I want to make a binary calculator and I have a problem with the subtraction part. Here is my code (I have tried to adapt one for sum that I've found on this website).
maxlen = max(len(s1), len(s2))
s1 = s1.zfill(maxlen)
s2 = s2.zfill(maxlen)
result = ''
carry = 0
i = maxlen - 1
while(i >= 0):
s = int(s1[i]) - int(s2[i])
if s <= 0:
if carry == 0 and s != 0:
carry = 1
result = result + "1"
else:
result = result + "0"
else:
if carry == 1:
result = result + "0"
carry = 0
else:
result = result + "1"
i = i - 1
if carry>0:
result = result + "1"
return result[::-1]
The program works fine with some binaries subtraction but it fails with others.
Can someone please help me because I can't find the mistake? Thanks a lot.
Short answer: Your code is wrong for the case when s1[i] == s2[i] and carry == 1.
Longer answer: You should restructure your code to have three separate cases for s==-1, s==0, and s==1, and then branch on the value of carry within each case:
if s == -1: # 0-1
if carry == 0:
...
else:
...
elif s == 0: # 1-1 or 0-0
if carry == 0:
...
else:
...
else: # 1-0
if carry == 0:
...
else:
...
This way you have a separate block for each possibility, so there is no chance of overlooking a case like you did on your first attempt.
I hope the answer below it helps.
def binarySubstration(str1,str2):
if len(str1) == 0:
return
if len(str2) == 0:
return
str1,str2 = normaliseString(str1,str2)
startIdx = 0
endIdx = len(str1) - 1
carry = [0] * len(str1)
result = ''
while endIdx >= startIdx:
x = int(str1[endIdx])
y = int(str2[endIdx])
sub = (carry[endIdx] + x) - y
if sub == -1:
result += '1'
carry[endIdx-1] = -1
elif sub == 1:
result += '1'
elif sub == 0:
result += '0'
else:
raise Exception('Error')
endIdx -= 1
return result[::-1]
normalising the strings
def normaliseString(str1,str2):
diff = abs((len(str1) - len(str2)))
if diff != 0:
if len(str1) < len(str2):
str1 = ('0' * diff) + str1
else:
str2 = ('0' * diff) + str2
return [str1,str2]
Related
I do not seem to understand the error of this code. I am trying to convert from hexadecimal to decimal. For example for a hexadecimal string that starts with 0x. Can someone help me, please? Thank you
def hex_char_decode(digit): # 0xbadf00d
if digit[0:2] == "0x":
digit = digit[2:]
bit_index = len(digit) - 1
sum = 0
i = 0
while bit_index >= 0:
if digit[len] >= "0" and digit[len] <= "9":
res = int(digit[len])
elif digit[len] >= "A" and digit[len] <= "F":
res = ord(digit[len]) - 55
elif hex[len] >= 'a' and hex[len] <= 'f':
res = ord(digit[len]) - 87
else:
sum = 1
break
sum += (res * (16 ** i))
bit_index -= 1
i += 1
return sum
if __name__ == "__main__":
# you should have all your main logic including inputs.
# function calls and output inside of if statement.
hexa = input()
result = hex_char_decode(hexa)
print(result)
I want to implement basic calculator only includes numbers, '+' and '-' in Python. But I found that I cannot change index i inside the while loop.
def basicCalculator(expression):
res = 0
sign = 1
i = 0
while i < len(expression): # using while loop here
if expression[i] == '+':
sign = 1
elif expression[i] == '-':
sign = -1
elif '0' <= expression[i] <= '9':
temp = int(expression[i])
while i + 1 < len(expression) and '0' <= expression[i + 1] <= '9':
i += 1 # want to increase i, but failed
temp = temp * 10 + int(expression[i])
res += sign * temp
return res
s = "2+3-999"
print(basicCalculator(s)) # ouput should be -994, but infinite
Then I tried another way: using for loop instead of while loop, get incorrect answer.
def basicCalculator(expression):
res = 0
sign = 1
for i in range(len(expression)): # using for loop
if expression[i] == '+':
sign = 1
elif expression[i] == '-':
sign = -1
elif '0' <= expression[i] <= '9':
temp = int(expression[i])
while i + 1 < len(expression) and '0' <= expression[i + 1] <= '9':
i += 1 # increase i
temp = temp * 10 + int(expression[i])
res += sign * temp
return res
s = "2+3-999"
print(basicCalculator(s)) # ouput should be -994, but got -1102
I don't know why I cannot change the index inside the while loop or for loop. It is not a good idea to change i in for loop. What is the best way to fix the bug?
