I am trying to make a function to round a floating point number to a defined length of digits. What I have come up with so far is this:
import Numeric;
digs :: Integral x => x -> [x] <br>
digs 0 = [] <br>
digs x = digs (x `div` 10) ++ [x `mod` 10]
roundTo x t = let d = length $ digs $ round x <br>
roundToMachine x t = (fromInteger $ round $ x * 10^^t) * 10^^(-t)
in roundToMachine x (t - d)
I am using the digs function to determine the number of digits before the comma to optimize the input value (i.e. move everything past the comma, so 1.234 becomes 0.1234 * 10^1)
The roundTo function seems to work for most input, however for some inputs I get strange results, e.g. roundTo 1.0014 4 produces 1.0010000000000001 instead of 1.001.
The problem in this example is caused by calculating 1001 * 1.0e-3 (which returns 1.0010000000000001)
Is this simply a problem in the number representation of Haskell I have to live with or is there a better way to round a floating point number to a specific length of digits?
I realise this question was posted almost 2 years back, but I thought I'd have a go at an answer that didn't require a string conversion.
-- x : number you want rounded, n : number of decimal places you want...
truncate' :: Double -> Int -> Double
truncate' x n = (fromIntegral (floor (x * t))) / t
where t = 10^n
-- How to answer your problem...
λ truncate' 1.0014 3
1.001
-- 2 digits of a recurring decimal please...
λ truncate' (1/3) 2
0.33
-- How about 6 digits of pi?
λ truncate' pi 6
3.141592
I've not tested it thoroughly, so if you find numbers this doesn't work for let me know!
This isn't a haskell problem as much as a floating point problem. Since each floating point number is implemented in a finite number of bits, there exist numbers that can't be represented completely accurately. You can also see this by calculating 0.1 + 0.2, which awkwardly returns 0.30000000000000004 instead of 0.3. This has to do with how floating point numbers are implemented for your language and hardware architecture.
The solution is to continue using your roundTo function for doing computation (it's as accurate as you'll get without special libraries), but if you want to print it to the screen then you should use string formatting such as the Text.Printf.printf function. You can specify the number of digits to round to when converting to a string with something like
import Text.Printf
roundToStr :: (PrintfArg a, Floating a) => Int -> a -> String
roundToStr n f = printf ("%0." ++ show n ++ "f") f
But as I mentioned, this will return a string rather than a number.
EDIT:
A better way might be
roundToStr :: (PrintfArg a, Floating a) => Int -> a -> String
roundToStr n f = printf (printf "%%0.%df" n) f
but I haven't benchmarked to see which is actually faster. Both will work exactly the same though.
EDIT 2:
As #augustss has pointed out, you can do it even easier with just
roundToStr :: (PrintfArg a, Floating a) => Int -> a -> String
roundToStr = printf "%0.*f"
which uses a formatting rule that I was previously unaware of.
I also think that avoiding string conversion is the way to go; however, I would modify the previous post (from schanq) to use round instead of floor:
round' :: Double -> Integer -> Double
round' num sg = (fromIntegral . round $ num * f) / f
where f = 10^sg
> round' 4 3.99999
4.0
> round' 4 4.00001
4.0
Related
I came across this question in a coding competition. Given a number n, concatenate the binary representation of first n positive integers and return the decimal value of the resultant number formed. Since the answer can be large return answer modulo 10^9+7.
N can be as large as 10^9.
Eg:- n=4. Number formed=11011100(1=1,10=2,11=3,100=4). Decimal value of 11011100=220.
I found a stack overflow answer to this question but the problem is that it only contains a O(n) solution.
Link:- concatenate binary of first N integers and return decimal value
Since n can be up to 10^9 we need to come up with solution that is better than O(n).
Here's some Python code that provides a fast solution; it uses the same ideas as in Abhinav Mathur's post. It requires Python >= 3.8, but it doesn't use anything particularly fancy from Python, and could easily be translated into another language. You'd need to write algorithms for modular exponentiation and modular inverse if they're not already available in the target language.
