Why does PRINT'ing a true boolean expression output -1? - basic

In Commodore 64 BASIC V2, PRINT'ing a true boolean expression outputs -1:
READY.
A=(5=5)
READY.
PRINT A
-1
Why -1 and not 1?

Commodore Basic doesn't have a boolean data type. A boolean expression evaluates to a number, where 0 means False and -1 means true.
As there is no boolean data type there are no boolean type expressions. You can use any numeric expression in an IF statement, and it will interpret any non-zero value as True.
Some languages where a boolean value is numeric or where it can be converted to a numeric value uses -1 to represent a true value. For an integer value 0 all bits are cleared, and for -1 all bits are set, so they can be seen as natural complements for each other.
Eventhough Commodore Basic doesn't use integer numbers but floating point numbers, the value -1 was supposedly chosen because some other languages uses it.

Short answer
Why -1 and not 1?
It's just a convention
Some speculation...
The reason underlying the choice for this convention may be related to the internal representation of integers numbers on most platforms:
Suppose a boolean value is stored in a 16 bits wide integer; false is 0 (every bit unset).
0 => 00000000 00000000
It makes sense to agree upon another convention where true is all bits set:
11111111 11111111
(a very reasonable choice since all 1 is the bitwise NOT of all 0)
The decimal representation of a signed integer whose bits are all set is -1
-1 => 11111111 11111111
While the binary representation of 1 is
1 => 00000001 00000000 (on a little endian platform).
So that's why -1 and not 1: is just a convention; but if you look at the binary representation of the value you may agree that the convention makes sense.
However this convention is far from being universally adopted despite all the above considerations. On many languages true casted to a numeric value is 1.
An important note about the code you posted:
You wrote
A=(5=5)
A is a real variable whose value is represented by 5 bytes (1 for exponent, 4 for mantissa).
The internal C64 representation of the real value 0 is all bits 0, but the representation of -1 is far from being all bits 1.
(all bits 1 would lead to the value -1.70141183e+38)
So again C64 Basic just sticks to a convention.
Finally let's explain the behaviuour of the IF statements in your code.
Any value different from 0 is evaluated as true in an IF statement (this happens on most languages, maybe all).
Bottom note:
If you want to take a look at the internal representation of a C64 variable (integer or real only, not strings) you may use the following code:
READY.
10 REM THE VARIABLE TO INSPECT
20 A=(5=5)
30 REM THE ADDR. OF THE FIRST VARIABLE
40 B=PEEK(45)+256*PEEK(46)
50 REM DISPLAY THE VAR'S 7 BYTES
60 FOR C=B TO B+6
70 PRINT C;": ";PEEK(C)
80 NEXT C
RUN
2227 : 65
2228 : 0
2229 : 129
2230 : 128
2231 : 0
2232 : 0
2233 : 0
Note the first two addresses (2227, 2228) store the variable's name (65, 0 for A)
You may try declaring the variable as integer and see the result with
20 A%=(5=5)

There is a way that this may be used, by customising a for loop, i.e., if you wanted something to happen until a key is pressed, one might:
0 for i=-1 to 0
1 rem logic here ...
10 get a$: i=(a$=""): next i
This is the same sort of logic as a do...while loop.
Edit - If you specifically wanted 0 to be false and 1 to be true, you could define a function as follows (I forgot about the ABS keyword as I've not used it in probably 20 years :-|) :
0 def fn b(x) = abs(x)
1 i = 7
2 a$ = "hello"
3 if fn b (i=6) then print "i is 6"
4 if fn b (i<10) then print "i is less than 10"
5 if fn b (a$="hi") then print "hey there!"
6 if fn b (a$="hello") then print a$

Related

Why does integer_decode in Rust work that way?

