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**CSCE 2100: Computing Foundations 1 Running Time of Programs**

Tamara Schneider Summer 2013

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**What is Efficiency? Time it takes to run a program? Resources**

Storage space taken by variables Traffic generated on computer network Amount of data moved to and from disk

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**Summarizing Running Time**

Benchmarking Use of benchmarks: small collection of typical inputs Analysis Group input based on size Running time is influenced by various factors Computer Compiler

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Running Time worst-case running time: maximum running time over all inputs of size 𝑛 average running time: average running time of all inputs of size 𝑛 best-case running time: minimum running time over all inputs of size 𝑛

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**Worst, Best, and Average Case**

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**Running Time of a Program**

𝑇(𝑛) is the running time of a program as a function of the input size 𝑛. 𝑇(𝑛) = 𝑐𝑛 indicates that the running time is linearly proportional to the size of the input, that is, linear time.

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**Running Time of Simple Statements**

We assume that “primitive operations” take a single instruction. Arithmetic operations (+, %, *, -, ...) Logical operations (&&, ||, ...) Accessing operations (A[i], x->y, ...) Simple assignment Calls to library functions (scanf, printf, ... )

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Code Segment 1 sum = 0; for(i=0; i<n; i++) sum++; 1

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Code Segment 1 sum = 0; for(i=0; i<n; i++) sum++; 1 1

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Code Segment 1 sum = 0; for(i=0; i<n; i++) sum++; 1 1 + (n+1)

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**Code Segment 1 1 1 + (n+1) + n = 2n+2 sum = 0; for(i=0; i<n; i++)**

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**Code Segment 1 1 1 + (n+1) + n = 2n+2 1 How many times? sum = 0;**

for(i=0; i<n; i++) sum++; 1 1 + (n+1) + n = 2n+2 1 How many times?

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**Code Segment 1 1 1 + (n+1) + n = 2n+2 1 How many times?**

sum = 0; for(i=0; i<n; i++) sum++; 1 1 + (n+1) + n = 2n+2 1 How many times? 1 + (2n+2) + n*1 = 3n + 3 Complexity?

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**Code Segment 2 sum = 0; for(i=0; i<n; i++) for(j=0; j<n; j++)**

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**Code Segment 2 1 sum = 0; for(i=0; i<n; i++) for(j=0; j<n; j++)**

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**Code Segment 2 1 2n+2 sum = 0; for(i=0; i<n; i++)**

for(j=0; j<n; j++) sum++; 1 2n+2

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**Code Segment 2 1 2n+2 2n+2 sum = 0; for(i=0; i<n; i++)**

for(j=0; j<n; j++) sum++; 1 2n+2 2n+2

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**Code Segment 2 1 2n+2 2n+2 1 sum = 0; for(i=0; i<n; i++)**

for(j=0; j<n; j++) sum++; 1 2n+2 2n+2 1

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**Code Segment 2 1 2n+2 2n+2 1 1 + (2n+2) + (2n+2)*n + n*n*1 Complexity?**

sum = 0; for(i=0; i<n; i++) for(j=0; j<n; j++) sum++; 1 2n+2 2n+2 1 1 + (2n+2) + (2n+2)*n + n*n*1 Complexity?

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**Code Segment 3 1 2n+2 ? 1 Complexity? sum = 0; for(i=0; i<n; i++)**

for(j=0; j<n*n; j++) sum++; 1 2n+2 ? 1 Complexity?

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**Code Segment 4 1 2n+4 ? 1 Complexity? sum = 0; for(i=0; i<=n; i++)**

for(j=0; j<i; j++) sum++; 1 2n+4 ? 1 Complexity? i=0 i=1 j=0 i=2 j=0 j=1 i=3 j=0 j=1 j=2 … i=n j=0 j=1 j=2 j= j=n-1

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**How Do Running Times Compare?**

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**Towards “Big Oh” t (time) c f(n), e.g. 5 x2 with c = 5, f(n)=x2**

T(n) describes the runtime of some program, e.g. T(n) = 2x2-4x+3 n (input size) n0 We can observe that for an input size n ≥ n0 , the graph of the function c f(n) has a higher time value than the graph for the function T(n). For n ≥ n0, c f(n) is an upper bound on T(n), i.e. c f(n) ≥ T(n).

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Big-Oh [1] It is too much work to use the exact number of machine instructions Instead, hide the details average number of compiler-generated machine instructions average number of instructions executed by a machine per second Simplification Instead of 4m-1 write O(m) O(m) ?!

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**Big-Oh [2] Restrict argument to integer 𝑛 ≥ 0**

𝑇(𝑛) is nonnegative for all 𝑛 Definition: 𝑇(𝑛) is 𝑂(𝑓(𝑛)) if ∃ an integer 𝑛0 and a constant 𝑐 > 0: ∀ 𝑛 ≥ 𝑛0, 𝑇 𝑛 ≤ 𝑐·𝑓(𝑛) ∃ “there exists” ∀ “for all”

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**Big-Oh - Example [1] Example 1: T(0) = 1 T(1) = 4 T(2) = 9**

Definition: T(n) is O(f(n)) if ∃ an integer n0 and a constant c > 0: ∀ n ≥ n0 T(n) ≤ cf(n) Example 1: T(0) = 1 T(1) = 4 T(2) = 9 in general : T(n) = (n+1)2 Is T(n) also O(n2) ???

