python中到任意给定数的路径复杂度(最快路由)
python中到任意给定数的路径复杂度(最快路由)
提问于 2016-01-06 01:14:16
今天我去参加一个数学竞赛，有一个问题是这样的：
你有一个给定的数字，现在你必须计算到那个数字的最短路径是什么，但这是有规则的。，
，
n，n。
当您到达n
- You时，可以通过将先前的数字增加一倍或将先前的两个数字相加来到达
- 。
示例：n = 25
最慢的路由：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 (只需不断添加1)
最快路由：1,2,4,8,16,24,25，复杂度=6
示例：n = 8最快路由：1,2,4,8，复杂度=3
示例：n = 15最快路由：1,2,3,6,9,15，复杂度=5
如何(用n <= 32)编写一个可以计算给定数目n的复杂度的程序？
我已经知道，对于任何给定的数字n(n <= 32 )，复杂度都低于1.45x2log(N)。所以现在我只需要计算复杂度低于1.45x2log(N)的所有路由，然后比较它们，看看哪一个是最快的“路由”。但是我不知道如何将所有的路由和所有这些放在python中，因为当给定的n发生变化时，路由的数量也会发生变化。
这就是我现在所拥有的：
number = raw_input('Enter your number here : ')
startnumber = 1
complexity = 0
while startnumber <= number回答 4
Stack Overflow用户
发布于 2016-01-06 05:00:30
我接受挑战:)
该算法相对较快。它在我的计算机上计算50ms内前32个数字的复杂度，而且我不使用多线程。(或前100个数字为370ms。)
这是一个递归的分支和切割算法。_shortest函数有3个参数:优化位于max_len参数中。例如，如果函数找到一个带有length=9的解决方案，它将停止考虑任何长度大于9的路径。找到的第一个路径总是非常好的路径，它直接跟随数字的二进制表示。例如，二进制: 111001 => 1,10,100,1000,10000,100000,110000,111000,111001.这并不总是最快的路径，但是如果你只搜索最快的路径，你可以减少大部分的搜索树。
#!/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))输出：
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]Stack Overflow用户
发布于 2016-01-06 02:11:57
你的问题有一个动态规划解，因为你可以将任意两个数相加，或者将一个数乘以2，我们可以尝试所有这些情况，并选择最小的一个。如果25的复杂度是5，路由包含9，那么我们知道9的解是4，我们可以使用9的解来生成25的解。我们还需要跟踪m的每个可能的最小解，以便能够使用它来求解n，其中m
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)
这最多可以解决n= 100
对于更大的数字，您可以通过添加几行代码来删除内部循环中的一些冗余计算，从而获得30%或更多的加速。为此，将在每次迭代中创建并修剪pos2变量：
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()))Stack Overflow用户
发布于 2016-01-06 04:39:41
蛮力逼迫
def solve(m, path):
if path[-1] == m:
return path
if path[-1] > m:
return False
best_path = [i for i in range(m)]
test_path = solve (m, path + [path[-1]*2])
if test_path and len(test_path) < len(best_path):
best_path = test_path
for k1 in path[:-1] :
for k2 in path[:-1] :
test_path = solve (m, path + [path[-1] + k1 + k2])
if test_path and len(test_path) < len(best_path):
#retain best
best_path = test_path
return best_path
print (solve(19, [1,2])) #[1, 2, 4, 8, 16, 19]
print (solve(25, [1,2])) #[1, 2, 4, 8, 16, 25]运行相当慢，我非常确定存在更智能的解决方案，但这看起来是正确的
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/34617451
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