Why do numbers with 17 and more digits turn EVEN automatically? - facebook

I'm testing a photo application for Facebook. I'm getting object IDs from the Facebook API, but I received some incorrect ones, which doesn't make sense - why would Facebook send wrong IDs? I investigated a bit and found out that numbers with 17 and more digits are automatically turning into even numbers!
For example, let's say the ID I'm supposed to receive from Facebook is 12345678912345679. In the debugger, I've noticed that Flash Player automatically turns it into 12345678912345678. And I even tried to manually set it back to an odd number, but it keeps changing back to even.
Is there any way to stop Flash Player from rounding the numbers? BTW the variable is defined as Object, I receive it like that from Facebook.

This is related to the implementation of data types:
int is a 32-bit number, with an even distribution of positive and
negative values, including 0. So the maximum value is
(2^32 / 2 ) - 1 == 2,147,483,647.
uint is also a 32-bit number, but it doesn't have negative values. So the
maximum value is
2^32 - 1 == 4,294,967,295.
When you use a numerical value greater than the maximum value of int or uint, it is automatically cast to Number. From the Adobe Doc:
The Number data type is useful when you need to use floating-point
values. Flash runtimes handle int and uint data types more efficiently
than Number, but Number is useful in situations where the range of
values required exceeds the valid range of the int and uint data
types. The Number class can be used to represent integer values well
beyond the valid range of the int and uint data types. The Number data
type can use up to 53 bits to represent integer values, compared to
the 32 bits available to int and uint.
53 bits have a maximum value of:
2^53 - 1 == 9,007,199,254,740,989 => 16 digits
So when you use any value greater than that, the inner workings of floating point numbers apply.
You can read about floating point numbers here, but in short, for any floating point value, the first couple of bits are used to specify a multiplication factor, which determines the location of the point. This allows for a greater range of values than are actually possible to represent with the number of bits available - at the cost of reduced precision.
When you have a value greater than the maximum possible integer value a Number could have, the least significant bit (the one representing 0 and 1) is cut off to allow for a more significant bit (the one representing 2^54) to exist => hence, you lose the odd numbers.
There is a simple way to get around this: Keep all your IDs as Strings - they can have as many digits as your system has available bytes ;) It's unlikely you're going to do any calculations with them, anyway.
By the way, if you had a value greater than 2^54-(1+2), your numbers would be rounded down to the next multiple of 4; if you had a value greater than 2^55-(1+2+4), they would be rounded down to the next multiple of 8, etc.

Related

What's Int.MaxValue between friends?

The max values of int, float and long in Scala are:
Int.MaxValue = 2147483647
Float.MaxValue = 3.4028235E38
Long.MaxValue = 9223372036854775807L
From the authors of Scala compiler, Keynote, PNW Scala 2013, slide 16 What's Int.MaxValue between friends?:
val x1: Float = Long.MaxValue
val x2: Float = Long.MaxValue - Int.MaxValue
println (x1 == x2)
// NO WONDER NOTHING WORKS
Why does this expression return true?
A Float is a 4-byte floating point value. Meanwhile a Long is an 8-byte value and an Int is also a 4-byte value. However, the way numbers are stored in 4-byte floating point values means that they have only around 8 digits of precision. Consequently, they do not have the capacity to store even the 4 most significant bytes (around 9-10 digits) of a Long regardless of the value of the least 4 significant bytes (another 9-10 digits).
Consequently, the Float representation of the two expressions is the same, because the bits that differ are below the resolution of a Float. Hence the two values compare equal.
Echoing Mike Allen's answer, but hoping to provide some additional context (would've left this as a comment rather than a separate answer, but SO's reputation feature wouldn't let me).
Integers have a maximum range of values defined as either 0 to 2^n (if it is an unsigned integer) or -2^(n-1) to 2^(n-1) (for signed integers) where n is the number of bits in the underlying implementation (n=32 in this case). If you wish to represent a number larger than 2^31 with a signed value, you can't use an int. A signed long will work up to 2^63. For anything larger than this, a signed float can go up to roughly 2^127.
One other thing to note is that these resolution issues are only in force when the value stored in the floating point number approaches the max. In this case, the subtraction operation causes a change in true value that is many orders of magnitude smaller than the first value. A float would not round off the difference between 100 and 101, but it might round off the difference between 10000000000000000000000000000 and 10000000000000000000000000001.
Same goes for small values. If you cast 0.1 to an integer, you get exactly 0. This is not generally considered a failing of the integer data type.
If you are operating on numbers that are many orders of magnitude different in size, and also not able to tolerate rounding errors, you will need data structures and algorithms that account for inherent limitations of binary data representation. One possible solution would be to use a floating point encoding with fewer bits of exponential, thereby limiting the max value but providing for greater resolution is less significant bits. For greater detail, check out:
look up the IEEE Standard 754 (which defines the floating point encoding)
http://steve.hollasch.net/cgindex/coding/ieeefloat.html
https://randomascii.wordpress.com/2012/02/25/comparing-floating-point-numbers-2012-edition/

How to get smallest value (stride) for particular number type in Swift?

