct?
Is this the distance from the South Pole?
So the first 1 is the distance traveled north/south? But 1/(2n* pi) is the radius from the Earth's axis isn't it and not distance traveled from the South Pole?
Sorry I am still a bit confused I guess.
thanks

Jim Yingst
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So the first 1 is the distance traveled north/south? But 1/(2n* pi) is the radius from the Earth's axis isn't it and not distance traveled from the South Pole?
You are correct. However they are very, very close to each other considering we're less than two miles from the south pole; the curvature of the earth is not that great. As I noted above:
Due to the curvature of the earth there's technically an extra x/(sin x) factor here, where x is the angle from the south pole (in radians) but 1/(2n*PI) is close enough, all things considered. Let's just flatten the pole out and not worry about it.

John Smith
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Posts: 2937
A hunter rises early one morning and walks one mile due south from his camp and spots a bear. He shoots the bear, wounding it and the bear goes due east one mile where the hunter finally catches up with him and finishes him off. The hunter drags his quarry exactly one mile due north back to his camp. What color was the bear?
Here is my variation:
A hunter rises early one morning on Jan 1st, 2002 and walks one mile due north (using a regular compass) and then one mile due east. The hunter stops to rest, and a hungry bear attacks and kills the hunter. Just before the hunter is killed, he phones his son at the camp and screams "Son, I walked one mile due north and then one mile due east, here is where you can find me if am killed". The bear tastes the hunter, but decides not to eat him. Assume that the body of the hunter stays highly visible and well-preserved for a year.
On Jan 1st, 2003, the hunter's son decided to find his father and give him a proper burial. So he leaves the same camp, walks one mile due north (using a regular compass) and then one mile due east, just like his father instructed him. But his father is not there!
Why?
[ July 06, 2003: Message edited by: Eugene Kononov ]