First, for testing purposes, let's define the slow and obvious version:
# Modulus that results are reduced by,
M = 10 ** 9 + 7
def slow_binary_concat(n):
"""
Concatenate binary representations of 1 through n (inclusive).
Reinterpret the resulting binary string as an integer.
"""
concatenation = "".join(format(k, "b") for k in range(n + 1))
return int(concatenation, 2) % M
Checking that we get the expected result:
>>> slow_binary_concat(4)
220
>>> slow_binary_concat(10)
462911642
Now we'll write a faster version. First, we split the range [1, n) into subintervals such that within each subinterval, all numbers have the same length in binary. For example, the range [1, 10) would be split into four subintervals: [1, 2), [2, 4), [4, 8) and [8, 10). Here's a function to do that splitting:
def split_by_bit_length(n):
"""
Split the numbers in [1, n) by bit-length.
Produces triples (a, b, 2**k). Each triple represents a subinterval
[a, b) of [1, n), with a < b, all of whose elements has bit-length k.
"""
a = 1
while n > a:
b = 2 * a
yield (a, min(n, b), b)
a = b
Example output:
>>> list(split_by_bit_length(10))
[(1, 2, 2), (2, 4, 4), (4, 8, 8), (8, 10, 16)]
Now for each subinterval, the value of the concatenation of all numbers in that subinterval is represented by a fairly simple mathematical sum, which can be computed in exact form. Here's a function to compute that sum modulo M:
def subinterval_concat(a, b, l):
"""
Concatenation of values in [a, b), all of which have the same bit-length k.
l is 2**k.
Equivalently, sum(i * l**(b - 1 - i)) for i in range(a, b)) modulo M.
"""
n = b - a
inv = pow(l - 1, -1, M)
q = (pow(l, n, M) - 1) * inv
return (a * q + (q - n) * inv) % M
I won't go into the evaluation of the sum here: it's a bit off-topic for this site, and it's hard to express without a good way to render formulas. If you want the details, that's a topic for https://math.stackexchange.com, or a page of fairly simple algebra.
Finally, we want to put all the intervals together. Here's a function to do that.
def fast_binary_concat(n):
"""
Fast version of slow_binary_concat.
"""
acc = 0
for a, b, l in split_by_bit_length(n + 1):
acc = (acc * pow(l, b - a, M) + subinterval_concat(a, b, l)) % M
return acc
A comparison with the slow version shows that we get the same results:
>>> fast_binary_concat(4)
220
>>> fast_binary_concat(10)
462911642
But the fast version can easily be evaluated for much larger inputs, where using the slow version would be infeasible:
>>> fast_binary_concat(10**9)
827129560
>>> fast_binary_concat(10**18)
945204784
You just have to note a simple pattern. Taking up your example for n=4, let's gradually build the solution starting from n=1.
1 -> 1 #1
2 -> 2^2(1) + 2 #6
3 -> 2^2[2^2(1)+2] + 3 #27
4 -> 2^3{2^2[2^2(1)+2]+3} + 4 #220
If you expand the coefficients of each term for n=4, you'll get the coefficients as:
1 -> (2^3)*(2^2)*(2^2)
2 -> (2^3)*(2^2)
3 -> (2^3)
4 -> (2^0)
Let the N be total number of bits in the string representation of our required number, and D(x) be the number of bits in x. The coefficients can then be written as
1 -> 2^(N-D(1))
2 -> 2^(N-D(1)-D(2))
3 -> 2^(N-D(1)-D(2)-D(3))
... and so on
Since the value of D(x) will be the same for all x between range (2^t, 2^(t+1)-1) for some given t, you can break the problem into such ranges and solve for each range using mathematics (not iteration). Since the number of such ranges will be log2(Given N), this should work in the given time limit.
