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bt3gl 2023-07-30 21:40:09 -07:00
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#!/usr/bin/env python
__author__ = "bt3"
''' Example of how to use a bit array in python as a "counter" dict'''
def print_dupl_ba(l1):
'''
>>> l1 = [0, 1, 2, 3, 4, 2, 6, 7, 8, 9]
>>> print_dupl_ba(l1)
2
'''
bs = bytearray(10)
for i in range(len(l1)):
if i == l1[i]:
bs[i] = 1
for index, bit in enumerate(bs):
if bit == 0:
return l1[index]
return None
if __name__ == '__main__':
import doctest
doctest.testmod()

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BIT-WISE
----------------------
1. To find a number:
11000101 is 2^0+2^2+2^6+2^7 = 197
2. Left shifting:
0010 1011 << 4 ---> 1011 000
3. Right shifting:
0010 1011 >> 4 ---> 0000 0010
or it can be filled with the copy of the first bit, instead of 0:
1011 0010 >> 4 ---> 1111 1011
4. XOR can cancels out:
15 ^ 12 ^ 15 = 12
5. 2^x:
left-shift 1 by x:
0000 0001 << x
so if x = 2, 2^2 = 4 -> 100
0000 0001 << 2 ---> 0000 0100
6. Is power of 2?
just do x&(x-1).
if 0 --> yes!

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#!/usr/bin/env python
__author__ = "bt3"
''' Clear a bit in a binary number.
Like the reverse of set bit:
1) first create a number filled of 1s,
with 0 at i (can create 0001000 and ~)
2) AND the number so it clears the ith bit
'''
def clear_bit(num, i):
mask = ~ (1 << i) # -0b10001
return bin(num & mask)
def clear_all_bits_from_i_to_0(num, i):
mask = ~ ( (1 << (i+1)) - 1)
return bin(num & mask)
def clear_all_bits_from_most_sig_to_1(num, i):
mask = ( 1 << i) -1
return bin(num & mask)
if __name__ == '__main__':
num = int('10010000', 2)
print clear_bit(num, 4) # '0b10000000'
num = int('10010011', 2)
print clear_all_bits_from_i_to_0(num, 2) # '0b10010000'
num = int('1110011', 2)
print clear_all_bits_from_most_sig_to_1(num, 2) #'0b11'

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#!/usr/bin/env python
__author__ = "bt3"
''' Find how many bits a int has:
1) Start with a mask of 1
2) Mask with AND
3) if result (if true): count += 1
(obs: to find the int of a bin do int('1001', 2)) and to show in bin
do bin(int))
'''
def find_bit_len(int_num):
lenght = 0
while int_num:
int_num >>= 1
lenght += 1
return lenght
if __name__ == '__main__':
for i in range(17):
print(find_bit_len(i))
print i.bit_length()

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#!/usr/bin/env python
__author__ = "bt3"
''' Find how many 1s in the binary:
1) Start with a mask of 1
2) Mask with AND
3) if result (if true): count += 1
(obs: to find the int of a bin do int('1001',
2)) and to show in bin do bin(int))
'''
def find_how_many_1_in_a_binary(n):
'''
>>> find_how_many_1_in_a_binary(9)
2
'''
counter = 0
while n:
if n & 1:
counter += 1
n >>= 1
return counter
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' Get a bit in a binary number:
1) Shifts 1 over by i bits
2) make an AND with the number
3) all the other than the bit at i are clean, now compare to 0
4) if the new value is not 0, bit i is 1
'''
def get_bit(num, i):
mask = 1 << i
return num & mask != 0
if __name__ == '__main__':
num = int('0100100', 2)
get_bit(num, 0) # 0
get_bit(num, 1) # 0
get_bit(num, 2) # 1
get_bit(num, 3) # 0
get_bit(num, 4) # 0
get_bit(num, 5) # 1
get_bit(num, 6) # 0

