I am totally new to programming.
How do I write a Python script to do the following without using bitwise operators, only allow to use logical and or operators. Thanks in advance!
I'm stuck.
& and
Sets result bit to 1 if both corresponding bits are 1.
E.g. 1010 & 1100 is 1000
| or
Sets result bit to 1 if one of the two corresponding bits is 1.
E.g 1010 | 1100 is 1110
^ xor
Sets result bit to 1 only if one of the corresponding bits is 1.
E.g. 1010 ^ 1100 is 0110
Output:
Enter binary expression: 110110 & 110011
Result: 110010
expr = input('Enter binary expression: ')
n1, op, n2 = expr.split()
n1 = int(n1)
n2 = int(n2)
enter image description here
If you're not allowed to use the bitwise operators, then probably the most straightforward way would be to go bit by bit and do the comparison yourself. This is easiest to do if we represent them as strings
We don't even need to convert the inputs to integers (which can be done by calling int(n1, 2) to specify base-2 instead of base-10), since they're given as strings in binary already. If they were given in base-10, we could read them in as integers and then use the built-in bin() method to represent them as binary strings.
Since there's a method bin() that can represent an integer as a binary string (prefaced with the characters 0b, which we can remove by slicing them off), we can use that, compare digits in the string, then convert the result back into an integer.
expr = input('Enter binary expression: ') # sample: enter "110110 & 110011"
n1, op, n2 = expr.split()
# -------------
result = ''
for (c1, c2) in zip(n1, n2):
next_bit = ""
if op == "&":
next_bit = "1" if (c1 == "1" and c2 == "1") else "0"
elif op == "|":
next_bit = "1" if (c1 == "1" or c2 == "1") else "0"
elif op == "^":
next_bit = "1" if (c1 != c2) else "0"
result += next_bit
print(result)
# 110010
This isn't the most efficient way (you can do some complicated iterative arithmetic and treat the numbers as integers, instead of treating them as strings and just comparing digit by digit), but it's probably the most straightforward and easiest to understand, especially given how the numbers are inputted here.
Related
I am doing a hackerrank example called Flipping bits where given a list of 32 bit unsigned integers. Flip all the bits (1->0 and 0->1) and return the result as an unsigned integer.
The correct code is:
def flippingBits(n):
seq = format(n, '032b')
return int(''.join(['0' if bit == '1' else '1' for bit in seq]), 2)
I dont understand the last line, what does the ''. part do? and why is there a ,2 at the end?
I have understood most of the code but need help in understanding the last part.
what does the ''. part do
'' represents an empty string which will be used as separator to join collection elements into string (some examples can be found here)
and why is there a ,2 at the end?
from int docs:
class int(x=0)
class int(x, base=10)
Return an integer object constructed from a number or string x
In this case it will parse the string provided in binary format (i.e. with base 2) into int.
I hope the below explanation helps:
def flippingBits(n):
seq = format(n, '032b') # change the format from base 10 to base 2 with 32bit size unsigned integer into string
return int(''.join(['0' if bit == '1' else '1' for bit in seq]), 2)
# ['0' if bit == '1' else '1' for bit in seq] => means: build a list of characters from "seq" string
# in which whenever there is 1 convert it to 0, and to 1 otherwise; then
# ''.join(char_list) => means: build string by joining characters in char_list
# without space between them ('' means empty delimiter); then
# int(num_string, 2) => convert num_string from string to integer in a base 2
Notice that you can do the bit flipping by using bit-wise operations without converting to string back and forth.
def flippingBits(n):
inverted_n = ~n # flip all bits from 0 to 1, and 1 to 0
return inverted_n+2**32 # because the number is a signed integer, the most significant bit should be flipped as well
I am a newby on this. I am trying to multiply every single element of the string below ('10010010') by 2 to the power of the position of the element in the string and sum all the multiplications. So far I am trying to do it like this, but I cannot achieve to figure out how to do it.
def decodingvalue(str1):
# read each character in input string
for ch in str1:
q=sum(2^(ch-1)*ch.isdigit())
return q
Function call
print(decodingvalue('10010010'))
Thanks a lot for your help!
