The position of the number is important ?

So is equal 135, 351, 531, 513 or do you count it like a diferent combination ?

So is equal 135, 351, 531, 513 or do you count it like a diferent combination ?

Solved

Posted on 2008-06-16

Is there a function or a library which has a function for this problem : I have x numbers (e.g 5) , and I want to list all possible combinations in of these numbers taken y (e.g 3) at a time. For example, let's say we have these 5 numbers: 1, 3, 5, 6, 7 , I want to print all the possible combinations taken 3 at time, which means 135, 356,567 etc etc.

If there is not such function, can you provide me the basic guidelines to build such a function? thanks a lot.

If there is not such function, can you provide me the basic guidelines to build such a function? thanks a lot.

4 Comments

So is equal 135, 351, 531, 513 or do you count it like a diferent combination ?

Have you checked out Codeproject et al?

e.g. http://www.codeguru.com/cp

The choosing 3 is pretty simple, you can do it by hard coding the stepping. Have an array of 3, lets call it y, and make the third one linked to an index which will step through the original list from the position of y[2] + 1 till the end. The second one is linked to the a position in the original list and it will increment by one once y[3]'s position hits the end of the original list. And y[1] goes to the next number once y[2] goes through all the numbers(that come after where y[1] is of course). You will end up having a bunch of loops within loops but this is the most straight foward way to do it. Making combinations of the 3 should be pretty straightfoward to do, just have 6 outputs where you hardcode the combinations since you have it stored in an array(like cout<<y[1]<<y[2]<<y[2]<<en

A, personally, cooler way to do it is to again have an array of 3 and to choose numbers for it just randomly choose 3, then have a map which has previously done numbers, cross check and if its a new number run the combination thing. Honestly, while the first way is messier and in my opinion less cool, its probaby faster despite having nested loops because the sizes are set but if speed is not an issue I say go for broke with random numbers :)

Combinations are surprisingly easy in C/C++. While it's really a recursion problem, there is a very easy non-recursive answer. For problems like 3 of 5 (n of y) it's not necessarily the most efficient, but it's very easy to understand.

1. Define the number of items in the source group. (in your example 5)

2. Initialize an integer (Counter) to 0.

3. Add 1 Counter.

4. If (Counter >> NumberOfSourceItems) break the loop

5. Count the number of 1 bits in Counter.

6. If the number of bits != the TargetCount, repeat from 3. (in your example 3)

7. Display the values at the corresponding bits.

(If Counter = 01011 binary, display 124. You start at the right, print values

when you encounter a 1, and skip values when you encounter a 0.)

8. repeat from 3.

If you can't get your head wrapped around the recursive solution, this will work quite well. :)

Good Luck,

Kent

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