I am interested in writing a function to convert a number to words in Julia. Such as f(1234) return to "one thousand two hundred and thirty-four".
I am getting held up on figuring out how to extract each digit and turn it back into a number. The following works, but it is pretty ugly. Thanks
x = 123456789
digits = split("$x","")
intx = zeros(length(digits))
for i=1:length(digits) intx[i]=int(digits[i]) end
intx
How about the digits function in Base?
In [23]: x = 123456789
Out [23]: 123456789
In [24]: digits(x)
Out [24]: 9-element Array{Int64,1}:
9
8
7
6
5
4
3
2
1
How about this:
x = 123456789
digits = Int[] # Empty array of Ints
while x != 0
rem = x % 10 # Modulo division - get last digit
push!(digits, rem)
x = div(x,10) # Integer division - drop last digit
end
println(digits) # [9,8,7,6,5,4,3,2,1]
To do it in the style you were doing it (which you probably shouldn't do), you can do
x = 123456789
digits = [int(string(c)) for c in string(x)]
println(digits) # {1,2,3,4,5,6,7,8,9}
Where c in string(x) iterates over the characters of the string, so c::Char. int(c) would give us the ASCII chracter code, so we need to change it back to a string then convert to integer.
Related
I am trying to fit 3 numbers inside 1 number.But numbers will be only between 0 and 11.So their (base) is 12.For example i have 7,5,2 numbers.I come up with something like this:
Three numbers into One number :
7x12=84
84x5=420
420+2=422
Now getting back Three numbers from One number :
422 MOD 12 = 2 (the third number)
422 - 2 = 420
420 / 12 = 35
And i understanded that 35 is multiplication of first and the second number (i.e 7 and 5)
And now i cant get that 7 and 5 anyone knows how could i ???
(I started typing this answer before the other one got posted, but this one is more specific to Arduino then the other one, so I'm leaving it)
The code
You can use bit shifting to get multiple small numbers into one big number, in code it would look like this:
int a, b, c;
//putting then together
int big = (a << 8) + (b << 4) + c;
//separating them again
a = (big >> 8) & 15;
b = (big >> 4) & 15;
c = big & 15;
This code only works when a, b and c are all in the range [0, 15] witch appears to be enough for you case.
How it works
The >> and << operators are the bitshift operators, in short a << n shifts every bit in a by n places to the left, this is equivalent to multiplying by 2^n. Similarly, a >> n shifts to to the right. An example:
11 << 3 == 120 //0000 1011 -> 0101 1000
The & operator performs a bitwise and on the two operands:
6 & 5 == 4 // 0110
// & 0101
//-> 0100
These two operators are combined to "pack" and "unpack" the three numbers. For the packing every small number is shifted a bit to the left and they are all added together. This is how the bits of big now look (there are 16 of them because ints in Arduino are 16 bits wide):
0000aaaabbbbcccc
When unpacking, the bits are shifted to the right again, and they are bitwise anded together with 15 to filter out any excess bits. This is what that last operation looks like to get b out again:
00000000aaaabbbb //big shifted 4 bits to the right
& 0000000000001111 //anded together with 15
-> 000000000000bbbb //gives the original number b
All is working exactly like in base 10 (or 16). Here after your corrected example.
Three numbers into One number :
7x12^2=1008
5*12^1=60
2*12^0=2
1008+60+2=1070
Now getting back Three numbers from One number :
1070 MOD 12 = 2 (the third number)
1070/12 = 89 (integer division) => 89 MOD 12 = 5
89 / 12 = 7
Note also that the maximum value will be 11*12*12+11*12+11=1727.
