I am trying to make a small program, that could do symbolic calculations. Nothing more complex than the five basic operations (sum, subtraction, multiplication, division and exponentiation). These things seem simple on paper, but I found that to get the computer to understand how to do these calculations is pretty hard.
And now the question is, how to represent a symbol so that it can be calculated on easily?
I have tried the following:
Numbers are rationals with numerical (integer) numerator and denominator. The rest are symbols with labels (x, y), and they have counts that consist of other symbols and rationals in the sense, that the symbols that are summed together are in an array, and symbols that are multiplied together are the counts of symbols. Thus (x+y)*z becomes symbol z with counts of [symbol x, symbol y], and x*y*z becomes symbol z with count of [symbol y with count of [symbol x]].
Addition is simple, if the other symbols label matches, then add the two counts together:
3x+yx => symbol x with count of [rational 3/1, symbol y]
Subtraction is as simple as addition, just add rational of negative number to the other:
3x+(-yx) => symbol x with count of [rational 3/1, symbol y with count of [rational -1/1]
Multiplication gets harder, but for every element of symbols count, add to their count the multiplier:
(1+y)x * z => symbol x with count of [symbol z, symbol y with count of [symbol z]]
Division is something I didn't get to, because how to find if something is divisible is actually hard, as there is need to find for every element of symbols count if it is divisible, including the original symbols count, and then find where can the eliminations be made.
So at this point I am a bit lost, and would like to know if there is a reasonably good way of representing symbols in symbolic calculations. Am I doing something right, or is this mess going to end up messier if I continue.
I also tried working with binary trees, but the division became too hard there as well.
Would it make more sense to represent the symbols as strings of characters, and try to do character by character computation? If yes, then how would division once again work, especially if there are fractions involved?
Related
I'm having trouble understanding a lecture slide in my school's machine learning course
why does the expected value of Y = f(X)? what does it mean
my understanding is that X, Y are vectors and f(X) outputs a vector of Y where each individual value (y_i) in the Y vector corresponds to a f(x_i) where x_i is the value in X at index i; But now it's taking the expected value of Y, which is going to be a single value, so how is that equal to f(X)?
X, Y (uppercase) are vectors
x_i,y_i (lowercase with subscript) are scalars at index i in X,Y
There is a lot of confusion here. First let's start with definitions
Definitions
Expectation operator E[.]: Takes a random variable as an input and gives a scalar/vector as an output. Let's say Y is a normally distributed random variable with mean Mu and Variance Sigma^{2} (usually stated as:
Y ~ N( Mu , Sigma^{2} ), then E[Y] = Mu
Function f(.): Takes a scalar/vector (not a random variable) and gives a scalar/vector. In this context it is an affine function, that is f(X) = a*X + b where a and b are fixed constants.
What's Going On
Now you can view linear regression from two angles.
Stats View
One angle assumes that your response variable-Y- is a normally distributed random variable because:
Y ~ a*X + b + epsilon
where
epsilon ~ N( 0 , sigma^sq )
and X is some other distribution. We don't really care how X is distributed and treat it as given. In that case the conditional distribution is
Y|X ~ N( a*X + b , sigma^sq )
Notice here that a,b and also X is a number, there is no randomness associated with them.
Maths View
The other view is the math view where I assume that there is a function f(.) that governs the real life process, that if in real life I observe X, then f(X) should be the output. Of course this is not the case and the deviations are assumed to be due to various reasons such as gauge error etc. The claim is that this function is linear:
f(X) = a*X + b
Synthesis
Now how do we combine these? Well, as follows:
E[Y|X] = a*X + b = f(X)
About your question, I first would like to challenge that it should be Y|X and not Y by itself.
Second, there are tons of possible ontological discussions over what each term here represents in real life. X,Y (uppercase) could be vectors. X,Y (uppercase) could also be random variables. A sample of these random variables might be stored in vectors and both would be represented with uppercase letters (the best way is to use different fonts for each). In this case, your sample will become your data. Discussions about the general view of the model and its relevance to real life should be made at random variable level. The way to infer the parameters, how linear regression algorithms works should be made at matrix and vectors levels. There could be other discussion where you should care about both.
I hope this overly unorganized answer helps you. In general if you want to learn such stuff, be sure you know what kind of math objects and operators you are dealing with , what do they take as input and what are their relevance to real life.
