×DROD Let's Play Youtube series? Shameless, indeed. What a punk!

Current Projects:

Aurora's Palace (complete)
Terrakept Resonant (complete)
Beethro's Brain (complete)
The City Beneath (completish)
Gigantic Jewel Lost (complete)
Magic Show (complete)
Cube of Memories (complete)
Castle Steele



Feel free to watch/insult/try to give advice to at:

http://www.youtube.com/user/Pearls13a (channel)

http://www.youtube.com/user/Pearls13a/videos?view=1 (playlists)

×You won't be able to find both R and r simultaneously with one measurement, since both variables can't be uniquely determined by a one dimensional measurement. So you'll want to solve for R^2 - r^2 (from the formula SA = pi(R^2 - r^2)

× There location of the measurement has to somehow relate to both radii still. There's only one logical place to measure.

×Measure across the annulus at a point tangent to the inner hole. Looking at the diagram attached to this post, you can create a right triangle by drawing a segment from the center of the annulus to the point of tangency and another from the center to the point where your line touches the outer circle. The segments have length r and R respectively. By the Pythagorean theorem, the length of half your measurement is sqrt(R^2 - r^2) So take your measurement, cut it in half, square it, then multiply by pi.

×Skell's idea generator: http://forum.caravelgames.com/ideasGenerator/view
Challenges Scoreboard: http://forum.caravelgames.com/challenges.php
Level map viewer: http://forum.caravelgames.com/map.php
Tactical DROD game: http://forum.caravelgames.com/tsspreview.php
(Note: Requires Flash)
Room Element Search tool http://forum.caravelgames.com/roomsearch.php
(Search through all published holds for rooms with various elements.)
Room position Heat Map http://forum.caravelgames.com/heatmap.php
DROD has a wiki! http://drod.wikia.com/wiki/Deadly_Rooms_Of_Death_Wiki
Skell's super cool list of turn sequence stuff https://docs.google.com/document/d/1QS9Vk3xsd4jlGF23J8p7c2IM5cmhwmoe6eCEaaAn8jY/edit#heading=h.ocovx14zokm4

It's not listed in "help", but [attachment] will embed the whatever image is attached to a post.

×
take a rule and offer it up to the hole in the washer so that it is just touching the edge of the hole.

measure the distance from the contact point to the outer edge of the washer, and call it h.

Then area = pi*h^2

1)  ______
2)  ______
3)  ______
4)  ______
5)  ______
6)  ______
7)  ______
8)  ______
9)  ______
10) ______

×1.61...

×11 x (value 7).

×But you chose the initial numbers. So how did I know?

× MAGIC

×But seriously, doing some simple arithmetic and multiplication reveals the secret.
Given numbers A and B at the start of a Fibonacci sequence, then you can solve easily for the seventh, ninth, and tenth number, as well as the sum of the ten numbers.
They are as follows, in terms of A and B:
7th#: 5A + 8B
9th#: 13A + 21B
10th#: 21A + 34B
sum: 55A + 88B

As can easily be seen, the sum is equal to 11*(7th#).

We then look at (10th#)/(9th#) = N, where N is some real number.
Then we get the following:
(21A + 34B)/(13A + 21B) = N
21A + 34B = N(13A + 21B)
21A + 34B = 13*N*A + 21*N*B

Solving for N in either case gives you N = 1.61.....

×You can see that the multiplicative terms in front of A and B follow the original Fibonacci sequence, which has any term equal to PHI*(previous term), PHI being 1.61.....

×An even harder challenge would be to determine if it is possible in 3 questions to determine who is who if you don't know what any words they say means.
If not possible in 3, what is the minimum needed?

×You know, let's just ignore the nature of these questions for the moment and tackle it from a more abstract point of view.

Let's ignore that we don't know what 'ya' and 'da' mean and assume there's some way for us to find out. Still, we only get three answers, all three of them being either y or d. This means there's a total of 2^3=8 possible outcomes to our questions.

Since we do need to find out this time excactly who's who, we will have to be able to tell the order in which we've asked the three guys, at least once we're done. Since there are three of them, there's 3!=6 possible orders.

