TripleM
Level: Smitemaster
Rank Points: 1373
Registered: 02-05-2003
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I've been trying a while, but haven't been successful. It's slightly out of memory-reach for the first program I tried writing to solve it by brute force..

halyavin
Level: Delver
Rank Points: 52
Registered: 02-20-2006
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My program says that player can't play on level 1 forever (even without next-level rule). There is a chance that the program contains an error though.

Nuntar
Level: Smitemaster

Rank Points: 4596
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Maybe it's time to admit that this puzzle is beyond the collective ability of the forum?

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50th Skywatcher

stigant
Level: Smitemaster

Rank Points: 1182
Registered: 08-19-2004
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That's fine with me... post another puzzle.

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Progress Quest Progress

aztcg7
Level: Master Delver

Rank Points: 104
Registered: 03-08-2005
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I've been contemplating this, and I'm almost entirely sure it's possible to break the game by staying on the first level. To demonstrate, I simplified the game by imagining a no next level on every square greyed, and a 2x2 grid with 2 colors. On this board, it's trivial to show that you can play forever. If the board is changed to 3 high, it's also trivial to show that you can continue to clear 4 colors. This is done by having the upper and lower pair have different tiles, with the middle pair a clearing ground.

For a minute, let's go back to the 2x2 with no next level. Can you play forever with 3 colors? You can, so long as you know what the next piece is going to be. By knowing, you can be sure to leave that one uncovered for clearing. Of course, you can't play forever with 4, as pure bad luck will have you with no more pairs to make.

Back to the 2x3. With this, our current lower and upper limits are 4 and 6. With the one lookahead, could we play forever with 5? Yes, although it will involve a change in strategy. Have the bottom 2 slots be filled with the first two different tiles that come along. For the third and fourth different tiles, have one on the other two rows, clearing as capable. Worst case, there are one of each of 4 of the 5 tiles, 2 on the bottom and one on the other two rows. from this, if the tile you next have to use is the 5th, you can decide where to put it such that on the next turn you can clear a pair.

I'm not sure how to use this strategy when you have to worry about the clearing rule, though. I'll think about it more tomorrow.

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In other news, is a considerably more stylish way to express sarcasm than , because everybody uses and I am /indie/. INDIE, I TELL YOU.

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
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Is it just me, or did someone post a puzzle and then delete it? (Something about double letter changing, I think?)

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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."

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
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Remember that puzzle that asked how many teeny tiny 2*2 DROD rooms there are? Here's a twist: How many moves can a 2*2 DROD room require? The room must contain Beethro and be possible in a finite amount of moves. Scripting isn't allowed, of course.
I'm pretty sure that the room I have in mind is optimal. If anyone comes up with a better one, though, they get two rank points instead of the usual one. Deal?
EDIT: You don't need exit stairs- just maximize the number of moves until the room is cleared.
EDIT 2: Press enter to see the number of moves you've taken in a room.

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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."

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[Last edited by lopsidation at 04-11-2009 07:44 PM]

Nuntar
Level: Smitemaster

Rank Points: 4596
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OK, here's a first stab.

SW square: Beethro, facing SE, on closed yellow door
NW square: roach
SE square: roach on on/off pressure plate that opens yellow door in SE square
NE square: bomb

This requires 11 moves.

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50th Skywatcher

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
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Good try, but 11 is not maximal.

____________________________
"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."

TFMurphy
Level: Smitemaster
Rank Points: 3118
Registered: 06-11-2007
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Nuntar wrote:
SW square: Beethro, facing SE, on closed yellow door
NW square: roach
SE square: roach on on/off pressure plate that opens yellow door in SE square
NE square: bomb

This requires 11 moves.

Actually, that requires 6 moves. You can simply turn clockwise and stab the bomb. If you replaced the Bomb with Halph, then it would indeed be 11 moves.

