I want to know if it's possible in MATLAB to discard overflow digits in MATLAB when I add two binary numbers.
I've only been able to find how to have a least number of binary digits, but how to I set a maximum number of digits?
For example:
e = dec2bin(bin2dec('1001') + bin2dec('1000'))
That gave me:
e =
10001
How do I get only '0001'?
dec2bin will always give you the minimum amount of bits to represent a number. If you would like to retain the n least significant digits, you have to index into the string and grab those yourself.
Specifically, if you want to retain only the n least significant digits, given that you have a base-10 number stored in num, you would do this:
out = dec2bin(num);
out = out(end-n+1:end);
Bear in mind that this performs no error checking. Should n be larger than the total number of bits in the string, you will get an out of bounds error. I'm assuming you're smart enough to use this and know what you're doing.
So for your example:
e = dec2bin(bin2dec('1001') + bin2dec('1000'));
n = 4, and so:
>> n = 4;
>> e = e(end-n+1:end)
e =
0001
Here is a more robust (but less efficient, I fear) way: What you describe is exactly the modulo operation. A 4-bit binary number is the remainder after a division by 0b10000 = 16. This can be done using the mod function in MATLAB.
>> e = dec2bin(mod(bin2dec('1001') + bin2dec('1000'),16),4)
e =
0001
Note: I added 4 as additional argument to the dec2bin function, so the output will always be 4-bit wide.
This can of course be generalized to any bit width: If you want to add 8-bit numbers, you will need the remainder of the division by 0b1'0000'0000 = 256, for example
>> e = dec2bin(mod(bin2dec('10011001') + bin2dec('10001000'),256),8)
e =
00100001
Or for shorter numbers, e.g. 2-bit wide, it is 0b100 = 4:
>> e = dec2bin(mod(bin2dec('10') + bin2dec('11'),4),2)
e =
01
Related
How to Increment a 3 digit Number by 1 in Brainfuck?
For Example. Getting a User input of 699 this code should output 700.
This is something i Tried but its not working .
,>,>,+
<<.>.>.
The reason it doesn't work is because you don't have a 3 digit number in memory. You have three ASCII values from 48-57 (characters 0-9). I will explain how to do what you need to, but won't provider the actual BF code for multiplication etc, you can find those elsewhere.
Let's call the characters X, Y and Z
First of all we need to convert them to a number between 0 and 9. Assuming that the user enters digits only, we can do this by subtracting 48 (ASCII character 0) from the ASCII value.
Ok so now we have three numbers from 0-9 as follows:
A = X - 48
B = Y - 48
C = Z - 48
Problem is they're still three separate digits and not one number. What's the actual number? In this case the number can be built in the following way:
N = 100A + 10B + C
So, you need to multiply A by 100, B by 10, C by 1, and then adding them all together. Could be done as follows:
N = A
N = (N * 10) + B
N = (N * 10) + C
After you've done that, you have an actual number in one cell, and you can increment it by doing N = N + 1 which in BF is a single +
Now this would work for numbers up to 255, which is the largest number that a BF cell can hold. You want to work with larger numbers? It gets more complicated, because now you have to split that number up into two or more cells, propagating the carries yourself etc. I won't go into that here because that gets much more complex (though there are algorithms you can find online to help as well) and I think this is enough to get you started.
EDIT: I realized that your code is also trying to print the incremented value. This requires more work, because the . command outputs the value of that cell as is as an ASCII character. But you need to output three digits. To do that you'd need to reverse the process above to split the number into three digits as follows:
C = N % 10
N = N / 10
B = N % 10
N = N / 10
A = N
Then you need to convert them from a number 0-9 to ASCII characters again as follows:
X = A + 48
Y = B + 48
Z = C + 48
and then finally you can output characters X, Y and Z in that order which would be the human readable incremented number.
Example: 6.321: I need it to be 6.322.
5.14875: I need it to be 5.14876.
How can I do this?
