I want to normalize an array so that each value is
in [0-1) .. i.e. "the max will never be 1 but the min can be 0."
This is not unlike the random function returning numbers in the same range.
While looking at this, I found that .99999999999999999===1 is true!
Ditto (1-Number.MIN_VALUE) === 1 But Math.ceil(Number.MIN_VALUE) is 1, as it should be.
Some others: Math.floor(.999999999999) is 0
while Math.floor(.99999999999999999) is 1
OK so there are rounding problems in JS.
Is there any way I can normalize a set of numbers to lie in the range [0,1)?
It may help to examine the steps that JavaScript performs of each of your expressions.
In .99999999999999999===1:
The source text .99999999999999999 is converted to a Number. The closest Number is 1, so that is the result. (The next closest Number is 0.99999999999999988897769753748434595763683319091796875, which is 1–2–53.)
Then 1 is compared to 1. The result is true.
In (1-Number.MIN_VALUE) === 1:
Number.MIN_VALUE is 2–1074, about 5e–304.
1–2–1074 is extremely close to one. The exact value cannot be represented as a Number, so the nearest value is used. Again, the nearest value is 1.
Then 1 is compared to 1. The result is true.
In Math.ceil(Number.MIN_VALUE):
Number.MIN_VALUE is 2–1074, about 5e–304.
The ceiling function of that value is 1.
In Math.floor(.999999999999):
The source text .999999999999 is converted to a Number. The closest Number is 0.99999999999900002212172012150404043495655059814453125, so that is the result.
The floor function of that value is 0.
In Math.floor(.99999999999999999):
The source text .99999999999999999 is converted to a Number. The closest Number is 1, so that is the result.
The floor function of 1 is 1.
There are only two surprising things here, at most. One is that the numerals in the source text are converted to internal Number values. But this should not be surprising. Of course text has to be converted to internal representations of numbers, and the Number type cannot perfectly store all the infinitely many numbers. So it has to round. And of course numbers very near 1 round to 1.
The other possibly surprising thing is that 1-Number.MIN_VALUE is 1. But this is actually the same issue: The exact result is not representable, but it is very near 1, so 1 is used.
The Math.floor function works correctly. It never introduces any error, and you do not have to do anything to guarantee that it will round down. It always does.
However, since you want to normalize numbers, it seems likely you are going to divide numbers at some point. When you divide, there may be rounding problems, because many results of division are not exactly representable, so they must be rounded.
However, that is a separate problem, and you have not given enough information in this question to address the specific calculations you plan to do. You should open a separate question for it.
Javascript will treat any number between 0.999999999999999994 and 1 as 1, so just subtract .000000000000000006.
Of course that's not as easy as it sounds, since .000000000000000006 is evaluated as 0 in Javascript, so you could do something like:
function trueFloor(x)
{
x = x * 100;
if(x > .0000000000000006)
x = x - .0000000000000006;
x = Math.floor(x/100);
return x;
}
EDIT: Or at least you'd think you could. Apparently JS casts .99999999999999999 to 1 before passing it to a function, so you'd have to try something like:
trueFloor("0.99999999999999999")
function trueFloor(str)
{
x=str.substring(0,9) + 0;
return Math.floor(x); //=> 0
}
Not sure why you'd need that level of precision, but in theory, I guess it works. You can see a working fiddle here
As long as you cast your insanely precise float as a string, that's probably your best bet.
Please understand one thing: this...
.999999999999999999
... is just a Number literal. Just as
.999999999999999998
.999999999999999997
.999999999999999996
...
... you see the pattern.
How JavaScript treats these literals is completely another story. And yes, this treatment is limited by the number of bits that can be used to store a Number value.
The number of possible floating point literals is infinite by definition - no matter how small is the range set for them. For example, take the ones shown above: how many of numbers very close to 1 you may express? Right, it's infinite: just keep appending 9 to the line.
But the container for each Number value is quite finite: it has 64 bits. That means, it can store 2^64 different values (Infinite, -Infinite and NaN among them) - and that's all.
You want to work with such literals anyway? Use Strings to store them, not Numbers - and some BigMath JS library (take your pick) to work with those values - as Strings, again.
But from your question it looks like you're not, as you talked about array of Numbers - Number values, that is. And in no way there can be .999999999999999999 stored there, as there is no such Number value in JavaScript.
Related
I'm trying to increase by one this number: 9223372036854775808:
var number = 9223372036854775808;
var plusOne = number + 1;
This should yield 9223372036854775809, but it instead yields 9223372036854776000.
