I am trying to have a solution for a pdf figure out a equation and round up to the nearest 1/8th of a yard. The equation I am using that I want rounded is (Text17/15)+.25 How can I make that round up everytime?
Assuming that Text17 is a text field, containing a numeric value, and you always want to round up, you could put somehting like this into the Calculate event of the field which is supposed to show the result:
event.value = util.printf("%.3f", Math.ceil(((this.getField("Text17").value*1 / 15) + 0.25) * 8) / 8 );
And that should do it.
Although both the functions perform the same operation, even they produce same o/p, what is basic difference between these two? Is there any performance related difference, if yes, then which one is better?
Thanks
The documentation is pretty clear on the difference:
NOTE
The REMAINDER function uses the round function in its formula, whereas
the MOD function uses the floor function in its formula.
In other words, when the arguments are positive integers, the mod function returns a positive number between 0 and the second argument. The remainder function returns a number whose absolute value is less than the second argument divided by 2.
The differences can be more striking for negative numbers. One example of a difference is:
REMAINDER(-15, 4)
MOD(-15, 4)
The first gives -3 and the second 1.
EDIT:
What is happening here? How many times does 4 go into -15. One method is "-4" times with a remained of 1. That is: -15 = 4*(-4) + 1. The other is "-3" times: -15 = 4*(-3) - 3.
The difference what is -15/4 expressed as an integer. Using floor, you get -4. Using round, you get -3.
The difference between MOD and REMAINDER function can be understood in the example below:
MOD(13,5): returns 3 whereas, REMAINDER (13,5) returns -2
A simple way to understand the difference is that MOD uses the floor function therefore it counts the occurrence of the second number within the first and returns what is left to complete the first number i.e. 2(5) gives 10 adding 3 gives 13 therefore MOD(13,5)=3
However, the REMAINDER uses a Round function it therefore gets the total number of the second number that could make up the first and then subtract the what makes it excess. i.e. 3(5) = 15 and subtracting 2 gives 13,therefore REMAINDER(13,5)= -2
REMAINDER (-15, 4): 4(-4) gives -16 and adding +1 gives -15 hence REMAINDER (-15,4)=+1
MOD (-15, 4): 3(-4) gives -12, adding -3 gives us -15 hence MOD(15,4)=-3
REMAINDER(-15, 4)--uses round without taking the sign of the number into
consideration.
hence -15/4= -3.75==> round(-3.75)= -4.--(ignore sign during round)
-4*4= -16
-15-(-16)=>1
There fore: REMAINDER(-15, 4)=1
MOD(-15, 4)----uses Floor without taking the sign of the number into
consideration.
-15/4= -3.75==> floor(-3.75)= -3.--(ignore sign in floor)
-3*4=-12
-15-(-12)=>-3
There fore: MOD(-15, 4)= -3
Mod(m,n) is so simple to understand.
Finding Value Mod() output value will always be the manually calculated remainder value when we divide m by n.
Finding Sign The sign of the output remains the same as the first parameter.
eg :
eg 1 : mod (11,3) is the remainder of 11/3 which is 2 and the same sign of 1st parameter. So output is 2
eg 2 : mod (-11,3) is the remainder of 11/3 which is 2 and the same sign of the 1st parameter. So the output is -2
Remainder(m,n) is a bit different
Finding Value You take two multiples of n, such that when multiplied gives the closest lower value and the closest upper value compared to the first parameter(m). The minimum difference between these values will be the output value
Finding Sign The sign of the output value will always be positive if the closest value is less than the first parameter and the if the closest value is greater than the 1st parameter then it will be negative.
eg :
eg 1 : remainder(10,3) The closest multiple values of 3 which are lesser and greater than 10 are 9(3x3) and 12(3x4). And the closest to 10 among 9 & 12 is 9. So the output will be the gap between 9 and 10 which is 1. The closest number is less than the 1st parameter. So the output is 1.
eg 2 : remainder(11,3) The closest multiple values of 11 which are lesser and greater than 11 are 9(3x3) and 12(3x4). And the closest to 12 among 9 & 12 is 12. So the output will be the gap between 12 and 11 which is 1. The closest number is greater than the 1st parameter. So the output is -1.