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#!/usr/bin/env python
__author__ = "bt3"
''' Given a real number between 0 and 1 (eg: 0.72), this method print the binary
representation. If the Number cannot be represented accurately in binary, with at
most 32 chars, print error:
'''
def get_float_rep(num):
'''
>>> get_float_rep(0.72)
('Error 2', '.1011100001010001111010111000010')
>>> get_float_rep(0.1)
('Error 2', '.0001100110011001100110011001100')
>>> get_float_rep(0.5)
'.1'
'''
if num >= 1 or num <= 0: return 'Error 1'
result = '.'
while num:
if len(result) >= 32: return 'Error 2', result
r = num*2
if r >= 1:
result += '1'
num = r - 1
else:
result += '0'
num = r
return result
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' Given two 32-bit numbers, N and M, and two bit positions, i and j, this
method insert M into N such that M starts at bit j and ends at bit i:
1) clear the bits j thru i in N'
2) shift M so that it lines up with bits j thru i
3) merge M and N
'''
def insert_small_bin_into_big_bin(M, N, i, j):
'''
>>> N = 0b10000000000
>>> M = 0b10011
>>> j = 6
>>> i = 2
>>> insert_small_bin_into_big_bin(M, N, i, j)
'0b10001001100'
'''
allOnes = ~0
left = allOnes << (j+1) # 1110000
right = ( (1 << i) - 1) # 0000111
mask = left | right # 1110111
N_cleared = N & mask
M_shifted = M << i
return bin( N_cleared | M_shifted)
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' Give a positive int, print the next smallest and next largest ints with
same number of 1 bits.
The brute force is:
1) find number of 1 bits
2) loop above and down until find same, checking for each
'''
def print_prev_same_1s(num):
n1s = find_num_1s(num)
# find prev
i = num-1
while True:
n1s_here = find_num_1s(i)
if n1s_here == n1s:
return bin(i)
i -= 1
if i < 0:
return None
def print_next_same_1s(num):
n1s = find_num_1s(num)
# find next
i = num+1
while True:
n1s_here = find_num_1s(i)
if n1s_here == n1s:
return bin(i)
i += 1
if i < 0:
return None
def find_num_1s(num):
counter = 0
while num:
if num & 1:
counter += 1
num >>= 1
return counter
if __name__ == '__main__':
num = 0b1001
n = '0b1010'
p = '0b110'
print_prev_same_1s(num) == p
print_next_same_1s(num) == n

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#!/usr/bin/env python
__author__ = "bt3"
''' This method returns the number of bits that are necessary to change to convert two
numbers A and B:
1) XOR
2) count 1s
'''
def count_bits_swap2(a, b):
count = 0
m = a^b
while m:
count +=1
m = m & (m-1)
return count
def count_bits_swap(a, b):
m = a^b
return count_1s(m)
def count_1s(m):
count = 0
while m:
if m& 1 :
count +=1
m >>= 1
return count
if __name__ == '__main__':
a = int('10010000', 2)
b = int('01011010', 2)
print count_bits_swap(a, b) #4
print count_bits_swap2(a, b) #4

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#!/usr/bin/env python
__author__ = "bt3"
''' Set a bit in a binary number:
1) Shifts 1 over by i bits
2) make an OR with the number, only the value at bit i will change and all the others bit
of the mask are zero so will not affect the num
'''
def set_bit(num, i):
mask = 1 << i
return bin( num | mask )
if __name__ == '__main__':
num = int('0100100', 2)
print set_bit(num, 0) #'0b100101'
print set_bit(num, 1) #'0b100110'
print set_bit(num, 2) # nothing change '0b100100'
print set_bit(num, 3) #'0b101100'
print set_bit(num, 4) #'0b110100'
print set_bit(num, 5) # nothing change '0b100100'

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#!/usr/bin/env python
__author__ = "bt3"
'''
swapping values in place without extra memory
'''
def swap_bit(a, b):
'''
>>> swap_bit(14, 73)
(73, 14)
'''
a = a^b
b = a^b
a = a^b
return a, b
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' Swap odd and even bits in a smart way in a binary:
1) first for odds, take n and move the odd:
(a) Mask all odd bits with 10101010 (0xAA)
(b) shift by right by 1
2) do the same to ints with 01010101
3) merge
'''
def swap_odd_even(num):
'''
>>> num = 0b11011101
>>> result = '0b1101110'
>>> swap_odd_even(num) == result
True
'''
mask_odd = 0xAA # 0b10101010
mask_even = 0x55 # 0b1010101
odd = num & mask_odd
odd >>= 1
even = num & mask_even
even >>= 1
return bin(odd | even)
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' This method merges set bit and clean bit:
1) first clear the bit at i using a mask such as 1110111
2) then shift the intended value v by i bits
3) this will create a number with bit i to v and all other to 0
4) finally update the ith bit with or
'''
def update_bit(num, i, v):
mask = ~ (1 << i)
return bin( (num & mask) | (v << i) )
if __name__ == '__main__':
num = int('10010000', 2)
print update_bit(num, 2, 1) # '0b10010100'