A little quiz ;-)

[RTADash]

In total, how many different color schemes are used throughout BD1 - BD2 (C64 version)? Don't forget amoeba caves!

[Dustin]

I come to a total of 22 different color schemes:

[Dustin]

1.Orange Grey1 White
2. LRed Purple W
3. Brown Orange White
4. Purple Orange W
5.LBlue Red W
6.Brown LRed W
7. LBlue LRed W
8.Brown LRed Green
9.Red LBlue W
10. Blue Grey3 W
11.Orange Purple W
12. Blue Orange W
13. Grey1 Orange W
14.Grey2 LRed W
15. Blue Orange Green
16. Grey2 Orange W
17. Orange LBlue W
The last five ones occur only in BD2:
18.Green Purple W
19.Green Orange W
20. Blue Grey2 W
21. Green Brown W
22.Blue LRed W

[RTADash]

Precisely - I dind't even realize that BD2E used a different gray than both int2's, but that is in fact correct!

The next question goes to you.

[Dustin]

A usual 40*22- cave made with the BD1 engine consists of the following items:
-in- and outbox (one each)
-one diamond
-one piece of magic wall
-fifteen pieces of normal brick wall
-the usual steelwall border
-one piece of amoeba.
The rest of the cave consists of nothing but dirt.
No other elements occur.

What is the maximum amount of diamonds you can get in such a cave?

EDIT: I modified the question a bit, since the original one was buggy.

[LogicDeLuxe]

EDIT: I modified the question a bit, since the original one was buggy.

I was just about to write the answer, which was of course 712 diamonds exactly. It's exploiting an overflow and completely predictable.
This is one reliable example:

I don't see anything buggy there.

[RTADash]

-one piece of amoeba.

I was able to get 172 using a fairly linear solution:

Edit: Stupid imageshack! - if anyone still wants to see my idea, i've reuploaded it here: http://www.file-upload.net/view-1170393 ... s.png.html

[BadDog73]

Heh, heh, heh! *rubs hands gleefully*

You can get 200. You have enough elements to reach from top to bottom of the cave minus one space. It turns out that if you are:

(please excuse the lack of picture)

AAA=
AAA*
AAAR
AAA_
====

A=amoeba, = is a wall or outbox, * is a diamond, R is Rockford, _ is dirt

You can move rockford down one space and quickly to the right, and the amoeba turns to diamonds...

So you get (amoeba max)-1 + the diamond you start with = 200.

[LogicDeLuxe]

If you get 200 boulders from the amoeba, you might be very lucky and able to push every one of them into the magic wall, which makes 201 diamonds in total. The odds that you get a suitable formation is slim, though, I think.
Due to the fact that amoebas can grow at more then one place in the last frame before converting to boulders, you might even get some more boulders which you could convert into diamonds with the magic wall. So theoretically you can get more than 201 diamonds in total.

[Dustin]

OK, I have to admit that I overestimated my knowledges about the BD1 engine a bit...
My original intention was as follows: There is one item too little to trap the amoeba into 199 cells. (Rory's solution was my mean trap ). So I wanted to let the diamond fall through the magic wall when the amoeba has grown to 199 cells, so that it converts into 199 diamonds.( That's why the BD1 engine was needed.) The diamond is lost then; there is no second magic wall to convert it back. So my intended solution was 199.

Even before I read Logic's second solution, I got aware that it's possible to let the amoeba convert into at least 200 boulders and then push them through the magic wall. If you use the brick walls and the diamond as a funnel and assume that the amoeba will grow exactly the way you want it to (the question was about the maximum amount, so such an assumption is allowed!), then it should be possible to push some few remaining boulders into the magic wall and get all the 201 or more diamonds. So it's your turn now again, Logic!

As for Logic's first 712-solution (which is related to the first version of my question, which was with "amoeba and dirt as much as you like" instead of "one piece of amoeba and the rest is dirt"): I admit that I don't understand it at all. I should have known that a living BD encyclopedia as you are would unpack a solution like this But please help me to understand it! Is the magic wall necessary for this solution? Why mustn't you take the diamond immediately and give the amoeba 712 cells? Why doesn't the amoeba actually convert into boulder immediately? Why notsett the brick walls at the very edge and give the amoeba even more space? Where does the limit of 711 cells for the amoeba come from????

As for BadDog's solution: I did not know that the BD1 amoeba could be framed like this, either.Nice!



OK, Logic, your question!

[LogicDeLuxe]

So I wanted to let the diamond fall through the magic wall when the amoeba has grown to 199 cells, so that it converts into 199 diamonds.

You did realize that you can not convert the amoeba by dropping a diamond into the magic wall, didn't you? The breakscan behavior only appears when a boulder is dropped into the amoeba.

Where does the limit of 711 cells for the amoeba come from????

That's easy. As I said, it's exploiting an overflow.
During each scan, every amoeba piece in the cave is counted and at the end checked against 200, which is $c8 in a hex byte. There is only one byte for this counter, so 255 ($ff) is the maximum countable amount, which would be turned to boulders immediately. 256 is $100 in hex, but it is truncated to $00, so the engine assumes that the amoeba is still small enough to grow further. for the magical 712, it is $2c8 in hexadecimal which is truncated to the well known $c8, ie. 200 pieces. Thus you can grow an amoeba which starts at least with 512 pieces up to 711 and will convert to boulders with the 712th piece.
To prevent this problem in PLCK, the original PLCK didn't allow you to place more than 200 pieces of amoeba. Though you could do this in hacked versions with this limit removed.
1stB and later engines increased this counter to 2 bytes, so this overflow won't happen, but you could set the maximum amoeba size to 712 pieces instead in those engines.


And now, the new question:
What's wrong in this picture?

[RTADash]

The eyes (butterflies) are in the wrong place at 105 seconds left?

[Dustin]

Another suggestion: In the "space" graphics, the background is full of stars, which are missing here (in the one cell of background that can be seen).

[RTADash]

Or maybe he removed an eye and forgot to replace the stars in the background. (If that was the case, would we both get to ask the next question?)

[LogicDeLuxe]

Another suggestion: In the "space" graphics, the background is full of stars, which are missing here (in the one cell of background that can be seen).

Exactly. You may ask the next question.