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TripleM
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Assuming '1 person' means '1 passenger', I think I get:

Click here to view the secret text


[Last edited by TripleM at 01-23-2014 05:29 AM]
01-23-2014 at 05:27 AM
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RyTracer
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Drat! You made that look easy. Okay, the puzzle is solved and you can post another, but also see if you can solve it if Lions take up 2 seats in the boat.

This is what comes of designing puzzles in Del Taco on lunch break.

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01-23-2014 at 06:58 AM
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TripleM
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Programming always makes things look easy :) Change a couple of characters and it's solvable with a minimum crossings of..

Click here to view the secret text


Will need a little time thinking up a new puzzle.
01-23-2014 at 07:11 AM
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TripleM
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I've done so many puzzles it's rare someone comes up with something unique, but I was surprised to have never heard this one before. It's pretty easy, but judge yourself by how long it takes you to come up with the right answer.

Secreted so you can start timing.. now:

Click here to view the secret text


[Last edited by TripleM at 01-23-2014 07:20 AM]
01-23-2014 at 07:19 AM
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Penumbra
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Again, late to the party :?

I also got
Click here to view the secret text
but in a different way

Click here to view the secret text


[Last edited by Penumbra at 01-23-2014 07:51 AM]
01-23-2014 at 07:50 AM
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Elfstone
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(PM'd an answer to TripleM, because if it is correct, I will never be able to think up another puzzle for here. :no )

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01-23-2014 at 12:56 PM
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RyTracer
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Yeah, for my puzzle I had intended there only be only one solution which involved never more than one Lion on the boat, so the Lions taking two seats works nicely. I also got the 21 trips solution. For curiosity's sake, could you post your solution, TripleM?

Also, in response to your puzzle, I'm going to have to go for "dos" (Spanish for two). Pretty sure that's wrong.

Or maybe "one." Reverse alphabetical order is still an order.

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[Last edited by RyTracer at 01-23-2014 05:53 PM]
01-23-2014 at 04:24 PM
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Jacob
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I go for 40.

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01-23-2014 at 08:49 PM
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stigant
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Doh... apparently I don't know how to spell ;-)

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01-23-2014 at 09:06 PM
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Someone Else
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You people and your weird spellings. I saw that and I thought "No, I'm pretty sure that U comes after R."
01-23-2014 at 10:02 PM
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TripleM
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Jacob wrote:
I go for 40.
Correct :) (Elfstone was also correct!)

RyTracer:

Click here to view the secret text

01-24-2014 at 12:07 AM
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RyTracer
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Good thing you said, "integer," or I would have put "e."

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01-24-2014 at 12:28 AM
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Elfstone
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TripleM wrote:
Jacob wrote:
I go for 40.
Correct :) (Elfstone was also correct!)
:D :thumbsup :hooray

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01-24-2014 at 09:37 PM
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Nuntar
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Someone needs to start this moving again....

What 4-digit square number gives another square number when all its digits are increased by 1? This can of course be solved by trial and error, but there is a more elegant way to find the solution.

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02-14-2014 at 01:54 PM
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stigant
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2025 and 3136:

Let k^2 and j^2 be the two numbers.
k^2 = j^2 + 1111
k^2 - j^2 = 1111
(k-j)(k+j) = 1111

Now, there are only 2 pairs of integers which multiply to 1111: 1111x1 and 11x101. Clearly (k+j) <> 1111 since k > 500 and k^2 would be 5+ digits. So k+j = 101 and k-j = 11. Solving this system gives k = 56 and j = 45.

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02-14-2014 at 03:05 PM
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stigant
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Sorry, I'll clean that up a little bit (and for some reason, I can't edit my post above). k and j need to be integers so k+j and k-j are also integers. Assume (without loss of generality*) that k > j > 0 so that k+j > k-j. Which means that once we find a pair of integers with the desired product (1111), k+j is the larger of those integers and k-j is the smaller. Since k+j = 1111 and k > j implies that k > 1111/2 and k^2 > 10000, we can reject the first pair.

* Assuming j > k leads to the same result, just switched, and assuming that j < 0 leads to the system k-j = 101 and k+j = 11 which has solution k = 56 and j = -45 but k^2 is still 3136 and j^2 is still 2025.

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02-14-2014 at 03:16 PM
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Nuntar
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That's it. Your turn! :thumbsup

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02-14-2014 at 03:39 PM
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stigant
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Ok, a number of years ago, my wife gave me a puzzle book by Martin Gardner (The Moscow Puzzles) which I basically never opened. But I happened to open it last night and thought this was an interesting puzzle. Gardner, of course tells a lovely little story to go with the puzzle which I have eliminated because I can't remember all the irrelevant details and, well, I don't care about the story.

So three people, a maid, a princess, and a suitor, who weigh 80, 100 and 180 pounds respectively, are trapped in a tower by the nefarious king, and need to descend from a window at the top of the tower. Outside the window is a two-bucket pulley system. That is, two buckets connected to a rope and hooked over a pulley just above their window. If you want to get something up to the window, you put it in one bucket at the bottom of the tower, put something heavier in the other bucket at the top of tower, and let gravity do its work. In order to descend safely, whatever is in the down-bucket must be within 10 pounds of whatever is in the up-bucket. There are 13 10-pound objects in the room at the top of the tower. How do the 3 prisoners escape the tower and live happily ever after?

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02-14-2014 at 04:17 PM
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Someone Else
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Well, first you start by lowering 1 object, then 2, and so on until there are 7 objects at the bottom and 6 in the bucket that just came up. The maid goes down, (bringing 7 objects up), and then 9 objects get sent down, bringing the maid back up.
The princess goes down, and the pattern repeats. Eventually, there's the princess and the maid at the bottom, and they leave the suitor because he was a jerk and the suitor comes down (bringing the princess and the maid back up) and then the whole thing is repeated.
02-14-2014 at 05:18 PM
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stigant
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yep... you're up

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02-14-2014 at 06:09 PM
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stigant
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Ok, looks like SE isn't going to post another puzzle, so I'll post one.

