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m994301009
(User) | Level 3 - 2008-05-27 22:13:08 |
I just can't seem to get level 3, I have been writing small programs to try to do it, but I can't seem to get it. I have tried:
1522 (which seems way to small)
1236950581248 (Which was my best answer)
and a whole bunch of other ones that I know were wrong because the was a few simple coding mistakes.
Please clarify: Does the amount of sides quadruple each time? Cause that is what I think the question is saying.
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thinkt4nk
(User) | Re: Level 3 - 2008-05-27 22:44:37 |
| look at the examples. DO the sides quadruple each time? |
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m994301009
(User) | Re: Level 3 - 2008-05-27 23:57:20 |
| All the example links are dead... |
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BourneJ
(User) | Re: Level 3 - 2008-05-28 10:28:30 |
| The links are dead, but I found a similar one. http://math.rice.edu/~lanius/frac/koch.html |
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m994301009
(User) | Re: Level 3 - 2008-05-30 02:20:14 |
| ok, so the amount of sides does increase four fold each time, i'm still going wrong somewhere :( |
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sajchurchey
(User) | Re: Level 3 - 2008-05-30 04:41:02 |
temporarily fixed using wayback machine.
http://web.archive.org/web/20070218122329/http://www.saltspring.com/brochmann/math/Fractals/fractal-1/fractal-1.00.html |
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BourneJ
(User) | Re: Level 3 - 2008-05-30 10:35:01 |
| Maybe you could try & see what the relation is between the perimeter every time, rather than just the number of sides. I think thats where you're going wrong. Just my 2 cents. |
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m994301009
(User) | Re: Level 3 - 2008-06-01 00:51:35 |
Well, what i'm trying to do is this,
Sides increase by 4 times each time, and side length becomes 1/3 of the original each, so if I just simulate this pattern, then multiply the 2 values together, than that should give me the answer shouldn't it? |
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m994301009
(User) | Re: Level 3 - 2008-06-01 00:53:28 |
| I see my problem now, each side is 1/6th of the original length, not 1/3rd. -_- |
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VforVendetta
(User) | Re: Level 3 - 2008-06-05 14:55:38 |
well, you can try this,
total perimeter=4*number of sides*length of each side
and create a loop to know the number of sides and length of each side in the 93rd minute and put those in the formula
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CodeX
(User) | Re: Level 3 - 2008-06-05 17:03:16 |
What this challange is based on is a koch snowflake, so you can look that up on wikipedia to see whats going on a bit better. Once you can see whats going on you may want to jott down a few steps of the snowflake, like (a=.5,p=3,l=1,n=0,v=3) so you have the area*, perimeter, length of a vertex, growth number and number of vertices. once you do two or 3 (might help to draw them too) you'll see that there is a pattern with v and l, as well as p which is worked out from v * l.
sorry if this was an overkill answer, but if not and you'd like some more help, repost |
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CodeX
(User) | Re: Level 3 - 2008-06-05 17:35:49 |
| forgot to mention the * next to area was to point out its not needed, if you were to work out area you'd a load of iteration so you could make a snowflake object in your oo language of choice, and then have it do steps, but programming it can result in missing accurecy and so you may want to just avoid working out the area (which won't go over a0*(8/5)). The best tool once you have figured out the pattern is to use a precision calculator like speedcrunch or http://ttmath.slimaczek.pl/online_calculator |
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m994301009
(User) | Re: Level 3 - 2008-06-09 05:10:09 |
| thanks for the advice, i'm going to start working on it now, its paper wasting time! |
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th3fr0g
(User) | Re: Level 3 - 2008-06-09 14:04:17 |
| what type of variables are you using. maybe your losing digits somewhere |
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alice_hoyden
(User) | Re: Level 3 - 2008-09-11 04:18:33 |
| Just use a for loop with minutes. Number of sides quadruples, length of sides gets divided by 3 with each iteration. Perimeter gets calculated after the loop. Make sure your variables are of a type that can hold the horrendous number you get. |
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utsab
(User) | Re: Level 3 - 2008-10-23 12:44:45 |
| 3*(4/3)^i [i is the no. of iteration ]is the ans i figured out...but dcant find the numeric soln |
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utsab
(User) | Re: Level 3 - 2008-10-23 12:53:59 |
| got it :) |
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coolmaddy
(User) | Re: Level 3 - 2009-01-08 22:28:03 |
from what i understud 3*((4/3)^k)
where k is the no of loops=no of minutes...
is this some problem with the approach |
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coolmaddy
(User) | Re: Level 3 - 2009-01-08 23:04:34 |
| i acted lyk a fool...taking 1hr 33 mins to be 99 mins rather than 93...sometimes i act lyk a idiot...sry guys |
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DKMFan
(User) | Re: Level 3 - 2009-01-27 17:41:36 |
| Ah, I remember solving Level 3. Damn that was a good day. |
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dokyriak
(User) | Re: Level 3 - 2011-03-29 03:32:34 |
| WTF. I thought it was 99 too!!! |
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