If you represent numbers as floating point or double precision floating point, this problem is a disaster.
If you can read in a number as a string (you mentioned get the number with the input command), you could do:
x = input('ENTER A NUMBER: ','s');
decimal_place = find(fliplr(x)=='.',1) - 1;
x_val = str2double(x);
if(~isempty(decimal_place))
y = x_val + 10 ^ -decimal_place;
else % if there is no decimal place, find first non-zero digit to get sigfig
warning('ambiguous number of significant digits');
first_nonzero_digit = find(fliplr(x)~='0',1);
if(~isempty(first_nonzero_digit))
y = x_val + 10 ^ (first_nonzero_digit - 1);
else
y = x_val + 1;
end
end
disp('your new number is');
disp(y);
Example test runs:
ENTER A NUMBER: 1.9
your new number is
2
ENTER A NUMBER: 3510
your new number is
3520
ENTER A NUMBER: 323.4374
your number is
323.4375
ENTER A NUMBER: 0
your number is
1
#AndrasDeak - I think you're right the first time. The hard part is not the rounding - it's defining the "last" decimal.
Since floating point numbers aren't exact, I can't think of a reliable way to find that "last" decimal place - in any language.
There is a very hacky way that comes to mind, though. You could "print" the number to a string, with 31 numbers after the decimal point, then working right from the dot, find the first place with 15 0s. (Since double precision numbers can only stably represent the first 14 decimal places and you get a 15th that varies, 31 decimal place will ALWAYS give you at least 15 0s after the last sig digit.)
>> a = 1.34568700030041234556
a =
1.3457
>> str = sprintf('%1.31', a)
str =
Empty string: 1-by-0
>> str = sprintf('%1.31f', a)
str =
1.3456870003004124000000000000000
>> idx = strfind(str, '000000000000000')
idx =
19
>> b = a*10^(idx(1)-3)
b =
1.3457e+16
>> sprintf('%10.20f', b)
ans =
13456870003004124.00000000000000000000
>> c = b+1
c =
1.3457e+16
>> sprintf('%10.20f', c)
ans =
13456870003004124.00000000000000000000
>> final = floor(c)/10^(idx(1)-3)
final =
1.3457
>> sprintf('%10.31f', final)
ans =
1.3456870003004124000000000000000
I think that's a relatively reliable implementation.
http://www.mathworks.com/matlabcentral/answers/142819-how-to-find-number-of-significant-figures-in-a-decimal-number
assuming your just trying to do regular rounding:
i'd use the round function built into matlab.
let's do your example above..
5.14875 has 5 decimal places and you want it to be converterd to 5.14876.
Lets assume you that the 6th decimal place was 9 (so your number is 5.148759)
%Step 1:changethe format so that your going to be able to see all of the
%decimal places
format long
%step2:now enter the original number
OriginalNumber=5.148755
%step 3 take the original number and round it to your new number
NewNumber=round(OriginalNumber,5)
this solution will not work if the 6th number (that you did not show) was a <5 because the computer will not know to round up
assuming your just trying to cut the numbers off...
You cannot do this in regular default matlab floating point numbers. To keep my explination simple I'll just state that without an explination. I'd do some review on the different ways matlab stores # (int vs floating point) on the matlab website. They have excellent documentation.
I wonder how to sum digits for a multi-digit number in Matlab.
For example 1241= 1+2+4+1 = 8
String-based answer:
>> n = 1241;
>> sum(int2str(n)-48)
ans =
8
The number is first converted to a string representation using int2str, then the ASCII code for '0' (i.e. 48) is subtracted from the ASCII code for each element of the string, producing a numeric vector. This is then summed to get the result.
