# Round And Round (12" Version) ##BEST##

The artwork by Peter Saville mirrors the Technique album cover, but uses a different statue with a red or blue (depending on which version is owned) background. With the single following the New Order tradition of not including the title within the lyrics, the line "The picture you see is no portrait of me" was stickered on varying sleeves.

## Round and Round (12" Version)

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Malu Halasa, reviewer of British music newspaper Record Mirror, found the track "flamboyant, eminently danceable" with sound of "traditional gem from the Joy Division / New Order period".[3] In turn Simon Williams New Musical Express chided the song for lack of novelty. He wrote: "New Order haven't seriously affected my tender yound hormones since "1963". And their latest evacuee from Technology unwittingly exacerbates this unsatisfying state of affairs, as the customary swathes of synths, rhythmic ribbons and bass bandages wrap themselves around a once-skeletal framework... I'm sure that it sounds great at le discotheque after 15 gallons of hyper-toxic, liver-sizzling Latvian lager, but what doesn't?".[4]

The int value to the closest multiple of 10 to the power minus ndigits, where ndigits is the precision after the decimal point. If two multiples are equally close, rounding is done toward the even choice.

Note: In python, if we round off numbers to floor or ceil without giving the second parameter, it will return 15.0 for example and in Python 3 it returns 15, so to avoid this we can use (int) type conversion in python. It is also important to note that the round ()function shows unusual behavior when it comes to finding the mean of two numbers.

Approach:The code takes a float number num and uses the math.floor() function to round it down to the nearest integer. It then prints the result. It then uses the math.ceil() function to round num up to the nearest integer, and prints the result.

Time Complexity:Both the math.floor() function and the math.ceil() function have a constant time complexity, which means that the time complexity of the original code is also constant. Similarly, the round() function also has a constant time complexity, so the time complexity of the alternative code is also constant.

Rounding involves converting a numeric value with a specified precision to a value with less precision. For example, you can use the Round(Double) method to round a value of 3.4 to 3.0, and the Round(Double, Int32) method to round a value of 3.579 to 3.58.

In a midpoint value, the value after the least significant digit in the result is precisely half way between two numbers. For example, 3.47500 is a midpoint value if it is to be rounded to two decimal places, and 7.500 is a midpoint value if it is to be rounded to an integer. In these cases, if the round-to-nearest strategy is used, the nearest value can't be easily identified without a rounding convention.

Midpoint values are rounded to the next number away from zero. For example, 3.75 rounds to 3.8, 3.85 rounds to 3.9, -3.75 rounds to -3.8, and -3.85 rounds to -3.9. This form of rounding is represented by the MidpointRounding.AwayFromZero enumeration member.

Midpoint values are rounded to the nearest even number. For example, both 3.75 and 3.85 round to 3.8, and both -3.75 and -3.85 round to -3.8. This form of rounding is represented by the MidpointRounding.ToEven enumeration member.

In .NET Core 3.0 and later versions, three additional rounding strategies are available through the MidpointRounding enumeration. These strategies are used in all cases, not just for midpoint values as MidpointRounding.ToEven and MidpointRounding.AwayFromZero are.

Rounding away from zero is the most widely known form of rounding, while rounding to nearest even is the standard in financial and statistical operations. It conforms to IEEE Standard 754, section 4. When used in multiple rounding operations, rounding to nearest even reduces the rounding error that is caused by consistently rounding midpoint values in a single direction. In some cases, this rounding error can be significant.

The following example illustrates the bias that can result from consistently rounding midpoint values in a single direction. The example computes the true mean of an array of Decimal values, and then computes the mean when the values in the array are rounded by using the two conventions. In this example, the true mean and the mean that results when rounding to nearest are the same. However, the mean that results when rounding away from zero differs by .05 (or by 3.6%) from the true mean.

In order to determine whether a rounding operation involves a midpoint value, the Round method multiplies the original value to be rounded by 10n, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. This is a slight variation on a test for equality, and as discussed in the "Testing for Equality" section of the Double reference topic, tests for equality with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.

The following example illustrates the problem. It repeatedly adds .1 to 11.0 and rounds the result to the nearest integer. 11.5 should round to 12 using either of the midpoint-rounding conventions (ToEven or AwayFromZero). However, as the output from the example shows, it does not. The example uses the "R" standard numeric format string to display the floating point value's full precision, and shows that the value to be rounded has lost precision during repeated additions, and its value is actually 11.499999999999998. Because .499999999999998 is less than .5, the midpoint-rounding conventions don't come into play and the value is rounded down. As the example also shows, this problem does not occur if you assign the constant value 11.5 to a Double variable.

If the rounding operation calls an overload that rounds a Double value, you can change the Double to a Decimal value and call an overload that rounds a Decimal value instead. Although the Decimal data type also has problems of representation and loss of precision, these issues are far less common.

Define a custom rounding algorithm that performs a "nearly equal" test to determine whether the value to be rounded is acceptably close to a midpoint value. The following example defines a RoundApproximate method that examines whether a fractional value is sufficiently near to a midpoint value to be subject to midpoint rounding. As the output from the example shows, it corrects the rounding problem shown in the previous example.

The Round method includes overloads that accept arguments of type Decimal and Double. There are no methods that round values of type Single. If you pass a Single value to one of the overloads of the Round method, it is cast (in C#) or converted (in Visual Basic) to a Double, and the corresponding Round overload with a Double parameter is called. Although this is a widening conversion, it often involves a loss of precision, as the following example illustrates. When a Single value of 16.325 is passed to the Round method and rounded to two decimal places using the rounding to nearest convention, the result is 16.33 and not the expected result of 16.32.

This unexpected result is due to a loss of precision in the conversion of the Single value to a Double. Because the resulting Double value of 16.325000762939453 is not a midpoint value and is greater than 16.325, it is always rounded upward.

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, Int32, MidpointRounding) method may not appear to round midpoint values as specified by the mode parameter. This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. This occurs because internally the method multiplies value by 10digits, and the multiplication operation in this case suffers from a loss of precision.

Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, MidpointRounding) method may not appear to round midpoint values to the nearest even integer. In the following example, because the floating-point value .1 has no finite binary representation, the first call to the Round(Double) method with a value of 11.5 returns 11 instead of 12.

Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, Int32) method may not appear to round midpoint values to the nearest even value in the digits decimal position. This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. This occurs because internally the method multiplies value by 10digits, and the multiplication operation in this case suffers from a loss of precision. 041b061a72