[SFW] Riddles / Puzzles / Brain Teasers - Page 36

To make this easier to talk about, let's make the lake the unit circle -- you're at (0,0) and the wolf is at some arbitrary point on the circle, say (0,1). Previously posters correctly said that if we want to get out at (0,1) safely, we need to somehow arrange for the wolf to be at the opposite point, (0,-1) when we're close enough to the circle to make a beeline for safety without the wolf catching up. Other posters just didn't correctly see how close "close enough" is.

First, notice that if we swim around the circle of radius r, for r near 1 (we're near the edge of the lake), the wolf will always easily keep up with us, staying right at the point on the circle closest to us, since for every distance d we swim in a circle, the wolf runs pi*d (arc length), which is less than the 4d he could cover going at top speed. However, if we stay close to the middle of the lake and swim in tiny circles around the origin, the situation is reversed... If we swim in circles of radius r, for r near 0, our angular velocity increases without limit.

So if we're faster near the center but the wolf is faster near the edge, where's the breakpoint? Other than the fact that 4>pi, this is the only significance of the wolf running at 4 times our speed. If we swim around the circle r=1/4 (a quarter of the way out to the edge), suppose we swim an arc of length theta. Then the distance we cover is s=r*theta=theta/4, so in the same time the wolf can cover 4 times that distance, or exactly theta distance. But the wolf is out at r=1, so for him, arc length s=r*theta=theta, and we see that you and the wolf cover exactly the same arc length.

Now we're nearly done. Anywhere inside the circle r=1/4, we can outpace the wolf's angular velocity, but anywhere outside, he can outpace ours. To get the wolf around to the opposite point, swim out to just slightly under radius 1/4 (within an arbitrarily small but nonzero distance). Start swimming in circles. Every time we go around the circle, the wolf lags behind us a little bit more, so keep doing this until he's half a circle behind. Depending on how close to 1/4 you are, this will take more or fewer trips around the circle, but our angular velocity is always higher so it's clear this happens eventually.

Now we're in the desired situation. The wolf is at (0,-1) opposite our exit point, and we're at (0,1/4), a quarter of the way there. Now stop circling and head to safety. We have to cover distance 3/4, so in that time the wolf can cover distance 3. But he's half a circle away -- he has to cover distance pi, which is more than the 3 he can manage at top speed. Huzzah!

Our angular velocity is higher than the wolf's if and only if we're less than a quarter of the way out from the center of the lake. Swim out slightly less than a quarter of the way and start circling. The wolf will gradually lag behind our position more and more -- when he's on the far point of the lake opposite our position, head straight for the closest point on the bank; you'll make it because 3<pi<4.

Weighing the water? This seems to not break any of the rules of the riddle.

Fill up the 5 gallon one
Pour 3 gallons into the 3 gallon tank
Empty 3 gallon tank
Pour remaining 2 gallons from 5 gallon tank into 3 gallon one
Fill up 5 gallon tank

You will now have a full 5 gallon tank + a 3 gallon tank with 2 gallons...

Fill 5 Gallon
Pour into three gallons in 3-gallon, so two gallons are remaining in the 5-gallon.
Empty the 3-gallon, and then pour the two gallons from the 5-gallon into the 3-gallon.
Fill the 5-gallon.
Pour one gallon into the 3-gallon, making the 3-gallon full.
There are four gallons left in the 5-gallon.

Lol I did this a longer way but got the same thing.
Fill the 3 gallon tank, pour into the 5 gallon tank.
Fill the 3 gallon tank again, pour 2 gallons into the 5 gallon tank (1 gallon left in 3 gallon tank).
Empty 5 gallon tank.
Pour 1 gallon from 3 gallon tank into 5 gallon tank.
Fill 3 gallon tank.
4 gallons.

My way is much more round about >.> same answer though!

If you imagine the pond has radius 'x', then the wolf has to run pi*x units to get to the opposite side. If you are swimming in a circle of radius x/4 or less from the center of the pond, then for you the distance to travel from one side to the other is less than 25% of the amount the wolf has to travel, so you can change angles faster than it.

So therefore if you swim in a circle at a radius a little less than x/4 for long enough, you'll end up across from the wolf. At this point, it's a little over 3/4*x for you to get to the center, whereas the wolf will need to travel pi*x. Even if he is going 4 times faster than you, he'll only make it 3*x of the way to where you are landing, so he won't make it in time.

Now if he's going 4.5 times your speed, this won't work. What a conundrum...

Edit: My instinct is that you would do something similar to what I posted to start, then spiral towards the pond edge to escape, but I have no idea how to show whether that would be better or how much better it would be.

Total area 1,2,3: obviously impossible
Total area 4: one 2x2
Total area 5,6,7: also obviously impossible
Total area 8: two 2x2s
Total area 9: one 3x3
Total area 10,11: impossible, slightly less obvious
Total area 12: three 2x2s
Total area 13: one 2x2 and one 3x3
Total area 14: impossible, slightly still less obvious
...and so on...

I think you should be able to leave the origin of the circle in a zig-zag type pattern like so (W= wolf, o = you, c = center)

             .
            (
          W(              o
            (
             .

             .
            (
           (              c
          W (              o
             .

             .
            (
           (              c
            (               o
            W.

             .
            (                 
           (              c o
           W(
            .

and so on until you reach the required distance out to apply the solution found for the 4 case

Using payments of 1, 2, 4, 8, 16, up to 2^n, you can make any number up to 2^(n+1)-1. So to get up to 79, the merchant can cut the chain up into pieces of length (1, 2, 4, 8, 16, 32), (1, 2, 4, 8), and 1 link left over. That's 79 total, cutting the circular chain into 11 pieces, so the optimal solution is at most 11.

Any price up to 63 can be covered with some combination of the first set of pieces listed above. Any price between 64 and 78 can be covered by the entire first set plus some combination of the second set. And 79 is obviously all 11 pieces.

23
use modulo 4(smallest 2x2 square). 3x3 is equivalent to 1. The biggest "impossible" is therefore the one that requires 3 3x3 squares to complete, and that happens at 27.

6

use modulo 3 instead of modulo 4 - and the 2x2 can be used as 3 (equivalent to 0) or 4 (equivalent to 1).

modulo 1 is covered at 4 (one 2x2)
module 2 is covered at 8 (two 2x2)
modulo 3 is covered at 9 (one 3x3)

last one to not be covered is at 6.