Chapter three exercises:
Minimum
function myMinimum(a,b){
if (a<b)
return a;
else
return b;
}
Recursion
function isEven(n)
{
if (n == 0) return true;
else if (n == 1) return false;
else if (n < 0) return isEven(-n);
else return isEven(n - 2);
}
Bean Counting
function countsB(n)
{
return countChar(n,“B”);
}
function countChar(a,b)
{
var count = 0;
for(var c = 0;c < a.length; c++)
{
if (a[c] == b) count++;
}
return count;
}1. Minimum
<html>
<body>
<script>
function min(a,b){
if(a<b){
document.write("a")
}
else{
document.write("b")
}
}
min(1809,1034)
</script>
</body>
</html>
2. Recursion
<html>
<body>
<script>
var one = "odd"
var zero = "even"
function isEven(n){
if(n>0){positive(n)
function positive(n){
var positiveReturn = n-2;
if(positiveReturn == 0){
document.write(zero);
}
else if(positiveReturn == 1){
document.write(one)
}
else{
positive(positiveReturn-2);
}
}
}
else if(n<0){negative(n)
function negative(n){
var negativeReturn = n+2;
if(negativeReturn == 0){
document.write(zero);
}
else if(negativeReturn == 1){
document.write(one)
}
else{
negative(negativeReturn+2);
}
}
}
}
isEven(-1)
</script>
</body>
</html>
3a. Bean Counting
<html>
<body>
<script>
function countBs(string){
let numberOfB = 0;
for(let n=0; n<string.length ; n++){
if(string[n] == "B"){
numberOfB = numberOfB+1;
}else {
numberOfB=numberOfB+0;
}
}
document.write(numberOfB);
}
countBs("BigBoxOfBalloonsInABakery")
</script>
</body>
</html>
3b. Bean Counting
<html>
<body>
<script>
function countChar(string, letter){
let numberOfLetter = 0;
for(let n=0; n<string.length ; n++){
if(string[n] === letter){
numberOfLetter = numberOfLetter+1;
}else {
numberOfLetter=numberOfLetter+0;
}
}
document.write(numberOfLetter);
}
function countBs(string){
return countChar(string, "a");
}
countBs("alabama bahama Mama")
</script>
</body>
</html>
MINIMUM
function min(x,y) {
if (x<y){
console.log(x);
}else if (y<x){
console.log(y);
}else{
console.log(“x and y are the same”);
}
};
min(3,3)
RECURSIVE
function isEven(num) {
if (num < 0) num = -num; // remove negative sign
switch (num) {
case 0: return true;
case 1: return false;
default: return isEven(num-2);
}
}
BEAN COUNTING
this one worked for me:
function countBs(string){
let resultB=0;
let resultA=0;
for (let count=0; count<string.length; count++){
string[count]==“b” ? resultA ++ : resultA=resultA;
string[count]==“B” ? resultB ++ : resultB=resultB;
}
return [resultA, resultB];
}
let answer=countBs(“BeansBeansBbbBeans”);
alert(answer);
Minimum
function min(a,b){
if (a < b) return a; else return b;}
console.log(min(0,10));
console.log(min(0,-10));
Recursion
function isEven (a){
if (a == 0) return true;
else if (a == 1) return false;
else return isEven (a - 2);
}
console.log(isEven(50));
// → true
console.log(isEven(75));
// → false
console.log(isEven(-1));
// → ??