I don't understand how integer_decode in num_traits works. For instance: we have
use num_traits::Float;
let num = 2.0f32;
// (8388608, -22, 1)
let (mantissa, exponent, sign) = Float::integer_decode(num);
But how we get those integers?
Binary representation of 2.0f32 has 0 sign bit, 1 bit as leading bit in exponent and mantissa consisting of zeros. How to get integer decode and why we choose this particular decomposition and not 8388608*2 as mantissa and -23 as exponent?
I didn't write the function, so take this answer with a grain of salt, as it's more of a gut feeling than knowledge. The rationale behind it is not explained in the comments in the function implementation, so unless the author of the code speaks up, we can't deliver more than educational guesses.
f32 is based on IEEE-754, which specifies that a 2.0 shall be represented as the following three parts:
the sign bit 0
indicates that 2.0 is positive
the exponent 128
it's one byte that indicates the exponent, with 127 representing 0. 128 means 1
the mantissa 0
the mantissa consists of 23 bits and has an implicit 1. in front of it. So 0 means 1.0.
To get the actual number, you need to do 0 * (-1) + 2 ^ (128 - 127) * 1.0, which is 2 ^ 1 * 1 = 2.
Now this is not the only way to compute that. You could also do:
map the sign bit to 1 and -1
instead of prefixing the mantissa with 1., add a 1 in front of it, making it an integer. (this avoids having to use a float to decode a float, which is nonsense for obvious reasons)
subtract 127 from the exponent, making it signed. Then, remove 23 from it to compensate that our mantissa is now shifted by 23 bits (because the mantissa is 23 bits long and we moved the comma all the way back to make it an integer).
This would, for 2.0 give us:
sign -1
mantissa 0b100000000000000000000000 = 8388608
exponent 128 - 127 - 23 = -22
Now we can do sign * mantissa * 2 ^ exponent, as specified in the documentation to get our value back.
Note how fast calculating those integers was: a binary decision for the sign, a binary or operation for the mantissa and a single u8 subtraction for the exponent (a single one because one can combine - 127 - 23 to - 150 beforehand).
why we choose this particular decomposition and not 8388608*2 as mantissa and -23 as exponent
The short version is that this guarantees that all possible mantissas can be treated the same way. It's 23 bits long and a 1 with the entire mantissa attached to it is always a valid integer. In the case of 0 this is a 1 with 23 0s, 0b100000000000000000000000, which is 8388608.
integer_decode() documentation is quite clear:
Returns the mantissa, base 2 exponent, and sign as integers, respectively. The original number can be recovered by sign * mantissa * 2 ^ exponent.
1 * 8388608 * 2^-22 == 2
Mostly IEEE 754 uses this description of finite floating-point numbers: A floating-point number has the form (−1)s×be×m, where:
s is 0 or 1.
e is any integer emin ≤ e ≤ emax.
m is a number represented by a digit string of the form d0.d1d2…dp−1 where di is an integer digit 0 ≤ di < b.
This form is described in IEEE-754 2019 clause 3.3. b is the base, p is the precision (number of digits in base b), and emin and emax are bounds on the exponent. This form is useful for certain things, such as describing the normalized form as starting with a leading “1.” or “0.” when the base is two. However, the standard also says, in the same clause:
It is also convenient for some purposes to view the significand as an 10 integer; in which case the finite floating-point numbers are described thus:
Signed zero and non-zero floating-point numbers of the form (−1)s×bp×c, where
s is 0 or 1.
q is any integer emin ≤ q+p−1 ≤ emax.
c is a number represented by a digit string of the form d0d1d2…dp−1 where di is an integer digit 0 ≤ di ≤ b (c is therefore an integer with 0 ≤ c < bp).
(My reproduction of the IEEE-754 text changes some of the typography slightly.) Note two things. First, this matches the results Float::integer_decode gives you. In the Float format, p is 24, so m should have 24 bits, not 25, so it can be 8,388,608 (223) and cannot be 16,777,216 (224).
Second, what makes this form useful is m is always an integer and can be any integer in that range—the low digit of m is immediately to the left of the radix point, so the consecutive values representable in this range of the floating-point format are consecutive integers, and we can analyze them and write proofs using number theory.
You could use alternate forms that are mathematically equivalent but let the significand be 224, but then the low digit in such a form would have to be zero (because there is no bit in the format to represent it having any other value), so it is not particularly useful.

bitwise operations in python (or,and, |) [duplicate]

