In NARS, fractions are represented using r example 23r45 will be the fraction 23⁄45. The problem is, if I have a variable that contains a rational number, how do I access the numerator? Does a "numerator" function exist such that numerator 23r45 returns 23?
There is no built-in "numerator" function, but you can easily construct one:
num←(⊣÷∨)∘1
num 23r45
23
For reference, the denominator can be found with:
den←(⊢÷∨)∘1
den 23r45
45
And so you can "split" a rational with:
n_d←(,÷∨)∘1
n_d 23r45
23 45
Related
I don't understand how floating point numbers are represented in hex notation in Swift. Apple's documentation shows that 0xC.3p0 is equal to 12.1875 in decimal. Can someone walk me through how to do that conversion? I understand that before the decimal hex value 0xC = 12. The 3p0 after the decimal is where I am stumped.
From the documentation:
Floating-Point Literals
...
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists of
an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 × 22, which evaluates
to 60. Similarly, 0xFp-2 represents 15 × 2-2, which evaluates to 3.75.
In your case
0xC.3p0 = (12 + 3/16) * 2^0 = 12.1875
Another example:
0xAB.CDp4 = (10*16 + 11 + 12/16 + 13/16^2) * 2^4 = 2748.8125
This format is very similar to the %a printf-format (see for example
http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html).
It can be used to specify a floating point number directly in its
binary IEEE 754 representation, see Why does Swift use base 2 for the exponent of hexadecimal floating point values?
for more information.
Interpret 0xC.3p0 using the place value system:
C (or 12) is in the 16^0 place
3 is in the 16^-1 place (and 3/16 == 0.1875)
p says the exponent follows (like the e in 6.022e23 in base 10)
0 is the exponent (in base 10) that is the power of 2 (2^0 == 1)
So putting it all together
0xC.3p0 = (12 + (3/16)) * 2^0 = 12.1875
In order to sum up what I've read, you can see those representations as follow:
0xC.3p0 = (12*16^0 + 3*16^-1) * 2^0 = 12.1875
From Martin R's example above :
0xAB.CDp4 = (10*16^1 + 11*16^0 + 12*16^-1 + 13*16^-2) * 2^4 = 2748.8125
The 0xC is 12, as you said. The decimal part is ((1/16)*3)*10^0.
So you need to take the decimal part and divide it by 16. Then you need to multiply it by 2 raised to the power of the number after the p
Hexadecimal -(0-9,A=10,B=11,C=12,D=13,E=14,F=15) and p0 means 2^0
ex: - 0xC = 12 (0x prefix represents hexadecimal)
After the decimal part as in 0xC.3p0 we divide the numbers with the power of 16
So here its 3/16 = 0.1875
so 0xC.3p0 = (12 + (3/16) ) 2^0
If it was 0xC.43p0 then for the 4 we would use 4/(16), for 3 we would use 3/(16 ^2) and similarly if the decimal part increases.
ex: 0xC.231p1 = (12 + 2/16 + 3/(256) + 1/(16^3)) 2^1 = 24.27392578125
I'm working through the first basic playground in https://github.com/nettlep/learn-swift using XCode
What exactly is happening with this expression?
0xC.3p0 == 12.1875
I've learned about hexadecimal literals and the special "p" notation that indicates a power of 2.
0xF == 15
0xFp0 == 15 // 15 * 2^0
If I try 0xC.3 I get the error: Hexadecimal floating point literal must end with an exponent.
I found this nice overview of numeric literals and another deep explanation, but I didn't see something that explains what .3p0 does.
I've forked the code and upgraded this lesson to XCode 7 / Swift 2 -- here's the specific line.
This is Hexadecimal exponential notation.
By convention, the letter P (or p, for "power") represents times two
raised to the power of ... The number after the P is decimal and
represents the binary exponent.
...
Example: 1.3DEp42 represents hex(1.3DE) × dec(2^42).
For your example, we get:
0xC.3p0 represents 0xC.3 * 2^0 = 0xC.3 * 1 = hex(C.3) = 12.1875
where hex(C.3) = dec(12.{3/16}) = dec(12.1875)
As an example, you can try 0xC.3p1 (equals hex(C.3) * dec(2^1)), which yields double the value, i.e., 24.375.
You can also study the binary exponent growth in a playground for hex-value 1:
// ...
print(0x1p-3) // 1/8 (0.125)
print(0x1p-2) // 1/4 (0.25)
print(0x1p-1) // 1/2 (0.5)
print(0x1p1) // 2.0
print(0x1p2) // 4.0
print(0x1p3) // 8.0
// ...