If this is really programming related, you will be using 16bits instead of 3*8 bits so sparing one byte. An easyer method not using base 12 would be fit each number into half a byte (better code efficiency and same transmission length):
7<<(4+4) + 5<<4 + 2 = 1874
1874 & 0x000F = 2
1874>>4 & 0x000F = 5
1874>>8 & 0x0F = 7
Because MOD(12) and division by 12 is much less efficient than working with powers of 2
you can use the principle of the positional notation to change from one or the other in any base
Treat yours numbers (n0,n1,...,nm) as a digit of a big number in the base B of your choosing so the new number is
N = n0*B^0 + n1*B^1 + ... + nm*B^m
to revert the process is also simple, while your number is greater than 0 find its modulo in respect to the base to get to get the first digit, then subtracts that digit and divide for the base, repeat until finish while saving each digit along the way
digit_list = []
while N > 0 do:
d = N mod B
N = (N - d) / B
digit_list.append( d )
then if N is N = n0*B^0 + n1*B^1 + ... + nm*B^m doing N mod B give you n0, then subtract it leaving you with n1*B^1 + ... + nm*B^m and divide by B to reduce the exponents of all B and that is the new N, N = n1*B^0 + ... + nm*B^(m-1) repetition of that give you all the digit you start with
here is a working example in python
def compact_num( num_list, base=12 ):
return sum( n*pow(base,i) for i,n in enumerate(num_list) )
def decompact_num( n, base=12):
if n==0:
return [0]
result = []
while n:
n,d = divmod(n,base)
result.append(d)
return result
example
>>> compact_num([2,5,7])
1070
>>> decompact_num(1070)
[2, 5, 7]
>>> compact_num([10,2],16)
42
>>> decompact_num(42,16)
[10, 2]
>>>
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.
I'm trying to understand the binary operators in C# or in general, in particular ^ - exclusive or.
For example:
Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time and constant space.
This can be done with ^ as follows: Do bitwise XOR of all the elements. Finally we get the number which has odd occurrences.
How does it work?
When I do:
int res = 2 ^ 3;
res = 1;
int res = 2 ^ 5;
res = 7;
int res = 2 ^ 10;
res = 8;
What's actually happening? What are the other bit magics? Any reference I can look up and learn more about them?
I know this is a rather old post but I wanted simplify the answer since I stumbled upon it while looking for something else.
XOR (eXclusive OR/either or), can be translated simply as toggle on/off.
Which will either exclude (if exists) or include (if nonexistent) the specified bits.
Using 4 bits (1111) we get 16 possible results from 0-15:
decimal | binary | bits (expanded)
0 | 0000 | 0
1 | 0001 | 1
2 | 0010 | 2
3 | 0011 | (1+2)
4 | 0100 | 4
5 | 0101 | (1+4)
6 | 0110 | (2+4)
7 | 0111 | (1+2+4)
8 | 1000 | 8
9 | 1001 | (1+8)
10 | 1010 | (2+8)
11 | 1011 | (1+2+8)
12 | 1100 | (4+8)
13 | 1101 | (1+4+8)
14 | 1110 | (2+4+8)
15 | 1111 | (1+2+4+8)
The decimal value to the left of the binary value, is the numeric value used in XOR and other bitwise operations, that represents the total value of associated bits. See Computer Number Format and Binary Number - Decimal for more details.
For example: 0011 are bits 1 and 2 as on, leaving bits 4 and 8 as off. Which is represented as the decimal value of 3 to signify the bits that are on, and displayed in an expanded form as 1+2.
