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J.M.S.S. Silva
Pigeon-hole Principle

              If n (> m)
            pigeons are put
                 into m
             pigeonholes,
            there's a hole
            with more than
              one pigeon.
Alternative Forms
• If n objects are to be allocated to m
  containers, then at least one container
  must hold at least ceil(n/m) objects.
• For any finite set A , there does not exist
  a bijection between A and a proper
  subset of A .
• Let |A| denote the number of elements in
  a finite set A. For two finite sets A and B,
  there exists a 1-1 correspondence f: A-
  >B iff |A| = |B|.
History
      The first statement of the
       principle is believed to
       have been made by
       Dirichlet in 1834 under the
       name Schubfachprinzip
       ("drawer principle" or
       "shelf principle")
      Also known as Dirichlet's
       box (or drawer) principle
General Problems
 There 750 students in the a batch at
 UOM. Prove that at least 3 of them have
 their birthdays on the same date ?
   ○ 366 * 2= 732 < 750
   ○ Thus at least 3 students have the birthday
    on the same date.
Problems on Relations
   There are 50 people in a room. some of them are
    friends. If A is a friend of B then B is also a friend of
    A. Prove that there are two persons in the room
    who have a same number of friends.

   In league T20 tournament of 16 cricket
    teams, every two teams have to meet in a game.
    Prove that at any time there are two teams which
    played equal number of matches.
Solution
   Case 1                             Case 2
     There exists a person              There does not exists
      with 49 friends (he is a            a person with 49
      friend of all other people)         friends
     Then there cannot be a             The no of friends vary
      person with 0 friends               between 0 – 48 .
     The no of friends vary             There are 50 people
      between 1 – 49 .                    and only 49 values.
     There are 50 people and
      only 49 values.
Problems On Divisibility
   Prove that there
    exists a multiple of
    2009 whose decimal
    expansion contains
    only digits 1 and 0.
Answer
   Consider 2010 numbers
          - 1,11,111,1111, … ,1111…111.

 Each of these numbers produce one of 2009
  remainders
        - 0,1,2,3 ,…,2008
 We have 2010 numbers and 2009 remainders
 By pigeon-hole principle some two numbers have the
  same remainder .Let those 2 numbers be A and B
  (A>B)
 Consider A-B. which is a multiple of 2009.
        - In the form of 11…1100…000
Problems On Divisibility
   Prove that of any 52 natural numbers one can find
    two numbers n and m such that either their sum m+n
    or difference m-n is divisible by 100.

     Consider sets {0},{1,99},{2,98}….{49,51},{50}
     There are 51 sets
     By pigeon hole principle at least 1 set should have 2
      members
     If we consider any set above if they have 2 members in the
      set, m+n or m-n is divisible by 100
Problems on Geometry
   51 points are placed, in a random way, into a
    square of side 1 unit. Can we prove that 3 of these
    points can be covered by a circle of radius 1/7 units
    ?
Answer
    To prove the result, we may divide
    the square into 25 equal smaller
    squares of side 1/5 units each.
    Then by the Pigeonhole Principle, at
    least one of these small squares
    should contain at least 3 points.
    Otherwise, each of the small
    squares will contain 2 or less points
    which will then mean that the total
    number of points will be less than
    50 , which is a contradiction to the
    fact that we have 51 points in the
    first case !
Answer - continue
   Now the circle
    circumvented around the
    particular square with the
    three points inside should
    have
   Radius=Sqrt(1/100+1/100
    )
           =Sqrt(1/50)
           <Sqrt(1/49)=1/7
                                 1/10
Applications
   Lossless data compression cannot guarantee
    compression for all data input files.
   The pigeonhole principle often arises in
    computer science. For example, collisions are
    inevitable in a hash table because the number
    of possible keys exceeds the number of
    indices in the array
   In probability theory, the birthday problem, or
    birthday paradox pertains to the probability
    that in a set of randomly chosen people some
    pair of them will have the same birthday
Applications
   The proof of Chinese Remainder Theorem
    is based on pigeon-hole principle
        Let m and n be relatively prime positive
        Integers. Then the system:
              x = a (mod m)
              x = b (mod n)
                    has a solution.
References
   http://en.wikipedia.org/wiki/Pigeonhole_principle
   http://en.wikipedia.org/wiki/Johann_Peter_Gustav_
    Lejeune_Dirichlet
   http://en.wikipedia.org/wiki/Lossless_data_compre
    ssion#Limitations
   Article on "What is Pigeonhole Principle?" by
    Alexandre V. Borovik, Elena V. Bessonova.
   Article on "Applications of the Pigeonhole
    Principle" by Edwin Kwek Swee Hee ,Huang
    Meiizhuo ,Koh Chan Swee ,Heng Wee Kuan ,
    River Valley High School
Thank You

