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February 09, 2012, 07:39:46 AM

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Author Topic: UDa ada patch resmi dari EA untuk fix story progression bug  (Read 938 times)

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SweetyTopic starter

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Complete list of changes:

    * Addresses some issues with Speed 2 and 3 moving too slowly. Some machines will have better results when using Speed 2 and 3 now.
    * Fix to story progression on/off selection toggle.
    * Fix for a possible crash with audio code.
    * Fix for babysitter routing off lots with babies.
    * Addresses some issues with Vsync and refresh rate problems.
    * Addresses some issues with DVD authentication errors and drive compatibility on startup.
    * This update deals with some issues on Mac systems that can crash the game while connecting to AFP servers.
    * Fix for a freeze that can occur when Sims attempt to clean out bad food from the fridge.

Source:
http://www.thesims3.com/game/patches/-1206416818
=======================================

tapiii, masalahnya, game kita kan bajakaaaan  :hua, sedngkan itu diupdatenya secara online.

Coba teman-teman katakan saya harus bagaimana supaya patchnya bisa dipake di game saya :shake

chucky_pakorem

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aduh punya saya juga bajakan lagi .. :Genit:
LALALALA~

treeag

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Download:

US ENGLISH
Worldwide

Kalau dulu pernah install patch 1.0.632, harus install ulang game nya :perhatian:



siapa mau coba pertama? :ups
« Last Edit: June 26, 2009, 09:17:32 PM by treeag »

caramelz

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  • Chammone! Who's BAD?!
Eh ud ada patchna? Ayo dx siapa mau coba, trus kasih tau hasilnya ke aku juga. Kan aku belum beli ts3 nih. Kalo berhasil aku mau beli nih..
I was born to never die
to live in bliss, to never cry

opshd

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emang udah ada cracknya yah ? :wew

Hey Mister
I am Mad Scientist
It's so cool!
Sonuvabich

gilang

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Gw ga bisa nginstall patchnya, soalnya versi gw pake Reloaded To Razor dimana Sims3GDF.dll nya ngga valid.. Ada yg bisa bantuin uplotin yg versi retailnya??
I don't have signature. But I admit that I actually have one. The sentence I wrote before was my signature. And so the sentence left to this one.

So we can conclude that this signature fulfill the mathematical induction:
Let f(n) be a function which returns string x that is a signature with length n sentences. We want to prove that f(n) is my signature with n sentences for every n >= 1
(1) Base case: f(1) = my signature with 1 sentence
(2) Hypothesis: f(n) is my signature
(3) Inductive case: f(n+1) = f(n)f(1). It is shown that f(n + 1) is a concatenation of two string x which both of it is a signature. So according to concatenation rule of regular expression, if the a and b is concatenated and both a and b have the same property, so ab must derives the same property of a and/or b.

Another proof:
  f(n+1) = f(n)f(1) = f(n-1)f(1)f(1) = f(n-2)f(1)f(1)f(1) = f(1)...f(1)f(1) = f(1)^(n+1)
It is shown that f(n+1) is a concatenation of any singleton string which is my signature. So because all of the n+1 strings are my signatures, so the concatenation of all n+i strings must be my signature.

So it is proven using mathematical induction that this is my signature.

Panjang ya? :hahak

SweetyTopic starter

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bedanya versi reloaded ama razor apa yaakk?? :wew:

Punya aku versinya 1.0.615.00107 . Aku blom pernah install patch apa2 sih.. masihh murni. Tapi aku uda punya hack satu.. masalah ga ya?

gilang

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Itu versi Reloaded.. Versi yg katanya Pre-release. Yg versi release serinya 1.0.631 dan itu dikeluarin oleh Razor

Yg versi Reloaded kaga bisa di-patch
I don't have signature. But I admit that I actually have one. The sentence I wrote before was my signature. And so the sentence left to this one.

So we can conclude that this signature fulfill the mathematical induction:
Let f(n) be a function which returns string x that is a signature with length n sentences. We want to prove that f(n) is my signature with n sentences for every n >= 1
(1) Base case: f(1) = my signature with 1 sentence
(2) Hypothesis: f(n) is my signature
(3) Inductive case: f(n+1) = f(n)f(1). It is shown that f(n + 1) is a concatenation of two string x which both of it is a signature. So according to concatenation rule of regular expression, if the a and b is concatenated and both a and b have the same property, so ab must derives the same property of a and/or b.

Another proof:
  f(n+1) = f(n)f(1) = f(n-1)f(1)f(1) = f(n-2)f(1)f(1)f(1) = f(1)...f(1)f(1) = f(1)^(n+1)
It is shown that f(n+1) is a concatenation of any singleton string which is my signature. So because all of the n+1 strings are my signatures, so the concatenation of all n+i strings must be my signature.

So it is proven using mathematical induction that this is my signature.

Panjang ya? :hahak

SweetyTopic starter

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jadi aku nga bisa pk patch yang dikasih tree??

gilang

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Sayang sekali tidak. Elu harus patch dari Reloaded to Razor dulu. Di IDWS ada. Tapi gedenya sampe 700 MB
I don't have signature. But I admit that I actually have one. The sentence I wrote before was my signature. And so the sentence left to this one.

So we can conclude that this signature fulfill the mathematical induction:
Let f(n) be a function which returns string x that is a signature with length n sentences. We want to prove that f(n) is my signature with n sentences for every n >= 1
(1) Base case: f(1) = my signature with 1 sentence
(2) Hypothesis: f(n) is my signature
(3) Inductive case: f(n+1) = f(n)f(1). It is shown that f(n + 1) is a concatenation of two string x which both of it is a signature. So according to concatenation rule of regular expression, if the a and b is concatenated and both a and b have the same property, so ab must derives the same property of a and/or b.

Another proof:
  f(n+1) = f(n)f(1) = f(n-1)f(1)f(1) = f(n-2)f(1)f(1)f(1) = f(1)...f(1)f(1) = f(1)^(n+1)
It is shown that f(n+1) is a concatenation of any singleton string which is my signature. So because all of the n+1 strings are my signatures, so the concatenation of all n+i strings must be my signature.

So it is proven using mathematical induction that this is my signature.

Panjang ya? :hahak