As for what's going on with the logic behind XOR here are some examples
From the original post
2^3 = 1
2 is a member of 1+2 (3) remove 2 = 1
2^5 = 7
2 is not a member of 1+4 (5) add 2 = 1+2+4 (7)
2^10 = 8
2 is a member of 2+8 (10) remove 2 = 8
Further examples
1^3 = 2
1 is a member of 1+2 (3) remove 1 = 2
4^5 = 1
4 is a member of 1+4 (5) remove 4 = 1
4^4 = 0
4 is a member of itself remove 4 = 0
1^2^3 = 0Logic: ((1^2)^(1+2))
(1^2) 1 is not a member of 2 add 2 = 1+2 (3)
(3^3) 1 and 2 are members of 1+2 (3) remove 1+2 (3) = 0
1^1^0^1 = 1 Logic: (((1^1)^0)^1)
(1^1) 1 is a member of 1 remove 1 = 0
(0^0) 0 is a member of 0 remove 0 = 0
(0^1) 0 is not a member of 1 add 1 = 1
1^8^4 = 13 Logic: ((1^8)^4)
(1^8) 1 is not a member of 8 add 1 = 1+8 (9)
(9^4) 1 and 8 are not members of 4 add 1+8 = 1+4+8 (13)
4^13^10 = 3 Logic: ((4^(1+4+8))^(2+8))
(4^13) 4 is a member of 1+4+8 (13) remove 4 = 1+8 (9)
(9^10) 8 is a member of 2+8 (10) remove 8 = 2
1 is not a member of 2+8 (10) add 1 = 1+2 (3)
4^10^13 = 3 Logic: ((4^(2+8))^(1+4+8))
(4^10) 4 is not a member of 2+8 (10) add 4 = 2+4+8 (14)
(14^13) 4 and 8 are members of 1+4+8 (13) remove 4+8 = 1
2 is not a member of 1+4+8 (13) add 2 = 1+2 (3)
To see how it works, first you need to write both operands in binary, because bitwise operations work on individual bits.
Then you can apply the truth table for your particular operator. It acts on each pair of bits having the same position in the two operands (the same place value). So the leftmost bit (MSB) of A is combined with the MSB of B to produce the MSB of the result.
Example: 2^10:
0010 2
XOR 1010 8 + 2
----
1 xor(0, 1)
0 xor(0, 0)
0 xor(1, 1)
0 xor(0, 0)
----
= 1000 8
And the result is 8.
The other way to show this is to use the algebra of XOR; you do not need to know anything about individual bits.
For any numbers x, y, z:
XOR is commutative: x ^ y == y ^ x
XOR is associative: x ^ (y ^ z) == (x ^ y) ^ z
The identity is 0: x ^ 0 == x
Every element is its own inverse: x ^ x == 0
Given this, it is easy to prove the result stated. Consider a sequence:
a ^ b ^ c ^ d ...
Since XOR is commutative and associative, the order does not matter. So sort the elements.
Now any adjacent identical elements x ^ x can be replaced with 0 (self-inverse property). And any 0 can be removed (because it is the identity).
Repeat as long as possible. Any number that appears an even number of times has an integral number of pairs, so they all become 0 and disappear.
Eventually you are left with just one element, which is the one appearing an odd number of times. Every time it appears twice, those two disappear. Eventually you are left with one occurrence.
[update]
Note that this proof only requires certain assumptions about the operation. Specifically, suppose a set S with an operator . has the following properties:
Assocativity: x . (y . z) = (x . y) . z for any x, y, and z in S.
Identity: There exists a single element e such that e . x = x . e = x for all x in S.
Closure: For any x and y in S, x . y is also in S.
Self-inverse: For any x in S, x . x = e
As it turns out, we need not assume commutativity; we can prove it:
(x . y) . (x . y) = e (by self-inverse)
x . (y . x) . y = e (by associativity)
x . x . (y . x) . y . y = x . e . y (multiply both sides by x on the left and y on the right)
y . x = x . y (because x . x = y . y = e and the e's go away)
Now, I said that "you do not need to know anything about individual bits". I was thinking that any group satisfying these properties would be enough, and that such a group need not necessarily be isomorphic to the integers under XOR.
But #Steve Jessup proved me wrong in the comments. If you define scalar multiplication by {0,1} as:
0 * x = 0
1 * x = x
...then this structure satisfies all of the axioms of a vector space over the integers mod 2.
Thus any such structure is isomorphic to a set of vectors of bits under component-wise XOR.
This is based on the simple fact that XOR of a number with itself results Zero.
and XOR of a number with 0 results the number itself.
So, if we have an array = {5,8,12,5,12}.
5 is occurring 2 times.
8 is occurring 1 times.
12 is occurring 2 times.
We have to find the number occurring odd number of times. Clearly, 8 is the number.
