Add solution for probelm_686 of project_euler (#5480)

* Added solution for probelm_686 of project_euler

* Changed documentation and formatting.

* Added ref link to optimization logic

* Update project_euler/problem_686/sol1.py

Co-authored-by: John Law <[email protected]>

Co-authored-by: John Law <[email protected]>

Author: srishtik2310

project_euler/problem_686/__init__.py

@@ -0,0 +1,160 @@
+"""
+Project Euler Problem 686: https://projecteuler.net/problem=686
+
+2^7 = 128 is the first power of two whose leading digits are "12".
+The next power of two whose leading digits are "12" is 2^80.
+
+Define p(L,n) to be the nth-smallest value of j such that
+the base 10 representation of 2^j begins with the digits of L.
+
+So p(12, 1) = 7 and p(12, 2) = 80.
+
+You are given that p(123, 45) = 12710.
+
+Find p(123, 678910).
+"""
+
+import math
+
+
+def log_difference(number: int) -> float:
+    """
+    This function returns the decimal value of a number multiplied with log(2)
+    Since the problem is on powers of two, finding the powers of two with
+    large exponents is time consuming. Hence we use log to reduce compute time.
+
+    We can find out that the first power of 2 with starting digits 123 is 90.
+    Computing 2^90 is time consuming.
+    Hence we find log(2^90) = 90*log(2) = 27.092699609758302
+    But we require only the decimal part to determine whether the power starts with 123.
+    SO we just return the decimal part of the log product.
+    Therefore we return 0.092699609758302
+
+    >>> log_difference(90)
+    0.092699609758302
+    >>> log_difference(379)
+    0.090368356648852
+
+    """
+
+    log_number = math.log(2, 10) * number
+    difference = round((log_number - int(log_number)), 15)
+
+    return difference
+
+
+def solution(number: int = 678910) -> int:
+    """
+    This function calculates the power of two which is nth (n = number)
+    smallest value of power of 2
+    such that the starting digits of the 2^power is 123.
+
+    For example the powers of 2 for which starting digits is 123 are:
+    90, 379, 575, 864, 1060, 1545, 1741, 2030, 2226, 2515 and so on.
+    90 is the first power of 2 whose starting digits are 123,
+    379 is second power of 2 whose starting digits are 123,
+    and so on.
+
+    So if number = 10, then solution returns 2515 as we observe from above series.
+
+    Wwe will define a lowerbound and upperbound.
+    lowerbound = log(1.23), upperbound = log(1.24)
+    because we need to find the powers that yield 123 as starting digits.
+
+    log(1.23) = 0.08990511143939792, log(1,24) = 0.09342168516223506.
+    We use 1.23 and not 12.3 or 123, because log(1.23) yields only decimal value
+    which is less than 1.
+    log(12.3) will be same decimal vale but 1 added to it
+    which is log(12.3) = 1.093421685162235.
+    We observe that decimal value remains same no matter 1.23 or 12.3
+    Since we use the function log_difference(),
+    which returns the value that is only decimal part, using 1.23 is logical.
+
+    If we see, 90*log(2) = 27.092699609758302,
+    decimal part = 0.092699609758302, which is inside the range of lowerbound
+    and upperbound.
+
+    If we compute the difference between all the powers which lead to 123
+    starting digits is as follows:
+
+    379 - 90 = 289
+    575 - 379 = 196
+    864 - 575 = 289
+    1060 - 864 = 196
+
+    We see a pattern here. The difference is either 196 or 289 = 196 + 93.
+
+    Hence to optimize the algorithm we will increment by 196 or 93 depending upon the
+    log_difference() value.
+
+    Lets take for example 90.
+    Since 90 is the first power leading to staring digits as 123,
+    we will increment iterator by 196.
+    Because the difference between any two powers leading to 123
+    as staring digits is greater than or equal to 196.
+    After incrementing by 196 we get 286.
+
+    log_difference(286) = 0.09457875989861 which is greater than upperbound.
+    The next power is 379, and we need to add 93 to get there.
+    The iterator will now become 379,
+    which is the next power leading to 123 as starting digits.
+
+    Lets take 1060. We increment by 196, we get 1256.
+    log_difference(1256) = 0.09367455396034,
+    Which is greater than upperbound hence we increment by 93. Now iterator is 1349.
+    log_difference(1349) = 0.08946415071057 which is less than lowerbound.
+    The next power is 1545 and we need to add 196 to get 1545.
+
+    Conditions are as follows:
+
+    1) If we find a power, whose log_difference() is in the range of
+    lower and upperbound, we will increment by 196.
+    which implies that the power is a number which will lead to 123 as starting digits.
+    2) If we find a power, whose log_difference() is greater than or equal upperbound,
+    we will increment by 93.
+    3) if log_difference() < lowerbound, we increment by 196.
+
+    Reference to the above logic:
+    https://math.stackexchange.com/questions/4093970/powers-of-2-starting-with-123-does-a-pattern-exist
+
+    >>> solution(1000)
+    284168
+
+    >>> solution(56000)
+    15924915
+
+    >>> solution(678910)
+    193060223
+
+    """
+
+    power_iterator = 90
+    position = 0
+
+    lower_limit = math.log(1.23, 10)
+    upper_limit = math.log(1.24, 10)
+    previous_power = 0
+
+    while position < number:
+        difference = log_difference(power_iterator)
+
+        if difference >= upper_limit:
+            power_iterator += 93
+
+        elif difference < lower_limit:
+            power_iterator += 196
+
+        else:
+            previous_power = power_iterator
+            power_iterator += 196
+            position += 1
+
+    return previous_power
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    print(f"{solution() = }")