I am trying to take a number and change it to something like this:
Number Change To
1 0001A
2 0002A
3 0003A
4 0001B
5 0002B
6 0003B
7 0001C
8 0002C
9 0003C
10 0001AA
11 0002AA
12 0003AA
13 0001AB
14 0002AB
15 0003AB
10 0001AC
11 0002AC
12 0003AC
13 0001BA
14 0002BA
15 0003BA
etc…
So Using the Letters 'A','B','C'
And Using the Numbers 1,2,3
For most of your inputs, the left-most 'digits' appears to be changing fastest, until you get to 12->13 transition, where the middle digit changes slowest:
1A
2A
3A
...
2AA
3AA
1AB
2AB
...
This is definitely awkward and will probably require a lot of extra effort to code than a scheme where the right-most or left-most digit changes fastest is strictly followed:
1A
2A
3A
...
2AA
3AA
1BA
2BA
...
or
1
2
3
a
b
c
11
12
13
1a
1b
1c
21
22
23
...
If you must follow exactly the scheme you've given here, then I'd recommend a lookup-table and simply hardcode this scheme into your source:
#include <stdio.h>
int main(int argc, char* argv[]) {
char table[][5]={"ZERO", "1A", "2A", "3A", "1B", "2B", "3B", "1C", "2C", "3C",
"1AA", "2AA", "3AA", "1AB", "2AB", "3AB", "1AC", "2AC", "3AC"};
int i;
for (i=1; i<argc; i++) {
int input=atoi(argv[i]);
printf("%d\t%s\n", input, table[input]);
}
return 0;
}
Please don't use this code directly -- I haven't done any input validation or array bounds checks.
Of course, I hope you pick a different approach entirely -- this simple codeword substitution works fine for a specific number of inputs, but some day someone might want to input a number greater than you've got in your table. (In this case, when someone wants to convert 19, it'll fail.)
A more generic approach, converting between bases, is an algorithm for converting between any two number bases. In your case, it could be converting into base 6 with the digits 1, 2, 3, A, B, C (if you strictly adhere to either a left-most changes fastest or right-most changes fastest approach).
Related
I have a boolean array (from previous computations) and I would like to select the related rows from several matrices. That is why I need the proper index array (to be reused later). This is easy in Matlab and python but I do not crock the correct julian way of doing it...
I am aware of DataFrames, but would like to find an orthodox matrix and array way of doing this.
In Matlab I would say:
n= 9; temp= 1:n; A= 1.0 + temp;
someTest= mod(temp,2) == 0; % just a substitute of a more complex case
% now I have both someTest and A!
inds= find(someTest); Anew= A(inds,:);
% I got inds (which I need)!
What I have got working is this:
n= 10; data= Array(1:n); A= 1.0 .+ data;
someTest= rem.(data,2) .== 0;
inds= [xy[2] for xy in zip(someTest,1:length(someTest)) if xy[1]]; # (*)
Anew= A[inds,:];
What I assumed is that there is some shorter way to express the above phrase. in v. 0.6 there was find() function, but I have not gotten good sense of the julia documentation yet (I am a very very newbie in this).
You can use the BitArray just directly to select the elements:
julia> A[someTest]
5-element Array{Float64,1}:
3.0
5.0
7.0
9.0
11.0
Fot your case:
julia> A[someTest,:] == A[inds,:]
true
find in 0.6 was renamed to findall in Julia 1.0.
To get inds, you can simply do the following:
inds = findall(someTest)
You do not have to compute the intermediate someTest first, which would allocate an array you do not intend to use. Instead, you can do the test with findall directly passing a predicate function.
inds = findall(x -> rem(x,2) == 0, data)
This will return indices of data for which the predicate rem(x,2) == 0 returns true. This will not allocate an intermediate array to find the indices, and should be faster.
As a side note, most of the time you do not need to materialize a range in Julia. Ranges are already iterable and indexable. They will automatically be converted to an Array when there is a need. Array(1:n) or collect(1:n) are usually redundant, and allocates more memory.
Your Matlab code doesn't work. A is just a row-vector (1x9 matrix), so when you try to do A(inds, :) you get an error:
>> Anew= A(inds,:)
Index in position 1 exceeds array bounds
(must not exceed 1).
But if you just fix that, you can solve the problem in exactly the same way in both Matlab and Julia, using either logical indices or regular ones:
Matlab (I'm making sure it's a matrix this time):
n = 9;
temp = (1:n).';
A = temp * (1:4);
inds = mod(temp,2) == 0;
>> A(inds, :) % using logical indices
ans =
2 4 6 8
4 8 12 16
6 12 18 24
8 16 24 32
>> A(find(inds), :) % using regular indices
ans =
2 4 6 8
4 8 12 16
6 12 18 24
8 16 24 32
And now, Julia:
n = 9;
temp = 1:n;
A = temp .* (1:4)'; # notice that we're transposing the opposite vector from Matlab
inds = mod.(temp, 2) .== 0; # you can use iseven.(temp) instead
julia> A[inds, :] # logical indices (BitArray)
4×4 Array{Int64,2}:
2 4 6 8
4 8 12 16
6 12 18 24
8 16 24 32
julia> A[findall(inds), :] # regular integer indices
4×4 Array{Int64,2}:
2 4 6 8
4 8 12 16
6 12 18 24
8 16 24 32
In this case, I would use the logical indices in both Julia and Matlab. In fact, the Matlab linter (in the editor) will tell that you should use logical indices here because it's faster. In Julia, however, there might be cases where it's more efficient to use inds = findall(iseven, temp), and just skip the logical BitArray, like #hckr says.
