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# List of integers & L.I.S.

 P: n/a Given a list of N arbitrarily permutated integers from set {1..N}. Need to find the ordering numbers of each integer in the LONGEST increasing sequence to which this number belongs. Sample: List: [4, 5, 6, 1, 2, 7, 3] Corresponding ordering numbers: [1, 2, 3, 1, 2, 4, 3] Details: e.g. number 7 belongs to increasing sequence 1, 2, 7; but this sequence is not the LONGEST sequence for "7". The longest sequence for the 7 is 4, 5, 6, 7. So, the 7's ordering number in this sequence is 4. The salt of the thing is to do this with an O(n*log(n)) algorithm! The straightforward O(n^2) algorithm is toooo slooow. Any ideas? Sep 7 '05 #1
39 Replies

 P: n/a On 7 Sep 2005 09:48:52 -0700, "n00m" wrote: Given a list of N arbitrarily permutated integers from set {1..N}. Need to find the ordering numbers of each integer in the LONGEST increasing sequence to which this number belongs. Sample: List: [4, 5, 6, 1, 2, 7, 3] Corresponding ordering numbers: [1, 2, 3, 1, 2, 4, 3] Details: e.g. number 7 belongs to increasing sequence 1, 2, 7; but this sequence is not the LONGEST sequence for "7". The longest sequence for the 7 is 4, 5, 6, 7. So, the 7's ordering number in this sequence is 4. The salt of the thing is to do this with an O(n*log(n)) algorithm! The straightforward O(n^2) algorithm is toooo slooow. Any ideas? I wrote something like that for the Discrete Mathematics I took last spring (we had to produce the longest sequence, though, rather than the indexes into the original list, our list was not necessarily a permutation of 1..N, and I "cheated" on the longest descending sequnce by changing one of the comparison operators and re-running the program). If push comes to shove, I suppose I could post the whole program (all 100 lines of it), but here's the comment near the top: # consider that there are 2 ** n subsequences (corresponding to the 2 ** # n permutations of the original sequence); the trick is to accumulate # them all at once while skipping the ones that are not strictly # ascending and also pruning sequences that obviously can't "win" It's not much to go on, but that bit after the ";" is the algorithm I use. I can't tell you how fast it runs in big-O notation, but my old 500 MHz PowerMac G4 digested lists of tens or hundreds of thousands of random integers while I waited. Regards, Dan -- Dan Sommers Sep 7 '05 #2

 P: n/a I coded a solution that can compute the ordering numbers for random.shuffle(range(1, 1000001)) in 2.5 seconds (typical, Win 2K Pro, Pentium 4 2.40GHz 785Meg RAM) Are you sure that this is not a homework problem? Sep 7 '05 #3

 P: n/a Thanks guys! Are you sure that this is not a homework problem? .... and let me reveal the secret: http://spoj.sphere.pl/problems/SUPPER/ Hardly it can be easily reduced to "standard" LIS problem (i.e. to find just a (any) Longest Increasing Sequence). I coded a solution that can compute the ordering numbers for random.shuffle(range(1, 1000001)) in 2.5 seconds (typical, Win 2K Pro, Pentium 4 2.40GHz 785Meg RAM) Sounds very impressive! It would be great to see your py solution got accepted on my link (yes! they accept Python 2.4.1 solutions! and of course the registration is free and fully anonymous). So far there is no any accepted Py solution for this problem (see link "Best Solutions" on the top of page of this problem). Note: their machines are PIII Xeon 700MHz (no limit. for memory usage). PS Thanks, Dan! but frankly speaking I'm not much interested just to get someone's ready-to-use code without understanding how it works. Sep 8 '05 #4

 P: n/a > ... and let me reveal the secret: http://spoj.sphere.pl/problems /SUPPER/ your question is different than the question on this website. also, what do you consider to be the correct output for this permutation? (according to your original question) [4, 5, 1, 2, 3, 6, 7, 8] Manuel Sep 8 '05 #5

 P: n/a So, this has no real world use, aside from posting it on a website. Thanks for wasting our time. You are making up an arbitrary problem and asking for a solution, simply because you want to look at the solutions, not because your problem needs to be solved. Clearly, this is a waste of time. Sep 8 '05 #6

