Question

Write algorithms that perform the operations \(u \times 10^{m} ; u\) divide \(10^{m} ; u\) rem \(10^{m}\) where \(u\) represents a large integer, \(m\) is a nonnegative integer, divide returns the quotient in integer division, and rem returns the remainder. Analyze your algorithms, and show that these operations can be done in linear time.

Short answer

Multiplication by \(10^{m}\) can be done by appending \(m\) zeroes to \(u\). Division by \(10^{m}\) involves discarding the last \(m\) digits of \(u\). The remainder when \(u\) is divided by \(10^{m}\) is simply the last \(m\) digits of \(u\). The time complexity for all these operations is linear.

Step-by-step solution

Algorithm for Multiplication by powers of 10

Multiplication by \(10^{m}\) can be done by simply appending \(m\) zeroes at the end of the number represented by \(u\). A python implementation might look as follows: \n\n ```python\n def multiply_by_power_of_ten(u, m): \n return u * 10**m \n ```

Algorithm for Division by powers of 10

Division by \(10^{m}\) can be achieved by discarding the last \(m\) digits of the number represented by \(u\). A python implementation could be as follows: \n\n ```python\n def divide_by_power_of_ten(u, m): \n if(m > len(str(u))): \n return 0 \n return int(str(u)[:-m]) \n ``` \n Here, we are checking whether \(m\) is larger than the number of digits in \(u\). If true, we simply return 0 being the result of integer division of any integer with a larger integer.

Algorithm for Remainder by powers of 10

Finding the remainder when \(u\) is divided by \(10^{m}\) equates to getting the last \(m\) digits of \(u\). Here is a possible python implementation: \n\n ```python\n def remainder_by_power_of_ten(u, m): \n if(m > len(str(u))): \n return u \n return int(str(u)[-m:]) \n ``` \n Here, we consider the case when \(m\) is larger than the number of digits in \(u\), in which case, \(u\) is the remainder itself.

Proof of linear time complexity

In the multiplication function, the time taken scales linearly with the value of \(m\). In the division and remainder functions, the time taken scales linearly with the number of digits of \(u\). So, complexity of all the three operations scales linearly with respect to the input size, which confirms that all three operations can be done in linear time.

Key concepts

Multiplication by Powers of Ten

Understanding how to multiply a number by powers of ten is a fundamental concept in arithmetic and computing. Given a large integer, denoted by u, and a nonnegative integer power, m, to multiply u by 10m means essentially shifting the decimal point m places to the right. This is equivalent to appending m zeros to the end of u if u were written down in decimal form.

For instance, if we have u = 123 and m = 2, multiplying u by 10m results in 12300. In a computer algorithm, this process does not require multiplicative operations for each digit; instead, the operation can be done by manipulating the digits directly in a single pass which is what affords it a linear time complexity.

Division by Powers of Ten

When dividing an integer by powers of ten, the action you're taking is quite straightforward. Similar to multiplication, division by 10m corresponds to shifting the decimal point m places to the left. In effect, this means removing m digits from the end of the number, as long as m is not greater than the number of digits in u.

If m exceeds the number of digits, the quotient is zero, since any number divided by a larger number in integer division yields a zero result. To put it into perspective, 12345 divided by 102 would give 123, as the last two digits are discarded. This is another operation that benefits from linear time complexity due to its direct manipulation of the number's digits without the need for iterative calculations.

Remainder by Powers of Ten

The remainder operation, often denoted by 'rem', determines what is left over after division. For the remainder by powers of ten, we are interested in the digits that get discarded during the division by 10m process - these discarded digits constitute the remainder. In practical terms, if you have an integer u and you want to find u rem 10m, you are simply retrieving the last m digits of u.

For example, the remainder of 12345 divided by 102 would be 45. It is important to note that if m is greater than the number of digits in u, then the remainder is u itself, as the division yields zero with u left over. This operation, like the others, can also be executed in linear time since it involves a single action of extracting a subset of digits from the number.

Linear Time Complexity

When analyzing algorithms, time complexity is a key factor. It refers to the amount of time an algorithm takes to complete, based on the size of the input. Linear time complexity, denoted as O(n), means that the time an algorithm takes to run is directly proportional to the size of the input. This is the most desirable time complexity for an operation because it suggests efficiency and scalability.

In the case of the operations discussed — multiplication, division, and remainder by powers of ten — they all can be performed in linear time. The key to their linear time complexity lies in their ability to operate directly on the digits of the number with a single pass through the number's length, rather than requiring repetitive, nested calculations. This results in the time taken being proportional to the number of digits of the input u, or the power m, ensuring an efficient algorithm for large-scale computations.