I was working on a problem where I'm finding all k-digit numbers whose digits sum up the given n.
I found how to do this and approach it as Integer Partitioning problem, however I would like to be able to only input n and k numbers (without the max_element) but when I try to delete it from the code it doesn't seem to work anymore.
How can I change that plus reverse it?
def c(n, k, max_element):
allowed = range(max_element, 0, -1)
def helper(n, k, t):
if k == 0:
if n == 0:
yield t
elif k == 1:
if n in allowed:
yield t + (n,)
elif 1 * k <= n <= max_element * k:
for v in allowed:
yield from helper(n - v, k - 1, t + (v,))
return helper(n, k, ())
for p in c(5, 3, 3):
print(p)
I tried using the reversed method but apparently it doesn't work in the generator.
Result:
(3, 1, 1)
(2, 2, 1)
(2, 1, 2)
(1, 3, 1)
(1, 2, 2)
(1, 1, 3)
Expected result:
113 122 131 212 221 311
There are a couple of problems here; the first is that you want the numbers in order and this code generates them in reverse order, because of range(max_element, 0, -1)
. The other problem is that since you're generating digits, the minimum element should be 0 and the maximum element should always be 9. We can fix both by changing that range to range(10)
.
We still need to be careful not to generate numbers starting with 0, so we'll make allowed
a parameter and use range(1, 10)
for just the first digit. I've also changed it to return the result as an integer instead of a tuple.
For reference, the code for this generator function comes from my answer to another question .
def c(n, k):
def helper(n, k, t, allowed):
if k == 0:
if n == 0:
yield t
elif k == 1:
if n in allowed:
yield 10*t + n
elif 0 <= n <= 9 * k:
for v in allowed:
yield from helper(n - v, k - 1, 10*t + v, range(10))
return helper(n, k, 0, range(1, 10))
Example:
>>> for p in c(5, 3):
... print(p)
...
104
113
122
131
140
203
212
221
230
302
311
320
401
410
500
This function should do the trick
def c(n, k, max_element):
allowed = range(max_element, 0, -1)
def helper(n, k, t):
if k == 0:
if n == 0:
yield t
elif k == 1:
if n in allowed:
yield t + (n,)
elif 1 * k <= n <= max_element * k:
for v in allowed:
yield from helper(n - v, k - 1, t + (v,))
return helper(n, k, ())
def reversed_iterator(iter):
return reversed(list(iter))
for p in reversed_iterator(c(5, 3, 3)):
print(p)
here is the output :
(1, 1, 3)
(1, 2, 2)
(1, 3, 1)
(2, 1, 2)
(2, 2, 1)
(3, 1, 1)
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