Solve PE 751: Concatenation Coincidence

Author: NonlinearFruit

README.md

@@ -13,6 +13,7 @@
 |-|-|-|
 |88|Product Sum Numbers|([src](project_euler/test_pe88_product_sum_numbers.py)) ([web](https://projecteuler.net/problem=88))|
 |700|Eulercoin|([src](project_euler/test_pe700_eulercoin.py)) ([web](https://projecteuler.net/problem=700))|
+|751|Concatenation Coincidence|([src](project_euler/test_pe751_concatenation_coincidence.py)) ([web](https://projecteuler.net/problem=751))|
 |800|Hybrid Integers|([src](project_euler/test_pe800_hybrid_integers.py)) ([web](https://projecteuler.net/problem=800))|
 |808|Reversible Prime Squares|([src](project_euler/test_pe808_reversible_prime_squares.py)) ([web](https://projecteuler.net/problem=808))|
 |816|Shortest Distance Among Points|([src](project_euler/test_pe816_shortest_distance_among_points.py)) ([web](https://projecteuler.net/problem=816))|

project_euler/test_pe751_concatenation_coincidence.py

@@ -0,0 +1,64 @@
+# <https://projecteuler.net/problem=751>
+# <p>A non-decreasing sequence of integers $a_n$ can be generated from any positive real value $\theta$ by the following procedure:
+# $$\begin{align}
+# \begin{split}
+# b_1 &amp;= \theta \\
+# b_n &amp;= \left\lfloor b_{n-1} \right\rfloor \left(b_{n-1} - \left\lfloor b_{n-1} \right\rfloor + 1\right)~~~\forall ~ n \geq 2 \\
+# a_n &amp;= \left\lfloor b_{n} \right\rfloor
+# \end{split}
+# \end{align}$$
+# Where $\left\lfloor \cdot \right\rfloor$ is the floor function.</p>
+# 
+# <p>For example, $\theta=2.956938891377988...$ generates the Fibonacci sequence: $2, 3, 5, 8, 13, 21, 34, 55, 89, ...$</p>
+# 
+# <p>The <i>concatenation</i> of a sequence of positive integers $a_n$ is a real value denoted $\tau$ constructed by concatenating the elements of the sequence after the decimal point, starting at $a_1$: $a_1.a_2a_3a_4...$</p>
+# 
+# <p>For example, the Fibonacci sequence constructed from $\theta=2.956938891377988...$ yields the concatenation $\tau=2.3581321345589...$ Clearly, $\tau \neq \theta$ for this value of $\theta$.</p>
+# 
+# <p>Find the only value of $\theta$ for which the generated sequence starts at $a_1=2$ and the concatenation of the generated sequence equals the original value: $\tau = \theta$. Give your answer rounded to $24$ places after the decimal point.</p>
+# Notes:
+# - Since it starts with 2, a1 is 2 and theta must round down to 2
+# - a2 is either 2 or 3, depending on whether or not the decimal places round up or down (since they get doubled)
+# - Since the tenths place in theta is a 2 or 3, it must round down. So a2 is 2.
+# - a3 is either 2 or 3, depending on whether or not the decimal places round up or down (since they get doubled)
+# - Theta is currently 2.2, so b1 = 2.2, b2 = 2.4 (or 2.5). In either case, a3 must be 2
+# - b3 must be greater than (or equal to) 2.5. So a4 must be 3
+# - Theta is now 2.223
+
+from math import floor
+from decimal import *
+
+def b(theta, n):
+    if n == 1: return theta
+    solution = theta
+    while n > 1:
+        solution = floor(solution) * (solution - floor(solution) + 1)
+        n -= 1
+    return solution
+
+def test_b_works():
+    assert 2 == b(2, 1)
+    assert Decimal('2.4') == b(Decimal('2.2'), 2)
+
+def a_sequence(theta):
+    solution = str(floor(b(theta,1))) + "."
+    for i in range(2,25):
+        solution += str(floor(b(theta,i)))
+    return solution
+
+def theta_generator(theta=2):
+    stringy_theta = str(theta)
+    if len(stringy_theta) >= 26:
+        return theta
+    solution = a_sequence(theta)
+    stringy_solution = str(solution)
+    if stringy_solution.startswith(stringy_theta):
+        next = stringy_solution[len(stringy_theta)]
+        if next == ".":
+            next += stringy_solution[len(stringy_theta) + 1]
+        return theta_generator(Decimal(stringy_theta + next))
+    else:
+        return -1
+
+def test_works():
+    assert Decimal('2.223561019313554106173177') == theta_generator()