A = 35356536576821;
A = abs(A);
xp = ceil(log10(A)):-1:1;
while ~isscalar(xp)
A = sum(fix(mod(A,10.^xp)./10.^[xp(2:end) 0]));
xp = ceil(log10(A)):-1:1;
end
this is the numeric approach
This one is the solution is character approach:
A = '35356536576821';
A = char(regexp(A,'\d+','match'));
while ~isscalar(A)
A = num2str(sum(A - '0'));
end
Both, first take the absolute number (strip the minus) then: the numeric one counts with log10() how many digits a number has and through modulus and divisions extracts the digits which are summed, while the char approach convert to numeric digits with implicit conversion of - '0', sums and converts back to string again.
Another all-arithmetic approach:
n = 1241; %// input
s = 0; %// initiallize output
while n>0 %// while there is some digit left
s = s + mod(n-1,10)+1; %// sum rightmost digit
n = floor(n/10); %// remove that digit
end
Youcan use this code
sum(int2str(n)-48)
where n, is your input number.
How can one find the last two digits of a decimal number using MATLAB?
Example:
59 for 1.23000659
35 for 56368.35
12 for 548695412
There will always be issues when you have a decimal number with many integer digits and fractional digits. In this case, the number of integer and decimal digits decide if we are correct or not in estimating the last two digits. Let's take at the code and the comments thereafter.
Code
%// num is the input decimal number
t1 = num2str(num,'%1.15e') %// Convert number to exponential notation
t1 = t1(1:strfind(t1,'e')-1)
lastind = find(t1-'0',1,'last')
out = str2num(t1(lastind-1:lastind)) %// desired output
Results and Conclusions
For num = 1.23000659, it prints the output as 59, which is correct thanks
to the fact that the number of integer and decimal digits don't add upto
more than 16.
For num = 56368.35, we get output as 35, which is correct again and the
reason is the same as before.
For num = 548695412, we are getting the correct output of 12 because of the
same good reason.
For an out of the question sample of num = 2736232.3927327329236576
(deliberately chosen a number with many integer and fractional digits),
the code run gives output as 33 which is wrong and the reason could be
inferred from the fact that integer and decimal digits add upto a much
bigger number than the code could handle.
One can look into MATLAB command vpa for getting more precision, if extreme cases like the 4th one are to dealt with.
Convert to string and then extract the last two characters:
x = 1.23; % x = 1.23
s = num2str(x); % s = "1.23"
t = s(end-1:end); % t = "23"
u = str2num(t); % u = 23
Note: depending on your specific needs you might want to supply a precision or formatSpec to num2str.
The other two answers are nice and straight forward, but here you have a mathematical way of doing it ;)
Assuming a as your number,
ashift=0.01*a; %shift the last two digits
afloor=floor(ashift); %crop the last two digits
LastDecimals=a-100*afloor; %substract the cropped number form the original, only the last two digits left.
Of course if you have non-natural numbers, you can figure those out too with the same "floor and subtract technique as above.
I have a strange problem with hex2dec function in Matlab.
I realized in 16bytes data, it omits 2 LSB bytes.
hex2dec('123123123123123A');
dec2hex(ans)
Warning: At least one of the input numbers is larger than the largest integer-valued floating-point
number (2^52). Results may be unpredictable.
ans =
1231231231231200
I am using this in Simulink. Therefore I cannot process 16byte data. Simulink interpret this as a 14byte + '00'.
You need to use uint64 to store that value:
A='123123123123123A';
B=bitshift(uint64(hex2dec(A(1:end-8))),32)+uint64(hex2dec(A(end-7:end)))
which returns
B =
1310867527582290490
An alternative way in MATLAB using typecast:
>> A = '123123123123123A';
>> B = typecast(uint32(hex2dec([A(9:end);A(1:8)])), 'uint64')
B =
1310867527582290490
And the reverse in the opposite direction:
>> AA = dec2hex(typecast(B,'uint32'));
>> AA = [AA(2,:) AA(1,:)]
AA =
123123123123123A
The idea is to treat the 64-integer as two 32-bit integers.
That said, Simulink does not support int64 and uint64 types as others have already noted..