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Combinatorial Problem Solving: Partitions and Generating Functions

Learn about partitions of integers and generating functions in combinatorial problem solving. Solve homework problems and explore examples.

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Combinatorial Problem Solving: Partitions and Generating Functions

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  1. MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 27, Wednesday, November 5

  2. 6.3. Partitions • Homework (MATH 310#9W): • Read 6.4. • Do 6.3: all odd numberes problems • Turn in 6.3: 2,4,16,20,22 • Volunteers: • ____________ • ____________ • Problem: 16.

  3. Partitions • A partition of a group of r identical objects divides the group into a collection of unordered subsets of various sizes. • Analogously, we define a partition of the interger r to be a collection of positive integers whose sum is r. Normally we write this object as a sum ans list the integers in increasing order. • 5 = 1 + 1 + 1 + 1 + 1 • 5 = 1 + 1 + 1 + 2 • 5 = 1 + 2 + 2 • 5 = 1 + 1 + 3 • 5 = 2 + 3 • 5 = 1 + 4 • 5 = 5

  4. The Generating Function • The generating function for partitions can be written as the infinite product • g(x) =1/[(1 – x)(1 – x2)... (1 – xr) ...]

  5. Example 1 • Find the generating function for ar, the number of ways to express r as a sum of distinct integers. • Answer: • g(x) = (1+x)(1 + x2) ... (1 + xk) ...

  6. Example 2 • Find a generating function for ar, the number of ways that we can choose 2¢, 3¢, and 5¢ stamps adding to the net value of r cents. • Answer: 1/[(1 – x2)(1 – x3)(1 – x5)]

  7. Example 3. • Show with generating functions that every positive integer can be written as a unique sum of distinct powers of 2. • Answer: The generating function g*(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8) ... • (1 – x) g*(x) = (1 – x)(1 + x)(1 + x2) ... = (1 – x2)(1 + x2)(1 + x4) ... = (1 – x4)(1 + x4)(1 + x8) ... = (1 – x8) (1 + x8) ... = ... = 1 + 0x + 0x2 + ... 0xk + ... = 1.

  8. Ferrers Diagram and Conjugate Partitions • Example: • 15 = 7 + 3 + 2 + 2 + 1 • Ferrers Diagram is shown on the left. • We we transpose the diagram we obtain the conjugate partition • 15 = 5 + 4 + 2 + 1 + 1 + 1+ 1

  9. Example 4 • Show that the number of partitions of an integer r as a sum of m positive integers is equal to the number of partitions of r as a sum of integers, the largest of which is equal to m.

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