This article assumes Ruby, but I think the same can be said for many languages.
There are many ways to find out how many digits a positive integer x will be in decimal.
One of them
Math.log10(x).floor + 1
However, is it really possible to get the correct answer? Even if you are not good at math, look at it as carefully as possible so that you can understand it even if you do not know much about Ruby.
However, if you just want to know the conclusion, [Is that okay? [#% E3% 81% 9D% E3% 82% 8C% E3% 81% A7% E3% 81% 84% E3% 81% 84% E3% 81% AE% E3% 81% 8B) ..
In this section, it is assumed that the local variable x is assigned a positive integer.
Ruby can handle any large integer if conditions such as memory allow it.
For Ruby integers (Integer class), generate a number string expressed in N-ary notation [Integer # to_s](https://docs.ruby-lang.org/ja/2.7.0/method/Integer/i There is a method called /inspect.html).
If you omit the argument, you get a sequence of numbers in decimal notation.
p (36 + 72).to_s # => "108"
On the other hand, in the String class, there is a method called String # length that counts the length (number of characters). There is.
p "Ruby".length # => 4
If you combine this
x.to_s.length
You can get the number of digits with. It's pretty easy.
However, I feel a little uneasy. Isn't it slow to create a huge string with a length of 100,000 for an integer of 100,000 digits when you just want to know the number of digits? It's actually pretty fast, but let's leave that story aside.
Ruby returns an array of the number of each digit when a non-negative integer (integer greater than or equal to 0) is expressed in N-ary notation Integer # digits There is a method called /2.7.0/method/Integer/i/digits.html).
If the argument is omitted, it is in decimal notation.
p 1234.digits # => [4, 3, 2, 1]
Since they are arranged in order from the bottom, the order of appearance is reversed.
If you use this and Array # length that returns the length of the array
x.digits.length
You can get the number of digits with.
Mystery thought, "It seems that integers are faster than strings, so why not be faster than x.to_s.length? ", But that was not the case. That's right. A huge array is created.
There are other ways, but the main subject
Math.log10(x).floor + 1
Let's go to. Why does this give me the number of digits in x?
Math.log10 is a method that returns the common logarithm value of the given argument.
Common logarithm function
y = \log_{10}x
Is an exponential function with 10 as base
x = 10^y
It was the inverse function of. This exponential function is easy to understand when $ y $ is an integer, but it is also defined for any real number.
Let's look at when $ y $ is an integer greater than or equal to 0
If you look at this, you can see that when $ y $ is an integer greater than or equal to 0, $ y + 1 $, that is, $ \ log_ {10} (x) + 1 $ is the number of digits.
However, $ x $, where $ y $ is an integer, is limited. What about the more common positive integer $ x $?
As a trial, consider $ x = 999 $. This is a little less than $ 1000
Let's use the floor
[^ fl]: Note that for negative numbers, -1.1.floor is -2, not -1 with the decimal part truncated.
By the way, in mathematical symbols, it seems that the floor function is written as $ \ lfloor a \ rfloor $, but from the above consideration, when $ x $ is $ 100 $ or more and $ 999 $ or less,
\lfloor \log_{10}x \rfloor
Turns out to be all $ 2 $. In other words, it looks like this.
Therefore, the number of digits can be obtained by adding 1 to $ \ floorloor \ log_ {10} x \ rfloor + 1 $. Mathematically.
Finally at the core.
If you write $ \ floorloor \ log_ {10} x \ rfloor + 1 $ in Ruby code
Math.log10(x).floor + 1
However, I feel a little uneasy. That was the point of this article.
That's because Math.log10 is a floating point operation of the Float class. Floating-point arithmetic usually involves errors.
Is it possible that the resulting digits will be out of order due to subtle errors?
What should I think about?
Looking at the positive integer x in the order of 1, 2, 3, ..., where does the digit change? Of course, it's 9 → 10 or 99 → 100 or 999 → 1000.
If an error in floating-point arithmetic gives an incorrect result, it is probably around this boundary.
It is possible that something like 10000000 is mistakenly made one less digit, or something like 9999999999 is mistakenly made one more digit.
Of these, numbers such as 1000000 are originally in the form of $ 10 ^ k $, so I feel that errors are unlikely to occur.
Then, let's experiment with the number of 9s.
Yes, I wrote this code.
1.upto(100) do |k|
puts "%3d %3d" % [k, Math.log10(10 ** k - 1).floor + 1]
end
Change $ k $ from 1 to 100 to calculate the digits of $ 10 ^ k -1 $ (that is, the number of 9s arranged in $ k $). Display this side by side with $ k $. It's a mathematical match, so see if the same numbers are lined up.
The result is as follows
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
50 51
51 52
52 53
53 54
54 55
55 56
56 57
57 58
58 59
59 60
60 61
61 62
62 63
63 64
64 65
65 66
66 67
67 68
68 69
69 70
70 71
71 72
72 73
73 74
74 75
75 76
76 77
77 78
78 79
79 80
80 81
81 82
82 83
83 84
84 85
85 86
86 87
87 88
88 89
89 90
90 91
91 92
92 93
93 94
94 95
95 96
96 97
97 98
98 99
99 100
100 101
It is off the middle. Here is an excerpt.
14 14
15 16
Up to $ 10 ^ {14} ―― 1 $ can be calculated correctly, but at $ 10 ^ {15} ―― 1 $, as expected, the number is one larger than the correct answer.
There was something that came to my mind here. "The precision of floating-point numbers is about 15 digits in decimal," he said. Ruby's Float is actually environment-dependent, so the accuracy cannot be unequivocally stated, but it is IEEE in a large number of environments. It seems that 754 "double precision" is used. In this case, the precision is about 15 digits in decimal.
So it's quite possible that the result of Math.log10 (x) .floor + 1 with an integer of 15 9s wasn't the correct number of digits.
For small integers, Math.log10 (x) .floor + 1 is fine, but for large integers there is an error.
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