Today I solved a rather interesting problem: To find (and display) all the combinations that can be formed from a word. Suppose the word entered is “abc,” the output would be “abc acb bac bca cab cba.” A little Combination theory in Mathematics would tell you that if the length of the word is n, the number of combinations that can be formed from that word is n! (i.e, factorial of n.) In the above case, it is 3! or 3x2x1 = 6.

I went about solving the problem in Pascal, and it certainly isn’t the language you want to program in for these kind of tasks. However, today it wasn’t my choice in the matter, so: Pascal it was. An explanation of the solution is given below (not the actual source code):

I used three functions for the task: combinationOf(word, stub), shuffle(word), and strip(position, word). combinationOf displays all the combinations of the word using recursion. shuffle (which shifts the word one position to the right, i.e, changes “abc” to “bca” and “bca” to “cab”) and strip (which strips the alphabet corresponding to the digit i.e, strip(1, “abc”) => “bc”) are helper functions called from combinationOf.

The crux of the code (which may not be valid Pascal) is given below:

function combinationOf(word: string, stub: string) : string;
var i: integer;
var prestub: string;
for i := 1 to length(word) do
prestub := word[i];
if(length(strip(i, word)) = 2) then
write(stub, preStub);
write(stub, preStub);
stub := stub + prestub;
combinationOf(shuffle(strip(i, word)), stub);

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