CentOS 8.0.1905 network installs on bare metal via External Repos succeeded

At the moment URLs  below worked for me as installation repositories, Hopefully this repositories will stay stable in a foreseeable future.  Particular source might be verified  when creating CentOS 8 virtual machine on any KVM Hypervisor, say on Ubuntu 18.05 Server . Even on UEFI box with high speed Internet connection network install runs faster then "dd" copying 7 GB file to USB pen and updating boot devices priority via UEFI interface.

http://mirror.corbina.net/pub/Linux/centos/8.0.1905/BaseOS/x86_64/os/
http://centos-mirror.rbc.ru/pub/centos/8.0.1905/BaseOS/x86_64/os/

At this point installer should be ready to start downloading packages  from the URL
pointed above after metatada gets obtained and the button "Begin installation" will become active on main panel. Go ahead and press it. Installer is supposed to download packages from repositories, printed above, preparing transaction from obtained source and run the transaction installing CentOS 8.0.1905 on bare metal. 

    

  Run time snapshots with Qemu-kvm VM running on KVM-Hypervisor  (CentOS 8.0.1905 based) .

    

 

LibreOffice install :-

Download LibreOffice_6.3.1_Linux_x86-64_rpm.tar.gz
 wget  http://download.documentfoundation.org/libreoffice/stable/6.3.1/rpm/x86_64/LibreOffice_6.3.1_Linux_x86-64_rpm.tar.gz
Set up as root
# tar -xvf LibreOffice_6.3.1_Linux_x86-64_rpm.tar.gz
# cd LibreOffice_6.3.1.2_Linux_x86-64_rpm
# yum localinstall RPMS/*.rpm

Predicates and technique 08.2016 for solution of systems of Boolean equations vs Classic approach (2013)

Consider a system somewhat more complex than 



######################
UPDATE as of 06/10/2109
########################
  I do have to notice that solving System 5 from http://kpolyakov.spb.ru/download/mea-2016-8.pdf  silently does what described down here without focusing attention on predicates (x3≡x4) and x3^x4  truth and false sets intersections. Actually four sets were obtained and theirs powers have been used to solve System 5. I sincerely apologize for missing this doing original post
END UPDATE
##########################
UPDATE as of 03/10/2019
##########################
See complete version of "update" here
Setting up a cross-reference table in 08/2016 approach of Helen Mironchick when moving to a new line of the system of Boolean equations  
The key place is a detailed description of building a cross-reference table 
when moving to a new line of the system. The chart generation is just a consequence.
First consider system
   (1) F1(x1,y1,z1)=>F2(x2,y2,z2) =1
   (2) F1(x2,y2,z2)=>F2(x3,y3,z3) =1
   (3) F1(x3,y3,z3)=>F2(x4,y4,z4) =1
   (4) F1(x4,y4,z4)=>F2(x5,y5,z5) =1
where F1 and F2 are triple predicates
Denote card(N) the power of set N
Denote n1,n2,m1,m2,s1,s2
    n1=card (falseSet_F2 ∩ falseSet_F1)
    n2=card (falseSet_F2 ∩ truthSet_F1)
    m1=card (truthSet_F2 ∩ falseSet_F1)
    m2=card (truthSet_F2 ∩ truthSet_F1)
    s1=card (falseSet_F1)
    s2=card (truthSet_F1)
Then following 08.2016 diagram takes place

###############
END UPDATE
###############
   ************
   System (1)
   ************
   (((x1=>y1)=>z1)⊕((z1=>y1)=>x1))=>((x2≡y2)≡z2)=1
   (((x2=>y2)=>z2)⊕((z2=>y2)=>x2)) v ((x3≡y3)≡z3)=1
   (((x3=>y3)=>z3)⊕((z3=>y3)=>x3))=>((x4≡y4)≡z4)=1
   (((x4=>y4)=>z4)⊕((z4=>y4)=>x4)) v ((x5≡y5)≡z5)=1

   Basic chart
   

Start 08.2016 run

    

  Polyakov's Control

   