The question has long had an answer, but I thought some context would be helpful.
https://en.wikipedia.org/wiki/Modulo_operation discusses the “modulo” operation, and the fact that it is based on the remainder after integer division. The problem is that the concept of “remainder” is not clearly defined where there are negative numbers.
Where negative numbers are involved, the difference between modulus and remainder is significant. Mathematically, the modulus operation maps an integer on to a range. This range begins at 0 and is cyclic. For example integer mod 3 is mapped on to the range:
0, 1, 2, 0, 1, 2, …
If the given integer is negative, the cycle is simply extended backwards. If, however, the modulus itself is negative, then the whole range is negative:
0, -2, -1, 0, -2, -1, …
https://rob.conery.io/2018/08/21/mod-and-remainder-are-not-the-same/ illustrates this as a clock with negative numbers.
The practical upshot of this is that the sign of the second number is the same as the sign of the result.
With a remainder, however, you attempt to divide the second number into the first number until you can’t go any more. The remaining difference is called the remainder. If the first number is negative, then, if you stop short, the remainder will be negative.
The practical upshot of which is that the sign of the first number is the sign of the result.
Most coding languages, including most variations of SQL, use a remainder calculation, but very often call it the modulus. And most languages use the notation a % b meaning a mod b.
Oracle, of course, has two functions, which should have been helpful. However, the first function, mod(), actually gives what everybody else calls the remainder. The second, remainder() gives a result which nobody else gives, due to the fact that it is focused on the nearest division. As per the documentation, the mod() functions uses the floor() function, while remainder() uses round().
If you want to compare the results, you can try:
SELECT
remainder(19,5) AS "Remainder ++",
remainder(-19,5) AS "Remainder -+",
remainder(19,-5) AS "Remainder +-",
remainder(-19,-5) AS "Remainder --",
mod(19,5) AS "Mod ++",
mod(-19,5) AS "Mod -+",
mod(19,-5) AS "Mod +-",
mod(-19,-5) AS "Mod --",
mod(mod(19,5) + 5,5) AS "Modulo ++",
mod(mod(-19,5) + 5,5) AS "Modulo ++",
mod(mod(19,-5) + -5,-5) AS "Modulo ++",
mod(mod(-19,-5) + -5,-5) AS "Modulo ++"
FROM DUAL
;
The modulo calculation gives the true modulus, according to https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Remainder
In all, you probably want the remainder as the most natural interpretation, and so you will use mod() to get this result.
I am currently trying to figure out how to multiply two numbers in fixed point representation.
Say my number representation is as follows:
[SIGN][2^0].[2^-1][2^-2]..[2^-14]
In my case, the number 10.01000000000000 = -0.25.
How would I for example do 0.25x0.25 or -0.25x0.25 etc?
Hope you can help!
You should use 2's complement representation instead of a seperate sign bit. It's much easier to do maths on that, no special handling is required. The range is also improved because there's no wasted bit pattern for negative 0. To multiply, just do as normal fixed-point multiplication. The normal Q2.14 format will store value x/214 for the bit pattern of x, therefore if we have A and B then
So you just need to multiply A and B directly then divide the product by 214 to get the result back into the form x/214 like this
AxB = ((int32_t)A*B) >> 14;
A rounding step is needed to get the nearest value. You can find the way to do it in Q number format#Math operations. The simplest way to round to nearest is just add back the bit that was last shifted out (i.e. the first fractional bit) like this
AxB = (int32_t)A*B;
AxB = (AxB >> 14) + ((AxB >> 13) & 1);
You might also want to read these
Fixed-point arithmetic.
Emulated Fixed Point Division/Multiplication
Fixed point math in c#?
