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Big-O n) 4n-2 ≤ cn for^0 ≥ n^ 1 so that ≥ n true for c = 4 and n 0 0 = 1 (^3) + 10n 2 + 4n +2 is O(n (^3) ) 5n 3 +10n^2 +4n +2^0 ≥^ 1 so that ≤ cn 3 for n ≥ n true for c = 21 and n 0 0 = 1 2 n + 3 is O(log^2 n) 2log^0 ≥^ 1 so that true for c = 5^2 n and^ + 3 n^ ≤^ c log^2 n^ for^ n^ ≥^ n^0 0 = 2
Big-O
- Given function f(n) and g(n), we say that f(n) is O(g(n)) if there are positive constants c and n 0 so that f(n) ≤ cg(n) for n ≥ n 0 f(n) is O(g(n)) g(n) is O(f(n))
g(n) grows more (^) Yes No
f(n) grows more (^) No Yes
g(n) and f(n) has same growth Yes^ Yes 5
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type Int_Array (Integer range <>) of Integer; procedure Measure (A : Int_Array ) is begin^ Sum^ : Integer := 0; for I in A' for (^) Sum := Sum + A(J);J in A' ; ; end Measure;
Inner loop Outer loop
Ex 1
is array
range loop range loop end loop end loop
Statement Runs in X time Executes
of times
Variable Sum is initialized Constant1 1 Array of size n is created Constant2 1 Variable I is created and initialized Constant3 1 I is tested against A’range (n) Constant4 n+ Variable J is created and initialized Constant5 n J is tested against A’range (n) Constant6 n(n+1) Sum is incremented by A(J) Constant7 n 2 J is incremented by 1 Constant8 n 2 I is incremented by 1 Constant9 n BigO.adb
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Ex 3
type Int_Array (Integer range <>) of Integer; procedure Measure (A : Int_Array ) is begin^ Sum^ : Integer := 0; for I in A' for (^) Sum := Sum + A(J in 1 .. 4 loopI ); (^) ---- only change to Ex 2only change to Ex 2 ; ; end Measure; BigO3.adb
is array
range loop
end loop end loop
CQ – Ex 3 Variable J is created and initialized Constant J is tested against I Constant Sum is incremented by A(J) Constant J is incremented by 1 Constant
- N, N(I+1), NI, N*I
- N, N5, N4, N*
- N, N*5, 4, 4
- I still don’t get it (^11)
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Ex 4
function Factorial (N : in Natural ) return Positive is beginif N = 0 then return 1; elsereturn N * Factorial (N-1); end Factorial;;
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How long time does executing the Factorial algorithm take?
(^2) )
end if
CQ – Ex 4
- O(n)
- O(n
- log n
- 42
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Merge
- Input : Array A and indices p, mid, r such that - p ≤ mid < r A(p .. mid) is sorted and subarray A(mid+1 .. r) is sorted
- Output : single sorted array A(p .. r)
- T(n) = O(n), where n=r-p+1 = # of elements being merged
Merge Sort Analysis
- The base case : when n =1 , T(n) =O(1) n ≥ 2, time for merge sort steps:
- Divide : Compute mid as the average of p, r ⇒ cost = O(1)
- Conquer : Solve 2 subproblems, each of size n/ 2 ⇒cost = 2 T( n/2 )
- Combine : merge to an n element subarray ⇒ cost = O (n)
T(n) = O(1) n = 1
2T(n/2) + O(n) + O(1) n> 1
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Solving Recurrences:
Iteration
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aT(n/b) + cn a(aT(n/b/b) + cn/b) + cn a^2 T(n/b^2 ) + cna/b + cn a^2 T(n/b^2 a^2 (aT(n/b^2 /b) + cn/b^2 ) + cn(a/b + 1) a^3 T(n/b^3 ) + cn(a^2 /b^2 ) + cn(a/b + 1) a^3 T(n/b^3 ) + cn(a^2 /b^2 + a/b + 1) … ak^ T(n/b k^ ) + cn(ak-1/b k-1^ + ak-2/b k-2^2 /b^2 + a/b + 1)
) + cn(a/b + 1)
( ) cn n b
n aT
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The Master Method
divide and conquer algorithm size n into a subproblems, each of size n / b
divide the problem + combine solved subproblems) be described by the function f(n) provides a simple “ cookbook
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Simplified Master Method
- T(n) = aT(n/b) + cn k , where and b > 1
( ) k
k
k
k
b
k
a
a b
a b
a b
O n
O n n
O n b
log
log
T(n) =
- Given: a
- An algorithm that divides the problem of
- Let the cost of each stage (i.e., the work to
- The master method ” solution
a,c > 0
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The Towers of Hanoi
- Goal : Move stack of rings to another peg
ring
For simplicity, suppose there were just 3 disks
the top disk from A to B.
A B C
The Towers of Hanoi
- May only move 1 ring at a time
- May never have larger ring on top of smaller
Since we can only move one disk at a time, we move
For simplicity, suppose there were just 3 disks
We then move the top disk from A to B.
A B C
The Towers of Hanoi
For simplicity, suppose there were just 3 disks
We then move the top disk from C to A.
A B C
The Towers of Hanoi
For simplicity, suppose there were just 3 disks
We then move the top disk from C to B.
A B C
The Towers of Hanoi
For simplicity, suppose there were just 3 disks
We then move the top disk from A to B.
A B C
The Towers of Hanoi
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Towers of Hanoi
- hanoi(from,to,other,number) number disks from to needle to if number=1 then move the top disk from needle from to needle to else hanoi(from,other,to, number-1) hanoi(from,to,other, 1) hanoi(other,to, from, number-1) end
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Some math that is good to know
b (xy) = log^ b b y b (x/y) = log^ b b y b b x b x a/log^ x b
- a(b+c)^ = ab^ ac
- abc^ = (ab^ )c
- ab^ /a c^ = a(b-c) log (^) ab
- b c^ = aclog^ ab
-- move the top -- from needle
- log x + log
- log x – log
- log xa = alog
- log a = log
- b = a