Consider this code:
x = 1 # 0001
x << 2 # Shift left 2 bits: 0100
# Result: 4
x | 2 # Bitwise OR: 0011
# Result: 3
x & 1 # Bitwise AND: 0001
# Result: 1
I can understand the arithmetic operators in Python (and other languages), but I never understood 'bitwise' operators quite well. In the above example (from a Python book), I understand the left-shift but not the other two.
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Bitwise operators are operators that work on multi-bit values, but conceptually one bit at a time.
AND is 1 only if both of its inputs are 1, otherwise it's 0.
OR is 1 if one or both of its inputs are 1, otherwise it's 0.
XOR is 1 only if exactly one of its inputs are 1, otherwise it's 0.
NOT is 1 only if its input is 0, otherwise it's 0.
These can often be best shown as truth tables. Input possibilities are on the top and left, the resultant bit is one of the four (two in the case of NOT since it only has one input) values shown at the intersection of the inputs.
AND | 0 1 OR | 0 1 XOR | 0 1 NOT | 0 1
----+----- ---+---- ----+---- ----+----
0 | 0 0 0 | 0 1 0 | 0 1 | 1 0
1 | 0 1 1 | 1 1 1 | 1 0
One example is if you only want the lower 4 bits of an integer, you AND it with 15 (binary 1111) so:
201: 1100 1001
AND 15: 0000 1111
------------------
IS 9 0000 1001
The zero bits in 15 in that case effectively act as a filter, forcing the bits in the result to be zero as well.
In addition, >> and << are often included as bitwise operators, and they "shift" a value respectively right and left by a certain number of bits, throwing away bits that roll of the end you're shifting towards, and feeding in zero bits at the other end.
So, for example:
1001 0101 >> 2 gives 0010 0101
1111 1111 << 4 gives 1111 0000
Note that the left shift in Python is unusual in that it's not using a fixed width where bits are discarded - while many languages use a fixed width based on the data type, Python simply expands the width to cater for extra bits. In order to get the discarding behaviour in Python, you can follow a left shift with a bitwise and such as in an 8-bit value shifting left four bits:
bits8 = (bits8 << 4) & 255
With that in mind, another example of bitwise operators is if you have two 4-bit values that you want to pack into an 8-bit one, you can use all three of your operators (left-shift, and and or):
packed_val = ((val1 & 15) << 4) | (val2 & 15)
The & 15 operation will make sure that both values only have the lower 4 bits.
The << 4 is a 4-bit shift left to move val1 into the top 4 bits of an 8-bit value.
The | simply combines these two together.
If val1 is 7 and val2 is 4:
val1 val2
==== ====
& 15 (and) xxxx-0111 xxxx-0100 & 15
<< 4 (left) 0111-0000 |
| |
+-------+-------+
|
| (or) 0111-0100
One typical usage:
| is used to set a certain bit to 1
& is used to test or clear a certain bit
Set a bit (where n is the bit number, and 0 is the least significant bit):
unsigned char a |= (1 << n);
Clear a bit:
unsigned char b &= ~(1 << n);
Toggle a bit:
unsigned char c ^= (1 << n);
Test a bit:
unsigned char e = d & (1 << n);
Take the case of your list for example:
x | 2 is used to set bit 1 of x to 1
x & 1 is used to test if bit 0 of x is 1 or 0
what are bitwise operators actually used for? I'd appreciate some examples.
One of the most common uses of bitwise operations is for parsing hexadecimal colours.
For example, here's a Python function that accepts a String like #FF09BE and returns a tuple of its Red, Green and Blue values.
def hexToRgb(value):
# Convert string to hexadecimal number (base 16)
num = (int(value.lstrip("#"), 16))
# Shift 16 bits to the right, and then binary AND to obtain 8 bits representing red
r = ((num >> 16) & 0xFF)
# Shift 8 bits to the right, and then binary AND to obtain 8 bits representing green
g = ((num >> 8) & 0xFF)
# Simply binary AND to obtain 8 bits representing blue
b = (num & 0xFF)
return (r, g, b)
I know that there are more efficient ways to acheive this, but I believe that this is a really concise example illustrating both shifts and bitwise boolean operations.
I think that the second part of the question:
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Has been only partially addressed. These are my two cents on that matter.
Bitwise operations in programming languages play a fundamental role when dealing with a lot of applications. Almost all low-level computing must be done using this kind of operations.
In all applications that need to send data between two nodes, such as:
computer networks;
telecommunication applications (cellular phones, satellite communications, etc).
In the lower level layer of communication, the data is usually sent in what is called frames. Frames are just strings of bytes that are sent through a physical channel. This frames usually contain the actual data plus some other fields (coded in bytes) that are part of what is called the header. The header usually contains bytes that encode some information related to the status of the communication (e.g, with flags (bits)), frame counters, correction and error detection codes, etc. To get the transmitted data in a frame, and to build the frames to send data, you will need for sure bitwise operations.
In general, when dealing with that kind of applications, an API is available so you don't have to deal with all those details. For example, all modern programming languages provide libraries for socket connections, so you don't actually need to build the TCP/IP communication frames. But think about the good people that programmed those APIs for you, they had to deal with frame construction for sure; using all kinds of bitwise operations to go back and forth from the low-level to the higher-level communication.
As a concrete example, imagine some one gives you a file that contains raw data that was captured directly by telecommunication hardware. In this case, in order to find the frames, you will need to read the raw bytes in the file and try to find some kind of synchronization words, by scanning the data bit by bit. After identifying the synchronization words, you will need to get the actual frames, and SHIFT them if necessary (and that is just the start of the story) to get the actual data that is being transmitted.
Another very different low level family of application is when you need to control hardware using some (kind of ancient) ports, such as parallel and serial ports. This ports are controlled by setting some bytes, and each bit of that bytes has a specific meaning, in terms of instructions, for that port (see for instance http://en.wikipedia.org/wiki/Parallel_port). If you want to build software that does something with that hardware you will need bitwise operations to translate the instructions you want to execute to the bytes that the port understand.