 P: n/a Working on this allowed me to avoid some _real_ (boring) work at my job. So clearly it served a very useful purpose! ;) Manuel Sep 8 '05 #7

 P: n/a [sp*****@gmail.com] So, this has no real world use, aside from posting it on a website. Thanks for wasting our time. You are making up an arbitrary problem and asking for a solution, simply because you want to look at the solutions, not because your problem needs to be solved. Clearly, this is a waste of time. If investigating algorithms irritates you, ignore it. The people writing papers on this topic don't feel it's a waste of time. For example, http://citeseer.ist.psu.edu/bespamya...umerating.html "Enumerating Longest Increasing Subsequences and Patience Sorting (2000)" Sergei Bespamyatnikh, Michael Segal That's easy to follow, although their use of a Van Emde-Boas set as a given hides the most challenging part (the "efficient data structure" part). Sep 8 '05 #8

 P: n/a > That's easy to follow, although their use of a Van Emde-Boas set as a given hides the most challenging part (the "efficient data structure" part). The "efficient data structure" is the easy part. Obviously, it is a dict of lists. ....or is it a list of dicts?... ....or is it a tuple of generators?... Anyway, "being aware of the literature", "thinking", "being smart", and "being Tim Peters" are all forms of cheating. I prefer not to cheat. ;) nOOm, I still need an answer... also, what do you consider to be the correct output for this permutation? (according to your original question) [4, 5, 1, 2, 3, 6, 7, 8] Manuel Sep 8 '05 #9

 P: n/a sp*****@gmail.com wrote: So, this has no real world use, aside from posting it on a website. I don't think you're quite right. We never know where we gain and where we lose. So clearly it served a very useful purpose! ;) Thanks, Manuel! your question is different than the question on this website. Not exactly so (maybe I'm wrong here). How I did it (but got TLE - Time Limit Exceeded (which is 9 seconds)). Firstly I find ordering numbers when moving from left to the right; then I find ord. numbers for backward direction AND for DECREASING subsequences: 4 5 1 2 3 6 7 8 << the list itself 1 2 1 2 3 4 5 6 << ordering numbers for forward direction 2 1 6 5 4 3 2 1 << ordering numbers for backward direction === 3 3 7 7 7 7 7 7 << sums of the pairs of ord. numbers Now those numbers with sum_of_ord.pair = max + 1 = 6 + 1 are the answer. So the answer for your sample is: 1 2 3 6 7 8 Btw, I did it in Pascal. Honestly, I don't believe it can be done in Python (of course I mean only the imposed time limit). http://spoj.sphere.pl/status/SUPPER/ Sep 8 '05 #10

 P: n/a PS: I've still not read 2 new posts. Sep 8 '05 #11

 P: n/a > 4 5 1 2 3 6 7 8 << the list itself 1 2 1 2 3 4 5 6 << ordering numbers for forward direction 2 1 6 5 4 3 2 1 << ordering numbers for backward direction === 3 3 7 7 7 7 7 7 << sums of the pairs of ord. numbers Oops! Sorry for miscounting in backward direction. Should be (anyway the answer stays the same): 4 5 1 2 3 6 7 8 << the list itself 1 2 1 2 3 4 5 6 << ordering numbers for forward direction 5 4 6 5 4 3 2 1 << ordering numbers for backward direction === 6 6 7 7 7 7 7 7 << sums of the pairs of ord. numbers Sep 8 '05 #12

 P: n/a n00m wrote: Firstly I find ordering numbers when moving from left to the right; then I find ord. numbers for backward direction AND for DECREASING subsequences: Sounds good. Btw, I did it in Pascal. Honestly, I don't believe it can be done in Python (of course I mean only the imposed time limit). http://spoj.sphere.pl/status/SUPPER/ Is there a platform specified? Below is an alleged solution in Python. -- --Bryan #!/user/bin/env python """ Python 2.4 solution to: http://spoj.sphere.pl/problems/SUPPER/ """ from sys import stdin def one_way(seq): n = len(seq) dominators = [n + 1] * (n * 1) # dominators[j] is lowest final value of any increasing sequence of # length j seen so far, as we left-right scan seq. score = [None] * n end = 0 for (i, x) in enumerate(seq): # Binary search for x's place in dominators low, high = 0, end while high - low > 10: mid = (low + high) >> 1 if dominators[mid] < x: low = mid + 1 else: high = mid + 1 while dominators[low] < x: low += 1 dominators[low] = x score[i] = low end = max(end, low + 1) return score def supernumbers(seq): forscore = one_way(seq) opposite = [len(seq) - x for x in reversed(seq)] backscore = reversed(one_way(opposite)) score = map(sum, zip(forscore, backscore)) winner = max(score) return sorted([seq[i] for i in range(len(seq)) if score[i] == winner]) _ = stdin.readline() sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, Sep 9 '05 #13

