List ADTs

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Algorithm Analysis
Capt Balazs Michelson 366
2
Recall

• We are typically concerned with worst case
  running time of algorithms as input grows
• Run time is function of input
• We count basic steps
• When we count steps we put in terms of n
• We can eliminate lower order terms

3
Recall

• Looked growth rates for alg_1 and alg_2
• We determined that alg_1 took k2 6k 19 steps
  and alg_2 took 5k 13 steps
• We are able to ignore constants and lower order
  terms
• alg_2 has a linear growth rate
• alg_1 grows like k2 (quadratic growth)

4
Technically Speaking

• Notation used to describe algorithms running
  time called Asymptotic Notation
• It is mathematic notation for functions that
  disregards constant factors and lower order terms
• Describes growth rate of a function in terms of
  input size

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? - Notation

• Function f(n) that has a quadratic growth rate is
  said to grow like ?(n2)
• Written as f(n) ?(n2)
• In other words
• f(n) has the same shape as n2
• f(n) grows at a rate proportional to n2

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More Generally

• f(n) ?(g(n))
• g(n) is n2 in our previous (specific) example
• alg_1 alg_2 example
• alg_1 f(n) ?(g(n)) g(n) is n2
• alg_2 f(n) ?(g(n)) g(n) is n

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f(n) ?(g(n))
After some cutoff point, n0 -where n0 is an
integer constant - f(n) is bound between constant
multiples of g(n). Therefore, f(n) grows like
g(n).
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Formal Definition

• ?(g(n)) f(n) ? (c1, c2, n0 gt0)
• where 0 ? c1g(n) ? f(n) ? c2g(n)
• ? n ? n0
• Less Formal
• A function (f(n)) belongs to the set ?(g(n)) if
  there exists c1 c2 gt 0 such that f(n) can be
  bound between c1g(n) and c2g(n) for all values
  of n greater than a sufficiently large value (n ?
  n0)

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Note!

• f(n) ?(g(n)) and f(n) ? ?(g(n)) are equivalent
• If f(n) ?(g(n)) then g(n) is an asymptotically
  tight bound for f(n)
• Key points
• ?(g(n)) is a tight bound
• ?(g(n)) is a set of functions

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Big-O Notation

• O(g(n)) f(n) ? (c, n0 gt0)
• where 0 ? f(n) ? c g(n)
• ? n ? n0

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Big-O
O(g(n)) describes a worst case upper bound. It
is an asymptotic upper bound for f(n). No
matter what, f(n) will be no worse than g(n).
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Big-O Example

• Use definition of big-O to show
• 2n2 22n -18 O(n2)

2n2 22n -18 ? 2n2 22n for all n gt 0 Want
n2 on right 22n ? n2 22 ? n 2n2 22n -18 ?
2n2 n2 for all n ? 22 2n2 22n -18 ? 3n2 for
all n ? 22 Thus, with c 3 and n0 22 we have
shown that 2n2 22n -18 O(n2)
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Example of Asymptotic Upper Bound
4g(n)4n2

• 4 g(n) 4n2
• 3n2 n2
• ? 3n2 9 for all n ? 3
• gt 3n2 5
• f(n)
• Thus, f(n)O(g(n))

f(n)3n25
g(n)n2
3
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Use Definition of Big-O to show that 6n3 is not
O(n2)
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? - Notation

• ?(g(n)) f(n) ? (c, n0 gt0)
• where 0 ? cg(n) ? f(n)
• ? n ? n0

Best case running time or lower bound
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Proving Tight Bound

• Can be difficult
• f(n) ?(g(n))
• iff f(n) O(g(n)) f(n) ?(g(n))

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f(n) ?(g(n)) iff f(n) O(g(n)) f(n) ?(g(n))

• Implies
• if f(n)O(g(n)) and g(n)O(f(n)) then f(n)?(g(n))

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f(n) ?(g(n)) iff f(n) O(g(n)) f(n) ?(g(n))

• Proof
• g(n) O(f(n)) implies that g(n) ? cf(n)
• 1/c g(n) ? f(n)
• let c 1/c
• cg(n) ? f(n)
• f(n) ? cg(n)
• ?f(n) ?(g(n))
• Since f(n) ?(g(n)) and g(n) O(f(n)), it
  follows that f(n) ?(g(n))

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Useful Properties of Asymptotic Notation

• if a is a constant, O(af(n)) O(f(n))
• Ex O(32n3) O(n3)
• if f(n) grows as fast as or faster than g(n)
  (remember the growth rate hierarchy), then O(f(n)
  g(n)) O(f(n))
• Ex O(5n4 3n2) ?(n4)
• O(f(n)) O(g(n)) O(f(n) g(n))
• Ex ?(n) ?(log n) ?(n log n)
• O(f(n)) O(g(n)) O(f(n) g(n))

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Show 8n3 - 5n2 10 O(n3)