As an example, the various ranges become:
1. 1 (D(x) = 1)
2. 2-3 (D(x) = 2)
3. 4-7 (D(x) = 3)
4. 8-15 (D(x) = 4)
i am trying to implement a recursive function which takes a float and returns a list of ints representing the continued fraction representation of the float (https://en.wikipedia.org/wiki/Continued_fraction) In general i think i understand how the algorithm is supposed to work. its fairly simply. What i have so far is this:
let rec float2cfrac (x : float) : int list =
let q = int x
let r = x - (float q)
if r = 0.0 then
[]
else
q :: (float2cfrac (1.0 / r ))
the problem is with the base case obviously. It seems the value r never does reduce to 0.0 instead the algorithm keeps on returning values which are the likes of 0.0.....[number]. I am just not sure how to perform the comparison. How exactly should i go about it. The algorithm the function is based on says the base case is 0, so i naturally interpret this as 0.0. I dont see any other way. Also, do note that this is for an assignment where i am explicitly asked to implement the algorithm recursively. Does anyone have some guidance for me? It would be much appreciated
It seems the value r never does reduce to 0.0 instead the algorithm keeps on returning values which are the likes of 0.0.....[number].
This is a classic issue with floating point comparisons. You need to use some epsilon tolerance value for comparisons, because r will never reach exactly 0.0:
let epsilon = 0.0000000001
let rec float2cfrac (x : float) : int list =
let q = int x
let r = x - (float q)
if r < epsilon then
[]
else
q :: (float2cfrac (1.0 / r))
> float2cfrac 4.23
val it : int list = [4; 4; 2; 1]
See this MSDN documentation for more.
You could define a helper function for this:
let withinTolerance (x: float) (y: float) e =
System.Math.Abs(x - y) < e
Also note your original solution isn't tail-recursive, so it consumes stack as it recurses and could overflow the stack. You could refactor it such that a float can be unfolded without recursion:
let float2cfrac (x: float) =
let q = int x
let r = x - (float q)
if withinTolerance r 0.0 epsilon then None
else Some (q, (1.0 / r))
4.23 |> Seq.unfold float2cfrac // seq [4; 4; 2; 1]
I need to write my own recursive function in ML that somehow uses ord to convert a string of numbers to integer type. I can use helper functions, but apparently I should be able to do this without using one (according to my professor).
I can assume that the input is valid, and is a positive integer (in string type of course).
So, the call str2int ("1234") should output 1234: int
I assume I will need to use explode and implode at some point since ord operates on characters, and my input is a string. Any direction would be greatly appreciated.
Given that you asked, I guess I can ruin all the fun for you. This will solve your problem, but ironically, it won't help you.
Well, the ordinal number for the character #'0' is 48. So, this means that if you subtract of any ordinal representing a digit the number 48 you get its decimal value. For instance
ord(#"9") - 48
Yields 9.
So, a function that takes a given character representing a number from 0-9 and turns it into the corresponding decimal is:
fun charToInt(c) = ord(c) - 48
Supposing you had a string of numbers like "2014". Then you can first explode the string into list of characters and then map every character to its corresponding decimal.
For instance
val num = "2014"
val digits = map charToInt (explode num)
The explode function is a helper function that takes a string and turn it into a list of characters.
And now digits would be a list of integers representing the decimal numbers [2,0,1,4];
Then, all you need is to apply powers of 10 to obtain the final integer.
2 * 10 ^ 3 = 2000
0 * 10 ^ 2 = 0
1 * 10 ^ 1 = 10
4 * 10 ^ 0 = 4
The result would be 2000 + 0 + 10 + 4 = 2014
You could define a helper function charsToInt that processes the digits in the string from left to right.
At each step it converts the leftmost digit c into a number and does addition with the 10x-multiple of n (which is the intermediary sum of all previously parsed digits) ...
fun charsToInt ([], n) = n
| charsToInt (c :: cs, n) = charsToInt (cs, 10*n + ord c - 48)
val n = charsToInt (explode "1024", 0)
Gives you: val n = 1024 : int
As you see the trick is to pass the intermediary result down to the next step at each recursive call. This is a very common technique when dealing with these kind of problems.
