TheAlgorithms · sudiptiwari · Oct 11, 2023 · Oct 11, 2023 · Oct 11, 2023 · Oct 11, 2023
diff --git a/bit_manipulation/binary_addition.py b/bit_manipulation/binary_addition.py
@@ -0,0 +1,133 @@
+# Information on Binary Addition:
+# https://www.tutorialspoint.com/addition-of-two-n-bit-binary-numbers
+def binary_and(input1: str, input2: str) -> str:
+    """
+    AND gate logic.
+    >>> binary_and('0','1')
+    '0'
+    >>> binary_and('0','0')
+    '0'
+    >>> binary_and('1','1')
+    '1'
+    >>> binary_and('1','0')
+    '0'
+    """
+    if input1 == "1" and input2 == "1":
+        return "1"
+    else:
+        return "0"
+
+
+def binary_or(input1: str, input2: str) -> str:
+    """
+    OR gate logic.
+    >>> binary_or('0','1')
+    '1'
+    >>> binary_or('0','0')
+    '0'
+    >>> binary_or('1','1')
+    '1'
+    >>> binary_or('1','0')
+    '1'
+    """
+    if input1 == "1" or input2 == "1":
+        return "1"
+    else:
+        return "0"
+
+
+def binary_xor(input1: str, input2: str) -> str:
+    """
+    XOR gate logic.
+    >>> binary_xor('0','1')
+    '1'
+    >>> binary_xor('0','0')
+    '0'
+    >>> binary_xor('1','1')
+    '0'
+    >>> binary_xor('1','0')
+    '1'
+    """
+    if input1 == input2:
+        return "0"
+    else:
+        return "1"
+
+
+def addition(number_1: str, number_2: str, number_of_bits: int) -> tuple[str, str]:
+    """
+    return tuple with ('sum','carry')
+    The number of bits in 'sum' = number_of_bits passed to the function.
+    (i.e, if number_of_bits = 5, the length of 'sum' will also be 5).
+
+    Explanation:
+    The formula of sum and carry for each bit in binary operations are:
+    carry:    C5 C4 C3 C2 C1 C0
+    number_1:    A4 A3 A2 A1 A0
+    number_2:  + B4 B3 B2 B1 B0
+    ----------------------------
+    answer:   C5 S4 S3 S2 S1 S0
+
+    The formula for sum is:
+    S0 = A0 XOR B0 XOR C0
+    S1 = A1 XOR B1 XOR C1
+    .
+    .
+    and so on.
+
+    The formula for carry is:
+    C1 = A0 AND B0 OR ((A0 XOR B0) AND C0)
+    C2 = A1 AND B1 OR ((A1 XOR B1) AND C1)
+    .
+    .
+    and so on.
+
+    The numbers are reversed so that individual bits are traversed from R to L.
+    Finally, the resultant sum is reversed again to retain the original format.
+
+    >>> addition('1010','1101', 4)
+    ('0111', '1')
+    >>> addition('11111','00000', 5)
+    ('11111', '0')
+    >>> addition('0011','1111', 5)
+    ('10010', '0')
+    >>> addition('10011','110001', 6)
+    ('000100', '1')
+    >>> addition('10011','110001', 7)
+    ('1000100', '0')
+    >>> addition('1001','111', 4)
+    ('0000', '1')
+    >>> addition('1001','111', 5)
+    ('10000', '0')
+    >>> addition('101','10', 3)
+    ('111', '0')
+
+    Do not perform an operation as this  >>> addition('101','10', 2).
+    Since adding 3-bit number with any other number results to at least 3 bit number
+    but you are expecting a 2 bit number which is not possible.
+    """
+    number_1 = number_1.zfill(number_of_bits)  # zero padding at front
+    number_2 = number_2.zfill(number_of_bits)  # zero padding at front
+    reversed_number_1 = number_1[
+        ::-1
+    ]  # reverse for right to left traversal of bits using for loop
+    reversed_number_2 = number_2[
+        ::-1
+    ]  # reverse for right to left traversal of bits using for loop
+    carry = "0"  # initial carry in (C0) = 0
+    binary_sum = ""
+    for i in range(number_of_bits):
+        binary_sum = binary_sum + binary_xor(
+            binary_xor(reversed_number_1[i], reversed_number_2[i]), carry
+        )
+        intermediate_xor = binary_xor(reversed_number_1[i], reversed_number_2[i])
+        intermediate_and = binary_and(reversed_number_1[i], reversed_number_2[i])
+        carry = binary_or(intermediate_and, binary_and(intermediate_xor, carry))
+    binary_sum = binary_sum[::-1]
+    return binary_sum, carry
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()