For example, if you have some physical buttons connected to the parallel port to control some other device, this is a line of code that you can find in the soft application:
read = ((read ^ 0x80) >> 4) & 0x0f;
Hope this contributes.
I didn't see it mentioned above but you will also see some people use left and right shift for arithmetic operations. A left shift by x is equivalent to multiplying by 2^x (as long as it doesn't overflow) and a right shift is equivalent to dividing by 2^x.
Recently I've seen people using x << 1 and x >> 1 for doubling and halving, although I'm not sure if they are just trying to be clever or if there really is a distinct advantage over the normal operators.
I hope this clarifies those two:
x | 2
0001 //x
0010 //2
0011 //result = 3
x & 1
0001 //x
0001 //1
0001 //result = 1
Think of 0 as false and 1 as true. Then bitwise and(&) and or(|) work just like regular and and or except they do all of the bits in the value at once. Typically you will see them used for flags if you have 30 options that can be set (say as draw styles on a window) you don't want to have to pass in 30 separate boolean values to set or unset each one so you use | to combine options into a single value and then you use & to check if each option is set. This style of flag passing is heavily used by OpenGL. Since each bit is a separate flag you get flag values on powers of two(aka numbers that have only one bit set) 1(2^0) 2(2^1) 4(2^2) 8(2^3) the power of two tells you which bit is set if the flag is on.
Also note 2 = 10 so x|2 is 110(6) not 111(7) If none of the bits overlap(which is true in this case) | acts like addition.
Sets
Sets can be combined using mathematical operations.
The union operator | combines two sets to form a new one containing items in either.
The intersection operator & gets items only in both.
The difference operator - gets items in the first set but not in the second.
The symmetric difference operator ^ gets items in either set, but not both.
Try It Yourself:
first = {1, 2, 3, 4, 5, 6}
second = {4, 5, 6, 7, 8, 9}
print(first | second)
print(first & second)
print(first - second)
print(second - first)
print(first ^ second)
Result:
{1, 2, 3, 4, 5, 6, 7, 8, 9}
{4, 5, 6}
{1, 2, 3}
{8, 9, 7}
{1, 2, 3, 7, 8, 9}
This example will show you the operations for all four 2 bit values:
10 | 12
1010 #decimal 10
1100 #decimal 12
1110 #result = 14
10 & 12
1010 #decimal 10
1100 #decimal 12
1000 #result = 8
Here is one example of usage:
x = raw_input('Enter a number:')
print 'x is %s.' % ('even', 'odd')[x&1]
Another common use-case is manipulating/testing file permissions. See the Python stat module: http://docs.python.org/library/stat.html.
For example, to compare a file's permissions to a desired permission set, you could do something like:
import os
import stat
#Get the actual mode of a file
mode = os.stat('file.txt').st_mode
#File should be a regular file, readable and writable by its owner
#Each permission value has a single 'on' bit. Use bitwise or to combine
#them.
desired_mode = stat.S_IFREG|stat.S_IRUSR|stat.S_IWUSR
#check for exact match:
mode == desired_mode
#check for at least one bit matching:
bool(mode & desired_mode)
#check for at least one bit 'on' in one, and not in the other:
bool(mode ^ desired_mode)
#check that all bits from desired_mode are set in mode, but I don't care about
# other bits.
not bool((mode^desired_mode)&desired_mode)
I cast the results as booleans, because I only care about the truth or falsehood, but it would be a worthwhile exercise to print out the bin() values for each one.
Bit representations of integers are often used in scientific computing to represent arrays of true-false information because a bitwise operation is much faster than iterating through an array of booleans. (Higher level languages may use the idea of a bit array.)
A nice and fairly simple example of this is the general solution to the game of Nim. Take a look at the Python code on the Wikipedia page. It makes heavy use of bitwise exclusive or, ^.
There may be a better way to find where an array element is between two values, but as this example shows, the & works here, whereas and does not.
import numpy as np
a=np.array([1.2, 2.3, 3.4])
np.where((a>2) and (a<3))
#Result: Value Error
np.where((a>2) & (a<3))
#Result: (array([1]),)
i didnt see it mentioned, This example will show you the (-) decimal operation for 2 bit values: A-B (only if A contains B)
this operation is needed when we hold an verb in our program that represent bits. sometimes we need to add bits (like above) and sometimes we need to remove bits (if the verb contains then)
111 #decimal 7
-
100 #decimal 4
--------------
011 #decimal 3
with python:
7 & ~4 = 3 (remove from 7 the bits that represent 4)
001 #decimal 1
-
100 #decimal 4
--------------
001 #decimal 1
with python:
1 & ~4 = 1 (remove from 1 the bits that represent 4 - in this case 1 is not 'contains' 4)..
Whilst manipulating bits of an integer is useful, often for network protocols, which may be specified down to the bit, one can require manipulation of longer byte sequences (which aren't easily converted into one integer). In this case it is useful to employ the bitstring library which allows for bitwise operations on data - e.g. one can import the string 'ABCDEFGHIJKLMNOPQ' as a string or as hex and bit shift it (or perform other bitwise operations):
>>> import bitstring
>>> bitstring.BitArray(bytes='ABCDEFGHIJKLMNOPQ') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
>>> bitstring.BitArray(hex='0x4142434445464748494a4b4c4d4e4f5051') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
the following bitwise operators: &, |, ^, and ~ return values (based on their input) in the same way logic gates affect signals. You could use them to emulate circuits.
To flip bits (i.e. 1's complement/invert) you can do the following:
Since value ExORed with all 1s results into inversion,
for a given bit width you can use ExOR to invert them.
In Binary
a=1010 --> this is 0xA or decimal 10
then
c = 1111 ^ a = 0101 --> this is 0xF or decimal 15
-----------------
In Python
a=10
b=15
c = a ^ b --> 0101
print(bin(c)) # gives '0b101'
You can use bit masking to convert binary to decimal;
int a = 1 << 7;
int c = 55;
for(int i = 0; i < 8; i++){
System.out.print((a & c) >> 7);
c = c << 1;
}
this is for 8 digits you can also do for further more.

