From 55a9a415cfcccad595054bf6a9d4a279abc0a3b2 Mon Sep 17 00:00:00 2001
From: mindaugl <mindelek@gmail.com>
Date: Tue, 8 Apr 2025 15:11:56 +0800
Subject: [PATCH 1/2] Add solution for the Euler project problem 164.

---
 project_euler/problem_164/__init__.py |  0
 project_euler/problem_164/sol1.py     | 58 +++++++++++++++++++++++++++
 2 files changed, 58 insertions(+)
 create mode 100644 project_euler/problem_164/__init__.py
 create mode 100644 project_euler/problem_164/sol1.py

diff --git a/project_euler/problem_164/__init__.py b/project_euler/problem_164/__init__.py
new file mode 100644
index 000000000000..e69de29bb2d1
diff --git a/project_euler/problem_164/sol1.py b/project_euler/problem_164/sol1.py
new file mode 100644
index 000000000000..97cb085e3508
--- /dev/null
+++ b/project_euler/problem_164/sol1.py
@@ -0,0 +1,58 @@
+"""
+Project Euler Problem 164: https://projecteuler.net/problem=164
+
+Three Consecutive Digital Sum Limit
+
+How many 20 digit numbers n (without any leading zero) exist such that no three
+consecutive digits of n have a sum greater than 9?
+
+Brute-force recursive solution with caching of intermediate results.
+
+>>> solution(10)
+21838806
+"""
+
+
+def solve(
+    digit: int, prev: int, prev2: int, sum_max: int, first: bool, cache: dict[str, int]
+) -> int:
+    """
+    Solve for remaining 'digit' digits, with previous 'prev' number, and
+    previous-previous 'prev2' number, total sum of 'sum_max'.
+    Pass around 'cache' to store/reuse intermediate results.
+
+    >>> solve(1, 0, 0, 9, True, {})
+    9
+    >>> solve(1, 0, 0, 9, False, {})
+    10
+    """
+    if digit == 0:
+        return 1
+    comb = 0
+    cache_str = f"{digit},{prev},{prev2}"
+    if cache_str in cache:
+        return cache[cache_str]
+    for v in range(sum_max - prev - prev2 + 1):
+        if first and v == 0:
+            continue
+        comb += solve(digit - 1, v, prev, sum_max, False, cache)
+    cache[cache_str] = comb
+    return comb
+
+
+def solution(n_digits: int = 20) -> int:
+    """
+    Solves the problem for n_digits number of digits.
+
+    >>> solution(2)
+    45
+    """
+    sum_max = 9
+    cache: dict[str, int] = {}
+    ans = solve(n_digits, 0, 0, sum_max, True, cache)
+
+    return ans
+
+
+if __name__ == "__main__":
+    print(f"{solution(10) = }")

From 2b883d1be4ad3594a567f51698890a8ff2a2b3a6 Mon Sep 17 00:00:00 2001
From: mindaugl <mindelek@gmail.com>
Date: Wed, 9 Apr 2025 16:18:45 +0800
Subject: [PATCH 2/2] Add solution for the Euler project problem 190.

---
 project_euler/problem_190/__init__.py |  0
 project_euler/problem_190/sol1.py     | 50 +++++++++++++++++++++++++++
 2 files changed, 50 insertions(+)
 create mode 100644 project_euler/problem_190/__init__.py
 create mode 100644 project_euler/problem_190/sol1.py

diff --git a/project_euler/problem_190/__init__.py b/project_euler/problem_190/__init__.py
new file mode 100644
index 000000000000..e69de29bb2d1
diff --git a/project_euler/problem_190/sol1.py b/project_euler/problem_190/sol1.py
new file mode 100644
index 000000000000..1cc0aa667d4a
--- /dev/null
+++ b/project_euler/problem_190/sol1.py
@@ -0,0 +1,50 @@
+"""
+Project Euler Problem 190: https://projecteuler.net/problem=190
+
+Maximising a Weighted Product
+
+Let S_m = (x_1, x_2, ..., x_m) be the m-tuple of positive real numbers with
+x_1 + x_2 + ... + x_m = m for which P_m = x_1 * x_2^2 * ... * x_m^m is maximised.
+For example, it can be verified that |_P_10_| = 4112
+(|__| is the integer part function).
+Find Sum_{m=2}^15 = |_P_m_|.
+
+Solution:
+- Fix x_1 = m - x_2 - ... - x_m.
+- Calculate partial derivatives of P_m wrt the x_2, ..., x_m. This gives that
+  x_2 = 2 * x_1, x_3 = 3 * x_1, ..., x_m = m * x_1.
+- Calculate partial second order derivatives of P_m wrt the x_2, ..., x_m.
+  By plugging in the values from the previous step, can verify that solution is maximum.
+
+>>> solution(5)
+10
+
+"""
+
+
+def solution(n: int = 15) -> int:
+    """
+    Calculate sum of P_m for m from 2 to n.
+
+    >>> solution(2)
+    1
+    >>> solution(3)
+    2
+    >>> solution(4)
+    4
+
+    """
+
+    ans = 0
+    for m in range(2, n + 1):
+        x1 = 2 / (m + 1)
+        capital_p = 1.0
+        for i in range(1, m + 1):
+            xi = i * x1
+            capital_p *= pow(xi, i)
+        ans += int(capital_p)
+    return ans
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")