the console.log for -1 returns error message “maximum stack call exceeded”, because the function recalls itself until 0 or 1 is reached, but a number less than one has no command to return “true” or “false”; this can be fixed by adding this line: else if (a < 0) return isEven(-a); before this line: else return isEven (a - 2);
Bean counting
function countBs(string) {
return countChar(string, “B”);
}
function countChar(string, c) {
let count = 0;
for (let i = 0; i < string.length; i++) {
if (string[i] == c) {
count += 1;
}
}
return count;
}
console.log(countBs(“BBC”));
// → 2
console.log(countChar(“kakkerlak”, “k”));
// → 4
Exercise 1 - Minimum
var x = prompt(“Please enter first value”);
var y = prompt(“Please enter second value”);
function Min(x,y){
if (x < y) return x;
else return y;
}
var answer = Min(x,y);
alert (answer);
Exercise 2 - recursion
var x = prompt(“Please enter value”);
function isEven( x ){
if (x == 0) return true;
else if (x == 1) return false;
else if (x < 0) return isEven(-x); //solution for maximum call stack size exceeded
else return isEven(x - 2);
}
var answer = isEven(x)
alert(answer)
Exercise 3 - counting bean
function countBs(word){
let count = 0;
for (i=0;i<word.length;i++){
if (word[i]===“B”) count += 1;
}
return count
}
var answer = countBs(“BbubbBbleBuBm”)
console.log(answer)
function countChar(word,char){
let count = 0;
for (i=0;i<word.length;i++){
if (word[i]===char) count += 1;
}
return count
}
var answer = countChar(“BbubbBbleBuBmuuuuu”,“u”)
console.log(answer)
Minimum:
function min(a,b){
if (a<b) return a;
else return b;
}
Recursion:
function isEven(n) {
if (n==0) return true;
else if (n==1) return false;
else if (n<0) return isEven(-n);
else return isEven(n-2);
}
Bean Counting:
function countChar(string, char) {
let count = 0;
for (let i = 0; i < string.length; i++) {
if (string[i] == char) {
count += 1;
}
}
return count;
}
function countBs(string) {
return countChar(string, “B”);
}
Exercise 1 Minimum:
function min(x,y) {
if (x>=y)
return y;
else return x;
}
Exercise 2 Recursion:
function isEven(x) {
if (x == 0)
return true;
else if (x == 1)
return false;
else if (x<0)
return isEven(-x);
else
return isEven(x-2);
}
Exercise 3 Bean Counting:
function countBs(x) {
let token = 0;
for (i = 0; i < x.length; i++)
{
if (x.charAt(i)==‘B’)
token++;
else
token = token + 0;
}
return token;
}
function countChar(x,y) {
let token = 0;
for (i = 0; i < x.length; i++)
{
if (x.charAt(i)==y)
token++;
else
token = token + 0;
}
return token;
}
if(x[i] === "B"){
token++
}
return token;
They both work the same.
When I learned Java in school I remember we had to use charAt() to pick a character from a String, so looked it up to see if it’s the same in Javascript and got the syntax from W3. I didn’t know you could just do x[i], I guess that’s an easier way to do it.
MINIMUM:
function min (a,b){
if (a < b){
return a
} else {return b}
};
console.log(min(20,30));
RECURSION
function min (a,b){
if (a < b){
return a
} else {return b}
};
console.log(min(20,30))
else if (x<0) return isEven(-x); solves the problem.
BEAN COUNTING
function countChar(str, chr){
let count = 0;
for (let i = 0; i < str.length; i+1){
if (str[i] == chr)
count+1;
}
return count;
}
function countBs(str){
return countChar(str, "B");
}
This is what i have so far and in console log it says even odd odd
did i do this right?
<h1> Title </h1>
<script>
function isEven(a){
if (a == 0){
return ("Even");}
else if (a == 1){
return ("Odd");}
else if (a < 0){
return isEven(-a);}
else{
return isEven(a-2);}}
console.log(isEven(50));
console.log(isEven(75));
console.log(isEven(-1));
</script>function isEven(a){
// % symbol called the Modulo Operator. It returns the remainder of dividing the left hand operand by right hand operand.
if(a % 2 === 0){
return ("Even");
}
else{
return ("odd");
}
}
<h1> Title </h1>
<script>
function isEven(a){
if (a % 2 === 0){
return ("Even");}
else if (a == 1){
return ("Odd");}
else if (a < 0){
return isEven(-a);}
else{
return isEven(a-2);}}
console.log(isEven(50));
console.log(isEven(75));
console.log(isEven(-1));
</script>
</body>
MINIMUM FUNCTION
function min(a,b){
if(a<b)
return a;
else
return b;
}
console.log(min(0, 10));
// → 0
console.log(min(0, -10));
// → -10
RECURSION FUNCTION
function isEven(a){
if (a%2==0)
return true;
if (a%2==1)
return false;
else
return isEven(a-2);
}
//If we change from a-2 to a+2 in the above line the program won’t crash and will work even if the given number is negative.
console.log(isEven(50));
// → true
console.log(isEven(75));
// → false
console.log(isEven(-1));
// ??
COUNTING FUNCTION
function countChar( s1 , s2){
let count=0;
for(var i=0 ; i< s1.length ;i++){
if (s1[i] == s2)
count++;
}
return count;
}
console.log(countChar(“BBC”,“B”));
// → 2
console.log(countChar(“kakkerlak”, “k”));
// → 4