Re: I passed a fizzbuzz test but failed at recursion.
Am 10.03.2010 16:55, schrieb Bill:
def fizzbuzz(num):
if num:
if num % 15 is 0: return fizzbuzz(num-1) + 'fizzbuzz \n'
elif num % 5 is 0: return fizzbuzz(num-1) + 'buzz \n'
elif num % 3 is 0: return fizzbuzz(num-1) + 'fizz \n'
else : return fizzbuzz(num-1) + ('%d \n' % num)
return ''
print fizzbuzz(100)
While I do understand that this is not a Perl group, and this probably
won't help the OP, I just could not resist:
for i in range(1,101):print('','fizz')[not i%3]+('','buzz')[not i%5]or i
However, when I try to decipher the logic of the function I imagine
the solution reversed as buzz fizz 98 97 ...etc... 4 fizz 2 1.
No, the function creates the string from right to left.
See:
fizzbuzz(100) = fizzbuzz(99) + buzz \n
= fizzbuzz(98) + fizz \n + buzz \n
...
= fizzbuzz(1) + 2 \n + fizz \n + 4 \n + buzz\n + fizz \n +
+ 97 \n + 98 \n + fizz \n + buzz \n
= fizzbuzz(0) + 1 \n + 2 \n + fizz \n + 4 \n + buzz\n + fizz
\n + + 97 \n + 98 \n + fizz \n + buzz \n
= + 1 \n + 2 \n + fizz \n + 4 \n + buzz\n + fizz \n +
+ 97 \n + 98 \n + fizz \n + buzz \n
Regards,
Michael
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Re: I passed a fizzbuzz test but failed at recursion.
On Mar 10, 10:55 am, Bill bsag...@gmail.com wrote:
Look at this recursive fizzbuzz function
fromhttp://www.codinghorror.com/blog/2007/02/why-cant-programmers-program...
def fizzbuzz(num):
if num:
if num % 15 is 0: return fizzbuzz(num-1) + 'fizzbuzz \n'
elif num % 5 is 0: return fizzbuzz(num-1) + 'buzz \n'
elif num % 3 is 0: return fizzbuzz(num-1) + 'fizz \n'
else : return fizzbuzz(num-1) + ('%d \n' % num)
return ''
print fizzbuzz(100)
This returns 1 2 fizz 4 ...etc... 97 98 fizz buzz which is correct.
However, when I try to decipher the logic of the function I imagine
the solution reversed as buzz fizz 98 97 ...etc... 4 fizz 2 1.
After all, the first num is 100 which decrements by one until a zero
stops the recursive loop.
My (faulty) reasoning is that fizzbuzz(100) would firstly print a
fizz and the last fizzbuzz(1) would finally print a 1.
My logic is wrong, but I don't know why.
There's only one print, it prints the string returned by fizzbuzz(100)
The string is constructed via recursion
ie:
fizzbuzz(6)
fizzbuzz(5)
fizzbuzz(4)
fizzbuzz(3)
fizzbuzz(2)
fizzbuzz(1)
fizzbuzz(0)
''+'1 \n'
'1 \n'+'2 \n'
'1 \n2 \n'+'fizz \n'
'1 \n2 \n fizz \n'+'4 \n'
'1 \n2 \n fizz \n4 \n'+'buzz \n'
'1 \n2 \n fizz \n4 \nbuzz \n'+'fizz \n'
print '1 \n2 \n fizz \n4 \nbuzz \nfizz \n'
clear as mud ;)
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Re: I passed a fizzbuzz test but failed at recursion.
On 3/10/2010 10:55 AM, Bill wrote:
Look at this recursive fizzbuzz function from
http://www.codinghorror.com/blog/2007/02/why-cant-programmers-program.html
def fizzbuzz(num):
if num:
if num % 15 is 0: return fizzbuzz(num-1) + 'fizzbuzz \n'
elif num % 5 is 0: return fizzbuzz(num-1) + 'buzz \n'
elif num % 3 is 0: return fizzbuzz(num-1) + 'fizz \n'
In all 3 branches, 'is' should be '=='. As written, this code depends on
the implementation treating 0 as a singleton, which CPython does as an
optimization, but which the language def does not require.
else : return fizzbuzz(num-1) + ('%d \n' % num)
return ''
print fizzbuzz(100)
This returns 1 2 fizz 4 ...etc... 97 98 fizz buzz which is correct.
However, when I try to decipher the logic of the function I imagine
the solution reversed as buzz fizz 98 97 ...etc... 4 fizz 2 1.
After all, the first num is 100 which decrements by one until a zero
stops the recursive loop.
My (faulty) reasoning is that fizzbuzz(100) would firstly print a
fizz and the last fizzbuzz(1) would finally print a 1.
If one reversed the string addition in each branch, it would.
As written, the 'word' for n is tacked on at the end.
My logic is wrong, but I don't know why.
Is this slightly revised version any clearer?
def fizzbuzz_rb(n):
if n:
previous = fizzbuzz_rb(n-1)
word = (not n % 15 and 'fizzbuzz \n' or
not n % 5 and 'buzz \n' or
not n % 3 and 'fizz \n' or
'%d \n' % n)
return previous + word
else:
return ''
or this equivalent tail-recursive version?
def fizzbuzz_rt(i,n,s):
if i = n:
word = (not i % 15 and 'fizzbuzz \n' or
not i % 5 and 'buzz \n' or
not i % 3 and 'fizz \n' or
'%d \n' % i)
return fizzbuzz_rt(i+1, n, s+word)
else:
return s
or this equivalent iterative version?
def fizzbuzz_it(n):
s = ''
for i in range(1,n+1):
s += (not i % 15 and 'fizzbuzz \n' or
not i % 5 and 'buzz \n' or
not i % 3 and 'fizz \n' or
'%d \n' % i)
return s
print (fizzbuzz_rb(100) == fizzbuzz_rt(1,100,'') == fizzbuzz_it(100))
# prints True
Terry Jan Reedy
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