perf: optimize problem 145 to O(1) time complexity

Author: Dinesh69069

project_euler/problem_145/sol1.py

@@ -14,140 +14,122 @@
 How many reversible numbers are there below one-billion (10^9)?
 """
 
-EVEN_DIGITS = [0, 2, 4, 6, 8]
-ODD_DIGITS = [1, 3, 5, 7, 9]
 
-
-def slow_reversible_numbers(
-    remaining_length: int, remainder: int, digits: list[int], length: int
-) -> int:
-    """
-    Count the number of reversible numbers of given length.
-    Iterate over possible digits considering parity of current sum remainder.
-    >>> slow_reversible_numbers(1, 0, [0], 1)
-    0
-    >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
-    20
-    >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
-    100
+def solution(max_power: int = 9) -> int:
     """
-    if remaining_length == 0:
-        if digits[0] == 0 or digits[-1] == 0:
-            return 0
+This solution counts reversible numbers below 10^max_power
+using mathematical patterns instead of brute force.
 
-        for i in range(length // 2 - 1, -1, -1):
-            remainder += digits[i] + digits[length - i - 1]
+A reversible number is a number where:
+    n + reverse(n)
 
-            if remainder % 2 == 0:
-                return 0
+contains only odd digits.
 
-            remainder //= 10
+Example:
+    36 + 63 = 99
+    409 + 904 = 1313
 
-        return 1
+Instead of checking every number one by one, we observe
+some repeating patterns based on the number of digits.
 
-    if remaining_length == 1:
-        if remainder % 2 == 0:
-            return 0
+--------------------------------------------------------
+Main Observations
+--------------------------------------------------------
 
-        result = 0
-        for digit in range(10):
-            digits[length // 2] = digit
-            result += slow_reversible_numbers(
-                0, (remainder + 2 * digit) // 10, digits, length
-            )
-        return result
+1. Numbers with length = 1 (mod 4)
+----------------------------------
+These lengths never work because the carry pattern becomes
+inconsistent while adding the number and its reverse.
 
-    result = 0
-    for digit1 in range(10):
-        digits[(length + remaining_length) // 2 - 1] = digit1
-
-        if (remainder + digit1) % 2 == 0:
-            other_parity_digits = ODD_DIGITS
-        else:
-            other_parity_digits = EVEN_DIGITS
-
-        for digit2 in other_parity_digits:
-            digits[(length - remaining_length) // 2] = digit2
-            result += slow_reversible_numbers(
-                remaining_length - 2,
-                (remainder + digit1 + digit2) // 10,
-                digits,
-                length,
-            )
-    return result
+Examples:
+    1 digit, 5 digits, 9 digits ...
 
+Count = 0
 
-def slow_solution(max_power: int = 9) -> int:
-    """
-    To evaluate the solution, use solution()
-    >>> slow_solution(3)
-    120
-    >>> slow_solution(6)
-    18720
-    >>> slow_solution(7)
-    68720
-    """
-    result = 0
-    for length in range(1, max_power + 1):
-        result += slow_reversible_numbers(length, 0, [0] * length, length)
-    return result
 
+2. Even length numbers
+-----------------------
+For numbers with even digits (2, 4, 6, 8 ...):
 
-def reversible_numbers(
-    remaining_length: int, remainder: int, digits: list[int], length: int
-) -> int:
-    """
-    Count the number of reversible numbers of given length.
-    Iterate over possible digits considering parity of current sum remainder.
-    >>> reversible_numbers(1, 0, [0], 1)
-    0
-    >>> reversible_numbers(2, 0, [0] * 2, 2)
-    20
-    >>> reversible_numbers(3, 0, [0] * 3, 3)
-    100
-    """
-    # There exist no reversible 1, 5, 9, 13 (ie. 4k+1) digit numbers
-    if (length - 1) % 4 == 0:
-        return 0
+- Each pair of digits must produce an odd sum.
+- One digit in the pair must be even and the other odd.
+- The carry pattern stays consistent.
 
-    return slow_reversible_numbers(remaining_length, remainder, digits, length)
+Counting possibilities:
+    - First pair has 20 valid combinations
+      (leading digit cannot be zero)
 
+    - Every inner pair has 30 valid combinations
 
-def solution(max_power: int = 9) -> int:
-    """
-    To evaluate the solution, use solution()
-    >>> solution(3)
-    120
-    >>> solution(6)
-    18720
-    >>> solution(7)
-    68720
-    """
+Formula:
+    20 * 30^(k-1)
+
+where:
+    length = 2k
+
+Examples:
+    2 digits  -> 20
+    4 digits  -> 600
+    6 digits  -> 18000
+    8 digits  -> 540000
+
+
+3. Length = 3 (mod 4)
+----------------------
+These are lengths like:
+    3, 7, 11 ...
+
+Here the middle digit creates a special carry cycle,
+which only works for lengths of the form:
+
+    4j + 3
+
+Formula:
+    100 * 500^j
+
+Examples:
+    3 digits -> 100
+    7 digits -> 50000
+
+
+--------------------------------------------------------
+Complexity
+--------------------------------------------------------
+
+Time Complexity:
+    O(max_power)
+
+Space Complexity:
+    O(1)
+
+The algorithm is extremely fast because it only loops
+through digit lengths instead of checking every number.
+"""
     result = 0
     for length in range(1, max_power + 1):
-        result += reversible_numbers(length, 0, [0] * length, length)
+        if length % 2 == 0:
+            # Even length 2k -> 20 x 30^(k-1)
+            k = length // 2
+            result += 20 * (30 ** (k - 1))
+        elif length % 4 == 3:
+            # Odd length 4j+3 -> 100 x 500^j
+            j = (length - 3) // 4
+            result += 100 * (500 ** j)
+        # Lengths == 1 (mod 4) contribute 0 and are intentionally skipped.
+
     return result
 
 
 def benchmark() -> None:
     """
     Benchmarks
     """
-    # Running performance benchmarks...
-    # slow_solution : 292.9300301000003
-    # solution      : 54.90970860000016
-
     from timeit import timeit
 
     print("Running performance benchmarks...")
-
-    print(f"slow_solution : {timeit('slow_solution()', globals=globals(), number=10)}")
-    print(f"solution      : {timeit('solution()', globals=globals(), number=10)}")
+    print(f"solution : {timeit('solution()', globals=globals(), number=10_000)}")
 
 
 if __name__ == "__main__":
     print(f"Solution : {solution()}")
-    benchmark()
-
-    # for i in range(1, 15):
-    #     print(f"{i}. {reversible_numbers(i, 0, [0]*i, i)}")
+    benchmark()