I have code like :
debt:=debt+(p_sum-p_addition-p_paid);
before this line
debt:=0;
p_sum:=36;
p_addition:=3.6;
p_paid:=32.4;
after this line debt variable gets value like : 1.3322676e-15;
You are using binary floating point arithmetic. Your non-integer values, are not representable in binary floating point. And hence you are subject to rounding errors.
You actually want to operate with decimal rather than binary representations. If you do that then your values will all be represented exactly and the arithmetic will be exact.
In Delphi, the decimal real valued data type is Currency. I suggest that you switch to Currency for these calculations.
Required reading on this topic is What Every Computer Scientist Should Know About
Floating-Point Arithmetic.
What you're seeing is simply scientific notation for a floating point value. This is just one of a few different ways of displaying the value in the Watch List. In this case the "normal format" would be: 0.0000000000000013322676 (Note there are 15 zeroes.) The Watch List has a limit of 18 digits to display the value. So in the interests of giving you the programmer a value as accurate as possible, it uses the scientific notation.
Whenever a floating point number is very small (as in this case) or very large, it is much more efficient to display it in scientific notation. The reason your calculation produces this very small result instead of 0 is due to rounding error as explained in David's Answer.
As for getting the value "back to normal format", there's no point. The value is simply a value in memory representing the result of a calculation. Basically a series of bytes (in this particular case $3B $AA $11 $F7 $FF $FF $D7 $3C). You don't want to fiddle with this value, otherwise you might introduce more rounding errors. Also, as a programmer you should be able to read the more accurate scientific notation.
However, at some point you may want to display this value to your users. In which case you will create an appropriate string representation of the value. The most appropriate format will depend on the nature of your application. (I.e. scientific notation will still be best in some cases.)
If you want to use fixed floating point notation, you could use FormatFloat('0.00', Value); In your case the very small number will be rounded to 0.00. (BTW, you can even put that into your Watch List.)
Read the help on FormatFloat for more information. You can choose to include/exclude thousand separators, or even use scientific notation in your own display format.
Reading Scientific Notation
Scientific notation is actually quite easy to read once you understand the format:
NumbereExponent E.g. 1.3322676e-15
This basically means: take the Number and multiply it by 10 to the power of Exponent. (This equates to shifting the decimal point left/right Exponent number of digits.) A negative exponent makes the number smaller and a positive exponent makes the number bigger.
Some examples:
1 nanosecond is 1.0e-9 seconds, or 0.000000001 seconds
1 second is 1.0e+9 nanoseconds, or 100000000 nanoseconds
1.5e0 = 1.5
1.5e1 = 15
1.5e-1 = 0.15
An important convention of scientific notation is that the number before the exponent always uses exactly 1 digit (non-zero) to the left of the decimal point (with the exponent having been adjusted accordingly). This makes it easy to compare the relative size of two numbers just by looking at the exponent. E.g.
1.01846e+7 is bigger than 9.999999999e+6
1.01846e-6 is bigger than 9.999999999e-7
1.234e+250 is much bigger than 9.876e-250
1.234e+3 is bigger than 1.1234e+3 (only when exponents are equal would you need to compare the actual numbers).
I have a program that solves equations and sometimes the solutions x1 and x2 are numbers with a lot of decimal numbers. For example when Δ = 201 (Δ = discriminant) the square root gives me a floating point number.
I need a good approximation of that number because I also have a function that converts it into a fraction. So I thought to do this:
Result := FormatFloat('0.#####', StrToFloat(solx1));
The solx1 is a double. In this way, the number '456,9067896' becomes '456,90679'.
My question is this: if I approximate in this way, the fraction of 456,9067896 will be correct (and the same) if I have 456,90679?
the fraction of 456,9067896 will be correct (and the same) if I have 456,90679?
No, because 0.9067896 is unequal to 0.90679.
But why do you want to round the numbers? Just let them be as they are. Shorten them only for visual representation.
If you are worried about complete correctness of the result, you should not use floating point numbers at all, because floating points are, by definition, a rounding of real numbers. Only the first 5-6 decimal digits of a 32-bit floating point are generally reliable, the following ones are unreliable, due to machine error.
If you want complete precision, you should be using symbolic maths (rational numbers and symbolic representation for irrational/imaginary numbers).
To compare two floating point values with a given precision, just use the SameValue() function from Math unit or its sibbling CompareValue().
if SameValue(456.9067896, 456.90679, 1E-5) then ...
You can specify the precision on which the comparision will take place.
