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nisargshah95
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 Joined: Sat Jan 07, 2012 7:50 pm
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by nisargshah95 » Thu Apr 26, 2012 6:32 am
Now I was working on this problem in school where I had to list out all possible permutations of a given three digit number. Here's my code 
Code: Select all
Output "Enter a threedigit number: "
Input no
hundreds_digit=no/100,
ones_digit=no%10,
tens_digit=no/1010*hundreds_digit
Now I've got the one's, ten's and the hundred's digit. The problem is, how do I display their different combinations other than manually write all six possible ways (that would be worse) like
Code: Select all
Output hundreds_digit, tens_digit, ones_digit
Output hundreds_digit, ones_digit, tens_digit
And so on...
Any help will be appreciated.

jaap
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 Joined: Tue Mar 25, 2008 3:57 pm

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by jaap » Thu Apr 26, 2012 6:56 am
You'll need to use some data structure like an array or a list (or even a string) so that you can access those digits individually by index number, rather than by specific variable names for each digit. That way the digits will be something like digit[1], digit[2], and digit[3] (or digit[0], digit[1], digit[2] depending on the programming language).
This allows you to use a variable for the index, i.e. digit[index] where index is a variable with the value 1, 2, or 3 (or 0, 1, 2 as noted above). It gives much more flexibility.To do what you want, you can then use three nested forloops, with if statements to avoid repeating a digit in one number.
If you then want to go on to make it work for numbers of any length, you will probably want to look at recursion.
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nisargshah95
 Posts: 5
 Joined: Sat Jan 07, 2012 7:50 pm
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by nisargshah95 » Thu Apr 26, 2012 7:14 am
Expand
[quote="jaap"]You'll need to use some data structure like an array or a list (or even a string) so that you can access those digits individually by index number, rather than by specific variable names for each digit. That way the digits will be something like digit[1], digit[2], and digit[3] (or digit[0], digit[1], digit[2] depending on the programming language).
This allows you to use a variable for the index, i.e. digit[index] where index is a variable with the value 1, 2, or 3 (or 0, 1, 2 as noted above). It gives much more flexibility.To do what you want, you can then use three nested forloops, with if statements to avoid repeating a digit in one number.
If you then want to go on to make it work for numbers of any length, you will probably want to look at recursion.[/quote]
I'm using C++. If I use an array to store each digit in each index then I'll have to input one digit and then other digit. What I mean to say is that I cannot input the whole number at a time. I'm in 12th grade so I haven't learnt advanced methods. Still I'll try considering your suggestion. Thanks.

jaap
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by jaap » Thu Apr 26, 2012 7:40 am
nisargshah95 wrote:I'm using C++. If I use an array to store each digit in each index then I'll have to input one digit and then other digit. What I mean to say is that I cannot input the whole number at a time. I'm in 12th grade so I haven't learnt advanced methods. Still I'll try considering your suggestion. Thanks.
You can adjust your code like this:
Code: Select all
int no;
// Input no  whatever code you use for inputting no goes here
// Assumption: At this point no is an integer between 100 and 999 inclusive.
int digits[3];
digits[0] = no/100;
digits[1] = no%10;
digits[2] = no/1010*digits[0];
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nisargshah95
 Posts: 5
 Joined: Sat Jan 07, 2012 7:50 pm
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by nisargshah95 » Fri Apr 27, 2012 5:12 am
Yeah! That's what I was exactly going to suggest before I saw your post. And now I've solved my problem. Did the following 
Code: Select all
seq[3]={ones_digit,tens_digit,hundreds_digit}
for (int i=0;i<=2;i++)
Output seq[i], seq[(i+1)%3], seq[(i+2)%3]
Output seq[i], swap(seq[(i+1)%3],seq[(i+2)%3]) //Where 'swap' will swap the last two digits and return a twodigit number.
Thanks for helping me out!