bitwise scripts fixed

src/bitwise/bit_array.py

@@ -1,17 +1,16 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/usr/bin/env python
 
-''' Example of how to use a bit array in python as a "counter" dict:
+__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
-'''
+    '''
 
-
-
-def print_dupl_ba(l1):
     bs = bytearray(10)
     for i in range(len(l1)):
         if i == l1[i]:
@@ -23,9 +22,6 @@ def print_dupl_ba(l1):
 
 
 
-    
-    
-
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/bitwise/bit_operations/clear_bits.py

@@ -1,40 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' Clear a bit in a binary number. It is almost 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
-        
-    >>> num = int('10010000', 2)
-    >>> clear_bit(num, 4)   
-    '0b10000000'
-    >>> num = int('10010011', 2)
-    >>> clear_all_bits_from_i_to_0(num, 2)  
-    '0b10010000'
-    >>> num = int('1110011', 2)
-    >>> clear_all_bits_from_most_sig_to_1(num, 2)  
-    '0b11'
-'''
-
-
-
-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__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/bit_operations/find_bit_len.py

@@ -1,28 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' 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))
-    
-    >>> for i in range(17): print(find_bit_len(i))
-'''
-
-
-
-def find_bit_len(int_num):
-    lenght = 0
-    while int_num:
-        int_num >>= 1
-        lenght += 1
-    return lenght
-    
-    
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/bit_operations/get_bit.py

@@ -1,41 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' 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
-    >>> 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
-'''
-
-
-
-def get_bit(num, i):
-    mask = 1 << i
-    return num & mask != 0
-
-
-    
-    
-    
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/bit_operations/num_bits_to_convert_2_nums.py

@@ -1,45 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' This method returns the number of bits that are necessary to change to convert two
-    numbers A and B:
-    1) XOR
-    2) count 1s
-        
-    >>> a = int('10010000', 2)
-    >>> b = int('01011010', 2)
-    >>> count_bits_swap(a, b)  
-    4
-    >>> count_bits_swap2(a, b)  
-    4
-'''
-
-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__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/bit_operations/set_bit.py

@@ -1,38 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' 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
-    >>> num = int('0100100', 2)
-    >>> set_bit(num, 0)   
-    '0b100101'
-    >>> set_bit(num, 1)   
-    '0b100110'
-    >>> set_bit(num, 2)    # nothing change
-    '0b100100'
-    >>> set_bit(num, 3)   
-    '0b101100'
-    >>> set_bit(num, 4)   
-    '0b110100'
-    >>> set_bit(num, 5)     # nothing change
-    '0b100100'
-'''
-
-
-
-def set_bit(num, i):
-    mask = 1 << i
-    return bin( num | mask ) 
-
-
-    
-    
-    
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/bit_operations/update_bit.py

@@ -1,26 +0,0 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
-
-''' 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
-        
-    >>> num = int('10010000', 2)
-    >>> update_bit(num, 2, 1)   
-    '0b10010100'
-'''
-
-
-
-def  update_bit(num, i, v):
-    mask = ~ (1 << i)
-    return bin( (num & mask) | (v << i) )    
-    
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/bitwise/find_bit_len.py

@@ -6,7 +6,8 @@ __author__ = "bt3"
     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))
+    (obs: to find the int of a bin do int('1001', 2)) and to show in bin
+    do bin(int))
 '''
 
 
@@ -22,4 +23,3 @@ if __name__ == '__main__':
     for i in range(17):
         print(find_bit_len(i))
         print i.bit_length()
-        print

src/bitwise/find_how_many_1_binary.py

@@ -1,21 +1,22 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/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))
-    
-    >>> find_how_many_1_in_a_binary(9)
-    2
-
+    (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:
@@ -25,9 +26,6 @@ def find_how_many_1_in_a_binary(n):
 
 
 
-    
-    
-
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/bitwise/get_float_rep_bin.py

@@ -1,20 +1,22 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/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'
-'''
+    '''
 
-def get_float_rep(num):
     if num >= 1 or num <= 0: return 'Error 1'
     result = '.'
     while num:

src/bitwise/insert_small_bin_into_big_bin.py

@@ -1,22 +1,25 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/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'
-'''
+    '''
 