With 2 bits you can represent the integer range of [-2, 1]. So using Q2.14 format, -0.25 would be stored as 11.11000000000000. Using 1 sign bit you can only represent -1, 0, 1, and it makes calculations more complex because you need to split the sign bit then combine it back at the end.
Multiply into a larger sized variable, and then right shift by the number of bits of fixed point precision.
Here's a simple example in C:
int a = 0.25 * (1 << 16);
int b = -0.25 * (1 << 16);
int c = (a * b) >> 16;
printf("%.2f * %.2f = %.2f\n", a / 65536.0, b / 65536.0 , c / 65536.0);
You basically multiply everything by a constant to bring the fractional parts up into the integer range, then multiply the two factors, then (optionally) divide by one of the constants to return the product to the standard range for use in future calculations. It's like multiplying prices expressed in fractional dollars by 100 and then working in cents (i.e. $1.95 * 100 cents/dollar = 195 cents).
Be careful not to overflow the range of the variable you are multiplying into. Your constant might need to be smaller to avoid overflow, like using 1 << 8 instead of 1 << 16 in the example above.
I have 2 inputs from 0-180 for x and y i need to add them together and stay in the range of 180 and 0 i am having some trouble since 90 is the mid point i cant seem to keep my data in that range im doing this in vb.net but i mainly need help with the logic
result = (x + y) / 2
Perhaps? At least that will stay in the 0-180 range. Are there any other constraints you're not telling us about, since right now this seems pretty obvious.
If you want to map the two values to the limited range in a linear fashion, just add them together and divide by two:
out = (in1 + in2) / 2
If you just want to limit the top end, add them together then use the minimimum of that and 180:
out = min (180, in1 + in2)
Are you wanting to find the average of the two or add them? If you're adding them, and you're dealing with angles which wrap around (which is what it sounds like) then, why not just add them and then modulo? Like this:
(in1 + in2) mod 180
Hopefully you're familiar with the modulo operator.
I am doing this small exercise.
declare #No decimal(38,5);
set #No=12345678910111213.14151;
select #No*1000/1000,#No/1000*1000,#No;
Results are:
12345678910111213.141510
12345678910111213.141000
12345678910111213.14151
Why are the results of first 2 selects different when mathematically it should be same?
it is not going to do algebra to convert 1000/1000 to 1. it is going to actually follow the order of operations and do each step.
#No*1000/1000
yields: #No*1000 = 12345678910111213141.51000
then /1000= 12345678910111213.141510
and
#No/1000*1000
yields: #No/1000 = 12345678910111.213141
then *1000= 12345678910111213.141000
by dividing first you lose decimal digits.
because of rounding, the second sql first divides by 1000 which is 12345678910111.21314151, but your decimal is only 38,5, so you lose the last three decimal points.
because when you divide first you get:
12345678910111.21314151
then only six decimal digits are left after point:
12345678910111.213141
then *1000
12345678910111213.141
because the intermediary type is the same as the argument's - in this case decimal(38,5). so dividing first gives you a loss of precision that's reflected in the truncated answer. multiplying by 1000 first doesn't give any loss of precision because that doesn't overload 38 digits.
It's probably because you lose part of data making division first. Notice that #No has 5-point decimal precision so when you divide this number by 1000 you suddenly need 8 digits for decimal part:
123.12345 / 1000 = 0.12312345
So the value has to be rounded (0.12312) and then this value is multiply by 1000 -> 123.12 (you lose 0.00345.
I think that's why the result is what it is...
The first does #No*1000 then divides it by 1000. The intermediates values are always able to represent all the decimal places. The second expression first divides by 1000, which throws away the last two decimal places, before multiplying back to the original value.
You can get around the problem by using CONVERT or CAST on the first value in your expression to increase the number of decimal places and avoid a loss of precision.
DECLARE #num decimal(38,5)
SET #num = 12345678910111213.14151
SELECT CAST(#num AS decimal(38,8)) / 1000 * 1000