So we have 8 possible outcomes, and we must be able to attribute each of them uniquely to one order. Since there's only 6 of them, this should be possible.

But wait! What about those cases where the guys choose randomly?

Let's abbreviate Past with <, Present with v and Future with > and consider the following three cases:

< v >
< > v
v < >

In case 2 and 3, one person will answer randomly, and in the first case even two. And, if I got it right, it's really a fifty-fifty-choice between y and d this time. This means that there must be two possible outcomes for case two and three and even four for case one. Which, in return, means that there are now, in some way, 11 cases.

Since we only have a set of 8 outcomes, though, it should be impossible to uniquely attribute one outcome to one order.

×You don't have to ask each person one question; you can ask the same person multiple questions.

×
Because if the last question is asked to the F, you get a random, unhelpful answer, we need to establish his position early and not ask him the third question. With trial and error, I came up with some questions to do that. Perhaps unelegantly, two of them require boolean logic operators. All directions are in my orientation, so if the sages are facing me, it is opposite for them.

The questions are*:
1. Does "ya" mean "no" XOR (are you OR the sage on the far left known as Current)?
2. Does "ya" mean "no" XOR are you known as Current?
3. Does "ya" mean "no" XOR (are you known as Current AND Past is standing somewhere to the left of Future)?

Ask the first question to the man on the left, the second to the man in the middle.

If they answer YY, possible outcomes are PCF and CPF. Ask the third question to the man on the left. Y=CPF, N=PCF.

If they answer YN, possible outcomes are PFC and CFP. Ask the third question to the man on the right*. Y=PFC, N=CFP.

If they answer NY, possible outcomes are PCF and FCP. Ask the third question to the man in the middle. Y=FCP, N=PCF.

If they answer NN, possible outcomes are PFC and FPC. Ask the third question to the man on the right. Y=PFC, N=FPC.

*edited slightly from original

×For the harder version, I believe it would take four questions, slightly modified of the ones above, but I have not yet worked on it.

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×They will answer your questions with "ya" or "da". If you knew specifically which one meant yes, your questions would work.

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×I can do it in 4 questions. They are:

1. (asked to left) Are you or the one on the left Current?
2. (asked to middle) Are you Past or Current?
3. (asked to right) Are you Current?
4. (asked to variable) Is Past somewhere to the left of Future?

After the first three questions (the literal answers followed by the cryptic answers, which is all we know; the first answer automatically becomes A):

PCF=YYY or NYY=AAA or ABB
PFC=YNY or NNY=ABA or AAB
CPF=YYY=AAA
CFP=YNY=ABA
FPC=NNY=AAB
FCP=NYY=ABB

If AAA, ask middle, A(Y)=PCF, B(N)=CPF

If ABB, ask middle, A(N)=FCP, B(Y)=PCF

If ABA, ask right, A(Y)=PFC, B(N)=CFP

If AAB, ask right, A(N)=FPC, B(Y)=PFC

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×To recap the solution of the previous version: The prisoners each give themselves a number. Prisoner #1 looks in box #1, then if that box contains prisoner #5's name, he looks in box #5, and so on. There must be a cycle of boxes leading to other boxes that eventually leads back to box #1. If the cycle is of length 50 or less, he will find his name in the required 50 tries and all other prisoners on the cycle will find their own names. So all prisoners succeed if the arrangement happens to contain no cycles of length greater than 50.

For this version, then, the first prisoner checks whether there is a cycle of length greater than 50 and if so (obviously there can only be one) he switches two boxes at maximum possible distance from each other in the cycle, which has the effect of splitting it into two cycles of half the length.

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.

×4, placed equidistant (so grown up!) beams from the center with increasing distances and radii.

Reasoning:

×Mastered:
King Dugan's Dungeon
The Choice
Journey to Rooted Hold
Perfection
Halph has a Bad Day
Beethro and the Secret Society
The City Beneath
Gunthro and the Epic Blunder

Working on:
Beethro's Teacher
Finding the First Truth
The Second Sky

I've never asked for help or hints. Yet.