Steps to join a Linux computer to Active Directory using AD
are as follows:
First, the benefits of using SSSD:
- · Reduced load on authentication servers.
- · Option for offline authentication
- · Single user account
Steps:
1) Make sure that
both the Active Directory and Linux systems have a properly configured
environment
2) On the Linux
client, add the Active Directory domain to the client's DNS configuration so
that it can resolve the domain's SRV records.
Search adserver.example.com
Nameserver 192.168.1.1
3) Set up the linux system as an AD client and enroll it
within the AD domain.
1) set up the
kerberos to use AD realm
1) vim
/etc/krb5.conf
2) configure
the logging and libdefaulrs section
[Logging]
FILE:
/var/logs/krb5libs.log
[Libdefaults]
Default_realm=example.com
Dns_lookup_realm=true
Dns_lookup_kdc = true
Ticket_lifetime=24h
Renew_lifetime=7d
Rdns =
false
Forwardable = yes
2) Configure the
samba server to connect to AD server
1) vim
/etc/samba/smb.conf
2)
configure [globals]
[global]
workgroup = EXAMPLE
client signing = yes
client use spnego = yes
kerberos method = secrets and
keytab log file = /var/log/samba/%m.log password server =
AD.EXAMPLE.COM
realm = EXAMPLE.COM
security = ads
3) add the linux
machine to the AD domain.
Kinit Adninistrator
Net ads join -k
Klist -k
4) configure sssd
[sssd]
config_file_version =
2
domains =
ad.example.com
services = nss, pam,
pac
Create a new domain section at the bottom of the file for
the Active Directory domain. This section has the format domain/NAME, such as
domain/ad.example.com. For each provider, set the value to ad, and give the
connection information for the specific Active Directory instance to connect
to.
[domain/ad.example.com]
id_provider = ad
auth_provider = ad
chpass_provider = ad
access_provider = ad
#codingexercise
Find the maximum width of a binary tree where the width is
the count of nodes at a given level
int GetMaxWidth(Node root)
{
int max = 0;
for (int i =l; i < height(root); i++){
int count =
GetWidth(root, i);
if (count >
max)
max =
count;
}
return root;
}
int GetWidth(Node root, int level)
{
if (root == null)
return 0;
if (level == 1)
return 1;
if (level > 1){
return
GetWidth(root.left, level -1) + GetWidth(root.right, level-1);
}
return 0;
}
#monty hall problem
A tv show host gives
you the option to open one of three doors behind one of which is a car and the
others have goats. You pick a door and the tv show host picks another. His door
has goat. Do you want to switch doors ?
Solution to the above problem is that you should switch
because the probability that the second door hides the car has increased to 2/3.
Previously, the two remaining doors haf a combined
probability of 2/3 but with the hosts choice, that probability has now become
entirely that of one unopened door to switch to
An employee works for an employer for 7 days. The employer
has a gold rod of 7 units. How does the employer pays to the employee so that
the employee gets 1 unit at the end of everyday.The employer can make at most 2
cuts in rod.
Solution : the rod should be cut in 4,2,1 length pieces.
A box contains n coins, of which 7 of them are counterfeit
with tails on both sides and the rest are fair coins. If one coin is selected
from the bag and tossed, the probability of getting a tail is 17/20. Find the
value of ‘n’.
N = 2. Working backwards from the last pèrson.
7/n x 1 + (n-7)/n x 1/2 = 17/20
N = 10
A boy goes to 20 of his friend’s houses with ‘n’ number of
newly purchased marbles in his hands. At every house he visits , he gives away
half of marbles he have and take one of his friend’s marble’s and adds it with
the one’s he is left with , he never had a problem of dividing an odd number of
marbles left and finally after leaving the his 20th friends house, he is left
with 2 marbles, can you guess the ‘n’ value?
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