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Division Rule - Discrete Structures - Lecture Slides, Slides of Discrete Structures and Graph Theory

English and Foreign Languages UniversityDiscrete Structures and Graph Theory

These solved exam paper are very easy to understand and very helpful to built a concept about the foundation of computers and discrete structures.The key points discuss in these notes are:Division Rule, Quotient of Size of Set, Partition Set, Equivalence Classes, Generalized Permutations, Unordering Portions, Permutation Theorem, Distinct Permutations, Combinations with Repetition

Typology: Slides

2012/2013

Uploaded on 04/27/2013

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The Division Rule
•Theorem: Suppose a set A has n elements and is
partitioned by the collection {A1, A2, ..., Ap},
where each partition set has m elements. Then:
p = n/m.
•In other words, if a set is partitioned into equal-
sized partition sets, then the number of partition
sets is the quotient of the size of the set with the
size of any partition set.
•For example, if a set has 100 elements and is
partitioned in 20-element subsets, then there
must be 5 subsets (equivalence classes).
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The Division Rule

  • Theorem: Suppose a set A has n elements and is partitioned by the collection {A 1 , A 2 , ..., A p }, where each partition set has m elements. Then: p = n / m.
  • In other words, if a set is partitioned into equal- sized partition sets, then the number of partition sets is the quotient of the size of the set with the size of any partition set.
  • For example, if a set has 100 elements and is partitioned in 20-element subsets, then there must be 5 subsets (equivalence classes).

Generalized Permutations

  • Permutations teach us how to count the number of orderings of the letters of COMPUTER (8!). What about the number of orderings of the letters of MISSISSIPPI?
  • In this case, we note that not all the letters are distinct. In particular, MISSISSIPPI → IIIISSSSPPM , so although we are still searching for an ordering structures, there are sub-unorderings present, induced by the repeated letters, for which we have to account.

Generalized Permutations Take 2

  • If we model this problem, purely as a combination, and not a permutation at all, we can reason the task as: 1. Choose 4 slots from 11 for the I ’s; 2. Choose 4 slots from the remaining 7 for the S ’s; 3. Choose 2 slots from the remaining 3 for the P ’s; 4. Place the M (only 1 way remaining).
  • This yields: C(11,4)C(7,4)C(3,2)C(1,1) = (11!7!3!)/(7!4!4!3!2!1!) = 11!/(4!4!2!).

Generalized Permutation Theorem

  • Theorem: Suppose a collection consists of n objects of which: n 1 are of type 1, indistinguishable from each other; n 2 are of type 2, indistinguishable from each other; ... nk are of type k , indistinguishable from each other; and n 1 + n 2 + ... + nk = n. Then the number of distinct permutations of the n objects is: C( n , n 1 )C( n − n 1 , n 2 )C( n − n 1 − n 2 , n 3 )...C( nk , nk ) = n! / ( n 1! n 2! n 3 !... nk !).

A Motivating Example

  • How many ways can I select 15 cans of soda from a cooler containing large quantities of Coke, Pepsi, Diet Coke, Root Beer and Sprite?
  • We have to model this problem using the chart: Coke Pepsi Diet Coke Root Beer Sprite A: 111 111 111 111 111 = B: 11 111111 111111 1 = C: 1111 1111111 1111 =
  • Here, we set an order of the categories and just count how many from each category are chosen.

A Motivating Example ( cont’d. )

  • Now, each event will contain fifteen 1’s, but we need to indicate where we transition from one category to the next. If we use 0 to mark our transitions, then the events become: A: 1110111011101110111 B: 1100111111011111101 C: 0011110111111101111
  • Thus, associated with each event is a binary string with #1’s = #things to be chosen and #0’s = #transitions between categories.

Another Example

  • How many ways can I fill a box holding 100 pieces of candy from 30 different types of candy? Solution: Here #slots = 100, #transitions = 30 − 1, so there are C(100+29,100) = 129!/(100!29!) different ways to fill the box.
  • How many ways if I must have at least 1 piece of each type? Solution: Now, we are reducing the #slots to choose over to (100 − 30) slots, so there are C(70+29,70) = 99!/70!29!

When to Use Generalized

Combinations

  • Besides categorizing a problem based on its order and repetition requirements as a generalized combination, there are a couple of other characteristics which help us sort: - In generalized combinations, having all the slots filled in by only selections from one category is allowed; - It is possible to have more slots than categories.

Integer Solutions with Restrictions

  • How many integer solutions are there to: a + b + c + d = 15, when a ≥ 3, b ≥ 0, c ≥ 2 and d ≥ 1?
  • Now, solution “strings” are 111 a 0 b 011 c 01 d , where the a,b,c,d are the remaining numbers of each category to fill in the remaining slots.
  • However, the number of slots has effectively been reduced to 9 after accounting for a total of 6 restrictions.
  • Thus there are C(9+3,9) = 12!/(9!3!) solutions.

More Integer Solutions & Restrictions

  • How many integer solutions are there to: a + b + c + d = 15, when a ≥ −3, b ≥ 0, c ≥ −2 and d ≥ −1?
  • In this case, we alter the restrictions and equation so that the restrictions “go away.” To do this, we need each restriction ≥ 0 and balance the number of slots accordingly.
  • Hence a ≥ − 3 +3 , b ≥ 0, c ≥ − 2 +2 and d ≥ − 1 +1 , yields a + b + c + d = 15 +3+2+1 = 21
  • So, there are C(21+3,21) = 24!/(21!3!) solutions.