How to express integer using symbols + * () and 1 with minimal cost? [closed]

Question

The task is to build integers from the symbols + * ( ) (addition, multiplication and brackets) and the digit 1. You are given an integer and must output an expression using the minimal number of characters. For example:

4    = 1+1+1+1  
23   = 11+11+1  
242  = (11+11)*11  
1000 = 1+(1+1+1)*(1+1+1)*111 
1997 = (1+1)*(1+1+1)*111+11*11*11 

Answer 1

You can use dynamic programming, and compute, for each number i < n, the shortest expression that computes i, and the shortest expression that computes i that can be used in a multiplicative context. In general, the second expression will be longer than the first: for example 2 can be constructed as '1+1' but if you want a '2' in a multiplication, then it'll be '(1+1)'.

Here's some unoptimized code that prints the shortest solutions for all numbers up to 2000. It runs in a fraction over 2 seconds on my laptop, but there's a lot of scope for removing all the string construction from the code. It runs in O(n^2) time.

def getbest(a, b):
    return a or b if not (a and b) else min((a, b), key=len)

def minconstruct(n):
    res = [[None, None] for _ in range(n + 1)]
    for i in xrange(1, n + 1):
        if set(str(i)) == set('1'):
            res[i][0] = res[i][1] = str(i)
        for j in xrange(1, i // 2 + 1):
            sol = '%s+%s' % (res[j][0], res[i-j][0])
            res[i][0] = getbest(res[i][0], sol)
            res[i][1] = getbest(res[i][1], '(' + sol + ')')
        for j in xrange(2, i):
            if i % j != 0:
                        continue
            sol = '%s*%s' % (res[j][1], res[i//j][1])
            res[i][0] = getbest(res[i][0], sol)
            res[i][1] = getbest(res[i][1], sol)             
    return res

r = minconstruct(2000)
for i, x in enumerate(r[1:]):
    print '%4d: %s' % (i, x[0])

Answer 2

Here's my recursive solution. It works for 2000 elements in about 1.4 sec on my tablet:

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 < k
        for d in divs:
            if d >= n: break
            k, r = divmod(n, d)
            if k < 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 < 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) < 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 <= n:
            p[j] = 1
            j = j * i

    return [i for (i, v) in enumerate(p) if v == 1 and i > 1]

Speed testing:

>>> timeit('to_onestr(2000)', 'from __main__ import to_onestr', number=1)
1.1375278780336457
>>> timeit('to_onestr(4000)', 'from __main__ import to_onestr', number=1)
3.6481025870678696
>>> timeit('to_onestr(6000)', 'from __main__ import to_onestr', number=1)
7.732885259577177

Also tested @Anonymous approach

>>> timeit('minconstruct(2000)', 'from __main__ import minconstruct', number=1)
12.012599471759474