A 38-move solution can result from the following:
NW: Roach Queen
NE: Oremite with NW-facing Force Arrow and Beethro
SW: Closed Yellow Door with SE-facing Arrow and Disarm Token
SE: Plate that opens the SW door

EDIT: And a 63-move solution (65 to clear but scored at 63) would be:
NW: Tar Mother, Oremites (NO TAR UNDERNEATH)
NE: Roach Queen, S-Facing Force Arrow, Conquer Token, Open Yellow Door
SW: Open Tarstuff Gate (will close as soon as play begins)
SE: Beethro (sword not facing NW or N), Orthosquare, Toggle Plate that toggles NE Door

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[Last edited by TFMurphy at 04-11-2009 09:41 PM]

mxvladi
Level: Smitemaster
Rank Points: 2505
Registered: 02-11-2008
IP: Logged

SE: Beethro, disarm token, a closed door and force arrow facing south
NW: Roach queen
NE: pressure plate that opens door at SE
SW: force arrow facing south

34 moves required

EDIT: Agrh, post collision

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[Last edited by mxvladi at 04-11-2009 09:05 PM]

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
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TFMurphy wrote:
EDIT: And a 63-move solution (65 to clear but scored at 63) would be:
NW: Tar Mother, Oremites (NO TAR UNDERNEATH)
NE: Roach Queen, S-Facing Force Arrow, Conquer Token, Open Yellow Door
SW: Open Tarstuff Gate (will close as soon as play begins)
SE: Beethro (sword not facing NW or N), Orthosquare, Toggle Plate that toggles NE Door

That's good- but it's still not the best possible. Everyone is missing a key element that will drastically improve scores...

____________________________
"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."

agaricus5
Level: Smitemaster
Rank Points: 1838
Registered: 02-04-2003
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lopsidation wrote:
That's good- but it's still not the best possible. Everyone is missing a key element that will drastically improve scores...

Might that...




____________________________
Resident Medic/Mycologist

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
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Yep! One point to you and one point to TFMurphy. The next puzzle goes to TFMurphy for figuring out how to make such a small room last two spawn cycles.

____________________________
"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."

TFMurphy
Level: Smitemaster
Rank Points: 3118
Registered: 06-11-2007
IP: Logged

'kay, sure.

In lopsidation's puzzle, I posted a setup that lasted two spawn cycles. So for my follow-up, I'll ask you to go one better. Same rules: but make it last *three* spawn cycles before it can be conquered. (And yes, this is possible -- I'm not asking you to do anything I haven't confirmed for myself.)

As a reminder, here's the constraints for lopsidation's puzzle: a 2x2 room, no scripting, no need to create an exit stair, must contain Beethro and be possible in a finite number of moves.

EDIT: I've since found a way for the room to last four spawn cycles, and if you succeed in doing so (or even more), I'll award more points (+2 for original puzzle, +2 for exceeding it), but I'll consider the puzzle solved if you reach three spawns. I will reveal my solutions to both sets at that time.

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[Last edited by TFMurphy at 04-13-2009 11:36 PM]

mxvladi
Level: Smitemaster
Rank Points: 2505
Registered: 02-11-2008
IP: Logged

puzzle which lasts 3 spawn cycles:

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[Last edited by mxvladi at 04-21-2009 12:26 PM]

TFMurphy
Level: Smitemaster
Rank Points: 3118
Registered: 06-11-2007
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That's it. You're up next, mxvladi.

My own original 3 spawn solution was as follows:

As for the 4 spawn solution... well, mxvladi was getting a little close:

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[Last edited by TFMurphy at 02-18-2012 06:15 PM : Had Arrow facing wrong direction in 4-spawn solution]

mxvladi
Level: Smitemaster
Rank Points: 2505
Registered: 02-11-2008
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Somebody else can have a go.

stigant
Level: Smitemaster

Rank Points: 1182
Registered: 08-19-2004
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Ok, I've got three relatively simple puzzles involving dodecahedrons (a regular dodecahedron is a 12-sided polyhedron made up of 12 regular pentagons). For the purposes of the puzzles, assume that the dodecahedron has edges of length 1.

The first two concern paths along the edges of the figure. There are 30 edges connecting the 20 vertices on a dodecahedron. Since each vertex has 3 edges, it's impossible to trace a path along every edge which doesn't traverse any of the edges twice.

Question 1: What is the minimum length path which traverses every edge at least once (note that the answer is obviously more than 30 since you can't traverse each edge once as noted above)?

Question 2: What is the minimum number of single traversal paths needed to include all edges? In other words, you are to put together a set of paths. Each edge is included in exactly one path. What is the minimum number of paths in the set?

The third question concerns paths on the surface of the figure.

