of g(x), c, 5 and k) in the next section. For the following code segments below, count the operations and produce a corresponding polynomial representation. f(x) = public static boolean isEmpty() {...


of g(x), c, 5 and k) in the next section. For the following code segments below, count the operations and<br>produce a corresponding polynomial representation.<br>f(x) =<br>public static boolean isEmpty() {<br>return head == null;<br>}<br>f(x)<br>public static int num_occurrences(int n) {<br>int count = 0;<br>for(int i = 0; i < n; i++) {<br>for(int j = 0; j < n; j++) {<br>if( i == j ) continue;<br>if(strings [i]<br>count++;<br>}<br>}<br>}<br>return count;<br>}<br>strings[j]) {<br>==<br>f(2)<br>public static void c(int n) {<br>for(int a = 0; a < n; a++) {<br>System.out.println( a<br>}<br>num_occurrences(n);<br>}<br>* a);<br>f(x)<br>public static boolean isPrime(int n) {<br>if(n<br>1) return false;<br>==<br>for(int i = 2; i <n; i++) {<br>if( n % i == 0 ) {<br>return false;<br>}<br>}<br>return true;<br>}<br>Demonstrating | f(x) | < c| g(x) | for all n > k<br>For each of the polynomials above, pick a reference function (g(x)) and constants c,k such that the<br>g(x) bounds f(x) in the most succinct terms possible for Big O.<br>

Extracted text: of g(x), c, 5 and k) in the next section. For the following code segments below, count the operations and produce a corresponding polynomial representation. f(x) = public static boolean isEmpty() { return head == null; } f(x) public static int num_occurrences(int n) { int count = 0; for(int i = 0; i < n;="" i++)="" {="" for(int="" j="0;" j="">< n;="" j++)="" {="" if(="" i="=" j="" )="" continue;="" if(strings="" [i]="" count++;="" }="" }="" }="" return="" count;="" }="" strings[j])="" {="=" f(2)="" public="" static="" void="" c(int="" n)="" {="" for(int="" a="0;" a="">< n;="" a++)="" {="" system.out.println(="" a="" }="" num_occurrences(n);="" }="" *="" a);="" f(x)="" public="" static="" boolean="" isprime(int="" n)="" {="" if(n="" 1)="" return="" false;="=" for(int="" i="2;" i="">< c|="" g(x)="" |="" for="" all="" n=""> k For each of the polynomials above, pick a reference function (g(x)) and constants c,k such that the g(x) bounds f(x) in the most succinct terms possible for Big O.
public static void main(String[] args) {<br>for(int a = 0; a < n; a++) //n<br>foo(); //n<br>}<br>public static foo() {<br>for(int a = 0; a < n; a++) //n<br>System.out.println(a);<br>}<br>Log Exponent Rule: Consider the following logarithm a(x) = log2 X°. Note that we can rewrite<br>this using the

Extracted text: public static void main(String[] args) { for(int a = 0; a < n;="" a++)="" n="" foo();="" n="" }="" public="" static="" foo()="" {="" for(int="" a="0;" a="">< n;="" a++)="" n="" system.out.println(a);="" }="" log="" exponent="" rule:="" consider="" the="" following="" logarithm="" a(x)="log2" x°.="" note="" that="" we="" can="" rewrite="" this="" using="" the="" "log="" roll"="" as="" c="" *="" log2="" x.="" since="" constants="" are="" factored="" out="" during="" asymptotic="" analysis,="" you="" can="" simply="" drop="" the="" constant="" multiplier="" (which="" on="" a="" log="" is="" it's="" exponent).="" •="" example="" f(x)="log2" x6="" f(x)="" becomes6="" *="" log2="" x,="" and="" o(log2="" x)="" log="" base="" rule:="" you="" may="" omit="" the="" base="" of="" the="" log="" and="" assume="" its="" base-2.="" transitivity:="" if="" a=""><>< c,="" then="" a="">< c.="" related="" to="" big="" ,="" if="" a="5x²" +="" 3xand="" i'm="" trying="" to="" prove="" that="" a="" is="" less="" than="" or="" equal="" to="" (i.e,="" bounded="" by)="" x2,="" we="" would="" write5x?="" +="" 3x="">< c="" *="" x2="" and="" find="" some="" pair="" of="" c,="" k="" such="" that="" this="" relationship="" holds.="" using="" transitivity,="" 5x?="" +="" 3x="">< 5a²="" +="" 3x?="">< 8x?="">
Jun 06, 2022
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