 P: n/a Bravo, Bryan! It's incredibly fast! But your code got WA (wrong answer). See my latest submission: http://spoj.sphere.pl/status/SUPPER/ Maybe you slipped a kind of typo in it? Silly boundary cases? Sep 9 '05 #14

 P: n/a Bravo, Bryan! Looks very neat! (pity I can't give it a try in my Py 2.3.4 because of reversed() and sorted() functions) And I've submitted it but got ... TLEs: http://spoj.sphere.pl/status/SUPPER/ Funnily, the exec.time of the best C solution is only 0.06s! PS In my 1st submission I overlooked that your code handles only 1 testcase (there are 10 of them); hence its 0.13s exec. time. PPS This is the code's text I submitted: #!/user/bin/env python from sys import stdin def one_way(seq): n = len(seq) dominators = [n + 1] * (n * 1) # dominators[j] is lowest final value of any increasing sequence of # length j seen so far, as we left-right scan seq. score = [None] * n end = 0 for (i, x) in enumerate(seq): # Binary search for x's place in dominators low, high = 0, end while high - low > 10: mid = (low + high) >> 1 if dominators[mid] < x: low = mid + 1 else: high = mid + 1 while dominators[low] < x: low += 1 dominators[low] = x score[i] = low end = max(end, low + 1) return score def supernumbers(seq): forscore = one_way(seq) opposite = [len(seq) - x for x in reversed(seq)] backscore = reversed(one_way(opposite)) score = map(sum, zip(forscore, backscore)) winner = max(score) return sorted([seq[i] for i in range(len(seq)) if score[i] == winner]) for tc in range(10): _ = stdin.readline() sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, Sep 9 '05 #15

 P: n/a n00m wrote: Bravo, Bryan! It's incredibly fast! Not compared to a good implementation for a compiled, low-level language. But your code got WA (wrong answer). See my latest submission: http://spoj.sphere.pl/status/SUPPER/ Maybe you slipped a kind of typo in it? Silly boundary cases? Hmmm ... wrong answer ... what could ... ah! Here's a problem: I bomb on the empty sequence. Correction below. I'm not a perfect programmer, but I like to think I'm a good programmer. Good programmers own their bugs. If there's another problem, I need more to go on. From what you write, I cannot tell where it fails, nor even what submission is yours and your latest. -- --Bryan #!/user/bin/env python """ Python solution to: http://spoj.sphere.pl/problems/SUPPER/ By Bryan Olson """ from sys import stdin def one_way(seq): n = len(seq) dominators = [n + 1] * (n * 1) score = [None] * n end = 0 for (i, x) in enumerate(seq): low, high = 0, end while high - low > 10: mid = (low + high) >> 1 if dominators[mid] < x: low = mid + 1 else: high = mid + 1 while dominators[low] < x: low += 1 dominators[low] = x score[i] = low end = max(end, low + 1) return score def supernumbers(seq): forscore = one_way(seq) opposite = [len(seq) - x for x in reversed(seq)] backscore = reversed(one_way(opposite)) score = map(sum, zip(forscore, backscore)) winner = max(score + [0]) return sorted([seq[i] for i in range(len(seq)) if score[i] == winner]) _ = stdin.readline() sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, Sep 9 '05 #16

 P: n/a Oops Bryan... I've removed my reply that you refer to... See my previous - CORRECT - reply. The code just times out... In some sense it doesn't matter right or wrong is its output. Btw, what is the complexity of your algorithm? Currently I'm at work and it's not easy for me to concentrate on our subject. Sep 9 '05 #17