In this case, the predicate G (x, y, z) = (((x => y) => z) ⊕ ((z => y) => x)) is given by a slightly more complicated function than in the previous post. In fact, any triple predicate F (x, y, z) with the power of the truth region “2” defined on 8 three-bit chains will give exactly the same number of solutions.
************
System (2)
************
Consider following system
(((x1=>y1)=>z1)⊕((z1=>y1)=>x1))=>((x2≡y2)≡z2)=1
(((x2=>y2)=>z2)⊕((z2=>y2)=>x2))=>((x3≡y3)≡z3)=1
(((x3=>y3)=>z3)⊕((z3=>y3)=>x3))=>((x4≡y4)=z4)=1
(((x4=>y4)=>z4)⊕((z4=>y4)=>x4))=>((x5≡y5)≡z5)=1
(((x5=>y5)=>z5)⊕((z5=>y5)=>x5))=>((x1≡y1)≡z1)=1
Perform two runs. First for system
(((x1=>y1)=>z1)⊕((z1=>y1)=>x1))=>((x2≡y2)≡z2)=1
(((x2=>y2)=>z2)⊕((z2=>y2)=>x2))=>((x3≡y3)≡z3)=1
(((x3=>y3)=>z3)⊕((z3=>y3)=>x3))=>((x4≡y4)≡z4)=1
(((x4=>y4)=>z4)⊕((z4=>y4)=>x4))=>((x5≡y5)≡z5)=1
(((x5=>y5)=>z5)⊕((z5=>y5)=>x5))=>((x1≡y1)≡z1)=0
Starting values for G(x1,y1,z1) are defined by
false triples ((x1≡y1)≡z1)
For 000  G(0,0,0) =0

For 101  G(1,0,1) =0

For 011  G(0,1,1) =1

For 110  G(1,1,0) =1
Thus  G(x1,y1,z1) starts  with 2/2
The result of first run is value G(x5,y5,z5) on line "1". It defines the number 
of false solutions - 1952 , which should be deducted from number of solutions of second system.
Second run :-

(((x1=>y1)=>z1)⊕((z1=>y1)=>x1))=>((x2≡y2)≡z2)=1

(((x2=>y2)=>z2)⊕((z2=>y2)=>x2))=>((x3≡y3)≡z3)=1
(((x3=>y3)=>z3)⊕((z3=>y3)=>x3))=>((x4≡y4)≡z4)=1
(((x4=>y4)=>z4)⊕((z4=>y4)=>x4))=>((x5≡y5)≡z5)=1

    Polyakov's Control
    

Предикаты и техника решения 08.2016 систем уравнений в булевых переменных

Рассмотрим систему несколько более сложную чем в 

    Например

    (((x1=>x2)=>x3)⊕((x1≡x2)≡x3))=>(x4 v x5 v x6)=1
    (((x4=>x5)=>x6)⊕((x4≡x5)≡x6))=>(x7 v x8 v x9)=1
    (((x7=>x8)=>x9)⊕((x7≡x8)≡x9))=>(x1 v x2 v x3)=1

    Решение в стиле 08.2016 
    

    Контроль по Полякову

    

В даном случае предикат F(x,y,z) =  (((x=>y)=>z)⊕((x≡y)≡z))  задан несколько более сложной функцией, чем в предыдущем посте.  В действительности, любой трехместный предикат G(x,y,z) с мощностью области истинности "3" определенный на 8 трех-битовых цепочках даст ровно тоже количество решений. Таким образом, все определяется мощностью множества истинности предиката стартующего имлликацию

Другой пример

((x1=>y1)=>z1)=>((x2≡y2)≡z2)=1

((x2=>y2)=>z2)=>((x3≡y3)≡z3)=1

((x3=>y3)=>z3)=>((x4≡y4)≡z4)=1

((x4=>y4)=>z4)=>((x5≡y5)≡z5)=1

Контроль по Полякову

Решение одной известной задачи с форума ЕГЭ Информатика в формате 08.2016

Исходная формулровка в группе Елены Мирончик
https://vk.com/club180658320?w=wall-180658320_64%2Fall

   Ссылка выше на мое первое решение по классике и решение Елены Александровны в новом стиле ( достаточно сложное для понимание ). Ниже следует решение той же задачи в формате 08.2016

 

Solution of one 23rd problem in 08/2016 format kind of Practice test №1 as of 02.09.2019

