Author Topic: TI - pisomka  (Read 2660 times)

bodik

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TI - pisomka
« on: 28.11.2006, 23:09:41 »
caute. uz ste niekto pisali zapoctovku z TI s plocicom? alebo aspon nemate nejake materialy na to? na hornade tam toho vela nie je :(

Johny Bravo

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Re: TI - pisomka
« Reply #1 on: 29.11.2006, 00:22:02 »

Ing. nemtom

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Re: TI - pisomka
« Reply #2 on: 29.11.2006, 00:52:08 »
thx man
brix will be shat

Shiro

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Re: TI - pisomka
« Reply #3 on: 29.11.2006, 13:54:24 »
jj, thx - aspon sa budem mat z coho ucit

woody

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Re: TI - pisomka
« Reply #4 on: 29.11.2006, 20:13:42 »
taakze u plocicu ...
1. nakresleny mealy treba transformovat do moora.
2. nekresleny automat treba redukovat
3. nakresleny treba determinovat redukovat nemusi
4. urobit zasobnikovy automat pre L={0^n 1^m 0^n | n,m>0}

vcelku lahke :)

VTI

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Re: TI - pisomka
« Reply #5 on: 30.11.2006, 04:03:19 »
naaahodou by sa nenasla dobra dusa co by tu hodila riesenie 4ky?

teda urobit zasobnikovy automat pre L={0^n 1^m 0^n | n,m>0} a druha skupina mala urobit zasobnikovy automat pre L={a^m b^n c^n | n,m>0}

mrtvyhubar

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Re: TI - pisomka
« Reply #6 on: 04.12.2006, 17:24:26 »
No ja by som tu 4ku riesil nejak takto,ale len boh a Hudak vedia ci to je dobre.
(q0,0,z,q0,0z)
(q0,0,0,q0,00)
(q0,1,0,q1,0)
(q1,1,0,q1,0)
(q1,0,0,q1,lambda)
(q1,lambda,z,qf,z)
Viditeľné škody na nábytku sú ťažko opraviteľné a kazia celkový vzhľad
bytu. Myslite na to pred tým než začnete búchať pesťou po stole kôli nefunkčnému zadaniu.
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mrkvy

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Re: TI - pisomka
« Reply #7 on: 07.12.2006, 06:41:28 »
Sa mi zjavil sef z nebies a povedal, ze toten Tvoj priklad je pre L={0^n 1^m 0^n | n,m>0} spravne.. ;) Inak super avatar...  :emot-drool:
Edit: Ale ak Ti pride retazec typu 000111111110010 thaq Ti to zoberie za spravny (v piatom riadku by si mal mat pouzite aj q2, potom by to nerobilo).. Otazne je ci taky retazec moze prist na vstup...  :emot-drunk:
« Last Edit: 07.12.2006, 06:46:08 by mrkvy »
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mrkvy :zuzka: svoj vodicak :D