backtracking n staircases at most k steps in a single jump - algorithm

Question

You need to climb a staircase that has n steps, and you decide to get some extra exercise by jumping up the steps. You can cover at most k steps in a single jump. Return all the possible sequences of jumps that you could take to climb the staircase, sorted.
My implementation is obviously giving me the wrong answer.
def climbingStaircase(n, k):
final_res=[]
final_res.append(CSR(n,k,[]))
return final_res
def CSR(n,k,res):
if n == 0:
return res
else:
for i in range(1,k+1):
if n-i>=0:
res.append(i)
n=n-i
res=CSR(n,i,res)
return res
For n = 4 and k = 2, the output should be
[[1, 1, 1, 1],
[1, 1, 2],
[1, 2, 1],
[2, 1, 1],
[2, 2]]
Actual output:
[[1,1,1,1,2,1]]
Can someone point out which part I'm missing?

Answer 1

One huge problem is in the code below: you deduct the quantity of steps for each possibility within the step range.
n=n-i
res=CSR(n,i,res)
When you're done exploring what you can do with a 1-step jump, you need to backtrack and try from the same starting point (this instance's original value of n) with a 2-step jump. Change the code to:
res = CSR(n-i, i, res)
This keeps the n value intact as you go through the loop.
In addition, you can't limit future jumps to the max of what you just took. Change that second parameter, too:
res = CSR(n-i, k, res)
That should get you moving. Also try this lovely debug blog for help. At least insert one or two tracing statements, such as
print n, k, res
at the top of your routine.
CAVEAT
This is not all of your trouble. The largest remaining problem is that CSR returns only one solution: every step you take is appended to the same list. You need a way to gather the completed solutions as separate lists; the append in climbingStaircase is executed only once, after CSR is entirely finished.
You need to recognize a completed solution at n==0.
DEBUGGING HELP
Here is a version of your program with the recursion parameters fixed, and debugging traces inserted.
indent = ""
def climbingStaircase(n, k):
final_res = []
final_res.append(CSR(n, k, []))
return final_res
def CSR(n, k, res):
global indent
indent += " "
print indent, n, k, res
if n == 0:
print "SOLUTION", res
else:
for i in range(1, k+1):
if n-i >= 0:
CSR(n-i, k, res + [i])
indent = indent[:-2]
print climbingStaircase(4, 2)
Note the use of "indent" to help visualize your recursion and backtracking. The critical part here is that, instead of updating res globally, I've left it as a local variable. I've also removed the return value for now, simply dumping to output the solutions as they're found. You can see how it works:
4 2 []
3 2 [1]
2 2 [1, 1]
1 2 [1, 1, 1]
0 2 [1, 1, 1, 1]
SOLUTION [1, 1, 1, 1]
0 2 [1, 1, 2]
SOLUTION [1, 1, 2]
1 2 [1, 2]
0 2 [1, 2, 1]
SOLUTION [1, 2, 1]
2 2 [2]
1 2 [2, 1]
0 2 [2, 1, 1]
SOLUTION [2, 1, 1]
0 2 [2, 2]
SOLUTION [2, 2]
[None]
With this stuff in place, I'm hopeful you can trace your logic and figure out how to capture the sequence of solutions at a level of your choosing.

Answer 2

Successfully implemented Prune's answer.
def climbingStaircase(n, k):
res=[]
CSR(n,k,[],res)
return res
def CSR(n,k,str_, res):
if n == 0:
res.append(str_)
else:
for i in range(1,k+1):
if n-i>=0:
CSR(n-i,k,str_+[i],res)

Answer 3

A quick Java version of this solution:
int[][] climbingStaircase(int n, int k) {
List<ArrayList<Integer>> list = new ArrayList<>();
climb(n, k, new ArrayList<Integer>(), list);
// convert to int[][]
int[][] result = new int[list.size()][];
for (int i=0; i<list.size(); i++) {
List<Integer> l = list.get(i);
int [] arr = new int[l.size()];
for (int j=0; j<l.size(); j++)
arr[j] = l.get(j);
result[i] = arr;
}
return result;
}
void climb(int n, int k, ArrayList<Integer> prev, List<ArrayList<Integer>> list) {
if (n==0) { // no more stairs, done climbing
list.add(prev);
} else {
for (int i=1; i<=k; i++) { // climb remaining stairs in intervals from 1 to k steps
if (i <= n) { // no need to test intervals larger than remaining # of stairs
ArrayList<Integer> branch = new ArrayList<>(prev);
branch.add(i);
climb(n-i, k, branch, list);
}
}
}
}

Answer 4

In Swift 5.5
func solution(n: Int, k: Int) -> [[Int]] {
var res_final = [[Int]]()
SRC(n: n, k: k, res: [], &res_final)
return res_final
}
var indent: String = ""
func SRC(n: Int, k: Int, res: [Int], _ res_final: inout [[Int]]) {
indent += " "
print(indent, n, k, res)
if n == .zero {
res_final.append(res)
print("Solution", res)
} else {
for i in 1...k {
if n-i >= .zero {
SRC(n: n-i, k: k, res: res + [i], &res_final)
}
}
}
indent = " "
}
solution(n: 4, k: 2)