[OT] stable algorithm with complexity O(n)
Duncan Booth
duncan.booth at invalid.invalid
Sat Dec 13 20:17:41 CET 2008
"Diez B. Roggisch" <deets at nospam.web.de> wrote:
> David HlÃ¡Äik schrieb:
>> Hi guys,
>>
>> i am really sorry for making offtopic, hope you will not kill me, but
>> this is for me life important problem which needs to be solved within
>> next 12 hours..
>>
>> I have to create stable algorithm for sorting n numbers from interval
>> [1,n^2] with time complexity O(n) .
>>
>> Can someone please give me a hint. Would be very very thankful!
>
> Unless I grossly miss out on something in computer science 101, the
> lower bound for sorting is O(n * log_2 n). Which makes your task
> impossible, unless there is something to be assumed about the
> distribution of numbers in your sequence.
>
> Who has given you that assignment - a professor? Or some friend who's
> playing tricks on you?
>
I think you must have fallen asleep during CS101. The lower bound for
sorting where you make a two way branch at each step is O(n * log_2 n), but
if you can choose between k possible orderings in a single comparison you
can get O(n * log_k n).
To beat n * log_2 n just use a bucket sort: for numbers with a known
maximum you can sort them digit by digit for O(n log_k n), and if you don't
restrict yourself to decimal then k can be as large as you want, so for the
problem in question I think you can set k=n and (log_n n)==1 so you get
O(n)
i.e. digit by digit bucket sort the numbers in base n.
Something like (untested):
n = len(data)
buckets1 = [[] for i in range(n)]
buckets2 = [[] for i in range(n)]
for x in data:
buckets1[x % n].append(x)
for x in (v for b in buckets1 for v in reversed(b)):
buckets2[x // n].append(x)
for x in (v for b in buckets2 for v in b):
print x
All loops are order n (the last two have about 2*n steps).
I lied about the untested:
>>> def f(data):
n = len(data)
buckets1 = [[] for i in range(n)]
buckets2 = [[] for i in range(n)]
for x in data:
buckets1[x % n].append(x)
for x in (v for b in buckets1 for v in reversed(b)):
buckets2[x // n].append(x)
for x in (v for b in buckets2 for v in b):
print x
>>> import random
>>> data = [ random.randint(1,100) for i in range(10)]
>>> f(data)
1
16
30
35
70
70
75
76
82
99
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