Sequence combinations of varying length using two 'alphabets' - objective-c

Question

I am trying to generate all possible combinations of a sequence of length k given a certain alphabet (this is to generate query sequences for a bioinformatics project).
The sequences are of the form:
A first character and last character that can be any of A C G U (call these Y) and k - 2 characters in between that can be any of A C G U or ? (call these X).
e.g. if k = 3 then the pattern is of the form YXY and if k = 5 then YXXXY.
Generating all possible sequences if k is known is easy, as you can just use k nested for loops. But if k is not known in advance, then this implementation does not fit.
The total number of possible sequences can be expressed by 4^2 * 5^(k-2). With k = 3 this only gives 80 combinations, but scale that up to k = 9 and you have 1,250,000!
Any tips, ideas or suggestions would be much appreciated.
I need to use every sequence generated, so they need to be either stored in an array, or passed at creation/generation to another function, it doesn't really matter which, although I would prefer not to have to store all of them.
Many Thanks.
N.B. I am writing in objective-c, but any c-style code, or psuedocode or just plain english descriptions of an algorithm would be helpful.
UPDATE:
Here is the objc code I wrote base on the brilliant answer by Analog File. Currently it just outputs one sequence per line to stdout, but I will modify it to produce an array of strings.
Many thanks to everyone who contributed.
NSArray *yAlphabet = [NSArray arrayWithObjects:#"A", #"C", #"G", #"U", nil];
NSArray *xAlphabet = [NSArray arrayWithObjects:#"A", #"C", #"G", #"U", #"?", nil];
int i, v;
int count = 0;
int numberOfCases = 16 * pow(5 , (k - 2));
for (int n = 0; n < (numberOfCases); n++) {
i = n;
v = i % 4;
i = i / 4;
count++;
printf("\n%s", [[yAlphabet objectAtIndex:v] cStringUsingEncoding:NSUTF8StringEncoding]);
for (int m = 1; m < (k - 1); m++) {
v = i % 5;
i = i / 5;
printf("%s", [[xAlphabet objectAtIndex:v] cStringUsingEncoding:NSUTF8StringEncoding]);
}
printf("%s", [[yAlphabet objectAtIndex:i] cStringUsingEncoding:NSUTF8StringEncoding]);
}
printf("\n");
NSLog(#"No. Sequences: %i", count);
UPDATE 2:
And here's the code, outputting the generated sequences to an array of strings. Note that k is the length of the desired sequences and is given as a parameter elsewhere. I have tested this up to k=9 (1,250,000 sequences). Also note that my code uses ARC, hence there is no memory deallocation shown.
NSArray *yAlphabet = [NSArray arrayWithObjects:#"A", #"C", #"G", #"U", nil];
NSArray *xAlphabet = [NSArray arrayWithObjects:#"A", #"C", #"G", #"U", #"?", nil];
NSMutableArray *sequences = [[NSMutableArray alloc] init];
int i, v;
int count = 0;
int numberOfCases = 16 * pow(5 , (k - 2));
for (int n = 0; n < (numberOfCases); n++) {
i = n;
v = i % 4;
i = i / 4;
count++;
NSMutableString *seq = [[NSMutableString alloc] initWithString:[yAlphabet objectAtIndex:v]];
for (int m = 1; m < (k - 1); m++) {
v = i % 5;
i = i / 5;
[seq appendString:[xAlphabet objectAtIndex:v]];
}
[seq appendString:[yAlphabet objectAtIndex:i]];
[sequences addObject:seq];
}
NSLog(#"No. Sequences looped: %i", count);
//print the array to confirm
int count1 = 0;
for (NSMutableString *str in sequences) {
fprintf(stderr, "%s\n", [str cStringUsingEncoding:NSUTF8StringEncoding]);
count1++;
}
NSLog(#"No. Sequences printed: %i", count1);
NSLog(#"Counts match? : %#", (count == count1 ? #"YES" : #"NO"));

Answer 1

You know how many cases you are going to get.
This is pseudocode (k is the sequence length)
for n = 0 to num_of_cases - 1
i = n
v = i % length_of_alphabeth_Y
i = i / length_of_alphabeth_Y
output vth char in alphabeth Y
for m = 1 to k-1
v = i % length_of_alphabeth_X
i = i / length_of_alphabeth_X
output vth char in alphabeth X
output ith char in alphabeth Y
output end of sequence
Each iteration of the outer loop generates a case.I wrote output but it's easy to instead "store" the data in a dynamically allocated structure (n indexes in the roes, first case is column 0 then m indexes in the columns and last case is column k-1. if you do that "end of sequence" need not be output as it's subsumed by the increment of n).
Note how we are effectively "counting" in base length_of_alphabeth, except we use different bases depending on the digit. Modulo gives you the least significant digit, and integer division gets rid of it and shifts next digit to least significant position.
If you can imagine n as being just a value, in no specific base, the logic is rather simple. You could probably write this yourself from scratch, once you understand it.

Answer 2

It sounds like k will be passed in as a parameter, so something like this (in Python-ish pseudo-code) should work
Y_alphabet = ['A','C','G','U']
X_alphabet = ['A','C','G','U','?']
outputs = []
for i in range(k):
if i == 0 or i == k-1:
current_alphabet = Y_alphabet
else:
current_alphabet = X_alphabet
last_outputs = outputs
outputs = []
for next_character in current_alphabet:
# This just replaces outputs with a new list that consists
# of all the possible sequences of length i appended with
# the current character
outputs += [seq + next_character for seq in last_outputs]

Answer 3

The basic form for doing this looks like this (Java-ish psuedocode)
char[] output = new char[k];
pubilc void go(cur_k){
if(cur_k>k) // do something - copy the array and store it, etc.
for( char letter : alphabet ){
output[cur_k]=char_letter;
go(cur_k+1);
}
}