How to get smallest value (stride) for particular number type in Swift? I mean, the shortest non-zero stride.
For example 1 for Int, 0.00...001 for Double etc...
This has been partially addressed in separate comments already, but to bring it all together...
For integer types:
The smallest possible increment between distinct values is 1. This is part of the definition of an integer, so it's such a foundational aspect of the type that there isn't (and needn't be) a special API for finding it.
(One can argue that successor constitutes such an API. But one can also argue that using successor where you could just use 1 makes your code much less readable.)
For floating-point types:
There is no one smallest possible increment. Because floating-point types use an exponent-based representation, the spacing between representable floating-point numbers varies with the exponent, as illustrated in this number line:
That image comes from What Every Computer Scientist Should Know About Floating-Point Arithmetic, an essential read for, well, everyone using floating-point numbers. You can read more about the concept in Wikipedia's pages for Unit in the Last Place and Machine Epsilon. Exploring Binary also has a good entry-level read on floating-point spacing.
Back to Swift — The Float and Double types conform to the FloatingPoint protocol (in Swift 3, currently in beta). This defines the features of IEEE 754 floating point formats, which includes both:
The Unit in the Last Place, or ulp, which tells you the increment between a number and the next greater representable number (but for some edge cases). This is related to, but not the same as, machine epsilon, since it scales with value. (ulpOfOne is the same as what other libraries call machine epsilon.)
nextUp and nextDown, which tell you the closest greater or lesser representable numbers.
Here's an example (conveniently showing that for 32-bit Float, the minimum increment gets bigger than one sooner than you might think):
let ichi: Float = 1.0
ichi.ulp // -> 1.192093e-07
ichi.nextUp // -> 1.00000012
let man: Float = 10_000
man.ulp // -> 10000
man.nextUp // -> 10000.001
let oku: Float = 100_000_000
oku.ulp // -> 8
oku.nextUp // -> 100000008.0
In Swift 2, there's no FloatingPoint protocol, but you can make use of the equivalent POSIX constants/functions imported from C: FLT_EPSILON and DBL_EPSILON are defined as the difference between 1.0 and the next representable value, and the nextafter functions find the increment at any value.
You can get these numbers by importing Darwin and the numbers you're looking for will be
DBL_MIN
FLT_MIN
Int.min
However these won't be 'nice' numbers like 0.00 ... 01 and Int will be signed so using Uint8.min will give you 0.

Efficiently Store Decimal Numbers with Many Leading Zeros in Postgresql

A number like:
0.000000000000000000000000000000000000000123456
is difficult to store without a large performance penalty with the available numeric types in postgres. This question addresses a similar problem, but I don't feel like it came to an acceptable resolution. Currently one of my colleagues landed on rounding numbers like this to 15 decimal places and just storing them as:
0.000000000000001
So that the double precision numeric type can be used which prevents the penalty associated with moving to a decimal numeric type. Numbers that are this small for my purposes are more or less functionally equivalent, because they are both very small (and mean more or less the same thing). However, we are graphing these results and when a large portion of the data set would be rounded like this it looks exceptionally stupid (flat line on the graph).
Because we are storing tens of thousands of these numbers and operating on them, the decimal numeric type is not a good option for us as the performance penalty is too large.
I am a scientist, and my natural inclination would just be to store these types of numbers in scientific notation, but it does't appear that postgres has this kind of functionality. I don't actually need all of the precision in the number, I just want to preserve 4 digits or so, so I don't even need the 15 digits that the float numeric type offers. What are the advantages and disadvantages of storing these numbers in two fields like this:
1.234 (real)
-40 (smallint)
where this is equivalent to 1.234*10^-40? This would allow for ~32000 leading decimals with only 2 bytes used to store them and 4 bytes to store the real value, for a total of maximally 6 bytes per number (gives me the exact number I want to store and takes less space than the existing solution which consumes 8 bytes). It also seems like sorting these numbers would be much improved as you'd need only sort on the smallint field first followed by the real field second.
You and/or your colleague seem to be confused about what numbers can be represented using the floating point formats.
A double precision (aka float) number can store at least 15 significant digits, in the range from about 1e-307 to 1e+308. You have to think of it as scientific notation. Remove all the zeroes and move that to the exponent. If whatever you have once in scientific notation has less than 15 digits and an exponent between -307 and +308, it can be stored as is.
That means that 0.000000000000000000000000000000000000000123456 can definitely be stored as a double precision, and you'll keep all the significant digits (123456). No need to round that to 0.000000000000001 or anything like that.
Floating point numbers have well-known issue of exact representation of decimal numbers (as decimal numbers in base 10 do not necessarily map to decimal numbers in base 2), but that's probably not an issue for you (it's an issue if you need to be able to do exact comparisons on such numbers).
What are the advantages and disadvantages of storing these numbers in
two fields like this
You'll have to manage 2 columns instead of one.
Roughly, what you'll be doing is saving space by storing lower-precision floats. If you only need 4 digits of precision, you can go further and save 2 more bytes by using smallint + smallint (1000-9999 + exponent). Using that format, you could cram the two smallint into one 32 bits int (exponent*2^16 + mantissa), that should work too.
That's assuming that you need to save storage space and/or need to go beyond the +/-308 digits exponent limit of the double precision float. If that's not the case, the standard format is fine.