 P: n/a > nor even what submission is yours and your latest. Oops.. my UserName there is ZZZ. Submissions in the html table are ordered by date DESC. Sep 9 '05 #18

 P: n/a n00m wrote: Oops Bryan... I've removed my reply that you refer to... See my previous - CORRECT - reply. The code just times out... In some sense it doesn't matter right or wrong is its output. If my code times out, then they are using an archaic platform. With respect to my code, you noted: Bravo, Bryan! It's incredibly fast! I myself did not claim 'incredibly fast'; but the code should beat the 9-second mark on any currently-viable platform, even if the machine were bought on special at Wal-Mart. For this kind of low-level challenge, Python cannot compete with the speed of C/C++, nor Ada, Pascal, ML, PL/1, nor even good Java implementations. Nevertheless, even in the rare cases in which efficiency of such computations is an issue, the Python solution is usually worthy contender, and often the superior solution. Human time is more valuable than machine time. Python emphasizes human-friendly code. Alas, I would be (and, since I did cite it, was) wrong on that. My code failed for the empty sequence. Wheter or not that was one of the test cases, and whether or not it was a consideration as the problem was defined, it was a bug. As I wrote: Good programmers own their bugs. Btw, what is the complexity of your algorithm? For a list of n items, time Theta(n ln(n)), space Theta(n). -- --Bryan Sep 9 '05 #19

 P: n/a Oh! Seems you misunderstand me! See how the last block in your code should look: for tc in range(10): _ = stdin.readline() sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, When I submitted it (for the 1st time) without for tc in range(10): the e-judge counted the output of your code as Wrong Answer; just because the e-judge got an answer for only the very 1st testcase (I think in it was not too large input data). Sep 9 '05 #20

 P: n/a n00m wrote: Oh! Seems you misunderstand me! See how the last block in your code should look: for tc in range(10): _ = stdin.readline() sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, Ah, I misunderstood the input. I thought they'd run it once for each of their test cases. Handling exactly ten inputs sucks, so how about: while True: if not stdin.readline().strip(): break sequence = [int(ch) for ch in stdin.readline().split()] supers = supernumbers(sequence) print len(supers) for i in supers: print i, print -- --Bryan Sep 9 '05 #21

 P: n/a It also timed out:( 241056 2005-09-09 20:11:19 ZZZ time limit exceeded - 7064 PYTH Btw, have a look at this nicest problem: http://spoj.sphere.pl/problems/COINS/ My py solution takes #64 place among its best solutions: http://spoj.sphere.pl/ranks/COINS/start=60 Sep 9 '05 #22

 P: n/a PS: ALL problems in problems.PDF file (weekly updated): http://spoj.sphere.pl/problems.pdf The friendliest online contester I've ever seen! JUST A NON-SUCH. Sep 9 '05 #23

 P: n/a n00m wrote: It also timed out:( Could be. Yet you did write: It's incredibly fast! I never intended to submit this program for competition. The contest ranks in speed order, and there is no way Python can compete with truly-compiled languages on such low-level code. I'd bet money that the algorithm I used (coded in C) can run with the winners. I also think I'd wager that the Python version outright trumps them on code size. My first version bombed for the zero-length sequence. That was a mistake, sorry, but it may not be one of their test-cases. I wonder how many of the accepted entries would perform properly. -- --Bryan Sep 9 '05 #24

 P: n/a Bryan Olson wrote: Could be. Yet you did write: > It's incredibly fast! I just was obliged to exclaim "It's incredibly fast!" because I THOUGHT your first version handled ALL TEN testcases from the input. But the code read from the *20-lines* input *ONLY 2* its first lines. Usually they place heavy data testcase(s) at the end of the (whole) input. Like this: 3 2 3 1 7 4 5 6 1 2 7 3 .... .... .... 100000 456 2 6789 ... ... ... ... ... 55444 1 ... 234 Surely producing an answer for list [2, 3, 1] will be "incredibly fast" for ANY language and for ANY algorithm. My first version bombed for the zero-length sequence. That was a mistake, sorry, but it may not be one of their test-cases. In my turn I can bet there's not an empty sequence testcase in the input. I wonder how many of the accepted entries would perform properly. Info of such kind they keep in secret (along with what the input data are). One more thing. They (the e-judge's admins) are not gods and they don't warrant that if they put 9 sec timelimit for a problem then this problem can be "solved" in all accepted languages (e.g. in Python). I never intended to submit this program for competition. "Competition" is not quite relevant word here. It just LOOKS as if it is a "regular" competetion. There nobody blames anybody. Moreover, judging on my own experience, there nobody is even interested in anybody. It's just a fun (but very useful fun). Sep 10 '05 #25