Question 3: A fly is standing in the exact center of one face of the dodecahedron, and a spider is standing in the exact center of the opposite face. If the spider must remain on the surface of the dodecahedron, what is the minimum distance he must traverse in order to have the fly for dinner? A decimal approximation will be fine, but I'd prefer an exact solution.

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Progress Quest Progress

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[Last edited by stigant at 04-21-2009 09:14 PM]

Nuntar
Level: Smitemaster

Rank Points: 4596
Registered: 02-20-2007
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1. A traversal path can have at most two odd vertices, the beginning and end. To reduce the number of odd vertices from twenty to two, the the minimum number of edges that must be added is clearly nine, since each edge connects two vertices. Therefore the minimum-length path is the number of edges of a dodecahedron (thirty) plus nine, which is thirty-nine.

2. Almost the same question. Each path can have at most two odd vertices, so to reach all twenty vertices you need ten paths. Note that an actual example can be the same for both questions: stand the dodecahedron on a table, and consider a path going around all five edges of the top pentagon, coming down to the middle and going all around the middle "loop", then coming down and going round all five edges of the bottom pentagon. Nine edges are left untouched. For question (1), each time you reach an endpoint of one of these nine edges, go along it and immediately back to add that edge to the path. For question (2), the ten paths are the path described and the nine single-edge paths.

3. This is messy, but from laying out the dodecahedron as a net and doing a bit of vector geometry and trig to work out the straight-line distance between the two points, I think the answer is (where phi is the golden ratio):

sqrt (1/4 tan^2 36 - 1/4 cos 18 tan 36 + 35/8 phi + 51/16)

Google calculator tells me this mess comes to 3.212, but simple measurement says it's much closer to 4, so I've obviously gone wrong

EDIT: Aha! I confused 1/(2 tan 36) with 1/2 tan 36. Now let me work this out again....

____________________________
50th Skywatcher

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[Last edited by Nuntar at 04-22-2009 12:05 AM]

Jutt
Level: Smitemaster
Rank Points: 863
Registered: 06-29-2007
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I get:

2√(5a² – 4a² cos(144°))
with a = 1/(2tan(36°))

which gives a result of approximately 3.9500

____________________________
Holds: An Architects Audition, Artful Architecture, Salamander, Elusive Exhibitions, Leftover Levels, Six Times Six
Collaborative: Way Forward, Advanced Concepts 2
Styles/Mods: Basalt, Sandstone, Garden, Clock using game elements

stigant
Level: Smitemaster

Rank Points: 1182
Registered: 08-19-2004
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Good... either of you feel free to continue the thread.

____________________________
Progress Quest Progress

12th Archivist
Level: Smitemaster

Rank Points: 789
Registered: 12-07-2008
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It's been well over a month since this thread was posted in, so this is alerting Nuntar and Jutt that they can post a new puzzle.

____________________________
It was going well until it exploded.
~Scott Manley

Check out the DROD Wikia project here!

B.J.Earles
Level: Goblin

Rank Points: 18
Registered: 04-29-2009
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If nobody else is interested in posting a new puzzle, I have one ready.

RoboBob3000
Level: Smitemaster

Rank Points: 1978
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B.J.Earles wrote:
If nobody else is interested in posting a new puzzle, I have one ready.

Go for it!

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http://beepsandbloops.wordpress.com/

B.J.Earles
Level: Goblin

Rank Points: 18
Registered: 04-29-2009
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Okay, here goes.

You are given two circles that share the same central axis. On each circle there are six colored lenses, two for each of the primary colors (red, yellow, and blue). You are then told to rotate the top circle clockwise so that each lens shifts two positions and the bottom circle counterclockwise so that each lens on it shifts three positions. The colors that you can now see are, starting form the 12 o clock position and moving clockwise, purple, green, red, yellow, green, and purple. You are then asked to rotate the circles again and in the same way. This time the colors are yellow, blue, red, blue, orange, and orange. If you rotate the circles this way a total of one hundred times, how many times will you see the color purple?

hint: purple = blue + red, orange = red + yellow, green = blue + yellow.

You should have all the information you need to solve it, but if you are confused go ahead and ask me questions.

Sillyman
Level: Smiter

Rank Points: 339
Registered: 09-08-2006
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First question: How in the world are blue and red subtractive primary colours?

Edit: It's not just confusing subtractive with additive, it's mixing the two together.