Original system

(x1=>x2)^((z1⊕y1) =>x1)=1
(x2=>x3)^((z2⊕y2) =>x2)=1
(x3=>x4)^((z3⊕z3) =>x3)=1
(x4=>x5)^((z4⊕y4) =>x4)=1
(x5=>y5)=>z5 =1

   Needless to say that original task 23 from mentioned Practice test  №1 as of 02.09.2019
    "190902_Informatika_-_Probny_Variant_1_S_Resheniem.pdf" may be solved via 08/2016 format in a couple of minutes

        

Solution of 23rd problem in 08/2016 format from VK's Informatics_100 News Wire as of 14/09/2019

Original code

Consider a bit more complicated system like below.
I believe it's understandable that handle two variables
case is significantly easier.

((x1=>y1)=>z1) =>((x2≡y2)≡z2)=1
((x2=>y2)=>z2) =>((x3≡y3)≡z3)=1
((x3=>y3)=>z3) =>((x4≡y4)≡z4)=1
((x4=>y4)=>z4) =>((x5≡y5)≡z5)=1
((x5=>y5)=>z5) =>((x6≡y6)≡z6)=1

Fork 08/2016 diagram

Polyakov's control

Solution Another 23rd problem in 08/2016 format from VK's Informatics_100 News Wire as of 12/09/2019

    https://vk.com/informatics_100?z=photo-40390768_457275192%2Fwall-40390768_196117
  
   (x1^y1)=>(¬x2 v ¬y2) ≡  ¬(x1^y1) v ¬(x2^y2) = ¬((x1^y1)^(x2^y2))
   ¬((x1^y1)^(x2^y2)) =1 ≡ (x1^y1)^(x2^y2)=0

   

   In particular case 08/2016 schema appears to be much easier
   then analysis of truth tables and arrows (pair) diagram

   

  Original system is equivalent.

   (x1^y1)^(x2^y2) =0
   (x2^y2)^(x3^y3) =0
   (x3^y3)^(x4^y4) =0
   (x4^y4)^(x5^y5) =0
   x2^y4=0
   Fork two 08/2016 diagrams ( free your mind )
     

Solution of 23rd problem in 08/2016 format from VK's Informatics_100 News Wire as of 12/09/2019

    Convert system to equivalent
   
    (x1 =>¬y1) => (x2 ≡ ¬y2) =1  ≡   (¬x1 v ¬y1) => (x2⊕y2) =1
    (¬x1 v ¬y1) => (x2⊕y2) =1 ≡  ¬(x1^y1) => (x2⊕y2) =1
   ¬(x1^y1) => (x2⊕y2) =1 ≡  (x1^y1) v (x2⊕y2)=1

    Conversion is done
    New system looks like

    (x1^y1) v (x2⊕y2)=1
    (x2^y2) v (x3⊕y3)=1
    (x3^y3) v (x4⊕y4)=1
    (x4^y4) v (x5⊕y5)=1
    (x5^y5) v (x6⊕y6)=1

    The core block to replicate per 08/2016 technique
    

  Now fork 08/2016 diagram

   

Solution of one system of Boolean equations based on Helen Mironchick's approach for treatment logical product of bit triples starting implication as of 12/09/2019

Original system

(x1^y1^z1)⊕(x3^y3^z3) =>((x2=>y2)=>z2)^((x2≡y2)≡z2)=1
(x2^y2^z2)⊕(x4^y4^z4) =>((x3=>y3)=>z3)^((x3≡y3)≡z3)=1
(x3^y3^z3)⊕(x5^y5^z5) =>((x4=>y4)=>z4)^((x4≡y4)≡z4)=1
(x4^y4^z4)⊕(x6^y6^z6) =>((x5=>y5)=>z5)^((x5≡y5)≡z5)=1

Notice that 2 is power of  truth_set\{1,1,1} of predicate F(x,y,z)=((x=>y)=>z)^((x≡y)≡z)

    Base diagram per Helen Mironchick has been set and Mapping Method table has been 
    forked  as well.