I think you trying convert binary to int. If that so you can do the following:
str = '101110101'
#length is counted 1 to n, decrementing by 1 changes to 0-(n-1)
c = len(str)-1
q = 0
for ch in str:
print(q,c,ch)
q = q + (int(ch)*(2**c)) #in python power is '**'
c = c-1
if c == -1:
break
print(q)
you can of course optimize it and finish in fewer lines.
In python ^ (caret operator) is a Bitwise XOR.
I am having some issues with some code I wrote for this problem:
“Write a function namedd calc that will evaluate a simple arithmetic expression. The input to your program will be a string of the form:
operand1 operator operand2
where operand1 and operand2 are non-negative integers and operator is a single-character operator, which is either +, -, or *. You may assume that there is a space between each operand and the operator. You may further assume that the input is a valid mathemat- ical expression, i.e. your program is not responsible for the case where the user enters gibberish.
Your function will return an integer, such that the returned value is equal to the value produced by applying the given operation to the given operands.
Sample execution:
calc("5 + 10") # 15
“You may not use the split or eval functions in your solution.
Hint: the hard part here is breaking the input string into its three component. You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters. Be careful of off-by-one errors in using the slice operator.
Hint: it’s best to test your code as you work. The first step should be to break the input string into its three components. Write a program that does that, have it print out the operator and the two operands on separate lines, and test it until you are convinced that it works. Then, modifying it to perform the desired mathematical operation should be straightforward. Test your program with several different inputs to make sure it works as you expect.”
Here is my code:
def calc(exp):
operand1 = int(exp[:1])
operand2 = int(exp[4:6])
operator = exp[2:3]
if(operator == "+"):
addition = operand1+operand2
return addition
if(operator == "-"):
subtraction = operand1-operand2
return subtraction
if(operator == "*"):
multiplication = operand1*operand2
return multiplication
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
My code does not fully meet the criteria of this question. It only works for single digit numbers. How can I make my code work for any number?
Like:
“504 + 507”
”5678 + 76890”
and so on?
Thank you. Any help is appreciated.
As the hint says, get the position of the first and last space of the expression, use it to extract the operand and the operators, and then evaluate accordingly.
def calc(exp):
#Get the position for first space with find
low_idx = exp.find(' ')
#Get the position for last space with rfind
high_idx = exp.rfind(' ')
#Extract operators and operand with slice, converting operands to int
operand1 = int(exp[0:low_idx])
operator = exp[low_idx+1:high_idx]
operand2 = int(exp[high_idx:])
result = 0
#Evaluate based on operator
if operator == '+':
result = operand1 + operand2
elif operator == '-':
result = operand1 - operand2
elif operator == '*':
result = operand1 * operand2
return result
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
print(calc("504 + 507"))
print(calc("5678 + 76890"))
#15
#-4
#12
#1011
#82568
The answer is in the specification:
You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters.
find and rfind are methods of string objects.
You could split it into three components using this code: (note: this doesn't use split or eval)
def splitExpression(e):
numbers = ["1","2","3","4","5","6","7","8","9","0"] # list of all numbers
operations = ["+","-","*","/"] # list of all operations
output = [] # output components
currentlyParsing = "number" # the component we're currently parsing
buildstring = "" # temporary variable
for c in e:
if c == " ":
continue # ignore whitespace
if currentlyParsing == "number": # we are currently parsing a number
if c in numbers:
buildstring += c # this is a number, continue
elif c in operations:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "operation" # we are expecting an operation now
else:
pass # unknown symbol!
elif currentlyParsing == "operation": # we are currently parsing an operation
if c in operations:
buildstring += c # this is an operation, continue
elif c in numbers:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "number" # we are expecting a number now
else:
pass # unknown symbol!
if buildstring: # anything left in the buffer?
output.append(buildstring)
buildstring = ""
return output
Usage: splitExpression("281*14") returns ["281","*","14"]
This function also accepts spaces between numbers and operations
You can simply take the string and use the split method for the string object, which will return a list of strings based on some separator.
For example:
stringList = "504 + 507".split(" ")
stringList will now be a list such as ["504", "+", "507"] due to the separator " " which is a whitespace. Then just use stringList[1] with your conditionals to solve the problem. Additionally, you can use int(stringList[0]) and int(stringList[2]) to convert the strings to int objects.
