From 206c2fce8db1de9b7f82f04a3276005a284b3c40 Mon Sep 17 00:00:00 2001
From: Paulo Custodio <pauloscustodio@gmail.com>
Date: Tue, 24 Sep 2024 22:23:16 +0100
Subject: Add Python solution to challenge 147

---
 challenge-147/paulo-custodio/perl/ch-1.pl   |  2 +-
 challenge-147/paulo-custodio/perl/ch-2.pl   |  2 +-
 challenge-147/paulo-custodio/python/ch-1.py | 36 ++++++++++++++++++++++++
 challenge-147/paulo-custodio/python/ch-2.py | 43 +++++++++++++++++++++++++++++
 4 files changed, 81 insertions(+), 2 deletions(-)
 create mode 100644 challenge-147/paulo-custodio/python/ch-1.py
 create mode 100644 challenge-147/paulo-custodio/python/ch-2.py

(limited to 'challenge-147')

diff --git a/challenge-147/paulo-custodio/perl/ch-1.pl b/challenge-147/paulo-custodio/perl/ch-1.pl
index 85b81d6210..76794b7c9b 100644
--- a/challenge-147/paulo-custodio/perl/ch-1.pl
+++ b/challenge-147/paulo-custodio/perl/ch-1.pl
@@ -2,7 +2,7 @@
 
 # Challenge 147
 #
-# TASK #1 › Truncatable Prime
+# TASK #1 > Truncatable Prime
 # Submitted by: Mohammad S Anwar
 # Write a script to generate first 20 left-truncatable prime numbers in base 10.
 #
diff --git a/challenge-147/paulo-custodio/perl/ch-2.pl b/challenge-147/paulo-custodio/perl/ch-2.pl
index f81f51987d..3f9ad91939 100644
--- a/challenge-147/paulo-custodio/perl/ch-2.pl
+++ b/challenge-147/paulo-custodio/perl/ch-2.pl
@@ -2,7 +2,7 @@
 
 # Challenge 147
 #
-# TASK #2 › Pentagon Numbers
+# TASK #2 > Pentagon Numbers
 # Submitted by: Mohammad S Anwar
 # Write a script to find the first pair of Pentagon Numbers whose sum and
 # difference are also a Pentagon Number.
diff --git a/challenge-147/paulo-custodio/python/ch-1.py b/challenge-147/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..f3982a0174
--- /dev/null
+++ b/challenge-147/paulo-custodio/python/ch-1.py
@@ -0,0 +1,36 @@
+#!/usr/bin/env python3
+
+# Challenge 147
+#
+# TASK #1 > Truncatable Prime
+# Submitted by: Mohammad S Anwar
+# Write a script to generate first 20 left-truncatable prime numbers in base 10.
+#
+# In number theory, a left-truncatable prime is a prime number which, in a given
+# base, contains no 0, and if the leading left digit is successively removed,
+# then all resulting numbers are primes.
+#
+# Example
+# 9137 is one such left-truncatable prime since 9137, 137, 37 and 7 are all
+# prime numbers.
+
+from sympy import isprime, nextprime
+
+def left_truncatable_prime_it():
+    prime = None
+    while True:
+        prime = nextprime(prime) if prime is not None else 2
+        if is_left_truncatable_prime(prime):
+            yield prime
+
+def is_left_truncatable_prime(p):
+    while True:
+        if not isprime(p):
+            return False
+        p = int(str(p)[1:]) if len(str(p)) > 1 else ''
+        if p == '':
+            return True
+
+it = left_truncatable_prime_it()
+out = [next(it) for _ in range(20)]
+print(", ".join(map(str, out)))
diff --git a/challenge-147/paulo-custodio/python/ch-2.py b/challenge-147/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..87ee1d7626
--- /dev/null
+++ b/challenge-147/paulo-custodio/python/ch-2.py
@@ -0,0 +1,43 @@
+#!/usr/bin/env python3
+
+# Challenge 147
+#
+# TASK #2 > Pentagon Numbers
+# Submitted by: Mohammad S Anwar
+# Write a script to find the first pair of Pentagon Numbers whose sum and
+# difference are also a Pentagon Number.
+#
+# Pentagon numbers can be defined as P(n) = n(3n - 1)/2.
+#
+# Example
+# The first 10 Pentagon Numbers are:
+# 1, 5, 12, 22, 35, 51, 70, 92, 117 and 145.
+#
+# P(4) + P(7) = 22 + 70 = 92 = P(8)
+# but
+# P(4) - P(7) = |22 - 70| = 48 is not a Pentagon Number.
+
+limit = 100_000_000
+
+pentagon = [1]
+is_pentagon = {}
+
+def find_pair():
+    is_pentagon_func(limit)  # build pentagon up to N
+    try_ = pentagon.copy()
+    for a in try_:
+        for b in try_:
+            if is_pentagon_func(a + b) and is_pentagon_func(abs(a - b)):
+                return a, b
+    raise Exception("No pair found")
+
+def is_pentagon_func(num):
+    while pentagon[-1] < num:
+        n = len(pentagon) + 1
+        p = n * (3 * n - 1) // 2
+        pentagon.append(p)
+        is_pentagon[p] = True
+    return is_pentagon.get(num, False)
+
+a, b = find_pair()
+print(f"({a},{b})")
-- 
cgit