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  1. PO
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    <p>Here's my recursive solution. It works for 2000 elements in about 1.4&nbsp;sec on my tablet:</p> <pre><code>import math def to_onestr(n, numbers=None, divs=None): if numbers is None: numbers = [None] * (n + 1) numbers[0] = ('', False) if divs is None: divs = get_divs(n) if numbers[n] is None: s = str(n) # Default representation is 11111 or 1+1+1+1 if s == '1'*len(s): res = (s, False) else: res = ("+".join(['1'] * n), True) # Find all representations d*k + r, d &lt; k for d in divs: if d &gt;= n: break k, r = divmod(n, d) if k &lt; d: d, k = k, d k_res, r_res, d_res = to_onestr(k, numbers, divs), to_onestr(r, numbers, divs), to_onestr(d, numbers, divs) res_str, res_bool = '', False if d != 1: res_str += '({})*'.format(d_res[0]) if d_res[1] else d_res[0] + '*' res_str += '({})'.format(k_res[0]) if k_res[1] else k_res[0] if d != 1 and len(k_res[0]) * d + d - 1 &lt; len(res_str): res_str = '+'.join([k_res[0]]*d) res_bool = True if r != 0: res_str += '+{}'.format(r_res[0]) res_bool = True if len(res_str) &lt; len(res[0]): res = (res_str, res_bool) numbers[n] = res return numbers[n] def get_divs(n): p = [1] * (n + 1) # Get all prime numbers + all numbers which contains only 1 + all numbers we could get from 11..1 by multiplication for i in range(2, int(math.ceil(math.sqrt(n)))): if p[i] == 1: for j in range(i * i, n, i): if j % i == 0: p[j] = 0 for x in xrange(2, len(str(n)) + 1): i = int('1'*x) j = i while j &lt;= n: p[j] = 1 j = j * i return [i for (i, v) in enumerate(p) if v == 1 and i &gt; 1] </code></pre> <p>Speed testing:</p> <pre><code>&gt;&gt;&gt; timeit('to_onestr(2000)', 'from __main__ import to_onestr', number=1) 1.1375278780336457 &gt;&gt;&gt; timeit('to_onestr(4000)', 'from __main__ import to_onestr', number=1) 3.6481025870678696 &gt;&gt;&gt; timeit('to_onestr(6000)', 'from __main__ import to_onestr', number=1) 7.732885259577177 </code></pre> <p>Also tested @Anonymous approach</p> <pre><code>&gt;&gt;&gt; timeit('minconstruct(2000)', 'from __main__ import minconstruct', number=1) 12.012599471759474 </code></pre>
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    1. POHow to express integer using symbols + * () and 1 with minimal cost?
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    1. PTAnswer
    1. USPeter Mortensen
    1. USRoman Pekar
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      1. VTUpMod
    1. COtest your code ,but something wrong with "len(str(n)) + 1]))"
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      1. UStcpiper
    2. CO@Pythoner had to go when submitted it, changed solution a bit, it now fills the list of all numbers from 0 to n.
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      1. USRoman Pekar
    3. COThis code gives "((11+1)*11+1)*(1+1+1+1+1)*(1+1+1)+1+1" for 1997 rather than "(1+1)*(1+1+1)*111+11*11*11". What's it minimizing?
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      1. USPaul Hankin
 

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