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**Big-Oh - Example [2] Definition:**

T(n) is O(f(n)) if ∃ an integer n0 and a constant c > 0: ∀ n ≥ n0 T(n) ≤ cf(n) T(n)=(n+1)2. We want to show that T(n) is O(n2). In other words, f(n) = n2 If this is true, there exist and integer n0 and a constant c > 0 such that for all n ≥ n0 : T(n) ≤ cn2

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**Big-Oh - Example [3] Definition:**

T(n) is O(f(n)) if ∃ an integer n0 and a constant c > 0: ∀ n ≥ n0 T(n) ≤ cf(n) T(n) ≤ cn2 ⇔ (n+1)2 ≤ cn2 Choose c=4, n0=1: Show that (n+1)2 ≤ 4n2 for n ≥ 1 (n+1)2 = n2 + 2n + 1 ≤ n2 + 2n2 + 1 = 3n2 + 1 ≤ 3n2 + n2 = 4n2 = cn2

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**Big-Oh - Example [Alt 3] Definition:**

T(n) is O(f(n)) if ∃ an integer n0 and a constant c > 0: ∀ n ≥ n0 T(n) ≤ cf(n) T(n) ≤ cn2 ⇔ (n+1)2 ≤ cn2 Choose c=2, n0=3: Show that (n+1)2 ≤ 2n2 for n ≥ 3 (n+1)2 = n2 + 2n + 1 ≤ n2 + n2 = 2n2 = cn2 For all n≥3: 2n+2 ≤ n2

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**Simplification Rules for Big-Oh**

Constant factors can be omitted O(54n2) = O(n2) Lower-oder terms can be omitted O(n4 + n2) = O(n4) O(n2) + O(1) = O(n2) Note that the highest-order term should never be negative. Lower order terms can be negative. Negative terms can be omitted since they do not increase the runtime.

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**Transitivity [1] What is transitivity? Is Big Oh transitive?**

if A☺B and B☺C, then A☺C example: a < b and b < c, then a < c e.g. 2 < 4 and 4 < 7, then 2 < since “<“ is transitive Is Big Oh transitive?

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Transitivity [2] if f(n) is O(g(n)) and g(n) is O(h(n)) then f(n) is O(h(n)) f(n) is O(g(n)): ∃ n1, c1 such that f(n) ≤ c1 g(n) ∀ n ≥ n1 g(n) is O(h(n)): ∃ n2, c2 such that g(n) ≤ c2 h(n) ∀ n ≥ n2 Choose n0 = max{n1,n2} and c = c1 c2 f(n) ≤ c1 g(n) ≤ c1 c2 h(n) ⇒ f(n) is O(h(n)) ≤ c2 h(n)

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**Tightness Use constant factor “1”**

Use tightest upper bound that we can proof 3n is O(n2) and O(n) and O(2n) Which one should we use?

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**Summation Rule [1] Consider a program that that contains 2 parts**

Part 1 takes T1(n) time and is O(f1(n)) Part 2 takes T2(n) time and is O(f2(n)) We also know that f2 grows no faster than f1 ⇒ f2(n) is O(f1(n)) What is the running time of the entire program? T1(n) + T2(n) is O(f1(n) + f2(n)) But can we simplify this?

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Summation Rule [2] T1(n) + T2(n) is O(f1(n)) since f2 grows no faster than f1 Proof: T1(n) ≤ c1 f1(n) for n ≥ n1 T2(n) ≤ c2 f2(n) for n ≥ n2 f2(n) ≤ c3 f1(n) for n ≥ n3 n0 = max{n1,n2,n3} T1(n) + T2(n) ≤ c1 f1(n) + c2 f2(n) = c1 f1(n) + c2 f2(n) ≤ c1 f1(n) + c2 c3 f1(n) = c1 +c2 c3 f1(n) = c f1(n) with c=c1+c2c ⇒ T1(n) + T2(n) is O(f1(n))

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**Summation Rule - Example**

//make A identity matrix scanf("%d", &d); for(i=0; i<n; i++) for(j=0; j<n; j++) A[i][j] = 0; A[i][i] = 1; 𝑂(1) 𝑂(𝑛) O(n2) 𝑂(1) 𝑂(𝑛) 𝑂(1) O(1) + O(n2) + O(n) = O(n2)

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**Summary of Rules & Concepts [1]**

Worst-case, average-case, and best-case running time are compared for a fixed input size n, not for varying n! Counting Instructions Assume 1 instruction for assignments, simple calculations, comparisons, etc. Definition of Big-Oh T(n) is O(f(n)) if ∃ an integer n0 and a constant c > 0: ∀ n ≥ n0 T(n) ≤ cf(n)

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**Summary of Rules & Concepts [2]**

Rule 1: Constant factors can be omitted Example: O(3n5) = O(n5) Rule 2: Low order terms can be omitted Example: O(3n5 + 10n4 - 4n3 + n + 1) = O(3n5) We can combine Rule 1 and Rule 2: Example: O(3n5 + 10n4 - 4n3 + n + 1) = O(n5)

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**Summary of Rules & Concepts [3]**

For O(f(n) + g(n)), we can neglect the function with the slower growth rate. Example: O(f(n) + g(n)) = O(n + nlogn) = O(nlogn) Transitivity: If f(n) is O(g(n)) and g(n) is O(h(n)) then f(n) is O(h(n)) Example: f(n)=3n, g(n)=n2, h(n)=n6 3n is O(n2) and n2 is O(n6) 3n is O(n6) Tightness: We try to find an upper bound Big-Oh that is as small as possible. Example: n2 is O(n6), but is O(n2) is a much tighter (and better) bound.

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**Solutions to Instruction Counts on Code Segments**

Instructions Big Oh Code Segment 1 3n + 3 O(n) Code Segment 2 3n2 + 4n + 3 O(n2) Code Segment 3 3n3 + 4n + 3 O(n3) Code Segment 4 Argh!

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