He wrote a set of instructions:

He included a map with those seven landmarks highlighted.

Then what he did was cut the column (in

*italics*in the instructions) out of the instructions and left it to Dick, and cut the column with the landmarks (in

**bold**) and left it to George. (Both sons could see the rest of the instructions: "Start at ___, Go ___ of the way from there to ___, Go ___ of the way from there to ___, etc.")

He knew (and knew Dick knew) that there are 7! or over 5000 ways the seven landmarks could be ordered, and digging an average of 2500 holes to find the "treasure" would be unfeasible. He also knew (and knew George knew) that without the fractions, George would just be wondering around aimlessly.

He figured his sons HAD to work together if they wanted the money. He was wrong. Which one of the sons was able to dig a single hole and get the money for himself without the help of his brother? How/why?

Originally posted by Ryan McGuire:

He knew (and knew Dick knew) that there are 7! or over 5000 ways the seven landmarks could be ordered, and digging an average of 2500 holes to find the "treasure" would be unfeasible.

Not true.

Reordering of the landmarks will not change the final location so Dick can get it without knowing the order of landmarks.

Originally posted by Jim Yingst:

The term "centroid" comes to mind...

Are you just guessing or are you trying to convince us you arrived at that logically/mathematically?

Besides, (now that I've looked it up, I see) there are a couple different "centroids" you could use. Which one is best here?

Sheriff

(Σx / N, Σy / N ), if you prefer.

[ May 02, 2006: Message edited by: Jim Yingst ]

"I'm not back." - Bill Harding, *Twister*

It does not work, and for some guy to read on wikipedia abotu centroids and to claim here that he is samrt and that this problem is easy , whithout ever giving any proof to his centroid ideea - is just plain imature.

Question 1: I took a piece of paper and drew 3 dots and tried to see if walking like in the problem gives the same result no matter what the order of the 3 dots is: IT DID NOT WORK.

Question 2: the guy that claimed any centroid is the solution (place to digg) to this problem: WHERE IS YOUR PROOF ? I mean , the greeks figured out 3000 years ago that you cannot simply state the answer to a question , you must prove it meaning give arguments on why it is so. Yet you come here writing a few words and claiming this is an easy problem.

Please prove that "any" (or at least one type of your choosing) centroid is the place to digg to get the money in this problem.

However, since you want a proof, try this. Using vectors, let's say the landmarks are at

**b**,

**t**,

**m**,

**d**,

**g**,

**w**,

**n**:

Start at

**b**.

Step 1:

**x1**=

**b**+ (

**t**-

**b**)/2

Step 2:

**x2**=

**x1**+ (

**m**-

**x1**)/3

Step 3:

**x3**=

**x2**+ (

**d**-

**x2**)/4

Step 4:

**x4**=

**x3**+ (

**g**-

**x3**)/5

Step 5:

**x5**=

**x4**+ (

**w**-

**x4**)/6

Step 6:

**x6**=

**x5**+ (

**n**-

**x5**)/7

So the final location is:

**n**/7 + 6/7(

**w**/6 + 5/6(

**g**/5 + 4/5(

**d**/4 + 3/4(

**m**/3 + 2/3(

**t**/2 +

**b**/2)))))

Reduces (because lots of the fractions cancel) to:

(

**n**+

**w**+

**g**+

**d**+

**m**+

**t**+

**b**)/7,

which is the centroid. And it's symmetric, so the ordering of the landmarks has no effect. And typing that up was much slower than solving it!

Given a sequence of points x_1, x_2, ...

Let's say we are at the centroid of the first n points: C_n = 1/n . SUM{i = 1, n}[x_i]

We then move 1/(n + 1) of the distance towards x_(n+1)

New position = C_n + 1/(n+1) . [x_(n+1) - C_n]

= n/(n+1) . C_n + 1/(n+1) . x_(n+1)

= 1/(n+1) . SUM{i = 1, n}[x_i] + 1/(n+1) . x_(n+1)

= 1/(n+1) . SUM{i = 1, n+1}[x_i]

= C_(n+1)

The case n = 1 is trivial - the centroid of one point is that point. After that, each step takes us to the centroid of all points considered so far.

Sheriff

Based on Myke's obvious respect for the traditions of Plato and Pythagoras, I'm sure a detailed proof of this will be forthcoming. It was probably just overlooked in the rush.

"I'm not back." - Bill Harding, *Twister*

Sheriff

- 1

Myke Enriq wrote:Centroids have nothing to do with it.

It does not work, and for some guy to read on wikipedia abotu centroids and to claim here that he is samrt and that this problem is easy , whithout ever giving any proof to his centroid ideea - is just plain imature.

Question 1: I took a piece of paper and drew 3 dots and tried to see if walking like in the problem gives the same result no matter what the order of the 3 dots is: IT DID NOT WORK.

Question 2: the guy that claimed any centroid is the solution (place to digg) to this problem: WHERE IS YOUR PROOF ? I mean , the greeks figured out 3000 years ago that you cannot simply state the answer to a question , you must prove it meaning give arguments on why it is so. Yet you come here writing a few words and claiming this is an easy problem.

Please prove that "any" (or at least one type of your choosing) centroid is the place to digg to get the money in this problem.

Myke,

You were able to disprove the "centroid" answer with a three-location counter-example? Can you provide a drawing of your example or at least a description of it? It seems to work just fine for the handful of triangle I tried. For instance...

With these three points, we end up at the same place regardless of whether the points are ordered A-B-C or C-A-B in the instructions. I worked out the other four possibilities for these three points as well and came up with the same final destination each time.

For the case of three points, the picture should like something like what's producible with the Java applet located here. The points that are labelled D,E and F on that page are the intermediate locations after the "Go half way from the first point to the second" step in the original instructions. If then go one third of the way from one of those side bisectors to the other triangle vertex, you should always end up at the point that's labelled centroid.

However, if you have a diagram that illustrates a different result I'd be interested in seeing it.

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