Add and fix Algorithms

.gitignore

@@ -22,3 +22,4 @@ docs/
 
 # Code coverage
 *.lst
+.aider*

src/math/ceil.d

@@ -1,20 +1,30 @@
+
 import std;
 
 /**
-     * Returns the smallest (closest to negative infinity)
-     *
-     * @param number the number
-     * @return the smallest (closest to negative infinity) of given
-     * {@code number}
-*/
+ * Computes the smallest integer value that is greater than or equal to a given floating-point number.
+ *
+ * ### Step-by-Step:
+ * 1. `rem` = for get remainder of number when divided by 1.
+ * 2. In first if = if number is Positive and `rem != 0` (it has remainder) then `number - rem + 1`.
+ * 3. In second if = if number is Negative and `rem != 0` (it has remainder) then `number - rem`.
+ * 4. If number is already an integer then return the number as is.
+ *
+ * Params:
+ * number = The floating-point value to evaluate.
+ *
+ * Returns:
+ * If input has no remainder, it returns `number`. Otherwise it removes remaining part, adds one to `number` and returns it.
+ */
 
-double ceil(double number) {
-    if (number - cast(int) number == 0) {
-        return number;
-    } else if (number - cast(int) number > 0) {
-        return cast(int)(number + 1);
-    } else {
-        return cast(int) number;
+double ceil(double number){
+    double rem = number % 1;
+    if( number > 0 && rem != 0) { 
+        return number - rem + 1;
+    } else if (number < 0 && rem != 0) { 
+        return number - rem;
+    } else { 
+        return number; 
     }
 }
 
@@ -23,6 +33,10 @@ unittest {
     assert(ceil(10.2) == 11);
     assert(ceil(9) == 9);
     assert(ceil(12.6) == 13);
+
+    assert(ceil(-10.2) == -10);
+    assert(ceil(-12.6) == -12);
+    assert(ceil(-5.0) == -5);
 
 }
 

src/math/floor.d

@@ -0,0 +1,39 @@
+
+/**
+ * Computes the largest integer value that is less than or equal to a given floating-point number.
+ *
+ * ### Step-by-Step:
+ * 1. `rem` = for get remainder of number when divided by 1.
+ * 2. In first if = if number is Positive and `rem != 0` (it has remainder) then `x - rem`.
+ * 3. In second if = if number is Negative and `rem != 0` (it has remainder) then `x - rem - 1`.
+ * 4. If number is already an integer then return the number as is.
+ *
+ * Params:
+ * x = The floating-point value to evaluate.
+ *
+ * Returns:
+ * If input has no remainder, return `x`. Otherwise if number positive remove remaining and returns it, if negative, subtract one and return it.
+ */
+
+double floor(double x) {    
+    double rem = x % 1;
+    if( x > 0 && rem != 0) {
+        return x - rem;
+    } else if (x < 0 && rem != 0) {
+        return x - rem - 1;
+    } else {
+        return x;
+    }
+}
+
+
+
+
+unittest {
+    assert(floor(3.5) == 3);
+    assert(floor(-3.5) == -4);
+    assert(floor(3.0) == 3);
+    assert(floor(-3.0) == -3);
+}
+
+void main() {}

src/math/gcd.d

@@ -0,0 +1,40 @@
+import std;
+
+/**
+ * Greatest Common Divisor (GCD) Calculation - Computes the largest positive integer that divides two integers without leaving a remainder.
+ *
+ * ### Step-by-Step:
+ * 1. If either input `a` or `b` is 0, return 0.
+ * 2. Loop continues until `b` is zero.
+ * 3. `temp` = store the value of `b` temporarily before update `b` to `a % b`.
+ * 4. `b = a % b` = update `b` to remainder of `a` divided by `b`.
+ * 5. `a = temp` = update `a` to previous value of `b`.
+ *
+ * Params:
+ * a = The first integer.
+ * b = The second integer.
+ *
+ * Returns:
+ * In the end return `a` which is the GCD of the original pair of integers.
+ */
+
+long gcd(long a, long b) {
+    if (a == 0 || b == 0) {
+        return 0;
+    }
+    while (b != 0) {
+        long temp = b;
+        b = a % b;
+        a = temp;
+    }
+    return a;
+}
+
+unittest {
+    assert(gcd(48, 18) == 6);
+    assert(gcd(56, 98) == 14);
+    assert(gcd(101, 10) == 1);
+    assert(gcd(54, 24) == 6);
+}
+
+void main(){}

src/math/is_prime.d

@@ -0,0 +1,39 @@
+import std;
+
+/**
+ * Prime Check - An algorithm to determine if an integer is a prime number.
+ *
+ * ### Step-by-Step:
+ * 1. First if input `n <= 1` return false because it is not prime number.
+ * 2. Second if input `n == 2` return true because 2 is a prime number.
+ * 3. Third if input `n` is even return false because even numbers greater than 2 are not prime.
+ * 4. Start a loop from 3 to square root of `n`, increment by 2. If `n` is divisible by any of these, return false.
+ *
+ * Params:
+ * n = The integer to check.
+ *
+ * Returns:
+ * In the end return true if `n` is not divisible by any number from 2 to square root of `n`.
+ */
+
+bool is_prime(int n){
+    if(n <= 1) return false;
+    if(n == 2) return true;
+    if(n % 2 == 0) return false;
+    for(int i = 3; i * i <= n; i += 2){
+        if(n % i == 0) return false;        
+    }
+    return true;
+}
+unittest {
+    assert(is_prime(2) == true);
+    assert(is_prime(3) == true);
+    assert(is_prime(4) == false);
+    assert(is_prime(5) == true);
+    assert(is_prime(10) == false);
+    assert(is_prime(13) == true);
+    assert(is_prime(17) == true);
+    assert(is_prime(20) == false);
+}
+
+void main() {}