Why does my UILabel start counting backward at 2 billion?

I have a UILabel in my iPhone app simulator. It displays a coin count and I have an action that adds 1 hundred million to the count. I want the number to keep going up but for some reason once the count hits 2 billion, it adds a minus sign and starts counting down, then counts back up to 2 billion and back down again and so on.
I want to be able to display a much greater number of digits ie trillions and so on... Does anyone know what's going on with this and how to fix it so the label digits will keep going up as high as I want.
I'm using Xcode and Interface Builder and running through the simulator. I'm storing the number in a int variable, if that matters.
You store your coin count in an int, that's the problem. A 4 byte int can't store numbers higher than 2,147,483,647. If you add 1 to 2,147,483,647 you will get −2,147,483,648, which is the smallest possible int.
If you want to store bigger numbers you have to use a long which can store numbers between −(2^63) and 2^63−1 (or −9,223,372,036,854,775,808 to 9,223,372,036,854,775,807).
See this for additional details.
This is occurring because, as #DrummerB pointed out, your int variable only has enough bits to store integral values in the range of -2,147,483,647 to 2,147,483,647. The reason this gets "reset" or "rolls over" back to a negative has to do with how computers store data, which is in binary.
For example, if you had an 8-bit integer (otherwise known as a byte) can store integral values from 0 to 255 if it is unsigned (meaning it can only store positive values) and -127 to 127 if it is signed (meaning it can store negative numbers). When an integer reaches its max value, it is represented in memory by all ones as you see here with the unsigned value 255:
255 = 11111111
So the maximum number that can be stored in an 8-bit int (byte) is 255. If you add 1 to this value, you end up flipping all the 1 values so that they are zeroes and since storing the value 256 would require a 9th bit you lose the 9th bit entirely and the integer value will appear to "roll over" to the minimum value.
Now.. As I stated above, the result of the addition above yields the value 256, but we only have 8 bits of storage in our integer so the most significant bit (9th bit) is lost. So you can picture it kinda like this with the pipes | marking your storage area:
only 8 bits of storage total
v
255 = 0|11111111|
+ 1 = 0|00000001|
-------------------
256 = 1|00000000|
^
9th bit is lost
In an unsigned int, the same is true, however the first bit is used to determine if the value is negative so you gain signing but you lose 1 bit of storage, resulting in your only having enough space to store the values 0 to 127 and 1 bit for signing.
Now that we understand what's going on, it should be noted that iOS is, at the time of this writing, a 32-bit operating system and while it can handle larger integers you probably don't want to use them all over the place as it's not optimized to work with these values.
If you just want to increase the range of values you can store in this variable, I would recommend changing it to an unsigned int, which can be done using the NSUInteger typedef.