EDIT:
Now I realized that your problem said to use find() instead of split(). Simply use the logic above but instead find(" ") the first whitespace. You will then need to find the second whitespace by slicing past the first whitespace using the two additional arguments available for find().
You need to split the string out instead of hard coding the positions of the indexes.
When coding you want to try to make your code as dynamic as possible, that generally means not hard coding stuff that could be a variable or in this case could be grabbed from the spaces.
Also in the if statements I modified them to elif as it is all one contained statement and thus should be grouped.
def calc(exp):
vals = exp.split(' ')
operand1 = int(vals[0])
operand2 = int(vals[2])
operator = vals[1]
if operator == '+':
return operand1+operand2
elif operator == '-':
return operand1-operand2
else:
return operand1*operand2
how to convert decimal to binary by using repeated division in python?
i know i have to use a while loop, and use modulus sign and others {%} and {//} to do this...but i need some kind of example for me to understand how its done so i can understand completely.
CORRECT ME, if I'm wrong:
number = int(input("Enter a numberto convert into binary: "))
result = ""
while number != 0:
remainder = number % 2 # gives the exact remainder
times = number // 2
result = str(remainder) + result
print("The binary representation is", result)
break
Thank You
Making a "break" without any condition, makes the loop useless, so the code only executes once no matter what.
-
If you don't need to keep the original number, you can change "number" as you go.
If you do need to keep the original number, you can make a different variable like "times".
You seem to have mixed these two scenarios together.
-
If you want to print all the steps, the print will be inside the loop so it prints multiple times.
If you only want to print the final result, then the print goes outside the loop.
while number != 0:
remainder = number % 2 # gives the exact remainder
number = number // 2
result = str(remainder) + result
print("The binary representation is", result)
-
The concatenation line:
Putting the print inside the loop might help you see how it works.
we can make an example:
the value in result might be "11010" (a string, with quotes)
the value in remainder might be 0 (an integer, no quotes)
str(remainder) turns the remainder into a string = "0" instead of 0
So when we look at the assignment statement:
result = str(remainder) + result
The right side of the assignment operator = is evaulated first.
The right side of the = is
str(remainder) + result
which, as we went over above has the values:
"0" + "11010"
This is string concatenation. It just puts one string on the end of the other one. The result is:
"0 11010"
"011010"
That is the value evaluated on the right side of the assignment statement.
result = "011010"
Now that is the value of result.
B_Number = 0
cnt = 0
while (N != 0):
rem = N % 2
c = pow(10, cnt)
B_Number += rem * c
N //= 2
# Count used to store exponent value
cnt += 1
return B_Number
Essentially, I have two strings of equal length, let's say 'AGGTCT' and 'AGGCCT' for examples sake. I want to compare them position by position and get a readout of when they do not match. So here I would hope to get 1 out because there is only 1 position where they do not match at position 4. If anyone has ideas for the positional comparison code that would help me a lot to get started.
Thank you!!
Use the following syntax to get the number of dissimilar characters for strings of equal size:
sum( str1 ~= str2 )
If you want to be case insensitive, use:
sum( lower(str1) ~= lower(str2) )
The expression str1 ~= str2 performs char-by-char comparison of the two strings, yielding a logical vector of the same size as the strings, with true where they mismatch (using ~=) and false where they match. To get your result simply sum the number of true values (mismatches).
EDIT: if you want to count the number of matching chars you can:
Use "equal to" == operator (instead of "not-equal to" ~= operator):
sum( str1 == str2 )
Subtract the number of mismatch, from the total number:
numel(str1) - sum( str1 ~= str2 )
You can compare all the element of the string:
r = all(seq1 == seq2)
This will compare char by char and return true if all the element in the resulting array are true. If the strings can have different sizes you may want to compare the sizes first. An alternative is
r = any(seq1 ~= seq2)
Another solution is to use strcmp:
r = strcmp(seq1, seq2)
Just would like to point out that you are asking to calculate the hamming distance (as you ask for alternatives - the article contains links to some). This is already discussed here. In short the builtin command pdist can do it.