Or you can use a currency value, which has fixed arithmetic precision of 4 digits. So, it won't have rounding issue any more. But you can not do all mathematic computation with it (huge or tiny numbers are not handled properly): its main use is for accounting computations.
You should better never use string representations to compare floats, since it may be very confusing, and do not have good rounding abilities.
If I compare two floating-point numbers, are there cases where a>=b is not equivalent to b<=a and !(a<b), or where a==b is not equivalent to b==a and !(a!=b)?
In other words: are comparisons always "symmetrical", such that I can get the same result on a comparison by swapping the operands and mirroring the operator? And are they always "negatable", such that negating an operator (e.g. > to <=) is equivalent to to applying a logical NOT (!) to the result?
Assuming IEEE-754 floating-point:
a >= b is always equivalent to b <= a.*
a >= b is equivalent to !(a < b), unless one or both of a or b is NaN.
a == b is always equivalent to b == a.*
a == b is equivalent to !(a != b), unless one or both of a or b is NaN.
More generally: trichotomy does not hold for floating-point numbers. Instead, a related property holds [IEEE-754 (1985) §5.7]:
Four mutually exclusive relations are possible: less than, equal, greater than, and unordered. The last case arises when at least one operand is NaN. Every NaN shall compare unordered with everything, including itself.
Note that this is not really an "anomaly" so much as a consequence of extending the arithmetic to be closed in a way that attempts to maintain consistency with real arithmetic when possible.
[*] true in abstract IEEE-754 arithmetic. In real usage, some compilers might cause this to be violated in rare cases as a result of doing computations with extended precision (MSVC, I'm looking at you). Now that most floating-point computation on the Intel architecture is done on SSE instead of x87, this is less of a concern (and it was always a bug from the standpoint of IEEE-754, anyway).
In Python at least a>=b is not equivalent to !(a<b) when there is a NaN involved:
>>> a = float('nan')
>>> b = 0
>>> a >= b
False
>>> not (a < b)
True
I would imagine that this is also the case in most other languages.
Another thing that might surprise you is that NaN doesn't even compare equal to itself:
>>> a == a
False
The set of IEEE-754 floating-point numbers are not ordered so some relational and boolean algebra you are familiar with no longer holds. This anomaly is caused by NaN which has no ordering with respect to any other value in the set including itself so all relational operators return false. This is exactly what Mark Byers has shown.
If you exclude NaN then you now have an ordered set and the expressions you provided will always be equivalent. This includes the infinities and negative zero.
Aside from the NaN issue, which is somewhat analogous to NULL in SQL and missing values in SAS and other statistical packages, there is always the problem of floating point arithmetic accuracy. Repeating values in the fractional part (1/3, for example) and irrational numbers cannot be represented accurately. Floating point arithmatic often truncates results because of the finite limit in precision. The more arithematic you do with a floating point value, the larger the error that creeps in.
Probably the most useful way to compare floating point values would be with an algorithm:
If either value is NaN, all comparisons are false, unless you are explicitly checking for NaN.
If the difference between two numbers is within a certain "fuzz factor", consider them equal. The fuzz factor is your tolerance for accumulated mathematical imprecision.
After the fuzzy equality comparison, then compare for less than or greater than.
Note that comparing for "<=" or ">=" has the same risk as comparison for precise equality.
Nope, not for any sane floating point implementation: basic symmetry and boolean logic applies. However, equality in floating point numbers is tricky in other ways. There are very few cases where testing a==b for floats is the reasonable thing to do.
Of course most languages have library functions for this, but suppose I want to do it myself.
Suppose that the float is given like in a C or Java program (except for the 'f' or 'd' suffix), for example "4.2e1", ".42e2" or simply "42". In general, we have the "integer part" before the decimal point, the "fractional part" after the decimal point, and the "exponent". All three are integers.
It is easy to find and process the individual digits, but how do you compose them into a value of type float or double without losing precision?
I'm thinking of multiplying the integer part with 10^n, where n is the number of digits in the fractional part, and then adding the fractional part to the integer part and subtracting n from the exponent. This effectively turns 4.2e1 into 42e0, for example. Then I could use the pow function to compute 10^exponent and multiply the result with the new integer part. The question is, does this method guarantee maximum precision throughout?
Any thoughts on this?
All of the other answers have missed how hard it is to do this properly. You can do a first cut approach at this which is accurate to a certain extent, but until you take into account IEEE rounding modes (et al), you will never have the right answer. I've written naive implementations before with a rather large amount of error.