src/math/lcm.d

@@ -0,0 +1,45 @@
+
+import std;
+
+/**
+ * Least Common Multiple (LCM) Calculation - Computes the smallest positive integer that is divisible by both of two given integers.
+ *
+ * ### Step-by-Step:
+ * 1. If any of inputs `a` or `b` are zero, return 0.
+ * 2. If any of inputs are negative number, we take absolute value of that.
+ * 3. `result` = store output and multiply two inputs.
+ * 4. In while loop, we take GCD of the two inputs.
+ *
+ * Params:
+ * a = The first integer.
+ * b = The second integer.
+ *
+ * Returns:
+ * In the end, multiply two inputs, divide by their GCD, and return this result.
+ */
+
+
+long lcm(long a , long b){
+    if (a == 0 || b == 0) {
+        return 0;
+    }
+    if(a < 0) a = -a;
+    if(b < 0) b = -b;
+
+    long result = a * b;
+     while (b != 0) {
+        long temp = b;
+        b = a % b;
+        a = temp;
+    }
+    return result / a;
+}
+
+unittest {
+    assert(lcm(4, 6) == 12);
+    assert(lcm(21, 6) == 42);
+    assert(lcm(8, 9) == 72);
+    assert(lcm(15, 20) == 60);
+}
+
+void main() {}

src/math/prime_sieve.d

@@ -0,0 +1,51 @@
+import std;
+
+/**
+ * Prime Sieve - Generates all prime numbers up to a specified limit N using the Sieve of Eratosthenes algorithm.
+ *
+ * ### Step-by-Step:
+ * 1. In the first, we check that the input cannot be less than 2 because it is not a prime number.
+ * 2. We create a boolean array `is_prime` (size `n + 1`) to save valid outputs.
+ * 3. First we set the entire array to true.
+ * 4. We set members 0 and 1 to false because 0 and 1 are not prime.
+ * 5. In the loop, we check whether the current number is prime or not, and if not, saved in `is_prime` array.
+ *
+ * Params:
+ * n = The upper bound limit.
+ *
+ * Returns:
+ * In the end, checking values inside our validator array and added in `primes` array, return `primes`.
+ */
+
+
+int[] prime_sieve(int n){
+    if(n < 2) return [];
+    bool[] is_prime = new bool[n + 1];
+    is_prime[] = true;
+    is_prime[0] = false;
+    is_prime[1] = false;
+    for(int i = 2; i * i <= n; i++){
+        if(is_prime[i]){
+            for(int j = i * i; j <= n; j += i){
+                is_prime[j] = false;
+            }
+        }
+    }
+    int[] primes = [];
+    for(int i = 2; i <= n; i++){
+        if(is_prime[i]){
+            primes ~= i;
+        }
+    }
+    return primes;
+}
+
+unittest {
+    assert(prime_sieve(10) == [2, 3, 5, 7]);
+    assert(prime_sieve(20) == [2, 3, 5, 7, 11, 13, 17, 19]);
+    assert(prime_sieve(1) == []);
+    assert(prime_sieve(2) == [2]);
+}
+
+void main() {}
+

src/math/round.d

@@ -0,0 +1,48 @@
+import std;
+
+/**
+ * Rounding Function - Rounds a floating-point number to the nearest integer. Numbers exactly halfway between two integers are rounded away from zero.
+ *
+ * ### Step-by-Step:
+ * 1. `rem` = for get remainder of number when divided by 1.
+ * 2. In first if = if number is positive and fractional part `>= 0.5`, round up by subtracting `rem` and adding 1.
+ * 3. In second if = if number is positive and fractional part `< 0.5`, round down by subtracting `rem`.
+ * 4. In third if = if number is negative and fractional part `> -0.5`, round towards zero by subtracting `rem`.
+ * 5. In fourth if = if number is negative and fractional part `<= -0.5`, round away from zero by subtracting `rem` and 1.
+ * 6. Else = if number is already an integer, return the number as is.
+ *
+ * Params:
+ * x = The floating-point value to round.
+ *
+ * Returns:
+ * The rounded integer value according to the step conditions.
+ */
+
+
+double round(double x){
+    double rem = x % 1;
+    if (rem >= 0.5 && x > 0){
+        return x - rem + 1;
+    }else if (rem < 0.5 && x > 0){
+        return x - rem;
+    }else if (rem > -0.5 && x < 0){
+        return x - rem;
+    }else if (rem <= -0.5 && x < 0){
+        return x - rem - 1;
+    } else {
+        return x;
+    }
+}
+
+
+unittest{
+    assert(round(10.2) == 10);
+    assert(round(9) == 9);
+    assert(round(12.6) == 13);
+
+    assert(round(-10.2) == -10);
+    assert(round(-12.6) == -13);
+    assert(round(-5.0) == -5);
+}
+
+void main(){}