What is the bitwise negation for an integer?

I have homework assignment with a piece to compute the bitwise negation of integer value. It say 512 go into -513.
I have a solution that does x = 512 y = 512*(-1)+(-1).
Is that correct way?
I think you need to first negate and add 1.
-x = ~x + 1
consequently
~x= -x -1
This property is based on the way negative number are represented in two's complement. To represent a negative number $A$ on n bits, one uses the complement of |A| to 2n, i.e. the number 2n-|A|
It is easy to see that A+~A=111...11 as bits in the addition will always be 0 and 1 and 111...111 is the number just before 2n, or 2n-1.
As -|A| is coded by 2n-|A|, and A +~A=2n-1, we can say that -A=~A+1 or equivalently ~A=-A-1
This is true for any number, positive or negative. And ~512=-512-1=-513
val = 512
print (~val)
output:
-513
~ bitwise complement
Sets the 1 bits to 0 and 1 to 0.
For example ~2 would result in -3.
This is because the bit-wise operator would first represent the number in sign and magnitude which is 0000 0010 (8 bit operator) where the MSB is the sign bit.
Then later it would take the negative number of 2 which is -2.
-2 is represented as 1000 0010 (8 bit operator) in sign and magnitude.
Later it adds a 1 to the LSB (1000 0010 + 1) which gives you 1000 0011.
Which is -3.
Otherwise:
y = -(512+1)
print (y)
output:
-513

How to count binary sequence in binary number in Python?

