Key words: Geometry, Unusual geometry, Math, Physics, Chemistry, High school, Geometry Olympiads, Free Geometry, Euclidean Geometry, Calculus, Geometric Construction. Oh yes, going-nowhere discussions, haikus, and poems too.
Saturday, March 26, 2011
Fox 336
Thursday, March 24, 2011
Fox 335
Monday, March 21, 2011
Fox 334
Monday, March 14, 2011
Fox 333
Red Fox: Happy Pi+1 Day!
Himalayan Fox: What is that exactly?
Red: Well, March 14 was the Pi Day, you know, 3.14, and today is March 15th.
Himalayan: But Pi+1 would be 4.14 which is April 14th, isn't it?
Red: Oh, I didn't think that way. Whatever it is, happy March 15th to you brother.
Himalayan: OK, I'll mechanically say "to you too", but can we really happy while thousands of souls swept away with the water?
Red: Well, we can not die with the dead.
Himalayan: But we can help the living.
Red: I hear you brother.
Himalayan: Then we can celebrate 2Pi Day a few months later.
Wednesday, March 9, 2011
Balanced Mancala Problem
We rarely ask pure math problems but here is one with a "very little" touch of geometry. This does not need to be defined as a mancala game, but here it goes:
We have stones coming in batches. Each stone has a color and a weight. If the color of a stone is:
Yellow: It must be placed every other pit (every 2 pits) Batch size is always: 6 Total batches: y Total number = 6y
Red: Must be placed every 3 pits Batch size: 4 Total batches: r Total number = 4r
Green: Must be placed every 4 pits Batch size: 3 Total batches: g Total numbers = 3g
Blue: Must be placed every 6 pits Batch size: 2 Total batches: b Total number = 2b
Purple: Must be placed only once in 12 pits Batch size: 1 Total batches: p Total numbers = p
So there are N=6y+4r+3g+2b+p many stones. The stones in the same batch have the same weight. Different batches may have different weights. WLOG, assume that all weights are integers. We have a proof that ending up with the best well-balanced mancala is very difficult (NP-Hard). Here "well-balanced" means that the pit with the maximum weight is minimized when all stones are distributed. Let's call this maximum pit weight as W.
Consider the following heuristic process:
Step 1. Sort the batches with respect to their weights (batches with the high-weight stones go first) Step 2. Insert the first batch starting from pit number 1. Step 3. Insert the next batch in a way that the total maximum weight throughout 12 pits remains minimum. Step 4. Repeat Step 3 until all batches are placed in the mancala. Let H be the maximum weight throughout 12 pits.
A simple Example: Suppose we have only 4 batches: Yellow (6 stones, each 45 grams) Blue (2 stones, each 40 grams) Yellow (6 stones, each 30 grams) Green (3 stones, each 20 grams) First batch (Yellow) goes to pits: 1, 3, 5, 7, 9, and 11. H=45. Second batch (Blue) goes to pits: 2 and 8. H=45. Third batch (Yellow) goes to pits: 2, 4, 6, 8, 10, and 12. H=70. Fourth batch (Green) goes to pits: 3, 7, and 11. H=70.
In this exercise, heuristic actually finds the optimum, i.e., H=W=70 grams, observed in pits 2 and 8.
And the question: Prove that the worst-case of the heuristic solution, H, is always less than 2W. In other words, W ≤ H ≤ 2W always holds. If you disagree, then try to generate a counter-example.
Good luck!