Finding the LCM of a range of numbers

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时光说笑
时光说笑 2020-12-08 05:03

I read an interesting DailyWTF post today, \"Out of All The Possible Answers...\" and it interested me enough to dig up the original forum post where it was submitted. This

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  •  星月不相逢
    2020-12-08 05:54

    There's a fast solution to this, so long as the range is 1 to N.

    The key observation is that if n (< N) has prime factorization p_1^a_1 * p_2^a_2 * ... p_k * a_k, then it will contribute exactly the same factors to the LCM as p_1^a_1 and p_2^a_2, ... p_k^a_k. And each of these powers is also in the 1 to N range. Thus we only need to consider the highest pure prime powers less than N.

    For example for 20 we have

    2^4 = 16 < 20
    3^2 = 9  < 20
    5^1 = 5  < 20
    7
    11
    13
    17
    19
    

    Multiplying all these prime powers together we get the required result of

    2*2*2*2*3*3*5*7*11*13*17*19 = 232792560
    

    So in pseudo code:

    def lcm_upto(N):
      total = 1;
      foreach p in primes_less_than(N):
        x=1;
        while x*p <= N:
          x=x*p;
        total = total * x
      return total
    

    Now you can tweak the inner loop to work slightly differently to get more speed, and you can precalculate the primes_less_than(N) function.

    EDIT:

    Due to a recent upvote I decideded to revisit this, to see how the speed comparison with the other listed algorithms went.

    Timing for range 1-160 with 10k iterations, against Joe Beibers and Mark Ransoms methods are as follows:

    Joes : 1.85s Marks : 3.26s Mine : 0.33s

    Here's a log-log graph with the results up to 300.

    A log-log graph with the results

    Code for my test can be found here:

    import timeit
    
    
    def RangeLCM2(last):
        factors = range(last+1)
        result = 1
        for n in range(last+1):
            if factors[n] > 1:
                result *= factors[n]
                for j in range(2*n, last+1, n):
                    factors[j] /= factors[n]
        return result
    
    
    def lcm(a,b):
        gcd, tmp = a,b
        while tmp != 0:
            gcd,tmp = tmp, gcd % tmp
        return a*b/gcd
    
    def EuclidLCM(last):
        return reduce(lcm,range(1,last+1))
    
    primes = [
     2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 
     31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 
     73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 
     127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 
     179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 
     233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 
     283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 
     353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 
     419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 
     467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 
     547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 
     607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 
     661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 
     739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 
     811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 
     877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 
     947, 953, 967, 971, 977, 983, 991, 997 ]
    
    def FastRangeLCM(last):
        total = 1
        for p in primes:
            if p>last:
                break
            x = 1
            while x*p <= last:
                x = x * p
            total = total * x
        return total
    
    
    print RangeLCM2(20)
    print EculidLCM(20)
    print FastRangeLCM(20)
    
    print timeit.Timer( 'RangeLCM2(20)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(20)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(20)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(40)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(40)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(40)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(60)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(60)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(60)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(80)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(80)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(80)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(100)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(100)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(100)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(120)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(120)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(120)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(140)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(140)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(140)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    
    print timeit.Timer( 'RangeLCM2(160)', "from __main__ import RangeLCM2").timeit(number=10000)
    print timeit.Timer( 'EuclidLCM(160)', "from __main__ import EuclidLCM" ).timeit(number=10000)
    print timeit.Timer( 'FastRangeLCM(160)', "from __main__ import FastRangeLCM" ).timeit(number=10000)
    

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