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Author SHA1 Message Date
duongoku
62dcbea943
Add power sum problem (#8832)
* Add powersum problem

* Add doctest

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Add more doctests

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Add more doctests

* Improve paramater name

* Fix line too long

* Remove global variables

* Apply suggestions from code review

* Apply suggestions from code review

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
2023-06-26 09:39:18 +02:00
Christian Clauss
d764eec655
Fix failing pytest quantum/bb84.py (#8838)
* Fix failing pytest quantum/bb84.py

* Update bb84.py test results to match current qiskit
2023-06-26 08:54:50 +05:30
3 changed files with 95 additions and 3 deletions

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@ -24,7 +24,6 @@ jobs:
- name: Run tests
# TODO: #8818 Re-enable quantum tests
run: pytest
--ignore=quantum/bb84.py
--ignore=quantum/q_fourier_transform.py
--ignore=project_euler/
--ignore=scripts/validate_solutions.py

93
backtracking/power_sum.py Normal file
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@ -0,0 +1,93 @@
"""
Problem source: https://www.hackerrank.com/challenges/the-power-sum/problem
Find the number of ways that a given integer X, can be expressed as the sum
of the Nth powers of unique, natural numbers. For example, if X=13 and N=2.
We have to find all combinations of unique squares adding up to 13.
The only solution is 2^2+3^2. Constraints: 1<=X<=1000, 2<=N<=10.
"""
from math import pow
def backtrack(
needed_sum: int,
power: int,
current_number: int,
current_sum: int,
solutions_count: int,
) -> tuple[int, int]:
"""
>>> backtrack(13, 2, 1, 0, 0)
(0, 1)
>>> backtrack(100, 2, 1, 0, 0)
(0, 3)
>>> backtrack(100, 3, 1, 0, 0)
(0, 1)
>>> backtrack(800, 2, 1, 0, 0)
(0, 561)
>>> backtrack(1000, 10, 1, 0, 0)
(0, 0)
>>> backtrack(400, 2, 1, 0, 0)
(0, 55)
>>> backtrack(50, 1, 1, 0, 0)
(0, 3658)
"""
if current_sum == needed_sum:
# If the sum of the powers is equal to needed_sum, then we have a solution.
solutions_count += 1
return current_sum, solutions_count
i_to_n = int(pow(current_number, power))
if current_sum + i_to_n <= needed_sum:
# If the sum of the powers is less than needed_sum, then continue adding powers.
current_sum += i_to_n
current_sum, solutions_count = backtrack(
needed_sum, power, current_number + 1, current_sum, solutions_count
)
current_sum -= i_to_n
if i_to_n < needed_sum:
# If the power of i is less than needed_sum, then try with the next power.
current_sum, solutions_count = backtrack(
needed_sum, power, current_number + 1, current_sum, solutions_count
)
return current_sum, solutions_count
def solve(needed_sum: int, power: int) -> int:
"""
>>> solve(13, 2)
1
>>> solve(100, 2)
3
>>> solve(100, 3)
1
>>> solve(800, 2)
561
>>> solve(1000, 10)
0
>>> solve(400, 2)
55
>>> solve(50, 1)
Traceback (most recent call last):
...
ValueError: Invalid input
needed_sum must be between 1 and 1000, power between 2 and 10.
>>> solve(-10, 5)
Traceback (most recent call last):
...
ValueError: Invalid input
needed_sum must be between 1 and 1000, power between 2 and 10.
"""
if not (1 <= needed_sum <= 1000 and 2 <= power <= 10):
raise ValueError(
"Invalid input\n"
"needed_sum must be between 1 and 1000, power between 2 and 10."
)
return backtrack(needed_sum, power, 1, 0, 0)[1] # Return the solutions_count
if __name__ == "__main__":
import doctest
doctest.testmod()

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@ -64,10 +64,10 @@ def bb84(key_len: int = 8, seed: int | None = None) -> str:
key: The key generated using BB84 protocol.
>>> bb84(16, seed=0)
'1101101100010000'
'0111110111010010'
>>> bb84(8, seed=0)
'01011011'
'10110001'
"""
# Set up the random number generator.
rng = np.random.default_rng(seed=seed)