Why? More important, how can I fix this?
The largest representable number in JavaScript is (2^53) - 1, or, written out, 9007199254740991. The number you have, 9223372036854775808, is more than 1024 times that quantity.
If you want to work with numbers larger than the one above, you should use a big integer library. JavaScript does not have one built in, however, so you'll need to grab it yourself. Personally, I use big-integer when I'm working on things that deal with really large numbers, e.g., Project Euler.
This is to do with JavaScript storing numbers internally as (double precision) floating point. As you go up the scale of floating point numbers, the numbers get more and more sparse until the point where you get incorrect results, because the next representable number is more that 1 away (As in your example). To properly handle large numbers, you will need to use a proper large number library such as javascript-bignum. If you only need integers, you can use BigInteger.js
So my problem is this, I'm writing a program that checks if number is even or odd without division. So I decided to take the number, turn it into a String with the
number.toString()
method. The problem I'm having is that if you put a number that is about 17 or more digits long the string is correct for about the first 17 digits then it's just 0's and sometimes 2's. For example,
function toStr (number)
{
return number.toString(10);
}
console.log(toStr(123456789123456789));
prints,
123456789123456780
any ideas?
The problem has nothing to do with strings or your function at all. Try going to your console and just entering the expression 123456789123456789 and pressing return.
You will likewise obtain 123456789123456780.
Why?
The expression 123456789123456789 within the JavaScript language is interpreted as a JavaScript number type, which can only be represented exactly to a certain number of base two significant figures. The input number happens to have more significant digits when expressed in base two than the number of base two significant figures available in JavaScript's representation of a number, and so the value is automatically rounded in base two as follows:
123456789123456789 =
110110110100110110100101110101100110100000101111100010101 (base two)
123456789123456780 =
110110110100110110100101110101100110100000101111100001100 (base two)
Note that you CAN accurately represent some numbers larger than a certain size in JavaScript, but only those numbers with no more significant figures in base two than JavaScript has room for. For instance, 2 times a very large power of 10, which would have only one significant figure in base two.
If you are designing this program to accept user input from a form or dialog box, then you will receive the input as a string. You only need to check the last digit in order to determine if the input number is odd or even (assuming it is indeed an integer to begin with). The other answer has suggested the standard way to obtain the last character of a string as well as the standard way to test if a string value is odd or even.
If you go beyond Javascript's max integer size (9007199254740992) you are asking for trouble: http://ecma262-5.com/ELS5_HTML.htm.
So to solve this problem, you must treat it as a string only. Then extract the last digit in the string and use it to determine whether the number is even or odd.
if(parseInt(("123456789123456789").slice(-1)) % 2)
//odd
else
//even
It's a 64-bit floating point number, using the IEEE 754 specification. A feature of this spec is that starting at 2^53 the smallest distance between two numbers is 2.
var x = Math.pow(2, 53);
console.log( x == x + 1 );
This difference is the value of the unit in the last place, or ULP.
This is similar in principle to trying to store fractional values in integral types in other languages; values like .5 can't be represented, so they are discarded. With integers, the ULP value is always 1; with floating point, the ULP value depends on how big or small the number you're trying to represent.
I play a game, and in the database we set 100663296 to be a GM Leader but also this field in the database gets written to for different things, so it changes that number to 100794368
i was told to possible use a bit-wise check to check whether the first number is the same as the second number, and I have googled on using bit-wise checks but got confused as to what to use for my check.
Here are some other numbers that change, including the one from above.
predefined number new changed number/ever changing number.
100663296 = 100794368
67108864 = 67239936
117440512 = 2231767040
so how should i go about checking these numbers?
And here is part of my code that i was using before i noticed the change in the numbers.
if (playerData[i].nameflags == 67108864)
{
playerRows += '<img src ="icons/GM-Icon.png" alt="GM" title="GM"></img>';
}
thx to Bergi, for the answer.
if (playerData[i].nameflags & 0x400000 /* === 0x400000 */)
this seams to work great.
also thx to vr1911428
and every one else for the help on this.
So let's convert those numbers to binary representation (unsigned integer):
> 100663296
00000110000000000000000000000000
> 100794368
00000110000000100000000000000000
> 67108864
00000100000000000000000000000000
> 67239936
00000100000000100000000000000000
> 117440512
00000111000000000000000000000000
> 2231767040
10000101000001100001000000000000
Notice that the last number is out of the scope of JavaScripts bitwise arithmetic, which only works with 32-bit signed integers - you won't be able to use the leftmost bit.