Here's what I came up with:
fun pow10 n =
if n = 0 then 1 else 10*pow10(n-1);
fun str2help (L,n) =
if null L then 0
else (ord(hd L)-48) * pow10(n) + str2help(tl L, n-1);
fun str2int (string) =
str2help(explode string, size string -1);
str2int ("1234");
This gives me the correct result, though is clearly not the easiest way to get there.
We want to program the gauss-elimination to calculate a basis (linear algebra) as exercise for ourselves. It is not homework.
I thought first of [[Int]] as structure for our matrix. I thought then that we can sort the lists lexicographically. But then we must calculate with the matrix. And there is the problem. Can someone give us some hints.
Consider using matrices from the hmatrix package. Among its modules you can find both a fast implementation of a matrix and a lot of linear algebra algorithms. Browsing their sources might help you with your doubts.
Here's a simple example of adding one row to another by splitting the matrix into rows.
import Numeric.Container
import Data.Packed.Matrix
addRow :: Container Vector t => Int -> Int -> Matrix t -> Matrix t
addRow from to m = let rows = toRows m in
fromRows $ take to rows ++
[(rows !! from) `add` (rows !! to)] ++
drop (to + 1) rows
Another example, this time by using matrix multiplication.
addRow :: (Product e, Container Vector e) =>
Int -> Int -> Matrix e -> Matrix e
addRow from to m = m `add` (e <> m)
where
nrows = rows m
e = buildMatrix nrows nrows
(\(r,c) -> if (r,c) /= (to,from) then 0 else 1)
Cf. Container, Vector, Product.
It will be easier if you use [[Rational]] instead of [[Int]] since you get nice division.
You probably want to start by implementing the elementary row operations.
swap :: Int -> Int -> [[Rational]] -> [[Rational]
swap r1 r2 m = --a matrix with r1 and r2 swapped
scale :: Int -> Rational -> [[Rational]] -> [[Rational]]
scale r c m = --a matrix with row r multiplied by c
addrow :: Int -> Int -> Rational -> [[Rational]] -> [[Rational]]
addrow r1 r2 c m = --a matrix with (c * r1) added to r2
In order to actually do guassian elimination, you need a way to decide what multiple of one row to add to another to get a zero. So given two rows..
5 4 3 2 1
7 6 5 4 3
We want to add c times row 1 to row 2 so that the 7 becomes a zero. So 7 + c * 5 = 0 and c = -7/5. So in order to solve for c all we need are the first elements of each row. Here's a function that finds c:
whatc :: Rational -> Rational -> Rational
whatc _ 0 = 0
whatc a b = - a / b
Also, as others have said, using lists to represent your matrix will give you worse performance. But if you're just trying to understand the algorithm, lists should be fine.
According to wiki shifts can be used to calculate powers of 2:
A left arithmetic shift by n is
equivalent to multiplying by 2^n
(provided the value does not
overflow), while a right arithmetic
shift by n of a two's complement value
is equivalent to dividing by 2^n and
rounding toward negative infinity.
I was always wondering if any other bitwise operators (~,|,&,^) make any mathematical sense when applied to base-10? I understand how they work, but do results of such operations can be used to calculate anything useful in decimal world?
"yep base-10 is what I mean"
In that case, yes, they can be extended to base-10 in several ways, though they aren't nearly as useful as in binary.
One idea is that &, |, etc. are the same as doing arithmetic mod-2 to the individual binary digits. If a and b are single binary-digits, then
a & b = a * b (mod 2)
a ^ b = a + b (mod 2)
~a = 1-a (mod 2)
a | b = ~(~a & ~b) = 1 - (1-a)*(1-b) (mod 2)
The equivalents in base-10 would be (note again these are applied per-digit, not to the whole number)
a & b = a * b (mod 10)
a ^ b = a + b (mod 10)
~a = 9-a (mod 10)
a | b = ~(~a & ~b) = 9 - (9-a)*(9-b) (mod 10)
The first three are useful when designing circuits which use BCD (~a being the 9's complement), such as non-graphing calculators, though we just use * and + rather than & and ^ when writing the equations. The first is also apparently used in some old ciphers.
