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#336245 brain puzzle
Posted by Anza Power on 14 October 2013  07:40 PM
#335077 Trianglia
Posted by Anza Power on 09 August 2013  06:58 PM
node: the node you are in (one of v_{0}...v_{n1})
from: which of the 3 edges you came from (0 1 or 2)
direction: weather your next move is to the left or right (0 or 1)
So if there were n nodes then there are 6*n possible states s_{0}...s_{6n1}.
Now the proof is trivial because each state has exactly 1 outgoing edge (because the move is deterministic) and exactly 1 incoming edge because given a state you can construct the previous state from it (it's too wordy to explain but just draw some state and see if you can figure out the 3 parameters of the previous state)
So you have a graph where each node has indegree=1 and outdegree=1, if you start at some node and go forward ofcourse you're eventually gonna end up back at the original node
 1
#332885 Probability of a rope cut exactly in half, in a single slice
Posted by Anza Power on 09 June 2013  07:13 PM
The who gave the answer 1/3 had the logic that there are only three possibilities of this experiment after the cut.
1. Left peice is bigger
2. right peice is bigger
3. left peice = right peice
so the probability he calculated is 1/3.
Ok, so by that logic if I go and fill out the lottery, there are 2 possibilities, either I win or I lose, so probability of me winning is 1/2...
But yeah as kingofpain said, the probability is 0, you can think of it like this: the rope is 1 meter long and you pick a number between 0 and 1 and slice there, now, how many numbers are there between 0 and 1?
(even if you think of the rope in a 3D space and give it a diameter the answer doesn't change)
 2
#326668 Amoeba evacuation puzzle
Posted by Anza Power on 07 December 2012  06:53 PM
Good and quick solve, it's not possible to evacuate the area.
Yeah I think that is correct, here's why:
Divide the board into diagonal lines like so:
Now let's define c{i} to be the number of amoebas at diagonal i, let's give weights to the diagonals w{i}=0.5^{i1}, so the weight of the first diagonal is 1 and the weight of every diagonal after that is half the weight of diagonal before it...
Now define board_sum to be the sum of the number of amoebas on each diagonal times the weight of that diagonal, boardsum_ = sum(c{i}w{i})...
At the beginning we have board_sum=1, whenever we make a move we split an amoaba at diagonal i into two amoebas at diagonal i+1, since the weight of diagonal i+1 is half that of diagonal i we can see that the board_sum has not changed...
So the board_sum is always constantly 1.
Now imagine the case where there is an infinite number of amoebas one in every cell, the board_sum in that case is:
(the convergence was calculated using the formula for the mean of a geometric distribution with parameter 0.5)
Now take away the amoebas at the cells we want to clear, their weight is:
1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.25 + 0.125 + 0.125 = 3.
So the board_sum of the board when there is an infinite number of amoebas filling every cell except the ones that we need to clear is 1, which means that if the number of amoebas was finite then the board_sum would be less than 1 which is impossible...
 1
#326667 The traveling ant
Posted by Anza Power on 07 December 2012  06:38 PM
Spoiler for
the answer is yes. Andy the ant is placed on the elastic band at the point of 1/100 an hour, or after 36 seconds.
as noted by rainman, (though i don't agree with his result) if the ant stands still the relative distance to the truck will remain the same.
so, after the next 36 seconds, the elastic band will double, and andy will be at slightly more than 1/10000 of the way. after the next 36 seconds, the elastic band will be 3/2 larger, putting andy at roughly 3/20000 +1/10000 = 5/20000. after the next 36 seconds, the elastic band will be 4/3 more, putting andy at roughly 20/60000 +1/10000 = 26/60000.
more precisely, we have... e^(0.01*x) 1 = 100; or that andy will reach the truck after 461.5 hours roughly.
I think your first sentence is wrong
You can imagine splitting the band to N points and marking them, now if the ant stood between any two points when the band stretches it will always be between those two points
But you can put it another way:
So if we take an N that is big enough so that v_{0}/N is less than the ant's walking speed then the ant could get from mark k to mark k+1 in finite time, and since there is a finite number of points it will get to the last point in finite time as well,,,
 1
#326633 Amoeba evacuation puzzle
Posted by Anza Power on 01 December 2012  01:27 PM
http://anzapower.web...moeba/game.html
It's not as easy as I expected...
 1
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