(I wrote the same code in Java using approach 2, everything is good, any difference?)
Rather than incrementing the index in a loop within the loop, I suggest just accumulating the current number in the main loop so you have less to keep track of:
def basicCalculator(expression):
res = 0
sign = 1
num = ""
for c in expression:
if c.isdecimal():
num += c
continue
res += int(num) * sign
num = ""
if c == "+":
sign = 1
elif c == "-":
sign = -1
else:
raise ValueError(f"Invalid operator {c}")
return res + int(num) * sign
print(basicCalculator("2+3-999")) # -994
I have a practice question that requires me to generate x number of alternating substrings, namely "#-" & "#--" using both recursion as well as iteration. Eg.string_iteration(3) generates "#-#--#-".
I have successfully implemented the solution for the iterative method,
but I'm having trouble getting started on the recursive method. How can I proceed?
Iterative method
def string_iteration(x):
odd_block = '#-'
even_block = '#--'
current_block = ''
if x == 0:
return ''
else:
for i in range(1,x+1):
if i % 2 != 0:
current_block += odd_block
elif i % 2 == 0:
current_block += even_block
i += 1
return current_block
For recursion, you almost always just need a base case and everything else. Here, your base case it pretty simple — when x < 1, you can return an empty string:
if x < 1:
return ''
After than you just need to return the block + the result of string_iteration(x-1). After than it's just a matter of deciding which block to choose. For example:
def string_iteration(x):
# base case
if x < 1:
return ''
blocks = ('#--', '#-')
# recursion
return string_iteration(x-1) + blocks[x % 2]
string_iteration(5)
# '#-#--#-#--#-'
This boils down to
string_iteration(1) + string_iteration(2) ... string_iteration(x)
The other answer doesn't give the same result as your iterative method. If you always want it to start with the odd block, you should add the block on the right of the recursive call instead of the left:
def string_recursion(x):
odd_block = '#-'
even_block = '#--'
if x == 0:
return ''
if x % 2 != 0:
return string_recursion(x - 1) + odd_block
elif x % 2 == 0:
return string_recursion(x - 1) + even_block
For recursive solution, you need a base case and calling the function again with some other value so that at the end you will have the desired output. Here, we can break this problem recursively like - string_recursive(x) = string_recursive(x-1) + string_recursive(x-2) + ... + string_recursive(1).
def string_recursion(x, parity):
final_str = ''
if x == 0:
return ''
if parity == -1: # when parity -1 we will add odd block
final_str += odd_block
elif parity == 1:
final_str += even_block
parity *= -1 # flip the parity every time
final_str += string_recursion(x-1, parity)
return final_str
odd_block = '#-'
even_block = '#--'
print(string_recursion(3, -1)) # for x=1 case we have odd parity, hence -1
# Output: #-#--#-
NB: You may not use built-in typecasting: code this yourself.
def str2int(s):
result = 0
if s[0] == '-':
sign = -1
i = 1
while i < len(s):
num = ord(s[i]) - ord('0')
result = result * 10 + num
i += 1
result = sign * result
return result
else:
i = 0
while i < len(s):
num = ord(s[i]) - ord('0')
result = result * 10 + num
i += 1
return result
NB: You may not use built-in str() or string template. Code this yourself.
def int2str(i):
strng = ""
if i > 0:
while i != 0:
num = i % 10
strng += chr(48+num)
i = i / 10
return strng[::-1]
else:
while i != 0:
num = abs(i) % 10
strng += chr(48+num)
i = abs(i) / 10
return '-' + strng[::-1]
I am a newbie and I have to write code based on basic. I write these function by myself but these look weird. Can you help me to improve code? Thank you
This maybe a better question for https://codereview.stackexchange.com/.
