Q&A Discuss: Stacking Boxes in A pyramid

[CD47-Bot]

Thread created automatically to discuss a question in the Q&A Forum.

Stacking Boxes in A pyramid

If you stack the boxes in a pyramid shape, 2 on bottom and one centered on top does it count as a stack of three because it would be taller than two? Or does it not count as a stack?

[RebAl]

Using that logic any boxes that touch a stack dont count and if all boxes are touching each other you dont get any points.

[AlbertW]

the word from jason morrella in the Q&A session after the san jose kickoff was that, if a pyramid stack was the highest one, then that stack's height would be the multiplier, and all bins supporting it (in the pile/pyramid) are not counted as points.

[Lee]

Forget, for the moment, the neatly stacked pyramid. Consider if a super robot tossed all 45 of the boxes into a pile within their scoring zone, with the resulting height of 6 box units. There is no practical way to determine which, if any, or all of the boxes form the stack.

[Greg Ross]

Originally posted by Lee
There is no practical way to determine which, if any, or all of the boxes form the stack.

Sounds like Jack Straws. Just start picking up bins at the outside edge until the top bin shifts.

[Madison]

Originally posted by gwross
Sounds like Jack Straws. Just start picking up bins at the outside edge until the top bin shifts.

But once the top bin shifts once, it may change how other bins do or do not support the stack.

If you should happen to take away one bin, and that bin happens to be the one that makes the top bin shift, you are ignoring the potential for others to exist that wouldn't cause that to happen.

Right? Or, maybe I'm misreading that.

[dlavery]

Originally posted by Lee
Forget, for the moment, the neatly stacked pyramid. Consider if a super robot tossed all 45 of the boxes into a pile within their scoring zone, with the resulting height of 6 box units. There is no practical way to determine which, if any, or all of the boxes form the stack.

Read Team Update #1. The clarification of Rule SC8 and Rule SC9 make it clear how this would be scored [(45-6)*6]

-dave

[Greg Ross]

Originally posted by M. Krass
But once the top bin shifts once, it may change how other bins do or do not support the stack.

If you should happen to take away one bin, and that bin happens to be the one that makes the top bin shift, you are ignoring the potential for others to exist that wouldn't cause that to happen.

Right? Or, maybe I'm misreading that.

I realize the question is now moot, but to answer your challenge: Just as in Jack Straws (Pick-Up-Sticks) you pick up the easy ones first. In this game, the judges would pick up the bins which obviously are not supporting the stack. Then they would evaluate the remaining bins, and remove the ones that in their best judgment are least likely to be supporting the stack. True, the judges could make a mistake, and attempt to move a bin which causes the pile to shift, when there were other bins that could have been removed with impunity. But remember -- the game isn't fair.

[petek]

Originally posted by dlavery
Read Team Update #1. The clarification of Rule SC8 and Rule SC9 make it clear how this would be scored [(45-6)*6]

-dave

I agree. Check Team Update 3 on Rule SC8:
"The height of the tallest stack located in the scoring zone (the “multiplier stack”), measured in whole Stack Height Units (as defined in SC9) is subtracted from the total number of containers to establish the “base score.” Containers in additional stacks of the same height will be scored normally;"

I interpret that to say it doesn't matter how many containers (or opposing robots) support the top container of a stack, it's "The height", not the number of containers, which determines the multiple and the number subtracted from the total container count.

Pete

[jskene]

Suppose there are 7 containers in a scoring zone, arranged as illustrated below, that is a 3-container high stack made up of a pyramid containing 5 containers, and 2 other containers in the zone:

0
0 0
0 0 0 0

Section 7.6 says no containers in the stack count as part of the base score, so the base score is 2, and the final score is 3 x 2 = 6

Rule SC8 says take the total number of containers (7), subtract the height of the stack (3) to get the base score. So the base score = 7-3 = 4. Then multiply this by the stack height. Score is 4 x 3 = 12.

Clearly one of these scores is incorrect.

I suspect that it is FIRST's intention that rule SC8 prevails, but it is confusing to have contradictory scoring information also in section 7.6

A prettier picture is shown at:

http://www.skene.org/robots/Final-Configuration.jpg