Fixed point vs Floating point number

I just can't understand fixed point and floating point numbers due to hard to read definitions about them all over Google. But none that I have read provide a simple enough explanation of what they really are. Can I get a plain definition with example?
A fixed point number has a specific number of bits (or digits) reserved for the integer part (the part to the left of the decimal point) and a specific number of bits reserved for the fractional part (the part to the right of the decimal point). No matter how large or small your number is, it will always use the same number of bits for each portion. For example, if your fixed point format was in decimal IIIII.FFFFF then the largest number you could represent would be 99999.99999 and the smallest non-zero number would be 00000.00001. Every bit of code that processes such numbers has to have built-in knowledge of where the decimal point is.
A floating point number does not reserve a specific number of bits for the integer part or the fractional part. Instead it reserves a certain number of bits for the number (called the mantissa or significand) and a certain number of bits to say where within that number the decimal place sits (called the exponent). So a floating point number that took up 10 digits with 2 digits reserved for the exponent might represent a largest value of 9.9999999e+50 and a smallest non-zero value of 0.0000001e-49.
A fixed point number just means that there are a fixed number of digits after the decimal point. A floating point number allows for a varying number of digits after the decimal point.
For example, if you have a way of storing numbers that requires exactly four digits after the decimal point, then it is fixed point. Without that restriction it is floating point.
Often, when fixed point is used, the programmer actually uses an integer and then makes the assumption that some of the digits are beyond the decimal point. For example, I might want to keep two digits of precision, so a value of 100 means actually means 1.00, 101 means 1.01, 12345 means 123.45, etc.
Floating point numbers are more general purpose because they can represent very small or very large numbers in the same way, but there is a small penalty in having to have extra storage for where the decimal place goes.
From my understanding, fixed-point arithmetic is done using integers. where the decimal part is stored in a fixed amount of bits, or the number is multiplied by how many digits of decimal precision is needed.
For example, If the number 12.34 needs to be stored and we only need two digits of precision after the decimal point, the number is multiplied by 100 to get 1234. When performing math on this number, we'd use this rule set. Adding 5620 or 56.20 to this number would yield 6854 in data or 68.54.
If we want to calculate the decimal part of a fixed-point number, we use the modulo (%) operand.
12.34 (pseudocode):
v1 = 1234 / 100 // get the whole number
v2 = 1234 % 100 // get the decimal number (100ths of a whole).
print v1 + "." + v2 // "12.34"
Floating point numbers are a completely different story in programming. The current standard for floating point numbers use something like 23 bits for the data of the number, 8 bits for the exponent, and 1 but for sign. See this Wikipedia link for more information on this.
The term ‘fixed point’ refers to the corresponding manner in which numbers are represented, with a fixed number of digits after, and sometimes before, the decimal point.
With floating-point representation, the placement of the decimal point can ‘float’ relative to the significant digits of the number.
For example, a fixed-point representation with a uniform decimal point placement convention can represent the numbers 123.45, 1234.56, 12345.67, etc, whereas a floating-point representation could in addition represent 1.234567, 123456.7, 0.00001234567, 1234567000000000, etc.
There's of what a fixed-point number is and , but very little mention of what I consider the defining feature. The key difference is that floating-point numbers have a constant relative (percent) error caused by rounding or truncating. Fixed-point numbers have constant absolute error.
With 64-bit floats, you can be sure that the answer to x+y will never be off by more than 1 bit, but how big is a bit? Well, it depends on x and y -- if the exponent is equal to 10, then rounding off the last bit represents an error of 2^10=1024, but if the exponent is 0, then rounding off a bit is an error of 2^0=1.
With fixed point numbers, a bit always represents the same amount. For example, if we have 32 bits before the decimal point and 32 after, that means truncation errors will always change the answer by 2^-32 at most. This is great if you're working with numbers that are all about equal to 1, which gain a lot of precision, but bad if you're working with numbers that have different units--who cares if you calculate a distance of a googol meters, then end up with an error of 2^-32 meters?
In general, floating-point lets you represent much larger numbers, but the cost is higher (absolute) error for medium-sized numbers. Fixed points get better accuracy if you know how big of a number you'll have to represent ahead of time, so that you can put the decimal exactly where you want it for maximum accuracy. But if you don't know what units you're working with, floats are a better choice, because they represent a wide range with an accuracy that's good enough.
It is CREATED, that fixed-point numbers don't only have some Fixed number of decimals after point (digits) but are mathematically represented in negative powers. Very good for mechanical calculators:
e.g, the price of smth is USD 23.37 (Q=2 digits after the point. ) The machine knows where the point is supposed to be!
Take the number 123.456789
As an integer, this number would be 123
As a fixed point (2), this
number would be 123.46 (Assuming you rounded it up)
As a floating point, this number would be 123.456789
Floating point lets you represent most every number with a great deal of precision. Fixed is less precise, but simpler for the computer..