C++: Generate combinations in new base - c++

Question

I am trying to essentially count in a new base, specified by a given alphabet. So for the following params:
#define ALPHABET "abc"
#define ALPHABET_LEN 3
I would get the following result:
0 | a
1 | b
2 | c
3 | aa
4 | ab
...
9 | ca
10| cb
I have tried to do this with the following code:
#include <iostream>
#include <string>
#define ALPHABET "abc"
#define ALPHABET_LEN 3
int main()
{
for (int i = 0; i <= 10; i++) {
int l = 0;
int n_ = i;
int n = i;
char secret[4];
while (n > 0) {
secret[l] = ALPHABET[n%ALPHABET_LEN];
n /= ALPHABET_LEN;
l++;
}
std::cout << i << " | " << secret << std::endl;
}
}
Unfortunately, this prints the following:
0 |
1 | b
2 | c
3 | ab
4 | bb
5 | cb
6 | ac
7 | bc
8 | cc
9 | aab
10 | bab
This is not the expected pattern. How can I fix this, and is there a better way than just peeling off the next character using mod?

Answer 1

This is a challenging algorithm problem. The description of the problem is misleading. A base conversion suggests that there will be a 'zero' value which would be something like this:
a, b, c, a0, aa, ab, ac, b0, b1, b2, c0, etc.
However, in this problem, 'a' represents 'zero', but 'a' also is the first digit.
Looking at this as a base conversion creates a rabbit hole of complexity.
Rather, one has to figure out the algorithm to calculate each successive digit and ignore the idea of base conversion.
The first approach is to come up with a formula for each digit, which looks like this:
int LEN = ALPHABET_LEN; // Use a shorter variable name
std::string = ALPHABET[i % LEN]; // first digit
if(i > LEN - 1) {
secret = ALPHABET[(i/LEN -1)%LEN] + secret;
}
if(i > LEN * (LEN+1) - 1) {
secret = ALPHABET[(i/(LEN *(LEN+1)) - 1)%LEN] + secret;
}
if(i > LEN * (LEN+1) *(LEN+1) - 1) {
secret = ALPHABET[(i/(LEN *(LEN+1) * (LEN+1) ) - 1)%LEN] + secret;
}
As you work out the formula for each successive digit, you realize that the base is really LEN+1 rather than LEN.
This approach works, but it always requires an additional if statement for each successive digit. Sure, for i = 1 .. 10, this works fine. But what if i = 10,000,000. It would require an endless successive series of if statements.
However, after discovering the pattern, it is now a little easier to create a while() statement that avoids the need for an endless series of if statements.
#include <iostream>
#include <string>
//works as #define, but I like string so I can easily test with
// longer strings such as "abcd"
std::string ALPHABET {"abc"};
int main()
{
const int LEN = ALPHABET.size(); // Shorten the var name
for (int i = 0; i <= 100; i++) { // use i <= 100 to verify the algorithm
int number = i; // save the number that we are working on for this row
std::string secret = "";
secret += ALPHABET[i % LEN];
while(number / LEN > 0){
number = number/LEN - 1; // the base is really 4 rather than 3
secret = ALPHABET[number%LEN] + secret;
}
std::cout << i << " | " << secret << std::endl;
}
}
The output will be:
0 | a
1 | b
2 | c
3 | aa
4 | ab
5 | ac
6 | ba
7 | bb
8 | bc
9 | ca
10 | cb
11 | cc
12 | aaa
13 | aab
14 | aac
15 | aba
16 | abb
17 | abc
18 | aca
19 | acb
20 | acc
21 | baa
22 | bab
23 | bac
24 | bba
25 | bbb
26 | bbc
27 | bca
28 | bcb
29 | bcc
30 | caa
31 | cab
32 | cac
33 | cba
34 | cbb
35 | cbc
36 | cca
37 | ccb
38 | ccc
39 | aaaa
40 | aaab
41 | aaac
42 | aaba
43 | aabb
44 | aabc
45 | aaca
46 | aacb
47 | aacc
48 | abaa
49 | abab
50 | abac
51 | abba
52 | abbb
53 | abbc
54 | abca
55 | abcb
56 | abcc
57 | acaa
58 | acab
59 | acac
60 | acba
61 | acbb
62 | acbc
63 | acca
64 | accb
65 | accc
66 | baaa
67 | baab
68 | baac
69 | baba
70 | babb
71 | babc
72 | baca
73 | bacb
74 | bacc
75 | bbaa
76 | bbab
77 | bbac
78 | bbba
79 | bbbb
80 | bbbc
81 | bbca
82 | bbcb
83 | bbcc
84 | bcaa
85 | bcab
86 | bcac
87 | bcba
88 | bcbb
89 | bcbc
90 | bcca
91 | bccb
92 | bccc
93 | caaa
94 | caab
95 | caac
96 | caba
97 | cabb
98 | cabc
99 | caca
100 | cacb
Process finished with exit code 0

Answer 2

int i = 0;
int n = i;
while (n > 0) {
Clearly, in first iteration, the inner loop will not run a single iteration, and thus secret will not contain a as it should.
P.S. You fail to null-terminate secret, so the behaviour of the program is undefined when you insert it into the output stream.