My task for homework is to generate binary sequence (sample input in range 1<=n<=30) without two zeros or three ones together. I made simple binary generator, which works as shown below. I need to modify it to obtain this output(and according outputs also):
n = int(input())
def gen(n):
if n == 0:
return ['']
l = gen(n-1)
start0 = []
start1 = []
for seq in l:
start0.append('0' + seq)
start1.append('1' + seq)
return start0 + start1
l1 = gen(n)
for elem in l1:
print(elem)
Sample Input
4
Sample Output(which I have right now)
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Sample Output I need to obtain:
0101
0110
1010
1011
1101
To put in simple, sequences like 1000, 1110, 1100, 0011, 0111 and so on(without two zeros OR three 1s in a row) should be excluded. Tried list methods for this, but I failed to work it properly at all range(1<=n<=30). Any ideas?
PS I have to do it without using itertools. Avoiding itertools is a part of the task.
Just add two next if-s into the loop.
These if-s literally mean next things:
n = int(input())
def gen(n):
if n == 0:
return ['']
l = gen(n-1)
start0 = []
start1 = []
for seq in l:
if not seq.startswith('0'):
start0.append('0' + seq)
if not seq.startswith('11'):
start1.append('1' + seq)
return start0 + start1
l1 = gen(n)
for elem in l1:
print(elem)
output for input 4:
0101
0110
1010
1011
1101
output for input 5:
01010
01011
01101
10101
10110
11010
11011
@Arty has the answer as asked. Here's a generator example for another solution:
def gen(n):
if n == 1:
yield from '01'
else:
for seq in gen(n-1):
if not seq.endswith('0'):
yield seq + '0'
if not seq.endswith('11'):
yield seq + '1'
n = int(input('Bits? '))
for elem in gen(n):
print(elem)
Output:
Bits? 8
01010101
01010110
01011010
01011011
01101010
01101011
01101101
10101010
10101011
10101101
10110101
10110110
11010101
11010110
11011010
11011011
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