So which bits do you want to compare now? There are lots of possibilities, the above scheme doesn't make it clear, yet it looks like you are looking for the 27th bit from the right (226 = 67108864). To match against it, you can apply a binary AND bitmask:
x & Math.pow(2, 26)
which should evaluate to 226 again or zero - so you can just check for truthiness. Btw, instead of using pow you could use hexadecimal notation: 0x4000000. With that, your condition will look like this:
if (playerData[i].nameflags & 0x400000 /* === 0x400000 */)
If you need to check for full bitwise equality of two integers, all you need is just '==' operator, but to use it, you should guarantee that both operands are integers:
left = 12323;
right = 12323;
if (left == right)
alert("Operands are binary equal; I'll guarantee that. :-)");
Be very careful though; if at least one of operands is string representing number, not a number, both operands will be considered strings and you can get confusing results:
left = "012323";
right = 12323;
if (left != right)
alert("Operands are not equal, even though they represent 'equal values', as they are compared as strings.");
In general, these days, the attempt to operate with strings representing data instead of data itself is a real curse of the beginners; and it's hard to explain to them. It is especially difficult to explain in JavaScript, with its loose-type typing concept, which is itself very complex and hard to understand, behind the illusory simplicity.
Finally, if you need to compare separate bits (and, from your question, I don't see this need), you can use binary operators:
https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Bitwise_Operators
That's it, basically.
...................
I'm caclulating the mean value of a function's request/sec, appearently the result number sometimes is too long so it displays as Infinity, is there a way to round it so it show a number only? Or make a sleep()/wait() while it's on Infinity?
well to be exactly, im monitoring req/sec on a graph, when it's infinity the line goes up not towards zero
It's not too long to display. If you get Inf then you can't do anything with it other than know that it is something larger than the maximum possible value. This is the behavior of IEEE floating point numbers that are used in JavaScript.
Probably the cause for this Infinity is a division by zero, not a big number.
You are most likely unintentionally dividing by zero.
var num = 1/0;
console.log(num);
//>Infinity
Conditionally check that the divisor is not null.
You can check the maximum value of an integer as follows:
console.log([Number.MAX_VALUE, Number.MIN_VALUE]);
//>[1.7976931348623157e+308, 5e-324]
See also the official ECMA Description on Numbers
I heard that you could right-shift a number by .5 instead of using Math.floor(). I decided to check its limits to make sure that it was a suitable replacement, so I checked the following values and got the following results in Google Chrome:
2.5 >> .5 == 2;
2.9999 >> .5 == 2;
2.999999999999999 >> .5 == 2; // 15 9s
2.9999999999999999 >> .5 == 3; // 16 9s
After some fiddling, I found out that the highest possible value of two which, when right-shifted by .5, would yield 2 is 2.9999999999999997779553950749686919152736663818359374999999¯ (with the 9 repeating) in Chrome and Firefox. The number is 2.9999999999999997779¯ in IE.
My question is: what is the significance of the number .0000000000000007779553950749686919152736663818359374? It's a very strange number and it really piqued my curiosity.
I've been trying to find an answer or at least some kind of pattern, but I think my problem lies in the fact that I really don't understand the bitwise operation. I understand the idea in principle, but shifting a bit sequence by .5 doesn't make any sense at all to me. Any help is appreciated.
For the record, the weird digit sequence changes with 2^x. The highest possible values of the following numbers that still truncate properly:
for 0: 0.9999999999999999444888487687421729788184165954589843749¯
for 1: 1.9999999999999999888977697537484345957636833190917968749¯
for 2-3: x+.99999999999999977795539507496869191527366638183593749¯
for 4-7: x+.9999999999999995559107901499373838305473327636718749¯
for 8-15: x+.999999999999999111821580299874767661094665527343749¯
...and so forth
Actually, you're simply ending up doing a floor() on the first operand, without any floating point operations going on. Since the left shift and right shift bitwise operations only make sense with integer operands, the JavaScript engine is converting the two operands to integers first:
2.999999 >> 0.5
Becomes:
Math.floor(2.999999) >> Math.floor(0.5)
Which in turn is:
2 >> 0
Shifting by 0 bits means "don't do a shift" and therefore you end up with the first operand, simply truncated to an integer.