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]
>>>
Im looking for a way to modify a binary byte value on Arduino.
Because of the Hardware, its neccesarry, to split a two digit number into 2 4-bit.
the code to set output is wire.write(byte, 0xFF) which sets all outputs on High.
0xFF = binary 1111 1111
the formula should be convert a value like this:
e.g nr 35 is binary 0010 0011
but for my use it should displayed as 0011 0101 which would be refer to 53 in reality.
The first 4 bits are for a BCD-Input IC which displays the 5 from 35, the second 4 bits are for a BCD-Input IC which displays the 3 from 35.
Does anybody has a idea how to convert this by code, or like a mathematical formula?
Possible numbers are from 00 to 59.
Thank you for your help
To convert a value n between 0 and 99 to BCD:
((n / 10) * 16) + (n % 10)
assuming n is an integer and thus / is doing integer division; also assumes this will be stored in an unsigned byte.
(If this is not producing the desired result, please either explain how it is incorrect for the example given, or provide a different example for which it is incorrect.)
#include <string.h>
int num = // Any number from 0 to 59
int tens = num/10;
int units = num-(tens*10);
// Make string array for binary
string tensbinary;
int quotient = tens;
char buffer[1];
// Convert numbers
for (int i = 0; i < 4; i++)
{
quotientint = quotientint % 2;
sprintf(buffer, 1, "%d", quotientint);
binary.append(buffer);
}
// Repeat above for the units
// Now join the two together
binarytens.append(binaryunits);
I don't know if this will work, but still, you might be able to extrapolate based on the available information in my code.
The last thing you need to do is convert the string to binary.
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.
How do I take a single ASCII character and convert it to its decimal equivelant in MIPs?
Do I simply have to have some conditions to subtract a certain amount from the ascii code to make it its decimal representation?
A single hex character should be checked if it's in the range
'0' thru '9' (48 thru 57),
'A' thru 'F' (65 thru 70), or
'a' thru 'f' (97 thru 102).
Anything else is an error. If it does fall within one of those ranges, perform the following:
Subtract 48 (brings '0'-'9' down to 0-9).
If it's still greater than 9, subtract 7 (brings 'A'-'F' down to 10-15).
If it's still greater than 15, subtract 32 (brings 'a'-'f' down to 10-15).
If you're certain that the character will always be uppercase for the non-decimal digits, you can skip the third step in each of those lists above but it doesn't require a lot of extra code to do it.
Here's a simplistic implementation of what Pax wrote (it assumes that hexadecimal digits - A to F are always upper case)
File hextodec.c
#include <stdio.h>
/*
*Converts an ASCII char to its decimal equivalent.
*Returns -1 on error.
*
*/
extern int hextodec(char* c);
int main(int argc,char **argv){
int i=0;
char digits[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','F'};
for (;i<16;i++){
printf("%c\t%d\n",digits[i],hextodec(digits+i));
}
return 0;
}
File hextodec.S
#include <mips/regdef.h>
/* int hextodec(char* c)
* first (and only) argument is set in register a0.
* return value is set in register v0.
* function calling convention is ignored.
*/
.text
.globl hextodec
.align 2
.ent hextodec
hextodec:
lbu t0,0(a0) #load byte from argument
li t1,0X30
li t2,0x39
andi t1,t1,0x000000ff #Cast to word for comparison.
andi t2,t2,0x000000ff
bltu t0,t1,ERROR #error if lower than 0x30
bgt t0,t2,dohex #if greater than 0x39, test for A -F
addiu t0,t0,-0x30 #OK, char between 48 and 55. Subtract 48.
b return
dohex: li t1,0x41
li t2,0x46
andi t1,t1,0x000000ff #Cast to word for comparison.
andi t2,t2,0x000000ff
/*is byte is between 65 and 70?*/
bltu t0,t1,ERROR #error if lower than 0x41
bgt t0,t2,ERROR #error if greater than 0x46
ishex: addiu t0,t0,-0x37 #subtract 55 from hex char ('A'- 'F')
b return
ERROR: addiu t0,zero,-1 #return -1.
return: move v0,t0 #move return value to register v0
jr ra
.end hextodec
test run
root#:~/stackoverflow# ./hextodec
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
A 10
B 11
C 12
D 13
E 14
F 15
root#:~/stackoverflow#
Yes subtracting 48 from the ASCII value will probably be easiest.