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Fix some typos in solution 1 of euler 686 (#6112)

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While reading this code I noticed some typos in the doc strings and wanted to fix them.
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KerimovEmil 2022-05-11 23:28:45 -04:00 committed by GitHub
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project_euler/problem_686/sol1.py
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@ -27,7 +27,7 @@ def log_difference(number: int) -> float:
Computing 2^90 is time consuming. Computing 2^90 is time consuming.
Hence we find log(2^90) = 90*log(2) = 27.092699609758302 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. 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. So we just return the decimal part of the log product.
Therefore we return 0.092699609758302 Therefore we return 0.092699609758302
>>> log_difference(90) >>> log_difference(90)
@ -57,14 +57,14 @@ def solution(number: int = 678910) -> int:
So if number = 10, then solution returns 2515 as we observe from above series. So if number = 10, then solution returns 2515 as we observe from above series.
Wwe will define a lowerbound and upperbound. We will define a lowerbound and upperbound.
lowerbound = log(1.23), upperbound = log(1.24) lowerbound = log(1.23), upperbound = log(1.24)
because we need to find the powers that yield 123 as starting digits. because we need to find the powers that yield 123 as starting digits.
log(1.23) = 0.08990511143939792, log(1,24) = 0.09342168516223506. 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 We use 1.23 and not 12.3 or 123, because log(1.23) yields only decimal value
which is less than 1. which is less than 1.
log(12.3) will be same decimal vale but 1 added to it log(12.3) will be same decimal value but 1 added to it
which is log(12.3) = 1.093421685162235. which is log(12.3) = 1.093421685162235.
We observe that decimal value remains same no matter 1.23 or 12.3 We observe that decimal value remains same no matter 1.23 or 12.3
Since we use the function log_difference(), Since we use the function log_difference(),
@ -87,7 +87,7 @@ def solution(number: int = 678910) -> int:
Hence to optimize the algorithm we will increment by 196 or 93 depending upon the Hence to optimize the algorithm we will increment by 196 or 93 depending upon the
log_difference() value. log_difference() value.
Lets take for example 90. Let's take for example 90.
Since 90 is the first power leading to staring digits as 123, Since 90 is the first power leading to staring digits as 123,
we will increment iterator by 196. we will increment iterator by 196.
Because the difference between any two powers leading to 123 Because the difference between any two powers leading to 123
@ -99,7 +99,7 @@ def solution(number: int = 678910) -> int:
The iterator will now become 379, The iterator will now become 379,
which is the next power leading to 123 as starting digits. which is the next power leading to 123 as starting digits.
Lets take 1060. We increment by 196, we get 1256. Let's take 1060. We increment by 196, we get 1256.
log_difference(1256) = 0.09367455396034, log_difference(1256) = 0.09367455396034,
Which is greater than upperbound hence we increment by 93. Now iterator is 1349. Which is greater than upperbound hence we increment by 93. Now iterator is 1349.
log_difference(1349) = 0.08946415071057 which is less than lowerbound. log_difference(1349) = 0.08946415071057 which is less than lowerbound.
@ -107,7 +107,7 @@ def solution(number: int = 678910) -> int:
Conditions are as follows: Conditions are as follows:
1) If we find a power, whose log_difference() is in the range of 1) If we find a power whose log_difference() is in the range of
lower and upperbound, we will increment by 196. lower and upperbound, we will increment by 196.
which implies that the power is a number which will lead to 123 as starting digits. 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, 2) If we find a power, whose log_difference() is greater than or equal upperbound,