If you're not scared of math, I highly recommend reading the following article by David Goldberg, What Every Computer Scientist Should Know About Floating-Point Arithmetic. You'll get a better understanding for what is going on under the hood, and why the bits are laid out as such.
My best advice is to start with a working atoi implementation, and move out from there. You'll rapidly find you're missing things, but a few looks at strtod's source and you'll be on the right path (which is a long, long path). Eventually you'll praise insert diety here that there are standard libraries.
/* use this to start your atof implementation */
/* atoi - christopher.watford#gmail.com */
/* PUBLIC DOMAIN */
long atoi(const char *value) {
unsigned long ival = 0, c, n = 1, i = 0, oval;
for( ; c = value[i]; ++i) /* chomp leading spaces */
if(!isspace(c)) break;
if(c == '-' || c == '+') { /* chomp sign */
n = (c != '-' ? n : -1);
i++;
}
while(c = value[i++]) { /* parse number */
if(!isdigit(c)) return 0;
ival = (ival * 10) + (c - '0'); /* mult/accum */
if((n > 0 && ival > LONG_MAX)
|| (n < 0 && ival > (LONG_MAX + 1UL))) {
/* report overflow/underflow */
errno = ERANGE;
return (n > 0 ? LONG_MAX : LONG_MIN);
}
}
return (n>0 ? (long)ival : -(long)ival);
}
The "standard" algorithm for converting a decimal number to the best floating-point approximation is William Clinger's How to read floating point numbers accurately, downloadable from here. Note that doing this correctly requires multiple-precision integers, at least a certain percentage of the time, in order to handle corner cases.
Algorithms for going the other way, printing the best decimal number from a floating-number, are found in Burger and Dybvig's Printing Floating-Point Numbers Quickly and Accurately, downloadable here. This also requires multiple-precision integer arithmetic
See also David M Gay's Correctly Rounded Binary-Decimal and Decimal-Binary Conversions for algorithms going both ways.
I would directly assemble the floating point number using its binary representation.
Read in the number one character after another and first find all digits. Do that in integer arithmetic. Also keep track of the decimal point and the exponent. This one will be important later.
Now you can assemble your floating point number. The first thing to do is to scan the integer representation of the digits for the first set one-bit (highest to lowest).
The bits immediately following the first one-bit are your mantissa.
Getting the exponent isn't hard either. You know the first one-bit position, the position of the decimal point and the optional exponent from the scientific notation. Combine them and add the floating point exponent bias (I think it's 127, but check some reference please).
This exponent should be somewhere in the range of 0 to 255. If it's larger or smaller you have a positive or negative infinite number (special case).
Store the exponent as it into the bits 24 to 30 of your float.
The most significant bit is simply the sign. One means negative, zero means positive.
It's harder to describe than it really is, try to decompose a floating point number and take a look at the exponent and mantissa and you'll see how easy it really is.
Btw - doing the arithmetic in floating point itself is a bad idea because you will always force your mantissa to be truncated to 23 significant bits. You won't get a exact representation that way.
You could ignore the decimal when parsing (except for its location). Say the input was:
156.7834e10... This could easily be parsed into the integer 1567834 followed by e10, which you'd then modify to e6, since the decimal was 4 digits from the end of the "numeral" portion of the float.
Precision is an issue. You'll need to check the IEEE spec of the language you're using. If the number of bits in the Mantissa (or Fraction) is larger than the number of bits in your Integer type, then you'll possibly lose precision when someone types in a number such as:
5123.123123e0 - converts to 5123123123 in our method, which does NOT fit in an Integer, but the bits for 5.123123123 may fit in the mantissa of the float spec.
Of course, you could use a method that takes each digit in front of the decimal, multiplies the current total (in a float) by 10, then adds the new digit. For digits after the decimal, multiply the digit by a growing power of 10 before adding to the current total. This method seems to beg the question of why you're doing this at all, however, as it requires the use of the floating point primitive without using the readily available parsing libraries.
Anyway, good luck!
Yes, you can decompose the construction into floating point operations as long as these operations are EXACT, and you can afford a single final inexact operation.
Unfortunately, floating point operations soon become inexact, when you exceed precision of mantissa, the results are rounded. Once a rounding "error" is introduced, it will be cumulated in further operations...
So, generally, NO, you can't use such naive algorithm to convert arbitrary decimals, this may lead to an incorrectly rounded number, off by several ulp of the correct one, like others have already told you.