We start with res=0 and XOR with all the elements of the array.
int res=0;
for(int i:array)
res = res ^ i;
1st Iteration: res = 0^5 = 5
2nd Iteration: res = 5^8
3rd Iteration: res = 5^8^12
4th Iteration: res = 5^8^12^5 = 0^8^12 = 8^12
5th Iteration: res = 8^12^12 = 8^0 = 8
The bitwise operators treat the bits inside an integer value as a tiny array of bits. Each of those bits is like a tiny bool value. When you use the bitwise exclusive or operator, one interpretation of what the operator does is:
for each bit in the first value, toggle the bit if the corresponding bit in the second value is set
The net effect is that a single bit starts out false and if the total number of "toggles" is even, it will still be false at the end. If the total number of "toggles" is odd, it will be true at the end.
Just think "tiny array of boolean values" and it will start to make sense.
The definition of the XOR (exclusive OR) operator, over bits, is that:
0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0
One of the ways to imagine it, is to say that the "1" on the right side changes the bit from the left side, and 0 on the right side doesn't change the bit on the left side. However, XOR is commutative, so the same is true if the sides are reversed.
As any number can be represented in binary form, any two numbers can be XOR-ed together.
To prove it being commutative, you can simply look at its definition, and see that for every combination of bits on either side, the result is the same if the sides are changed. To prove it being associative, you can simply run through all possible combinations of having 3 bits being XOR-ed to each other, and the result will stay the same no matter what the order is.
Now, as we proved the above, let's see what happens if we XOR the same number at itself. Since the operation works on individual bits, we can test it on just two numbers: 0 and 1.
0 XOR 0 = 0
1 XOR 1 = 0
So, if you XOR a number onto itself, you always get 0 (believe it or not, but that property of XOR has been used by compilers, when a 0 needs to be loaded into a CPU register. It's faster to perform a bit operation than to explicitly push 0 into a register. The compiler will just produce assembly code to XOR a register onto itself).
Now, if X XOR X is 0, and XOR is associative, and you need to find out what number hasn't repeated in a sequence of numbers where all other numbers have been repeated two (or any other odd number of times). If we had the repeating numbers together, they will XOR to 0. Anything that is XOR-ed with 0 will remain itself. So, out of XOR-ing such a sequence, you will end up being left with a number that doesn't repeat (or repeats an even number of times).
This has a lot of samples of various functionalities done by bit fiddling. Some of can be quite complex so beware.
What you need to do to understand the bit operations is, at least, this:
the input data, in binary form
a truth table that tells you how to "mix" the inputs to form the result
For XOR, the truth table is simple:
1^1 = 0
1^0 = 1
0^1 = 1
0^0 = 0
To obtain bit n in the result you apply the rule to bits n in the first and second inputs.
If you try to calculate 1^1^0^1 or any other combination, you will discover that the result is 1 if there is an odd number of 1's and 0 otherwise. You will also discover that any number XOR'ed with itself is 0 and that is doesn't matter in what order you do the calculations, e.g. 1^1^(0^1) = 1^(1^0)^1.
This means that when you XOR all the numbers in your list, the ones which are duplicates (or present an even number of times) will XOR to 0 and you will be left with just the one which is present an odd number of times.
As it is obvious from the name(bitwise), it operates between bits.
Let's see how it works,
for example, we have two numbers a=3 and b=4,
the binary representation of 3 is 011 and of 4 is 100, so basically xor of the same bits is 0 and for opposite bits, it is 1.
In the given example 3^4, where "^" is a xor symbol, will give us 111 whose decimal value will be 7.
for another example, if you've given an array in which every element occurs twice except one element & you've to find that element.
How can you do that? simple xor of the same numbers will always be 0 and the number which occur exactly once will be your output. because the output of any one number with 0 will be the same name number because the number will have set bits which zero don't have.
