Fast Sorting and PatternAvoiding Permutations

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Fast Sorting andPattern-Avoiding Permutations

• David Arthur
• Stanford University

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Sorting

• A classical problem
• Sort a list of n distinct items
• A classical theorem
• At least O(n log n) comparisons are required
• Proof
• Need to distinguish between n! input orderings
• Takes lg(n!) O(n log n) comparisons

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The Challenge

• What if not every input ordering is possible?
• Example 1
• Maintain a sorted list
• Some items have changed since list was last
  sorted
• But the list is still almost sorted
• What is the sorted order now?
• No decreasing subsequence of length k
• Can sort in O(n log k) time

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The Challenge

• What if not every input ordering is possible?
• Example 2
• Profile a computer program
• Store data every time a function call finishes
• In what order were the functions called?
• No subsequence ordered like (3,1,2)
• Can sort in O(n) time

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The Common Thread

• Examples of pattern-avoiding permutations
• A permutation p contains s if p has a
  subsequence ordered in the same way as s
• Otherwise p avoids s
• Example (4,2,5,1,3) contains (1,3,2)

Sortedorder
Given order
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Why this?

• Continuum of complexity on permutations
• Every length n permutation avoids some pattern of
  length O(n0.5)
• For fixed s, permutations avoiding s are simple
• Theorem Marcus, Tardos 04At most Cn
  permutations of length n avoid s
• Theoretical possibility Can sort any s-avoiding
  permutation in lg Cn O(n) time.

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The Big Question

• Suppose the input ordering p is known to avoid
  some s
• Can p be sorted in less than O(n log n) time?
• Hard question!
• This talk
• For many s, you can sort in O(n log log log n)
  time
• If so, call s good
• Approach find operations preserving goodness

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Symmetry

• Reverse pattern
• r(s1, s2,, ss) (ss, ss-1,, s1)
• e.g., r(1,3,2) (2,3,1)
• Fact If s is good, then r(s) is good
• If p avoids r(s), then r(p) avoids s
• Algorithm Reverse p, then sort it

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Symmetry

• Complement pattern
• (s1, s2,, ss) (s1s1, s1s2,, s1ss)
• e.g., (1,3,2) (3,1,2)
• Fact If s is good, then s is good
• If p avoids s under lt ordering,then p avoids s
  under gt ordering
• Algorithm Sort p under gt ordering, then reverse
  it

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Our Result

• Direct sums
• (s1, s2,, ss) (t1, t2,, tt) (s1, s2,,
  ss, st1, st2,, stt)
• e.g., (1,3,2) (2,1) (1,3,2,5,4)
• Theorem If s and t are good, then s t is good
• This talk If s is good, then (1) s is good

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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 1 Preprocess p to find minimal elements
  (nothing below and left of them)

O(n) time
Minimal elementsSet k of min. elements
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 2 Partition p into columns around min.
  elements

O(n) time
Columns of columns k
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Remark Each column avoids s

If a column contains a sequence ordered like s
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Remark Each column avoids s

then p contains a sequence ordered like (1) s,
which is impossible.
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 3 Recursively sort each column

O(n log log log n) time
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 4 Merge each sorted column with minimal
  elements to partition p into rows

O(n k2) time
Rows of rows of columns k
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 5 Recursively sort each row(each row
  avoids s)

O(n log log log n) time
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Step 6 Concatenate the sorted rows

O(n) time
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Proof

• Goal if s is good, then (1) s is good
• Algorithm 1
• Summary O(n log log log n k2) time
• What if k is large?

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Proof

• Goal if s is good, then (1) s is good
• Algorithm 2
• Partition p into O(log log n) layers, each with a
  small number of minimal elements
• Sort each layer with Algorithm 1
• Merge the results
• O(n log log log n) time

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Summary
All length 4 patterns (by symmetry class)
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Summary

• Just a start on a difficult problem!