The SpiderMonkey source code has:
switch (op) {
case JSOP_LSH:
case JSOP_RSH:
if (!js_DoubleToECMAInt32(cx, d, &i)) // Same as Math.floor()
return JS_FALSE;
if (!js_DoubleToECMAInt32(cx, d2, &j)) // Same as Math.floor()
return JS_FALSE;
j &= 31;
d = (op == JSOP_LSH) ? i << j : i >> j;
break;
Your seeing a "rounding up" with certain numbers is due to the fact the JavaScript engine can't handle decimal digits beyond a certain precision and therefore your number ends up getting rounded up to the next integer. Try this in your browser:
alert(2.999999999999999);
You'll get 2.999999999999999. Now try adding one more 9:
alert(2.9999999999999999);
You'll get a 3.
This is possibly the single worst idea I have ever seen. Its only possible purpose for existing is for winning an obfusticated code contest. There's no significance to the long numbers you posted -- they're an artifact of the underlying floating-point implementation, filtered through god-knows how many intermediate layers. Bit-shifting by a fractional number of bytes is insane and I'm surprised it doesn't raise an exception -- but that's Javascript, always willing to redefine "insane".
If I were you, I'd avoid ever using this "feature". Its only value is as a possible root cause for an unusual error condition. Use Math.floor() and take pity on the next programmer who will maintain the code.
Confirming a couple suspicions I had when reading the question:
Right-shifting any fractional number x by any fractional number y will simply truncate x, giving the same result as Math.floor() while thoroughly confusing the reader.
2.999999999999999777955395074968691915... is simply the largest number that can be differentiated from "3". Try evaluating it by itself -- if you add anything to it, it will evaluate to 3. This is an artifact of the browser and local system's floating-point implementation.
If you wanna go deeper, read "What Every Computer Scientist Should Know About Floating-Point Arithmetic": https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
Try this javascript out:
alert(parseFloat("2.9999999999999997779553950749686919152736663818359374999999"));
Then try this:
alert(parseFloat("2.9999999999999997779553950749686919152736663818359375"));
What you are seeing is simple floating point inaccuracy. For more information about that, see this for example: http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems.
The basic issue is that the closest that a floating point value can get to representing the second number is greater than or equal to 3, whereas the closes that the a float can get to the first number is strictly less than three.
As for why right shifting by 0.5 does anything sane at all, it seems that 0.5 is just itself getting converted to an int (0) beforehand. Then the original float (2.999...) is getting converted to an int by truncation, as usual.
I don't think your right shift is relevant. You are simply beyond the resolution of a double precision floating point constant.
In Chrome:
var x = 2.999999999999999777955395074968691915273666381835937499999;
var y = 2.9999999999999997779553950749686919152736663818359375;
document.write("x=" + x);
document.write(" y=" + y);
Prints out: x = 2.9999999999999996 y=3
The shift right operator only operates on integers (both sides). So, shifting right by .5 bits should be exactly equivalent to shifting right by 0 bits. And, the left hand side is converted to an integer before the shift operation, which does the same thing as Math.floor().
I suspect that converting 2.9999999999999997779553950749686919152736663818359374999999
to it's binary representation would be enlightening. It's probably only 1 bit different
from true 3.
Good guess, but no cigar.
As the double precision FP number has 53 bits, the last FP number before 3 is actually
(exact): 2.999999999999999555910790149937383830547332763671875
But why it is
2.9999999999999997779553950749686919152736663818359375
(and this is exact, not 49999... !)
which is higher than the last displayable unit ? Rounding. The conversion routine (String to number) simply is correctly programmed to round the input the the next floating point number.
2.999999999999999555910790149937383830547332763671875
.......(values between, increasing) -> round down
2.9999999999999997779553950749686919152736663818359375
....... (values between, increasing) -> round up to 3
3
The conversion input must use full precision. If the number is exactly the half between
those two fp numbers (which is 2.9999999999999997779553950749686919152736663818359375)
the rounding depends on the setted flags. The default rounding is round to even, meaning that the number will be rounded to the next even number.
Now
3 = 11. (binary)
2.999... = 10.11111111111...... (binary)
All bits are set, the number is always odd. That means that the exact half number will be rounded up, so you are getting the strange .....49999 period because it must be smaller than the exact half to be distinguishable from 3.
I suspect that converting 2.9999999999999997779553950749686919152736663818359374999999 to its binary representation would be enlightening. It's probably only 1 bit different from true 3.
And to add to John's answer, the odds of this being more performant than Math.floor are vanishingly small.
I don't know if JavaScript uses floating-point numbers or some kind of infinite-precision library, but either way, you're going to get rounding errors on an operation like this -- even if it's pretty well defined.
It should be noted that the number ".0000000000000007779553950749686919152736663818359374" is quite possibly the Epsilon, defined as "the smallest number E such that (1+E) > 1."