I would like to count '01' sequence in 5760 binary bits.
First, I would like to combine several binary numbers then count # of '01' occurrences.
For example, I have 64 bits integer. Say, 6291456. Then I convert it into binary. Most significant 4 bits are not used. So I'll get 60 bits binary 000...000011000000000000000000000
Then I need to combine(just put bits together since I only need to count '01') first 60 bits + second 60 bits + ...so 96 of 60 bits are stitched together.
Finally, I want to count how many '01' appears.
s = binToString(5760 binary bits)
cnt = s.count('01');
num = 6291226
binary = format(num, 'b')
print(binary)
print(binary.count('01'))
If I use number given by you i.e 6291456 it's binary representation is 11000000000000000000000 which gives 0 occurrences of '01'.
If you always want your number to be 60 bits in length you can use
binary = format(num,'060b')
It will add leading 0 to make it of given length
Say that nums is your list of 96 numbers, each of which can be stored in 64 bits. Since you want to throw away the most 4 significant bits, you are really taking the number modulo 2**60. Thus, to count the number of 01 in the resulting string, using the idea of #ShrikantShete to use the format function, you can do it all in one line:
''.join(format(n%2**60,'060b') for n in nums).count('01')

How to replace a digit in a string of numbers with a random number

I have a string which is a sequence of numbers between 0-9, I would like to replace some of the numbers in that string with random numbers, but I'm unable to do with with either rand or randi as I keep getting the following error:
Conversion to char from cell is not possible.
Error in work (line 58)
mut(i,7:14) = {mute_by};
Here's what I'm currently doing to try and alter some digits of my string:
% mutate probability 0.2
if (rand < 0.2)
% pick best chromosome to mutate
mut = combo{10};
mute_by = rand([0,9]);
for i = 1:5
mut(i) = {mute_by};
end
end
mut represents the string 110202132224154246176368198100
How would I go about doing this? I assumed it would be fairly simple but I've been going over the documentation for a while now and I can't find the solution.
What I would do is generate a logical array that represents true if you want to replace a particular position in your string and false otherwise. How you're determining this is by generating random floating pointing point numbers of the same size as your string, and checking to see if their values are < 0.2. For any positions in your string that satisfy this constraint, replace them with another random integer.
The best way to do this would be to convert your string into actual numbers so that you can actual modify their positions numerically, create our logical array, then replace these values with random integers at these co-ordinates.
Something like this:
rng(123); %// Set seed for reproducibility
mut = '110202132224154246176368198100';
mut_num = double(mut) - 48;
n = numel(mut_num);
vec = rand(1, n) < 0.4;
num_elem = sum(vec);
mut_num(vec) = randi(10,1,num_elem) - 1;
mut_final = char(mut_num + 48);
Let's go through this code slowly. rng is a random seed generator, and I set it to 123 so that you're able to reproduce the same results that I have made as random number generation... is of course random. I declare the number you have made as a string, then I have a nifty trick of turning each character in your string into a single element of a numeric array. I do this by casting to double, then subtracting by 48. By casting to double, each element gets converted into its ASCII code, and so 0 would be 48, 1 would be 49 and so on. Therefore, we need to subtract the values by 48 so that we can bring this down to a range of [0-9]. I then count how long your string is with numel, then figure out which values I want to replace by generating that logical vector we talked about.
We then count up how many elements we need to change, then generate a random integer vector that is exactly this size, and we use this logical vector to index into our recently converted string (being stored as a numeric array) with random integers. Note that randi generates values from 1 up to whatever maximum you want. Because this starts at 1, I have to generate up to 10, then subtract by 1. The output of randi gets placed into our numeric array, and then we convert our numeric array into a string with char. Note that we need to add by 48 to convert the numbers into their ASCII equivalents before creating our string.
I've changed the probability to 0.4 to actually see the changes better. Setting this to 0.2 I could barely notice any changes. mut_final would contain the changed string. Here is what they look like:
>> mut
mut =
110202132224154246176368198100
>> vec
vec =
Columns 1 through 13
0 1 1 0 0 0 0 0 0 1 1 0 0
Columns 14 through 26
1 1 0 1 1 0 0 0 0 0 0 0 1
Columns 27 through 30
1 1 1 0
>> mut_final
mut_final =
104202132444143248176368195610
vec contains those positions in the string you want to change, starting from the 2nd position, 3rd position, etc. The corresponding positions in mut change with respect to vec while the rest of the string is untouched and is finally stored in mut_final.

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