Examples of Recurrence Equations

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Examples of Recurrence Equations

• CS463
• Irena Pevac
• CCSU

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Recurrence Equations

• Example 1)
• T(0 ) 0
• T(n) T(n-1) 1 for ngt1
• ______________________
• Solution T(n) n
• Example 2)
• T(0 ) 0
• T(n) T(n-1) n for ngt1
• ______________________
• Solution T(n) n(n1)/2

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Recurrence Equations

• Example 3)
• T(1) 0
  
• T(n) T(n/2) 1 for
  ngt2
• ______________________
  ___________________
• Solution T(n) lg n
• Example 4)
• T(0) 0 T(1) 0
  
• T(n) 2 T(n/2) n for
  ngt2
• ______________________
  ___________________
• Solution T(n) n lg n

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Recurrence Equations

• Example 5)
• T(0)0
• T(n)2T(n-1) 1 for ngt1
• ________________
• Solution T(n) 2n-1
• Example 6)
• T(1) 1
• T(n) 3T(n/2) n for ngt2 n
  is power of 2
• _________________________________________
• Solution T(n) 3nlg3 2n

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T(n)T(n-1) 1

• T(n) T(n-1) 1 T(n-2) 11
• T(n-3)111
• .
• T(n-n) 11 1
• T(0) n
• 0 n n
• ________________________________
• Solution is T(n) n

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Test the solution

• T(0) 0
• T(n) T(n-1) 1 for ngt1
  ______________________
• Solution T(n) n
• 00
• nn-11
• ___________
• nn Correct!!!

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T(n) T(n-1) n

• T(n) T(n-1) n
• T(n-2) (n-1)n
• T(n-3)(n-2)(n-1)n
• .
• T(n-n) 12 n
• T(0) 123n
• n(n1)/2
• ________________________________
• Solution is T(n) n(n1)/2

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Test the Solution

• T(0) 0
• T(n) T(n-1) n for ngt1
  ______________________
• Solution T(n) n(n1)/2
• 01/2 0
• n(n1)/2 (n-1)n/2 n
• ___________
• (n1)/2 (n-1)/2 1
• n/21/2n/2 -1/2 1
• 1/2 1/2 Correct!!!

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T(n) T(n/2) 1

• T(1) 0
• T(n) T(n/2) 1 for ngt2
• ______________________
  ___________________
• Solution T(n) lg n
• T(n) T(n/2) 1 T(n/22) 11
• T(n/23) 111 ..
• T(n/2k) k1 n/2k 1, k
  lg n
• T(n) 0 lg n lg n

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Test the solution

• T(1) 0
• T(n) T(n/2) 1 for ngt2
• ____________________
• lg1
• lg n lg(n/2) 1
• ____________________
• lg n lg n - lg 2 1
• lg n lg n -11 Correct!!!

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T(n) 2 T(n/2) n

• T(0) 0 T(1) 0
• T(n) 2 T(n/2) n for ngt2
• ________________________
• Assume that n 2k. If n 1, k 0.
• T(20) 0
• T(2k)2 T(2k-1)2k
• ___________________ multiply by 1/2k
• T(1)0
• T(2k)/ 2k 2 T(2k-1)/ 2k 2k/ 2k

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T(n) 2T(n/2) n

• Substitute T(2k)/ 2k with U(k)
• U(0) 0
• U(k)U(k-1) 1
• _____________________________________________
• Solution determined earlier U(k) k
• T(2k)/ 2k k T(2k) 2k k
• Finally, since 2kn
• T(n) n lg n

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Solution Testing

• Solution T(n) n lg n should satisfy equations
• T(1)0
• T(n) 2 T(n/2) n
• ________________________
• T(1) 1 lg 1 1 00 base case is correct.
• n lg n 2 (n/2) lg(n/2) n
• n lg n n ( lg n- lg 2) n
• n lg n n lg n n n
• n lg n n lg n Obviously correct!!!
• Recursive step is correct.

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T(n)2T(n-1) 1

• T(0)0
• T(n)2T(n-1) 1 for ngt1
• ________________
• Solution T(n) 2n-1
• T(0)0
• T(1)2 T(0) 1 1
  1 21-1
• T(2) 2 T(1) 1 211 3 3
  22-1
• T(3) 2 T(2) 1 231 7 7
  23-1
• T(4) 2 T(3) 1 2 7 1 15 15
  24-1
• T(5) 2 T(4) 1 215 1 31 31
  25-1

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Test the solution

• T(0)0
• T(n)2T(n-1) 1 for ngt1
• ________________
• T(n) 2n-1
• ________________
• 20-1 0
• 2n-1 2 (2n-1-1 ) 1
• 2n-1 2n-2 1
• 2n-1 2n-1 Correct!!!

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T(n) 3T(n/2) n

• T(1) 1
• T(n) 3T(n/2) n
• T(n) 3T(n/2) n 3(3T(n/22) n/2) n
• 32T(n/22) 3n/2 n
• 32(3T(n/23) n/22 ) 3n/2 n
• 33T(n/23) 32n/22 3n/2 n
• 33(3T(n/24) n/23) 32n/22 3n/2
  n
• 33(3T(n/24) n/23) 32n/22 3n/2
  n
• 34T(n/24) 33n/23 32n/22 3n/2
  n
• . . .
• 3kT(n/2k) (3/2)k-1n (3/2)k-2n
  (3/2)2n (3/2)n n
• where n/2k 1, n 2k, k lg n

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T(n) 3T(n/2) n
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T(n) 3T(n/2) n

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Test the Solution

• T(n) 3T(n/2) n