Conceptual problems

<div class='quotetop'>QUOTE(eagle33199 @ May 2 2007, 07:11 PM) [snapback]434431[/snapback]</div>

You make a "T" out of the planks, with the horizontal stem of the T on the outside of the moat, each end resting on land with the center over water, and the vertical stem of the T being the part that goes across the moat, with one end resting on land and one end resting on the center of the other plank.

This fails for large outside diameters of the moat (assuming it's circular in the first point), as the center will not be more than .5 meters from the edge of the moat beyond a certain point.

Part 2: What is the maximum diameter of the moat that this will work for? You may ignore any overlap for simplification.

 

<div class='quotetop'>QUOTE(priusenvy @ May 2 2007, 09:00 PM) [snapback]434514[/snapback]</div>

It'd be a weird moat for this to work. Assuming no overlap (which itself won't work - you gotta have SOME overlap, however small - but for simplicity, we'll go with no overlap) - the largest radius of moat that would make this configuration workable is 5.3125 meters - barely bigger than it is across! That would leave a patch of ground .625 meters across for the castle to occupy - a castle not even a miniature Barbie could cram herself into as a studio!

Perhaps the moat has more reasonable dimensions but has a tiny cove with this radius, and the opposite bank a matching point 5 meters away, but the puzzle would have to specify such a topography explicitly to be a legitimate puzzle, and not leave something like that ambiguous.

I have a feeling the answer lies somewhere else, perhaps using a cantilever, but the puzzle's landscape is utterly barren - there's no one else around to lend a hand, no rocks, no trees, nothing to make a cantlever work either, so if the TEE answer is "correct", I'll wager the puzzle's inventor didn't do his trigonometry to see if it was even legitimate.

Mark Baird
Alameda CA

 

<div class='quotetop'>QUOTE(airportkid @ May 2 2007, 10:14 PM) [snapback]434554[/snapback]</div>

You must be thinking of a different diagram than I am, because just eyeballing it I get a much larger number.

I'm asking for the largest circle where, if I place one 4.5 meter plank so each end touches the circle, the distance between the middle of the plank and the circle is at least 0.5 meters. That 0.5 meters, plus the 4.5 meters of the other plank add up to to the five meter width of the moat that the planks need to span.

Damn! You're right. I got a radius of 5.31. Guess I can't draw worth crap.

 

I will assume the planks are a standard 3"h x 8"w (height and width).

If the planks are made of redwood, then the pair would weigh 138 lbs.
If Balsa wood, 50 lbs.

In fresh water these planks would displace 307 lbs if fully submerged.
In salt water, 315 lbs.


With redwood, our person would have to weigh less than 169 lbs if the moat is fresh water . . .. 177 lbs if salt.
With balsa, our person would have to weigh less than 257 lbs if the moat is fresh water . . . 265 lbs if salt.

Under the above conditions, our person can float across and stay dry.

 

<div class='quotetop'>QUOTE(eagle33199 @ May 3 2007, 05:35 AM) [snapback]434640[/snapback]</div>

All of the loads are vertical. The "T" shape is seen when viewing from above. The ends of one plank rest on the ground on the outside of the moat, one end of the other rests on the first plank, the other end on the ground on the inside of the moat.

I don't think you asked the question correctly. You should have said the moat was square. Then you don't run into the ridiculous limitation of the 5.31 meter outside radius moat when you assume it's circular (and that's allowing for no overlap).

 

<div class='quotetop'>QUOTE(eagle33199 @ May 3 2007, 12:16 PM) [snapback]434921[/snapback]</div>

This worked out faster than I thought it would.

Seven locks will do the trick. Line your generals up. Give the 1st four the key to the 1st lock. There is now no possible quartet of generals that won't possess key #1. Tell the 1st general to go to the other end of the line (shifting the line by one general). Give the 1st four generals in this line the key to lock #2. There is now no possible quartet of generals that won't possess key #2. Again, have the 1st general go to the other end of line, shifting the line). Give the 1st four generals key #3. Shift the line. Keep handing out each lock's 4 keys to the 1st four generals in each shifted line. When you're done, every general will possess 4 keys, and no possible quartet will not possess a key to all 7 locks.

While I can't explain WHY, no possible trio will have all 7 keys (as revealed by my spreadsheet).

Mark Baird
Alameda CA

 

<div class='quotetop'>QUOTE(eagle33199 @ May 3 2007, 02:08 PM) [snapback]434997[/snapback]</div>

I see I screwed up in describing how to hand out the keys. This is the matrix I came up with:

A 1246
B 1346
C 1356
D 1357
E 2357
F 2457
G 2467

No general has a consecutive sequence of keys greater than 2 (and some have no consecutive sequence at all). Can't explain why, but somehow that prevents any trio from having all 7.

My shifting line results in your distribution - I should have thought that one out a little more. I was trying to make it clear - but wound up distorting the result!

MB

EDIT - Nope - still wrong - I just double checked and A, B & D have all 7. Damn. I thought this was too easy.

 

<div class='quotetop'>QUOTE(eagle33199 @ May 3 2007, 04:59 PM) [snapback]435117[/snapback]</div>

Drat, you beat me - I worked out the identical answer.

I don't see how it can be done with fewer locks or copies of keys. At least 4 generals must have a copy of each key to assure that every possible quartet has at least one of that key. To blockade the triplets, every possible quartet should have its own lock. There are 35 quartets and (interestingly) 35 triplets -of necessity, of course, each triplet is a subset of several quartets, so won't have the full complement of keys that a quartet has. According to my spreadsheet, each triplet will be missing exactly one of the keys.

MB

 

<div class='quotetop'>QUOTE(efusco @ May 3 2007, 08:21 PM) [snapback]435252[/snapback]</div>

"H" is great except it only produces 5 lines, not 6.

I think the secret is a line of four somewhere that does double duty as two overlapped lines of 3. Maybe. I haven't worked this one out yet. The answer's in that damn safe and I can't find enough generals who remembered to bring their twenty key keyrings to open it and find out!

MB

 

<div class='quotetop'>QUOTE(eagle33199 @ May 3 2007, 06:34 PM) [snapback]435167[/snapback]</div>

Triangles are our friend!

Draw a triangle.
Now plant a tree on each corner.
Now plant three more trees, one on each line midway between the corner trees.
The news trees also form a triangle, but don't draw lines between these new trees - instead, draw lines between the corner trees and the new tree directly across.
Now plant the seventh tree where these lines intersect in the middle.

You now have six rows of three trees each.

 

Here's an easy one:

Without lifting your pencil from the paper, can you draw four lines that pass through all nine dots below:

Code:

*   *   *

*   *   *

*   *   *

 

<div class='quotetop'>QUOTE(priusenvy @ May 4 2007, 02:36 PM) [snapback]435675[/snapback]</div>

Put the pencil on the lower right dot. Draw up diagonally through the center dot to the top left dot. Draw straight down through all the left dots to an imaginary point on a line that would pass upward diagonally through the bottom center and right center dots. Draw that line through those two dots and keep going to another imaginary point in line with the upper dots. Draw the last line back through the upper dots and remove the pencil at the upper left dot and go celebrate with a beer.

MB