A fun trick to swap two integers without a temporary variable is by using bitwise XOR:
void swap(int &a, int &b) {
a = a ^ b;
b = b ^ a; //b now = a
a = a ^ b; //knocks out the original a
}
This works because XOR is a commutative so a ^ b ^ b = a.
Yes, there are other useful operations, but they tend to be oriented towards operations involving powers of 2 (for obvious reasons), e.g. test for odd/even, test for power of 2, round up/down to nearest power of 2, etc.
See Hacker's Delight by Henry S. Warren.
In every language I've used (admittedly, almost exclusively C and C-derivatives), the bitwise operators are exclusively integer operations (unless, of course, you override the operation).
While you can twiddle the bits of a decimal number (they have their own bits, after all), it's not necessarily going to get you the same result as twiddling the bits of an integer number. See Single Precision and Double Precision for descriptions of the bits in decimal numbers. See Fast Inverse Square Root for an example of advantageous usage of bit twiddling decimal numbers.
EDIT
For integral numbers, bitwise operations always make sense. The bitwise operations are designed for the integral numbers.
n << 1 == n * 2
n << 2 == n * 4
n << 3 == n * 8
n >> 1 == n / 2
n >> 2 == n / 4
n >> 3 == n / 8
n & 1 == {0, 1} // Set containing 0 and 1
n & 2 == {0, 2} // Set containing 0 and 2
n & 3 == {0, 1, 2, 3} // Set containing 0, 1, 2, and 3
n | 1 == {1, n, n+1}
n | 2 == {2, n, n+2}
n | 3 == {3, n, n+1, n+2, n+3}
And so on.
You can calculate logarithms using just bitwise operators...
Finding the exponent of n = 2**x using bitwise operations [logarithm in base 2 of n]
You can sometime substitute bitwise operations for boolean operations. For example, the following code:
if ((a < 0) && (b < 0)
{
do something
{
In C this can be replaced by:
if ((a & b) < 0)
{
do something
{
This works because one bit in an integer is used as the sign bit (1 indicates negative). The and operation (a & b) will be a meaningless number, but its sign will be the bitwise and of the signs of the numbers and hence checking the sign of the result will work.
This may or may not benefit performance. Doing two boolean tests/branches will be worse on a number of architectures and compilers. Modern x86 compilers can probably generate a single branch using a some of the newer instruction even with the normal syntax.
As always, if it does result in a performance increase... Comment the code - i.e. put the "normal" way of doing it in a comment and say it's equivalent but faster.
Likewise, ~ | and ^ can be used in a similar way it all the conditions are (x<0).
For comparison conditions you can generally use subtraction:
if ((a < b) | (b < c))
{
}
becomes:
if (((a-b) | (b-c)) < 0)
{
}
because a-b will be negative only if a is less than b. There can be issues with this one if you get within a factor of 2 of max int - i.e. arithmetic overflow, so be careful.
These are valid optimizations in some cases, but otherwise quite useless. And to get really ugly, floating point numbers also have sign bits... ;-)
EXAMPLE:
As an example, lets say you want to take action depending on the order of a,b,c. You can do some nested if/else constructs, or you can do this:
x = ((a < b) << 2) | ((b < c) << 1) | (c < a);
switch (x):
I have used this in code with up to 9 conditions and also using the subtractions mentioned above with extra logic to isolate the sign bits instead of less-than. It's faster than the branching equivalent. However, you no longer need to do subtraction and sign bit extraction because the standard was updated long ago to specify true as 1, and with conditional moves and such, the actual less-than can be quite efficient these days.