two brothers puzzle and facebook

1. Puzzle | Two Brothers and
Facebook
Hup university
5-2-2018

2. Puzzle
• Sahil and Ritik are brothers. One day they were discussing who is
smarter. But gradually the discussion turned into an argument.
Mother came and tried to handle the situation. She gave them a
problem to solve and the one who will solve the problem first,
would be considered smarter than the other. The problem is
• :
In a group of 6 people, you might find some people are friends on
Facebook, or you might find that no one is friend on Facebook. The
brothers are supposed to prove that there is always a group of 3
people where either :
•
–All3 people are mutual friends on Facebook.
• –All3 people are strangers (i.e, no one is friend on Facebook).
• Can you help the brothers to reach to the solution ?

3. Analyses
• Fact:-6 people, (A,B,C,D,E,F)
• Rules
1-people are friends
2-no one is friend
• Goal
– The brothers are supposed to prove that there is
always a group of 3 people where either :
– All3 people are friends
– All3 people are strangers

4. Set A:-ABC, ABD, ABE, ABF, ACD, ACE,
ACF, ADE, ADF, AEF
A
AE AF
ABC
ABD
ABE
ABF
AB
ACD
ACE
ACF
AC
ADE
ADF
AD
AEF
Relation of A
{A} with {B, C,D,E,F}
If the new child is not a child
on old father , if it is delete
the new child
{AB} with {C,D,E,F}
{AC} with {D,E,F}
{AD} with {E,F}
{AE} with {F}

5. {BE} with {F}
{BD} with {E,F}
Relation of B
B
BE BF
BCD
BCE
BCF
BC
BDE
BDF
BD
BEF
Set B:-BCD, BCE, BCF, BDE, BDF, BEF
{B} with {C,D,E,F}
If the new child is not a child
on old father , if it is delete
the new child
{BC} with {D,E,F}

6. Relation of C
C
CE CF
CDE
CDF
CD
CEF
Set C:-CDE, CDF, CEF
{C} with {D,E,F}
If the new child is not a child
on old father , if it is delete
the new child
{CD} with {E,F}
{CE} with {F}

7. Relation of D
D
DE DF
DEF
Set D:-DEF
{D} with {E,F}
If the new child is not a child
on old father , if it is delete
the new child
{DE} with {F}

8. 3
friends
3
none
6 PEOPLE
1 ABC DEF A B C D E F
2 ABD CEF A B C D E F
3 ABE CDF A B C D E F
4 ABF CDE A B C D E F
5 ACD BEF A B C D E F
6 ACE BDF A B C D E F
7 ACF BDE A B C D E F
8 ADE BCF A B C D E F
9 ADF BCE A B C D E F
10 BCD AEF A B C D E F
11 BCE ADF A B C D E F
12 BCF ADE A B C D E F
13 BDE ACF A B C D E F
14 BDF ACE A B C D E F
15 BEF ACD A B C D E F
3
friends
3
none
6 PEOPLE
16 CDE ABF ABCDEF
17 CDF ABE ABCDEF
18 CEF ABD ABCDEF
19 DEF ABC ABCDEF

9. Steps
• Start with {A} friends with {B,C,D,E,F}
• Each two friend {AB, AC, AD, AE,AF} will be a
friend with {B, C, D, E, F}, delete repeated
friends.
• Next, start with {B} friends with {C,D,E,F},
• A is not in the set because of the repetition
• Each two friend {BC,BD,BE,BF} will be a friend
with {C, D,E, F} (A & B is not in the set of
repetition)

10. steps
• Next, start with {C} friends with {D,E,F},
• A&B is not in the set because of the repetition
• Each two friend {CD, CE, CF} will be a friend
with {D,E, F} (A, B & C is not in the set of
repetition) .
• Every new child will be check with the
previous explored child because of the
repetition.

11. Steps
• Next, start with {D} friends with {E,F},
• A,B,C is not in the set because of the
repetition
• Each two friend {DE,DF} will be a friend with
{E, F} (A B C D is not in the set)