the path complexity (fastest route) to any given number in python

Question

Today I went to a math competition and there was a question that was something like this:

You have a given number n, now you have to like calculate what's the shortest route to that number, but there are rules.

1. You start with number 1
2. You end when you reach n
3. You can get to n either by doubling your previous number, or by adding two previous numbers.

Example: n = 25

Slowest route : 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25 (You just keep adding 1)

Fastest route : 1,2,4,8,16,24,25, complexity = 6

Example: n = 8 Fastest route : 1,2,4,8, complexity = 3

Example : n = 15 Fastest route : 1,2,3,6,9,15, complexity = 5

How do I make a program that can calculate the complexity of a given number n (with n <= 32)?

I already know that for any given number n ( n <= 32 ) that the complexity is lower that 1.45 x 2log(n). So now I only need to calculate all the routes with complexity below 1.45 x 2log(n) and than compare them and see which one is the fastest 'route'. But I have no idea how to put all the routes and all this in python, because the number of routes changes when the given number n changes.

This is what I have for now:

number = raw_input('Enter your number here : ')
startnumber = 1
complexity = 0

while startnumber <= number

Answer 1

I accept the challenge :)

The algorithm is relatively fast. It calculates the complexity of the first 32 numbers in 50ms on my computer, and I don't use multithreading. (or 370ms for the first 100 numbers.)

It is a recursive branch and cut algorithm. The _shortest function takes 3 arguments: the optimization lies in the max_len argument. E.g. if the function finds a solution with length=9, it stops considering any paths with a length > 9. The first path that is found is always a pretty good one, which directly follows from the binary representation of the number. E.g. in binary: 111001 => [1,10,100,1000,10000,100000,110000,111000,111001]. That's not always the fastest path, but if you only search for paths that are as least as fast, you can cut away most of the search tree.

#!/usr/bin/env python

# Find the shortest addition chain...
# @param acc List of integers, the "accumulator". A strictly monotonous
#        addition chain with at least two elements.
# @param target An integer > 2. The number that should be reached.
# @param max_len An integer > 2. The maximum length of the addition chain
# @return A addition chain starting with acc and ending with target, with
#         at most max_len elements. Or None if such an addition chain
#         does not exist. The solution is optimal! There is no addition
#         chain with these properties which can be shorter.
def _shortest(acc, target, max_len):
    length = len(acc)
    if length > max_len:
        return None
    last = acc[-1]
    if last == target:
        return acc;
    if last > target:
        return None
    if length == max_len:
        return None
    last_half = (last / 2)
    solution = None
    potential_solution = None
    good_len = max_len

    # Quick check: can we make this work?
    # (this improves the performance considerably for target > 70)
    max_value = last
    for _ in xrange(length, max_len):
        max_value *= 2
        if max_value >= target:
            break
    if max_value < target:
        return None

    for i in xrange(length-1, -1, -1):
        a = acc[i]
        if a < last_half:
            break

        for j in xrange(i, -1, -1):
            b = acc[j]
            s = a+b
            if s <= last:
                break

            # modifying acc in-place has much better performance than copying
            # the list and doing
            #   new_acc = list(acc)
            #   potential_solution = _shortest(new_acc, target, good_len)

            acc.append(s)
            potential_solution = _shortest(acc, target, good_len)
            if potential_solution is not None:
                new_len = len(potential_solution)
                solution = list(potential_solution)
                good_len = new_len-1

            # since we didn't copy the list, we have to truncate it to its
            # original length now.
            del acc[length:]

    return solution

# Finds the shortest addition chain reaching to n.
# E.g. 9 => [1,2,3,6,9]
def shortest(n):
    if n > 3:
        # common case first
        return _shortest([1,2], n, n)
    if n < 1:
        raise ValueError("n must be >= 1")
    return list(xrange(1,n+1))

for i in xrange(1,33):
    s = shortest(i)
    c = len(s) - 1
    print ("complexity of %2d is %d: e.g. %s" % (i,c,s))