I think a while back we had the puzzle about dropping balls out of windows and determining the maximum floor from which you could drop a ball in the least number of drops. To refresh your memory:

You have 2 identical balls and a building with 36 floors. Your job is to determine the maximum floor from which you can drop a ball without it breaking in as few drops as possible. Create a strategy for minimizing the worst case scenario.

The solution is (my discussion below spoils some of this, so solve it now if you want to):
Click here to view the secret text


So I was thinking today that minimizing the total number of drops isn't really what you'd want to do. You'd want to minimize the total number of flights of stairs that you had to climb. Going down stairs isn't that bad, so we won't worry about that. So for example, in the solution above, suppose that you dropped the first ball from floors 8, 15 and 21 where it broke, and the second ball from floors 16, 17, and 18 where it broke, you would have gone up 8 floors, down to the bottom to retrieve ball 1, up to floor 15, down to the bottom to retrieve ball 1, up to 21, then down to 16, then up to 17, down to bottom, and then up to 18. This gives a total of 8 + 15 + 21 + 17 + 18 = 79 floors climbed. Of course, you could be a little more efficient by dropping the second ball from floor 15 instead of retrieving ball 1, and then retrieving both for a total of 15 + 21 + 17 + 18 = 71 floors climbed. The worst case scenario for the strategy above is dropping from floors 8, 15, 21, 26, 30, 33(break), 31, 32 which requires climbing a total of 137 floors.

Can you find a strategy that, in the worst case, does better than that? I will note that, in contrast to the original problem, this one requires some running of numbers, but that I was able to find the optimal solution by hand in a reasonable amount of time (and then verify that it is optimal by running all the possible combinations on a computer).

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04-01-2014 at 10:12 PM
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lopsidation
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OK, here we go.
Click here to view the secret text


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07-19-2014 at 12:29 AM
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lopsidation
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If that answer was correct, then here's the next puzzle. It's based on a true story.

You're playing DROD at 2 AM. You're on a quest to master every user-made hold in existence. Others have called this a fool's errand. You're going to prove them wrong. You only have one hold left: Bad Evil Maze.

You start playing the hold. The entire hold is a maze that takes up all of one room. The only game elements present are walls, floors, and a single staircase. So that's why it was rated 2 fun and 0 brains.

Suddenly, your monitor, speakers, and most of your keyboard crumble to dust.

Your hard drive still works, as do the 11 DROD movement keys (123456789QW). But you can't see or hear anything in the level. Worse, you don't remember anything about the layout of the maze. You'll have to navigate blind.

Rats. You won't be able to sleep until you are 100% sure you have beaten this room. Is there any sequence of moves that guarantees victory?

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07-20-2014 at 07:19 PM
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Nuntar
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Yes - provided of course that the maze is solvable.

This is taking de-optimisation to an extreme, but you could in theory draw a map of every possible maze, with different starting positions counting as different mazes (1216 x 1215 x 2^1214, a finite number). Place them in some order. Start with the first, and input a sequence of commands that will get you to the exit if that is the correct map. Then take the second, work out where the keys you have entered so far will get you, then input a sequence that reaches the exit from there. Eventually you will either hit the staircase by luck, or reach the correct map.

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07-21-2014 at 01:44 PM
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lopsidation
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You got it! Your turn.

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07-21-2014 at 02:39 PM
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lopsidation
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OK, I'm going to call two weeks enough time for a new puzzle. Feel free to post yours anytime, though, Nuntar!

Deep in the woods, you see a row of ten entrances to ten dark caves. Your goblin guide tells you that a hidden passage connects two of the caves. Since you are a smitemaster, this is important information. However, a goblin cannot always be trusted, so you demand evidence.

"Show me the hidden passage."
"Noz! Delver cannot enter cave. Goblin tribe tear to pieces."
"Well, then, you go in one entrance, and leave by another. That would convince me."
"Let me think... noz! Goblinz value privacy. If delver learns where passage is, delver must be killed."
"Gee, you're a toughie. Suppose there was a way for you to make me believe that a hidden passage exists, such that I wouldn't learn where it is, or anything about which two entrances it connects. Would you do it?"
"Iz surely impossible! But I'z would do it, yess."

Well? How about it?

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[Last edited by lopsidation at 08-09-2014 02:21 AM : Clarification: You can't learn *anything* about which two entrances are connected.]
08-09-2014 at 12:30 AM
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blorx1
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I could do it if I had an arbitrarily long piece of string and maybe I could do something similar. I don't otherwise quite know what I'd do and I don't have another puzzle to post so I won't say anything more.

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08-10-2014 at 02:23 AM
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lopsidation
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I'll accept alternate solutions to this puzzle, as long as they're fairly reasonable. A long piece of string? Sure!

Also, you may assume that the passage can be traversed in some reasonable amount of time, e.g. ten minutes.

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"Happiness is like a cat. If you pay no attention to it and go about your business, you'll find it rubbing against your legs and jumping into your lap."

[Last edited by lopsidation at 08-10-2014 01:42 PM]
08-10-2014 at 01:34 PM
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blorx1
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I'll post my solution using string for now, since it likely requires some similar thinking to the actual solution.

Click here to view the secret text


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[Last edited by blorx1 at 08-10-2014 03:19 PM]
08-10-2014 at 03:18 PM
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Tuttle
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I had the same idea as soon as you said string. But:
Click here to view the secret text

08-10-2014 at 03:33 PM
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