Finally, this is also explained in Apple`s Language Reference - Lexical Types: Floating-Point Literals:
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists
of an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 x 2^2, which
evaluates to 60. Similarly, 0xFp-2 represents 15 x 2^-2, which
evaluates to 3.75.
I don't understand how floating point numbers are represented in hex notation in Swift. Apple's documentation shows that 0xC.3p0 is equal to 12.1875 in decimal. Can someone walk me through how to do that conversion? I understand that before the decimal hex value 0xC = 12. The 3p0 after the decimal is where I am stumped.
From the documentation:
Floating-Point Literals
...
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists of
an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 × 22, which evaluates
to 60. Similarly, 0xFp-2 represents 15 × 2-2, which evaluates to 3.75.
In your case
0xC.3p0 = (12 + 3/16) * 2^0 = 12.1875
Another example:
0xAB.CDp4 = (10*16 + 11 + 12/16 + 13/16^2) * 2^4 = 2748.8125
This format is very similar to the %a printf-format (see for example
http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html).
It can be used to specify a floating point number directly in its
binary IEEE 754 representation, see Why does Swift use base 2 for the exponent of hexadecimal floating point values?
for more information.
Interpret 0xC.3p0 using the place value system:
C (or 12) is in the 16^0 place
3 is in the 16^-1 place (and 3/16 == 0.1875)
p says the exponent follows (like the e in 6.022e23 in base 10)
0 is the exponent (in base 10) that is the power of 2 (2^0 == 1)
So putting it all together
0xC.3p0 = (12 + (3/16)) * 2^0 = 12.1875
In order to sum up what I've read, you can see those representations as follow:
0xC.3p0 = (12*16^0 + 3*16^-1) * 2^0 = 12.1875
From Martin R's example above :
0xAB.CDp4 = (10*16^1 + 11*16^0 + 12*16^-1 + 13*16^-2) * 2^4 = 2748.8125
The 0xC is 12, as you said. The decimal part is ((1/16)*3)*10^0.
So you need to take the decimal part and divide it by 16. Then you need to multiply it by 2 raised to the power of the number after the p
Hexadecimal -(0-9,A=10,B=11,C=12,D=13,E=14,F=15) and p0 means 2^0
ex: - 0xC = 12 (0x prefix represents hexadecimal)
After the decimal part as in 0xC.3p0 we divide the numbers with the power of 16
So here its 3/16 = 0.1875
so 0xC.3p0 = (12 + (3/16) ) 2^0
If it was 0xC.43p0 then for the 4 we would use 4/(16), for 3 we would use 3/(16 ^2) and similarly if the decimal part increases.
ex: 0xC.231p1 = (12 + 2/16 + 3/(256) + 1/(16^3)) 2^1 = 24.27392578125
I have a, very long, integer. The integer is represented by a array of unsigned chars.
Example: the integer 1234 with base 10 is represented in the array as [4,3,2,1], [2,2,3,2] (base 8) and [2,13,4] (base 16)
Now I want to convert my integer with base n to another integer with base m. In my persued for a answer I came accross Wallar's algorithm, originally from here.
from math import *
def baseExpansion(n,c,b):
j = 0
base10 = sum([pow(c,len(n)-k-1)*n[k] for k in range(0,len(n))])
while floor(base10/pow(b,j)) != 0: j = j+1
return [floor(base10/pow(b,j-p)) % b for p in range(1,j+1)]
At first I thought this was my answer but unfortunately it is not. The problem I have is that the algorithm computes the sum. In my case this is a problem because the variable base10 is of type unsigned integer of 32 bits. Therefore when my integer, represented as a array, has more then 10 digits it can not convert the number anymore. Anyone has a solution?
Here's the school-book algorithm for doing what you're trying. You start with a representation for zero and call it a running total. Then, for each digit of the number to be converted, starting with the most significant and going to the least significant, 1) multiply the running total by the base of the source number and 2) add the digit to the running total. Now all you need is algorithms to do the multiplication and addition (and you can actually do both at once). Here's how to do that: 1) set the current digit to a variable, call it "carry", 2) for each digit in your new number, starting with the least significant and going to the most significant: 2a) set carry to the current digit in the new number times the output base plus carry, 2b) set the current digit to carry mod the output base, 2c) set carry to carry divided by the output base. And that should do it. There is an implementation of what you are trying to do somewhere here: http://www.cis.ksu.edu/~howell/calculator/comparison.html
I've got a calculation for example 57 / 30 so the solution will be 1,766666667..
How do i first of all get the 1,766666667 i only get 1 or 1.00 and then how do i round the solution (to be 2)?
thanks a lot!
57/30 performs integer division. To obtain a float (or double) result you should make 1 of the operands a floating point value:
result = 57.0/30;
To round result have a look at standard floor and ceil functions.