Verification of Helen Mironchick recent idea based on the logical product of bit triples on several Boolean systems having three variables as of 09/09/2019

Original systems 

(x1^y1^z1≡x3^y3^z3)=>(x2 v y2 v z2)=1
(x2^y2^z2≡x4^y4^z4)=>(x3 v y3 v z3)=1
(x3^y3^z3≡x5^y5^z5)=>(x4 v y4 v z4)=1

(x1^y1^z1⊕x3^y3^z3)=>(x2 v y2 v z2)=1
(x2^y2^z2⊕x4^y4^z4)=>(x3 v y3 v z3)=1
(x3^y3^z3⊕x5^y5^z5)=>(x4 v y4 v z4)=1

Second diagram forced to work ( for XOR update of original one )

    At this point we start testing with implication target different from considered by Helen 

    Mironchick ( for 2 variables case )
    (x1^y1^z1⊕x3^y3^z3)=>((x2=>y2)=>z2)=1      (3/5)
    (x2^y2^z2⊕x4^y4^z4)=>((x3=>y3)=>z3)=1      (3/5)
    (x3^y3^z3⊕x5^y5^z5)=>((x4=>y4)=>z4)=1      (3/5)

      (x1^y1^z1≡x3^y3^z3)=>((x2=>y2)=>z2)=1      (3/5)

      (x2^y2^z2≡x4^y4^z4)=>((x3=>y3)=>z3)=1      (3/5)       
      (x3^y3^z3≡x5^y5^z5)=>((x4=>y4)=>z4)=1      (3/5)    

    (x1^y1^z1⊕x3^y3^z3)=>((x2≡y2)≡z2)=1      (4/4)
    (x2^y2^z2⊕x4^y4^z4)=>((x3≡y3)≡z3)=1      (4/4)
    (x3^y3^z3⊕x5^y5^z5)=>((x4≡y4)≡z4)=1      (4/4)

Solution of onе System in Boolean Variables in 08/2016 format from Informatics_100 VK's News Wire as of 09/09/2019

Original Source

Fork 08/2016 diagram bits {0,0} fo xj&&yj are obviously denied 

Advantages of predicates implementation when solving 23rd problem of USE in Informatics

Original problem


On right hand side of equations we put 1 instead 0 to avoid 
building quite simple Mapping Method arrows diagrams.
The intend is to build more complicated ones  
   (x1^y1≡x2^y2) =>x3^y3=1
   (x3^y3=>x2^y2) =>x4^y4=1
   (x3^y3≡x4^y4) =>x5^y5=1
   (x5^y5=>x4^y4) =>x6^y6=1
Introduce double predicates zj=(xj)^(yj) . The power of zj truth set is 1 . 
Just one pair of bits {1,1}, the power of zj  false set is 3 . 
False set contains three pairs of bits {1,0; 0,1; 0,0}. 
In general utilizing of predicates for solving 23rd problem is described here
https://mapping-metod.blogspot.com/2019/07/blog-post.html

System itself
   (z1≡z2) =>z3 =1
   (z3=>z2) =>z4 =1
   (z3≡z4)=>z5 =1
   (z5=>z4) =>z6 =1
Notice that outgoing numbers should match powers of truth and false sets
of predicate zj 

    Advantages of predicates approach when solving 23rd problem.
    Consider a bit more complicated system kind of

    (x1^y1^w1≡x2^y2^w2) =>x3^y3^w3=1
    (x3^y3^w3=>x2^y2^w2) =>x4^y4^w4=1
    (x3^y3^w3≡x4^y4^w4) =>x5^y5^w5=1
    (x5^y5^w5=>x4^y4^w5) =>x6^y6^w6=1

    Introduce triple predicates zj=(xj)^(yj)^(wj) . The power of zj truth set is 1 . 
   Just one triple of bits {1,1,1}, the power of zj  false set is 7 . Triples
   {0,0,0};{0,0,1};{0,1,0};{0,1,1};{1,0,0};{1,0,1};{1,1,0} build a false set for zj
   in particular case  

 Regarding predicates zj system keeps to stay the same
   (z1≡z2) =>z3 =1
   (z3=>z2) =>z4 =1
   (z3≡z4)=>z5 =1
   (z5=>z4) =>z6 =1
Notice that outgoing numbers should match powers of truth and false sets
of predicate zj  which would differ from ones been used for double predicate zj