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Pigeon hole principle

  • 2. Pigeon-hole Principle  If n (> m) pigeons are put into m pigeonholes, there's a hole with more than one pigeon.
  • 3. Alternative Forms • If n objects are to be allocated to m containers, then at least one container must hold at least ceil(n/m) objects. • For any finite set A , there does not exist a bijection between A and a proper subset of A . • Let |A| denote the number of elements in a finite set A. For two finite sets A and B, there exists a 1-1 correspondence f: A- >B iff |A| = |B|.
  • 4. History  The first statement of the principle is believed to have been made by Dirichlet in 1834 under the name Schubfachprinzip ("drawer principle" or "shelf principle")  Also known as Dirichlet's box (or drawer) principle
  • 5. General Problems  There 750 students in the a batch at UOM. Prove that at least 3 of them have their birthdays on the same date ? ○ 366 * 2= 732 < 750 ○ Thus at least 3 students have the birthday on the same date.
  • 6. Problems on Relations  There are 50 people in a room. some of them are friends. If A is a friend of B then B is also a friend of A. Prove that there are two persons in the room who have a same number of friends.  In league T20 tournament of 16 cricket teams, every two teams have to meet in a game. Prove that at any time there are two teams which played equal number of matches.
  • 7. Solution  Case 1  Case 2  There exists a person  There does not exists with 49 friends (he is a a person with 49 friend of all other people) friends  Then there cannot be a  The no of friends vary person with 0 friends between 0 – 48 .  The no of friends vary  There are 50 people between 1 – 49 . and only 49 values.  There are 50 people and only 49 values.
  • 8. Problems On Divisibility  Prove that there exists a multiple of 2009 whose decimal expansion contains only digits 1 and 0.
  • 9. Answer  Consider 2010 numbers - 1,11,111,1111, … ,1111…111.  Each of these numbers produce one of 2009 remainders - 0,1,2,3 ,…,2008  We have 2010 numbers and 2009 remainders  By pigeon-hole principle some two numbers have the same remainder .Let those 2 numbers be A and B (A>B)  Consider A-B. which is a multiple of 2009. - In the form of 11…1100…000
  • 10. Problems On Divisibility  Prove that of any 52 natural numbers one can find two numbers n and m such that either their sum m+n or difference m-n is divisible by 100.  Consider sets {0},{1,99},{2,98}….{49,51},{50}  There are 51 sets  By pigeon hole principle at least 1 set should have 2 members  If we consider any set above if they have 2 members in the set, m+n or m-n is divisible by 100
  • 11. Problems on Geometry  51 points are placed, in a random way, into a square of side 1 unit. Can we prove that 3 of these points can be covered by a circle of radius 1/7 units ?
  • 12. Answer  To prove the result, we may divide the square into 25 equal smaller squares of side 1/5 units each. Then by the Pigeonhole Principle, at least one of these small squares should contain at least 3 points. Otherwise, each of the small squares will contain 2 or less points which will then mean that the total number of points will be less than 50 , which is a contradiction to the fact that we have 51 points in the first case !
  • 13. Answer - continue  Now the circle circumvented around the particular square with the three points inside should have  Radius=Sqrt(1/100+1/100 ) =Sqrt(1/50) <Sqrt(1/49)=1/7 1/10
  • 14. Applications  Lossless data compression cannot guarantee compression for all data input files.  The pigeonhole principle often arises in computer science. For example, collisions are inevitable in a hash table because the number of possible keys exceeds the number of indices in the array  In probability theory, the birthday problem, or birthday paradox pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday
  • 15. Applications  The proof of Chinese Remainder Theorem is based on pigeon-hole principle Let m and n be relatively prime positive Integers. Then the system: x = a (mod m) x = b (mod n) has a solution.
  • 16. References  http://en.wikipedia.org/wiki/Pigeonhole_principle  http://en.wikipedia.org/wiki/Johann_Peter_Gustav_ Lejeune_Dirichlet  http://en.wikipedia.org/wiki/Lossless_data_compre ssion#Limitations  Article on "What is Pigeonhole Principle?" by Alexandre V. Borovik, Elena V. Bessonova.  Article on "Applications of the Pigeonhole Principle" by Edwin Kwek Swee Hee ,Huang Meiizhuo ,Koh Chan Swee ,Heng Wee Kuan , River Valley High School