Growing Bacteria

jeetee · Post by **jeetee** » Tue Mar 24, 2009 7:10 pm

I think I don't get the description of the algorithm correctly.

When I calculate the first 8 days starting with one bacteria, I find that there are 48 bacterias after 8 days, not 47.
I wrote the algorithm in C++, but of course my answer isn't correct, which is to expect when I can't even work out the first 8 days manually...

Could anyone hint me out ?

gfoot · Post by **gfoot** » Tue Mar 24, 2009 10:19 pm

I agree with the 8-day answer of 47, but that's through reasoning rather than simulation. Maybe we interpreted the rules differently, but I can't see how.

bsguedes · Post by **bsguedes** » Tue Mar 24, 2009 10:27 pm

Each bacterium is able to generate ONLY two new bacteria, one in the next day to its birth and the second in the second day after is birth. Each one of the generated bacteria follow these rules too. It is also necessary to remove from population the dead bacteria (I think that jeetee has forgotten to kill the first bacterium to get 48 instead of 47 in the 8th day).

Cheers,
Bruno.

E2YnDlEMXiU · Post by **E2YnDlEMXiU** » Wed Mar 25, 2009 6:16 pm

even if your counts are slightly off for the 8 day example (mine was as well), simply execute your program and see when it reaches 10**12... you can easily figure out the answer then.

Bierdeckel · Post by **Bierdeckel** » Fri Mar 27, 2009 3:15 pm

Its realy simple.

You could finde the answer without any coding. Math is your friend

Even with coding its realy simple. My vb-code is just 13 lines long.

horst · Post by **horst** » Mon Apr 20, 2009 10:28 am

I get 51 bacteria for day 8

1 => 1
2 => 3
3 => 6
4 => 10
5 => 15
6 => 25
7 => 34
8 => 51

This challenge drives me crazy...

lukas · Post by **lukas** » Mon Apr 20, 2009 11:41 am

horst wrote:I get 51 bacteria for day 8

1 => 1
2 => 3
3 => 6
4 => 10
5 => 15
6 => 25
7 => 34
8 => 51

This challenge drives me crazy...

If one bacteria splits at day 1 you have two on day 2 and not three.

My Ruby Code was also just 8 lines long. Although a nice challenge.

horst · Post by **horst** » Mon Apr 20, 2009 1:55 pm

Ouch, thank you, I thought a bit too complicated. Got it now =)

homeas · Post by **homeas** » Wed Apr 29, 2009 2:04 pm

finally a challenge to relax,

why writing any code at all? - just use the windows (or pocket or xxx) calculator ...

bsguedes · Post by **bsguedes** » Wed Apr 29, 2009 2:54 pm

homeas wrote:finally a challenge to relax,

why writing any code at all? - just use the windows (or pocket or xxx) calculator ...

The main idea was to be an easy challenge =]

contagious · Post by **contagious** » Sun May 31, 2009 1:54 pm

fibonacci sequence... again

ellarso · Post by **ellarso** » Wed Jun 24, 2009 7:52 pm

contagious wrote:fibonacci sequence... again

Not quite! In a fibonacci sequence each number is the sum of the previous two.
0,1,1,2,3,5,8,13,21,34,55,89,...
Especially, the hint for day 8 (47) does not work!

gfoot · Post by **gfoot** » Thu Jun 25, 2009 2:09 am

It is related, but discussion of exactly how should probably go in the "solved" forum...

psycore · Post by **psycore** » Wed Jul 01, 2009 5:45 pm

Damn. I tried over weeks to create some java code for the bacteria cycle. I got the cycle of the new bacteria kind in code. Then I remembered the current topic of our maths lessons. Two minutes of calculator action. But hey, at least I learned a lot about coding java.

And to the guys wondering about the cycle. It says that there is that cycle for a bacteria and that scientist have found a new kind of that bacteria. But really, don't focus on the cycle so much.

Mosher · Post by **Mosher** » Fri Jan 22, 2010 9:59 pm

Found a rather simple solution via Access

Code excerpt as follows (don´t wonder about german words, i´m german ^^)

Code: Select all

x = 1
a = 1
b = 0
c = 0
d = 0
e = 0
...

s is the amount of bacteria after day x

a, b, c, d, and e are the amount of bacteria which are in stadium 1, 2, 3, 4, 5 accordig to challange´s description

a1, b1, c1, d1 are auxiliary variables to save the old data for preventing circle-referrence between variables