-def insert_small_bin_into_big_bin(M, N, i, j):
     allOnes = ~0
     left = allOnes << (j+1) # 1110000
     right = ( (1 << i) - 1) # 0000111
@@ -27,8 +30,6 @@ def insert_small_bin_into_big_bin(M, N, i, j):
     return bin( N_cleared | M_shifted)
 
 
-    
-
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/bitwise/next_with_same_num_1s.py

@@ -1,21 +1,13 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/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
-    
-    >>> num = 0b1001
-    >>> next = '0b1010'
-    >>> prev = '0b110'
-    >>> print_prev_same_1s(num) == prev
-    True
-    >>> print_next_same_1s(num) == next
-    True
-
 '''
 
 
@@ -59,6 +51,9 @@ def find_num_1s(num):
 
 
 if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
+    num = 0b1001
+    n = '0b1010'
+    p = '0b110'
+    print_prev_same_1s(num) == p
+    print_next_same_1s(num) == n
 

src/bitwise/set_bit.py

@@ -8,18 +8,14 @@ __author__ = "bt3"
     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'

src/bitwise/swap_in_place.py

@@ -8,6 +8,10 @@ __author__ = "bt3"
 
 
 def swap_bit(a, b):
+    '''
+    >>> swap_bit(14, 73)
+    (73, 14)
+    '''
     a = a^b
     b = a^b
     a = a^b
@@ -15,7 +19,5 @@ def swap_bit(a, b):
 
 
 if __name__ == '__main__':
-    a = 14
-    b = 73
-    a2, b2 = swap_bit(a, b)
-    print "a was {0}, now it is {1}. \nb was {2}, now it is {3}".format(a, a2, b, b2)
+    import doctest
+    doctest.testmod()

src/bitwise/swap_odd_even.py

@@ -1,6 +1,7 @@
-#!/usr/bin/python3
-# mari von steinkirch @2013
-# steinkirch at gmail
+#!/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:
@@ -8,17 +9,18 @@
         (b) shift by right by 1
     2) do the same to ints with 01010101
     3) merge
-    
-    >>> num = 0b11011101
-    >>> result = '0b1101110'
-    >>> swap_odd_even(num) == result
-    True
-
 '''
 
 
 
 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
@@ -29,9 +31,6 @@ def swap_odd_even(num):
 
 
 
-    
-    
-
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/searching_and_sorting/sorting/bubble_sort.py

@@ -1,8 +1,6 @@
-#!/usr/bin/python
-
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+#!/usr/bin/env python
 
+__author__ = "bt3"
 
 
 def bubble_sort(seq):

src/searching_and_sorting/sorting/count_sort.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 

src/searching_and_sorting/sorting/gnome_sort.py

@@ -1,8 +1,6 @@
-#!/usr/bin/python
-
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+#!/usr/bin/env python
 
+__author__ = "bt3"
 
 
 def gnome_sort(seq):

src/searching_and_sorting/sorting/heap.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 class Heap(object):

src/searching_and_sorting/sorting/heap_sort.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 ''' Heapsort using Pythons libraries'''

src/searching_and_sorting/sorting/insertion_sort.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 

src/searching_and_sorting/sorting/merge_sort.py

@@ -1,7 +1,7 @@
-#!/usr/bin/python
+#!/usr/bin/env python
+
+__author__ = "bt3"
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
 
 
 

src/searching_and_sorting/sorting/quick_sort.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 ''' Some examples of how to implement Quick Sort in Python

src/searching_and_sorting/sorting/selection_sort.py

@@ -1,7 +1,6 @@
-#!/usr/bin/python
+#!/usr/bin/env python
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
+__author__ = "bt3"
 
 
 def selection_sort(seq):

src/searching_and_sorting/sorting/sort_anagrams_together.py

@@ -1,7 +1,7 @@
-#!/usr/bin/python
+#!/usr/bin/env python
+
+__author__ = "bt3"
 
-__author__ = "Mari Wahl"
-__email__ = "marina.w4hl@gmail.com"
 
 ''' A method to sort an array so that all the anagrams are together. Since we only
     want the anagrams to be grouped, we can use a dictionary for this task. This