 P: n/a Bryan; My own version also timed out. And now I can tell: it's incredibly SLOW. Nevertheless it would be interesting to compare speed of my code against yours. I can't do it myself because my Python is of 2.3.4 version. Just uncomment "your" part. import bisect def oops(w,a,b): for m in w: j=bisect.bisect_left(a,m) a.insert(j,m) b.insert(j,max(b[:j]+[0])+1) def n00m(n,w): a,b=[],[] oops(w,a,b) v=map(lambda x: -x, w[::-1]) c,d=[],[] oops(v,c,d) e=map(sum, zip(b, d[::-1])) mx=max(e) f=[] for i in xrange(n): if e[i]==mx: f.append(i+1) print len(f) def one_way(seq): n = len(seq) dominators = [n + 1] * (n * 1) score = [None] * n end = 0 for (i, x) in enumerate(seq): low, high = 0, end while high - low > 10: mid = (low + high) >> 1 if dominators[mid] < x: low = mid + 1 else: high = mid + 1 while dominators[low] < x: low += 1 dominators[low] = x score[i] = low end = max(end, low + 1) return score def supernumbers(seq): forscore = one_way(seq) opposite = [len(seq) - x for x in reversed(seq)] backscore = reversed(one_way(opposite)) score = map(sum, zip(forscore, backscore)) winner = max(score + [0]) return sorted([seq[i] for i in range(len(seq)) if score[i] == winner]) def b_olson(sequence): supers = supernumbers(sequence) print len(supers) import random, time n=50000 w=range(1,n+1) random.shuffle(w) t=time.time() n00m(n,w) print 'n00m:',time.time()-t """ t=time.time() b_olson(w) print 'b_olson:',time.time()-t """ Sep 10 '05 #26

 P: n/a [Bryan Olson, on the problem at http://spoj.sphere.pl/problems/SUPPER/ ] I never intended to submit this program for competition. The contest ranks in speed order, and there is no way Python can compete with truly-compiled languages on such low-level code. I'd bet money that the algorithm I used (coded in C) can run with the winners. I also think I'd wager that the Python version outright trumps them on code size. Oh, it's not that bad . I took a stab at a Python program for this, and it passed (3.44 seconds). It just barely made it onto the list of "best" solutions, which I also guess is ranked by elapsed runtime. The Java program right above it took 2.91 seconds, but ate more than 27x as much RAM ;-) I didn't make any effort to speed this, beyond picking a reasonable algorithm, so maybe someone else can slash the runtime (while I usually enjoy such silliness, I can't make time for it at present). I'll include the code below. Alas, without access to the input data they use, it's hard to guess what might be important in their data. On my home box, chewing over random 100,000-element permutations took less than a second each (including the time to generate them); I'm pretty sure they're using slower HW than mine (3.4 GHz P5). My first version bombed for the zero-length sequence. That was a mistake, sorry, but it may not be one of their test-cases. I wonder how many of the accepted entries would perform properly. No idea here, and didn't even think about it. Notes: the `all` returned by crack() is a list such that all[i] is list of all (index, value) pairs such that the longest increasing subsequence ending with `value` is of length i+1; `value` is at index `index` in the input permutation. The maximal LISs thus end with the values in all[-1]. findall() iterates backwards over `all`, to accumulate all the values that appear in _some_ maximal LIS. There's clearly wasted work in findall() (if someone is looking for an algorithmic point to attack). Curiously, no use is made of that values are integers, outside of input and output; any values with a total ordering would work fine in crack() and findall(). """ # http://spoj.sphere.pl/problems/SUPPER/ def crack(xs): from bisect import bisect_right as find smallest = [] all = [] n = 0 for index, x in enumerate(xs): i = find(smallest, x) if i == n: smallest.append(x) all.append([(index, x)]) n += 1 else: all[i].append((index, x)) if x < smallest[i]: smallest[i] = x return all def findall(all): constraints = all[-1] allints = [pair[1] for pair in constraints] for i in xrange(len(all) - 2, -1, -1): survivors = [] for pair in all[i]: index, value = pair for index_limit, value_limit in constraints: if index < index_limit and value < value_limit: survivors.append(pair) allints.append(value) break constraints = survivors return sorted(allints) def main(): import sys while 1: n = sys.stdin.readline() if not n: break n = int(n) perm = map(int, sys.stdin.readline().split()) assert n == len(perm) supers = findall(crack(perm)) perm = None # just to free memory print len(supers) print " ".join(map(str, supers)) if __name__ == "__main__": main() """ Sep 11 '05 #27