BUT LET'S SEE HOW FAR WE CAN GO:
If you carefully reconstruct the float like this:
if(biasedExponent >= 0)
return integerMantissa * (10^biasedExponent);
else
return integerMantissa / (10^(-biasedExponent));
there is a risk to exceed precision both when cumulating the integerMantissa if it has many digits, and when raising 10 to the power of biasedExponent...
Fortunately, if first two operations are exact, then you can afford a final inexact operation * or /, thanks to IEEE properties, the result will be rounded correctly.
Let's apply this to single precision floats which have a precision of 24 bits.
10^8 > 2^24 > 10^7
Noting that multiple of 2 will only increase the exponent and leave the mantissa unchanged, we only have to deal with powers of 5 for exponentiation of 10:
5^11 > 2^24 > 5^10
Though, you can afford 7 digits of precision in the integerMantissa and a biasedExponent between -10 and 10.
In double precision, 53 bits,
10^16 > 2^53 > 10^15
5^23 > 2^53 > 5^22
So you can afford 15 decimal digits, and a biased exponent between -22 and 22.
It's up to you to see if your numbers will always fall in the correct range... (If you are really tricky, you could arrange to balance mantissa and exponent by inserting/removing trailing zeroes).
Otherwise, you'll have to use some extended precision.
If your language provides arbitrary precision integers, then it's a bit tricky to get it right, but not that difficult, I did this in Smalltalk and blogged about it at http://smallissimo.blogspot.fr/2011/09/clarifying-and-optimizing.html and http://smallissimo.blogspot.fr/2011/09/reviewing-fraction-asfloat.html
Note that these are simple and naive implementations. Fortunately, libc is more optimized.
My first thought is to parse the string into an int64 mantissa and an int decimal exponent using only the first 18 digits of the mantissa. For example, 1.2345e-5 would be parsed into 12345 and -9. Then I would keep multiplying the mantissa by 10 and decrementing the exponent until the mantissa was 18 digits long (>56 bits of precision). Then I would look the decimal exponent up in a table to find a factor and binary exponent that can be used to convert the number from decimal n*10^m to binary p*2^q form. The factor would be another int64 so I'd multiply the mantissa by it such that I obtained the top 64-bits of the resulting 128-bit number. This int64 mantissa can be cast to a float losing only the necessary precision and the 2^q exponent can be applied using multiplication with no loss of precision.
I'd expect this to be very accurate and very fast but you may also want to handle the special numbers NaN, -infinity, -0.0 and infinity. I haven't thought about the denormalized numbers or rounding modes.
For that you have to understand the standard IEEE 754 in order for proper binary representation. After that you can use Float.intBitsToFloat or Double.longBitsToDouble.
http://en.wikipedia.org/wiki/IEEE_754
If you want the most precise result possible, you should use a higher internal working precision, and then downconvert the result to the desired precision. If you don't mind a few ULPs of error, then you can just repeatedly multiply by 10 as necessary with the desired precision. I would avoid the pow() function, since it will produce inexact results for large exponents.
It is not possible to convert any arbitrary string representing a number into a double or float without losing precision. There are many fractional numbers that can be represented exactly in decimal (e.g. "0.1") that can only be approximated in a binary float or double. This is similar to how the fraction 1/3 cannot be represented exactly in decimal, you can only write 0.333333...
If you don't want to use a library function directly why not look at the source code for those library functions? You mentioned Java; most JDKs ship with source code for the class libraries so you could look up how the java.lang.Double.parseDouble(String) method works. Of course something like BigDecimal is better for controlling precision and rounding modes but you said it needs to be a float or double.
Using a state machine. It's fairly easy to do, and even works if the data stream is interrupted (you just have to keep the state and the partial result). You can also use a parser generator (if you're doing something more complex).
I agree with terminus. A state machine is the best way to accomplish this task as there are many stupid ways a parser can be broken. I am working on one now, I think it is complete and it has I think 13 states.
The problem is not trivial.
I am a hardware engineer interested designing floating point hardware. I am on my second implementation.
I found this today http://speleotrove.com/decimal/decarith.pdf
which on page 18 gives some interesting test cases.
Yes, I have read Clinger's article, but being a simple minded hardware engineer, I can't get my mind around the code presented. The reference to Steele's algorithm as asnwered in Knuth's text was helpful to me. Both input and output are problematic.
All of the aforementioned references to various articles are excellent.
I have yet to sign up here just yet, but when I do, assuming the login is not taken, it will be broh. (broh-dot).
Clyde