Output:

complexity of  1 is 0: e.g. [1]
complexity of  2 is 1: e.g. [1, 2]
complexity of  3 is 2: e.g. [1, 2, 3]
complexity of  4 is 2: e.g. [1, 2, 4]
complexity of  5 is 3: e.g. [1, 2, 4, 5]
complexity of  6 is 3: e.g. [1, 2, 4, 6]
complexity of  7 is 4: e.g. [1, 2, 4, 6, 7]
complexity of  8 is 3: e.g. [1, 2, 4, 8]
complexity of  9 is 4: e.g. [1, 2, 4, 8, 9]
complexity of 10 is 4: e.g. [1, 2, 4, 8, 10]
complexity of 11 is 5: e.g. [1, 2, 4, 8, 10, 11]
complexity of 12 is 4: e.g. [1, 2, 4, 8, 12]
complexity of 13 is 5: e.g. [1, 2, 4, 8, 12, 13]
complexity of 14 is 5: e.g. [1, 2, 4, 8, 12, 14]
complexity of 15 is 5: e.g. [1, 2, 4, 5, 10, 15]
complexity of 16 is 4: e.g. [1, 2, 4, 8, 16]
complexity of 17 is 5: e.g. [1, 2, 4, 8, 16, 17]
complexity of 18 is 5: e.g. [1, 2, 4, 8, 16, 18]
complexity of 19 is 6: e.g. [1, 2, 4, 8, 16, 18, 19]
complexity of 20 is 5: e.g. [1, 2, 4, 8, 16, 20]
complexity of 21 is 6: e.g. [1, 2, 4, 8, 16, 20, 21]
complexity of 22 is 6: e.g. [1, 2, 4, 8, 16, 20, 22]
complexity of 23 is 6: e.g. [1, 2, 4, 5, 9, 18, 23]
complexity of 24 is 5: e.g. [1, 2, 4, 8, 16, 24]
complexity of 25 is 6: e.g. [1, 2, 4, 8, 16, 24, 25]
complexity of 26 is 6: e.g. [1, 2, 4, 8, 16, 24, 26]
complexity of 27 is 6: e.g. [1, 2, 4, 8, 9, 18, 27]
complexity of 28 is 6: e.g. [1, 2, 4, 8, 16, 24, 28]
complexity of 29 is 7: e.g. [1, 2, 4, 8, 16, 24, 28, 29]
complexity of 30 is 6: e.g. [1, 2, 4, 8, 10, 20, 30]
complexity of 31 is 7: e.g. [1, 2, 4, 8, 10, 20, 30, 31]
complexity of 32 is 5: e.g. [1, 2, 4, 8, 16, 32]

Answer 2

There is a dynamic programming solution to your problem since you either add any two numbers or multiply a number by 2 we can try all those cases and choose the minimum one also if the complexity of 25 was 5 and the route contains 9 then we know the solution for 9 which is 4 and we can use the solution for 9 to generate the solution for 25.We also need to keep track of every possible minimum solution for m to be able to use it to use it to solve n where m < n

def solve(m):
    p = [set([frozenset([])]) for i in xrange(m+1)] #contains all paths to reach n
    a = [9999 for _ in xrange(m+1)]
    #contains all complexities initialized with a big number
    a[1] = 0
    p[1] = set([frozenset([1])])
    for i in xrange(1,m+1):
        for pos in p[i]:
            for j in pos: #try adding any two numbers and 2*any number
                for k in pos:
                    if (j+k <= m):
                        if a[j+k] > a[i]+1:
                            a[j+k] = a[i] + 1
                            p[j+k] = set([frozenset(list(pos) + [j+k])])
                        elif a[j+k] == a[i]+1:
                            p[j+k].add(frozenset(list(pos) + [j+k]))
    return a[m],sorted(list(p[m].pop()))

n = int(raw_input())
print solve(n)

this can solve up to n = 100

For larger numbers you can get a 30% or more speedup by adding a couple lines to remove some redundant calculations from the inner loop. For this the pos2 variable is created and trimmed on each iteration:

def solve(m):
    p = [set([frozenset([])]) for i in xrange(m+1)] #contains all paths to reach n
    a = [9999 for _ in xrange(m+1)]
    #contains all complexities initialized with a big number
    a[1] = 0
    p[1] = set([frozenset([1])])
    for i in xrange(1,m+1):
        for pos in p[i]:
            pos2 = set(pos)
            for j in pos: #try adding any two numbers and 2*any number
                for k in pos2:
                    if (j+k <= m):
                        if a[j+k] > a[i]+1:
                            a[j+k] = a[i] + 1
                            p[j+k] = set([frozenset(list(pos) + [j+k])])
                        elif a[j+k] == a[i]+1:
                            p[j+k].add(frozenset(list(pos) + [j+k]))
                pos2.remove(j)
    return a[m],sorted(list(p[m].pop()))