 P: n/a Tim Peters; INCREDIBLE~ 241433 2005-09-11 04:23:40 Tim Peters accepted 3.44 7096 PYTH BRAVO! I just wonder have I grey cells enough for to understand how your algo works... and hopefully it's not your last solved problem on the contester. I'm pretty sure they're using slower HW than mine (3.4 GHz P5). As I mentioned before their 4 identical machines are PIII Xeon 700MHz. PS: each accepted solution automatically gets into "Best Solutions" list. Sep 11 '05 #28

 P: n/a [Tim Peters, on the problem at http://spoj.sphere.pl/problems/SUPPER/ ] ... [n0**@narod.ru] INCREDIBLE~ 241433 2005-09-11 04:23:40 Tim Peters accepted 3.44 7096 PYTH BRAVO! It's different now ;-) I added the two lines import psyco psyco.full() and time dropped to 2.29, while memory consumption zoomed: 2005-09-11 18:44:57 Tim Peters accepted 2.29 42512 PYTH That surprised me: my own test cases on Windows using Python 2.4.1 enjoyed the same order of speedup, but barely increased RAM usage. Perhaps they installed an older (or newer <0.9 wink>) version of psyco. I just wonder have I grey cells enough for to understand how your algo works... You do! Work out some small examples by hand, and it will become clear; be sure to read the comments before the code, because they explain it. and hopefully it's not your last solved problem on the contester. I have to stop now, else I wouldn't do anything else <0.3 wink>. ... Sep 11 '05 #29

 P: n/a Tim Peters wrote: [Bryan Olson, on the problem at http://spoj.sphere.pl/problems/SUPPER/ ]I never intended to submit this program for competition. Thecontest ranks in speed order, and there is no way Python cancompete with truly-compiled languages on such low-level code.I'd bet money that the algorithm I used (coded in C) can runwith the winners. I also think I'd wager that the Python versionoutright trumps them on code size. Oh, it's not that bad . I took a stab at a Python program for this, and it passed (3.44 seconds). [...] I didn't make any effort to speed this, beyond picking a reasonable algorithm, so maybe someone else can slash the runtime Hmmm ... I used the Theta(n lg n) algorithm ... how the heck... Aha! The 'bisect' module changed since last I looked. It still has the Python implementation, but then the last few lines say: # Overwrite above definitions with a fast C implementation try: from _bisect import bisect_right, bisect_left, insort_left, insort_right, insort, bisect except ImportError: pass Binary-search is the inner-loop in this algorithm. I wrote my own binary-search, so I was executing Theta(n lg n) Python statements. Tim's use of bisect means that his inner-loop is in C, so he does Theta(n) Python statements and Theta(n lg n) C statement. The key to fast Python: use a good algorithm, and keep the inner loop in C. -- --Bryan Sep 11 '05 #30

 P: n/a Tim Peters wrote: The chance that Raymond Hettinger is going to recode _your_ functions in C is approximately 0 ;-) Who is Raymond Hettinger? Sep 14 '05 #32

 P: n/a n00m wrote: Tim Peters wrote: The chance that Raymond Hettinger is going to recode _your_ functions in C is approximately 0 ;-) Who is Raymond Hettinger? See python-dev and, wrt this thread, http://docs.python.org/whatsnew/node12.html. Reinhold Sep 14 '05 #33