____________________________
Who, me?
FNORD

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[Last edited by Sillyman at 06-30-2009 08:41 PM]

Nuntar
Level: Smitemaster

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The question, as stated, is ambiguous, since it's not clear whether the first rotation described is the first of the 100, or whether the 100 start from the second position. It's also unclear whether we are counting purples that appear after each of the 100 rotations or whether the position before the first rotation is also included. However, I'll try to give as full an answer as I can.

(1) Rotating the top circle five steps clockwise relative to the bottom circle is equivalent to rotating it one step anticlockwise, so we can imagine the bottom circle as fixed and the top one as rotating one step anticlockwise each time.

(2) In every six steps, we see purple four times, since each blue on the inner circle aligns with each red on the outer circle once and vice versa. (For the sake of the puzzle I'll ignore your mistake, as Sillyman pointed out, of confusing subtractive with additive primaries.) So in 102 steps we would see purple 68 times.

(3) Now we just need to subtract any purples that appear after the 101st and 102nd steps. Every six steps completes a cycle, so the 102nd step returns us to the starting position, which either contains 2 or 0 purples depending on the interpretation of the question. The 101st step is equivalent to doing one rotation clockwise. If the 100 rotations start from the second position, this returns us to the first position and the final answer is 68 - 0 - 2 = 66.

(4) However, if the 100 rotations were meant to start from the first position (the more likely interpretation, as the puzzle is more interesting this way) then we have to work out the configuration of the circles to work out what position results from doing one rotation clockwise. The first position contains red flanked by two colours that don't contain red, so the red in the second position must arise from the other two red circles. Hence the reds are at 12 o'clock and 4 o'clock in the top circle and at 10 and 4 in the bottom circle. To get the first position as described, the top circle must have blue at 10 o'clock and the bottom circle blue at 12.

(5) Then the first rotation brings the blue on the top circle to 8 o'clock, which is green in the first position and blue in the second. Hence the bottom circle has blue at 8 o'clock and yellow at 2 o'clock, the top circle being the other way round. Finally, both circles start with yellow at 6 o'clock (since the first position has yellow there). In this configuration, the position after one clockwise rotation has exactly one purple, at 4 o'clock. So the answer is 65, or 67 if the two purples in the initial position are also counted.

____________________________
50th Skywatcher

B.J.Earles
Level: Goblin

Rank Points: 18
Registered: 04-29-2009
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Sillyman wrote:
First question: How in the world are blue and red subtractive primary colours?

D'oh! I guess I messed that up. I must have been thinking of paint pigments, which is different from light. Oh well, at least the puzzle is still solvable.

Nuntar wrote:
The question, as stated, is ambiguous, since it's not clear whether the first rotation described is the first of the 100, or whether the 100 start from the second position. It's also unclear whether we are counting purples that appear after each of the 100 rotations or whether the position before the first rotation is also included.

The one hundred rotations do indeed begin with the first rotation stated in the problem (sorry, I didn't mean for that to be obscure), and you must include the position that you receive the circles in (so all in all, you look at the circles 101 times).

Nuntar wrote:
So the answer is 65, or 67 if the two purples in the initial position are also counted.

Sorry Nuntar. That's not the answer I have. You are on the right track though.

brian_s
Level: Smitemaster

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In the second position, the two disks are in some order Y-B-R-B-R-Y and Y-B-R-B-Y-R (starting at 12:00 clockwise).

In order to match the first position Y-B-R-B-R-Y is the top disk and Y-B-R-B-Y-R is the bottom disk.

There will be six distinct positions, which repeat as the disks are rotated.

Position 1 - 2 purple
R-B-R-Y-Y-B
B-Y-R-Y-B-R

Position 2 - 0 purple
Y-B-R-B-R-Y
Y-B-R-B-Y-R

Position 3 - 3 purple
R-Y-Y-B-R-B
B-Y-R-Y-B-R

Position 4 - 1 purple
R-B-R-Y-Y-B
Y-B-R-B-Y-R

Position 5 - 1 purple
Y-B-R-B-R-Y
B-Y-R-Y-B-R

Position 6 - 1 purple
R-Y-Y-B-R-B
Y-B-R-B-Y-R

Positions 1-5 will each occur 17 times and position 6 will occur 16 times, for a total of 17*(2+0+3+1+1)+16*(1)=135 occurances of purple.