03.01
[4**3,2+3.4**2,complex(1,1)**2]
Or even simpler
[4**3,2+3.4**2,(1+1j)**2]
03.02
a = "Hello, roof world!"
word = "roof"
num = a.index(word)
chars = len(word)
print(a[:num]+a[num+chars:])
03.03
s = "CAGTACCAAGTGAAAGAT"
s.count('A')
a=[1,2,3]
b=[4,5,6]
c=a+b
# or
d=a
d.extend(b)
print(c,d)
04.01
x=3.7
(x>3.4 and x<=6.6) or (x==2)
04.02
dict={'snake':'Schlange',
'jungle':'Dschungel',
'lives':'lebt',
'in':'in',
'the':'der/die/das'}
s="The snake lives in the jungle"
for word in s.split():
print(dict[word.lower()])
05.01
for i in range(2,1000):
is_prime=True
for j in range(2,i):
if (i%j==0):
is_prime=False
if is_prime:
print (i)
05.02: Simply break the loop since it is not necessary to search further if one divisor is found.
for i in range(2,1000):
is_prime=True
for j in range(2,i):
if (i%j==0):
is_prime=False
break
if is_prime:
print (i)
05.03
# very terse:
a, b = 0, 1
while a < 100000:
print(a)
#c = a + b
#a = b
#b = c
a, b = b, a + b
# Fibonacci sequence in a list
l=[0]
idx=0
fib=1
while fib<100000:
l.append(fib) # add x_i
fib=l[idx]+l[idx+1] # work out x_i+1
idx=idx+1
print(l)
And the square Fibonaccis:
# determine square Fibonaccis
l=[0]
fib=1
while fib<100000:
l.append(fib)
fib=l[-2]+l[-1]
# print square Fibonacci numbers
for fib in l:
is_square=False
for j in range(fib):
if j*j==fib:
is_square=True
break
if is_square:
print("square number =",fib,"(="+str(j)+"*"+str(j)+")")
s='pbatenghyngvbaf lbh unir fhpprrqrq va qrpelcgvat gur fgevat'
base=ord('a') # ascii code for 'a'
decode=-13
res=''
for character in s.lower():
if character==' ':
t=character
else:
ascii=ord(character)
t=chr(((ascii+decode)-base)%26+base) # (ascii-97)%26 in range 0 ... 25
res+=t
print(decode,res)
The translation of ascii can also be done explicitly:
s='pbatenghyngvbaf lbh unir fhpprrqrq va qrpelcgvat gur fgevat'
base=ord('a') # ascii code for 'a'
decode=-13
res=''
for character in s.lower():
if character==' ':
tt=ord(' ')
else:
ascii=ord(character)
tt=ascii+decode
if tt>=base+26:
tt-=26
elif tt<base:
tt+=26
res+=chr(tt)
print(decode,res)
This is the solution when first loading the alphabet into a list:
s='pbatenghyngvbaf lbh unir fhpprrqrq va qrpelcgvat gur fgevat'
decode=-13
alphabet=[]
for i in range(ord('a'), ord('z')+1):
alphabet.append(chr(i))
# or a very pythonic way
#alphabet=[chr(i) for i in range(ord('a'),ord('z')+1)]
decode=-13
res=''
for character in s.lower():
if character==' ':
tt=' '
else:
idx=alphabet.index(character)+decode
if idx>=26:
idx-=26
elif idx<0:
idx+=26
tt=alphabet[idx]
res+=tt
print(decode,res)
Determine the unkown cipher shift trying out all
s='fn jyxuxprin oxa cqn rwlxwenwrnwln'
base=ord('a') # ascii code for 'a'
for decode in range(-10,20):
#decode=-13
res=''
for character in s.lower():
if character==' ':
t=character
else:
ascii=ord(character)
t=chr(((ascii+decode)-base)%26+base)
res+=t
print(decode,res)
Note solution -7 and 17 are equivalent.
We can also try to be smart: the most common letter used in English words is the "e" (Most common letter from Oxford dictionary). So let's try to get a guess of the required decode parameter by counting letters first (will not work good for short text sequences though - why?):
s='fn jyxuxprin oxa cqn rwlxwenwrnwln'
max=0
max_letter=''
for letter in s:
count=s.count(letter)
if count>max:
max=count
max_letter=letter
print('The letter that occured most often in the string is "'+max_letter+'" with '+str(max)+' counts.')
decode=ord('e')-ord(max_letter)
print('The most likely "decode" parameter is thus '+str(decode))
To obtain better statistics, one could also include the second most letter and so on.