This is about how to do number conversion between binary to octal, octal to hexadecimal, binary to hexadecimal.. ( in all these decimal is no where there, either in source or destination )
Whenever decimal is involved either in source or destination i have a general methodology as,
if decimal is source, do the mod operation of that number with the base of destination. ( and continue the same way, and get the result ),
Example: convert decimal 20 to octal
20 mod 8 = 4
2 = 2
So it is 24 in Octal. This way you can do for anything ( binary, hex ) from Decimal.
if decimal is destination, do the multiplication operation with the source's base from 0 to N which starts from left to right. And add up to get the decimal.
Example: convert binary 1010 to decimal
0 : 0 x 2**0 = 0
1 : 1 x 2**1 = 2
0 : 0 x 2**2 = 0
1: 1 x 2**3 = 8
The sum is 10 in Decimal. This way you can do for anything ( octal, hex ) to decimal.
Questions
Is there any similar general logic, which i can use for conversion between octal, hex and binary ?
Hope the above is clear.. Else let me know.
The conversion between binary and either octal or hex is even easier than decimal. Just break the binary digits up into groups of 3 or 4 (depending on oct or hex), then convert each group to a digit. (If you don't have a multiple of 3 or 4 binary digits, just left-pad the number with zeros.)
Example:
111101111010 binary
1111 0111 1010
F 7 A
F7A hex
111101111010
111 101 111 010
7 5 7 2
7572 oct
Converting back is just the opposite. Each digit in octal or hex gets converted directly to either 3 or 4 binary digits.
For ex.
double size = 10.35;
i should get
value = 1035;
exponent = -2;
so when i re calculate i will get 10.35.
i.e 1035 * 10^-2 = 10.35;
Please help me.
Thanks in advance
In general this is not possible since the fractional part of a double is stored in powers-of-2, and might or might not match powers-of-10.
For example: When looking at powers-of-2 vs powers-of-3: Just like 1/2 == 2^-1 == 5 * 10^-1 has a match, 1/3 == 3^-1 == ?? does not have a match.
However, you can approximate it.
It would have an answer if you would ask for powers-of-2. In that case you can just look at the double representation (see IEEE-754 here) and extract the right bits.
Very simplistically (in C#):
double size = 10.36;
int power = 0;
while (size != (int)size)
{
size *= 10.0;
power--;
}
Console.WriteLine("{0} * 10 to the {1}", size, power);
Though I'm sure with a bit more thought a more elegant solution can be found.
This doesn't go the other way where you've got a large number (103600 say) and want to get the smallest value to some power (1036 * 10^2).
I had to do something very similar. Here's a solution in Python (it hasn't been tested very well):
def normalize(value, fdigits=2):
"""
Convert a string representing a numerical value to value-digit/exponent form.
Round the fractional portion to the given number of digits.
value the value (string)
fdigits the number of digits to which to round the fractional
portion
"""
# if empty string, return error
if not value:
return None
# split value by decimal
v = value.split('.')
# if too many decimals, return error
if len(v) > 2:
return None
# add empty string for fractional portion if missing
elif len(v) == 1:
v.append('')
# assign whole and fractional portions
(w, f) = v
# pad fractional portion up to number of significant digits if necessary
if len(f) < fdigits:
f += ('0' * (fdigits - len(f)))
# if the number of digits in the fractional portion exceeds the
# number of digits allowed by fdigits
elif len(f) > fdigits:
# convert both portions to integers; use '0' for whole portion if missing
(wi, fi) = (int(w or '0'), int(f[:fdigits]))
# round up if first insignificant digit is gteq 5
if int(f[fdigits]) >= 5:
fi += 1
# roll whole value up if fractional portion rounds to a whole
if len(str(fi)) > fdigits:
wi += 1
fi = 0
# replace the whole and fractional strings
(w, f) = (str(wi), ("%0" + str(fdigits) + "d") % fi)
# derive value digits and exponent
n = w.lstrip() + f
l = len(n)
x = -fdigits
n = n.rstrip('0')
x += (l - len(n))
# return value digits and exponent
return (int(n), x)