Not withstanding there is no error checking, one obvious comment is you have common code that can be factored out. Only capture in the if, else what is unique rather than repeat the while loop:
def str2int(s):
if s[0] == '-':
sign = -1
i = 1
else:
sign = 1
i = 0
result = 0
while i < len(s):
num = ord(s[i]) - ord('0')
result = result * 10 + num
i += 1
return sign * result
It is generally considered better form in python to iterate over list rather than indices:
def str2int(s):
sign = 1
if s[0] == '-':
sign = -1
s = s[1:]
result = 0
for c in s:
num = ord(c) - ord('0')
result = result * 10 + num
return sign * result
These last lines are equivalent to a standard map and reduce (reduce is in functools for py3). Though some would argue against it:
from functools import reduce # Py3
def str2int(s):
sign = 1
if s[0] == '-':
sign = -1
s = s[1:]
return sign * reduce(lambda x,y: x*10+y, map(lambda c: ord(c) - ord('0'), s))
There are similar opportunities to do the same for int2str().
I am asked to binary search a list of names and if these names start with a particular letter, for example A, then I am to print that name.
I can complete this task by doing much more simple code such as
for i in list:
if i[0] == "A":
print(i)
but instead I am asked to use a binary search and I'm struggling to understand the process behind it. We are given base code which can output the position a given string. My problem is not knowing what to edit so that I can achieve the desired outcome
name_list = ["Adolphus of Helborne", "Aldric Foxe", "Amanita Maleficant", "Aphra the Vicious", "Arachne the Gruesome", "Astarte Hellebore", "Brutus the Gruesome", "Cain of Avernus"]
def bin_search(list, item):
low_b = 0
up_b = len(list) - 1
found = False
while low_b <= up_b and found == False:
midPos = ((low_b + up_b) // 2)
if list[midPos] < item:
low_b = midPos + 1
elif list[midPos] > item:
up_b = midPos - 1
else:
found = True
if found:
print("The name is at positon " + str(midPos))
return midPos
else:
print("The name was not in the list.")
Desired outcome
bin_search(name_list,"A")
Prints all the names starting with A (Adolphus of HelBorne, Aldric Foxe .... etc)
EDIT:
I was just doing some guess and check and found out how to do it. This is the solution code
def bin_search(list, item):
low_b = 0
up_b = len(list) - 1
true_list = []
count = 100
while low_b <= up_b and count > 0:
midPos = ((low_b + up_b) // 2)
if list[midPos][0] == item:
true_list.append(list[midPos])
list.remove(list[midPos])
count -= 1
elif list[midPos] < item:
low_b = midPos + 1
count -= 1
else:
up_b = midPos - 1
count -= 1
print(true_list)
Not too sure if this is what you want as it seems inefficient... as you mention it seems a lot more intuitive to just iterate over the entire list but using binary search i found here i have:
def binary_search(seq, t):
min = 0
max = len(seq) - 1
while True:
if max < min:
return -1
m = (min + max) // 2
if seq[m][0] < t:
min = m + 1
elif seq[m][0] > t:
max = m - 1
else:
return m
index=0
while True:
index=binary_search(name_list,"A")
if index!=-1:
print(name_list[index])
else:
break
del name_list[index]
Output i get:
Aphra the Vicious
Arachne the Gruesome
Amanita Maleficant
Astarte Hellebore
Aldric Foxe
Adolphus of Helborne
You just need to found one item starting with the letter, then you need to identify the range. This approach should be fast and memory efficient.
def binary_search(list,item):
low_b = 0
up_b = len(list) - 1
found = False
midPos = ((low_b + up_b) // 2)
if list[low_b][0]==item:
midPos=low_b
found=True
elif list[up_b][0]==item:
midPos = up_b
found=True
while True:
if found:
break;
if list[low_b][0]>item:
break
if list[up_b][0]<item:
break
if up_b<low_b:
break;
midPos = ((low_b + up_b) // 2)
if list[midPos][0] < item:
low_b = midPos + 1
elif list[midPos] > item:
up_b = midPos - 1
else:
found = True
break
if found:
while True:
if midPos>0:
if list[midPos][0]==item:
midPos=midPos-1
continue
break;
while True:
if midPos<len(list):
if list[midPos][0]==item:
print list[midPos]
midPos=midPos+1
continue
break
else:
print("The name was not in the list.")
the output is
>>> binary_search(name_list,"A")
Adolphus of Helborne
Aldric Foxe
Amanita Maleficant
Aphra the Vicious
Arachne the Gruesome
Astarte Hellebore