 P: n/a Got it! He is a kind of pythonic monsters. Btw, why it's impossible to reply to old threads? Namely, there're no more "Reply" link in them. Only "Reply to author" etc. Sep 14 '05 #34

 P: n/a "n00m" wrote in message news:11********************@f14g2000cwb.googlegrou ps.com... Who is Raymond Hettinger? The Python developer who, in the last few years, has perhaps been the most active in coding or recoding library modules, such as itertools and sets, in C. He also partipates in development discussions, helps with SourceForge tracker items (RFEs, bugs, and patches) and occasionally posts here, especially about his modules, for which he is the expert. Terry J. Reedy Sep 14 '05 #35

 P: n/a n00m wrote: Got it! He is a kind of pythonic monsters. Btw, why it's impossible to reply to old threads? Namely, there're no more "Reply" link in them. Only "Reply to author" etc. Perhaps because you are not using a real Usenet client? Reinhold Sep 14 '05 #36

 P: n/a Thank you both for your replies. And my personal "Thank you!" to Mr. Hettinger for all his tremendous work! Perhaps because you are not using a real Usenet client? Yes! And I don't even know what is the beast - Usenet client. I just keep in Favorites of my browser (IE 6.0) this link: http://groups.google.com/group/comp.lang.python/ Sep 15 '05 #37

 P: n/a I hope nobody have posted similar solution (it's tested, but I didn't submit it to contest): from bisect import bisect_right as find def supernumbers(ls): indices = [0]*len(ls) for i, e in enumerate(ls): indices[e - 1] = i result = [None]*len(ls) borders = [] for i in indices: ind = find(borders, i) result[i] = ind + 1 if ind == len(borders): borders.append(i) else: borders[ind] = i return result At least, it looks shorter than other solutins. Of course, it allows number of optimizations like prefetching length and caching borders.append. If I'm correct, it takes O(n*log (max sequence length)) time that can be a win for huge datasets. With the best regards, anton. Sep 15 '05 #38

 P: n/a Anton, it simply does not work! Try supernumbers([2,1,4,5,3]). Sep 16 '05 #39

 P: n/a Hellom n00m! I beg your pardon for not having replied for such long time---I was really busy. Yes, the posted code doesn't solve the supernumbers problem, but solves the problem you posted first as I understood it :) --- for each number find a position of this number in the longest sequence it belongs to. Ok, anyway, I hope that I can suggest a solution to the supernumbers problem (see below). Actually, as I find out today testing it against Tim Peters's solution, they are of the same kind, but I use one additional optimimization. And the last note: I didn't persue the maximal performace, rather clarity of the algorithm. With the best regards, anton. from bisect import bisect as find def supernumbers(l): indices = [0]*len(l) for i, e in enumerate(l): indices[e - 1] = i # Now we can find out index of each element in the longest ascending sequence it belongs to # I call this index 'generation' # ess is an array indexed by generations and holding # ascending list of generation's elements # Note: suffix s stands for a list, therefore ess---list of lists of elements # Note: ess holds not the elements itself, but elements decreased by 1 ess = [] # Borders i.e. positions that separate generations borders = [] for e, i in enumerate(indices): ind = find(borders, i) if ind < len(borders): borders[ind] = i ess[ind].append(e) else: borders.append(i) ess.append([e]) # Now we should filter out those elements that don't # belong to the longest sequences gens = reversed(ess) # walk generations in reverse order fes = gens.next() # fes stands for elements' filter r = fes # r is a list of supernumbers; obvioulsy all the elements of the latest generation belong to r for es in gens: new_fes = [] s = 0 # As elements are sorted ascendingly we can check lesser and lesser lists for e in es: s = find(fes, e, s) if s == len(fes): # There is no more elements in filter that are greater # than elements in the current generation break if indices[fes[s]] > indices[e]: # Nice---element e belongs to the longest sequence new_fes.append(e) # Note: the following invariant holds: len(new_fes) > 0 fes = new_fes r += new_fes return [e + 1 for e in r] # As we used numbers from 0 internally Sep 26 '05 #40

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