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# Permutations of multiple string arrays' elements

Posted on 2003-12-10
Medium Priority
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I am working on an insurance system web app. The system allows users to set up price quotes based on a number of factors. The quotes must be made for all combinations of selected factors. So, if I’m making a quote, the factors that I choose to base it on are:

1.       Age

2.       Sex

3.       Smoker

Now, for each factor, I must define options, like so:

1. Age
a.       21-34
b.       35-64
c.       65-99

2. Sex
a.       Male
b.       Female

3. Smoker
a.       Non-Smoker
b.       Less than 1 pack per day
c.       More than 1 pack per day

OK, now comes the hard part. I need to generate every possible permutation of these three factors individual options, like so:

1.       Male, 21-34, Non-Smoker

2.       Male, 35-64, Non-Smoker

3.       Male, 65-99, Non-Smoker

4.       Male, 21-34, Less than 1 pack per day

5.       Male, 35-64, Less than 1 pack per day

6.       Male, 65-99, Less than 1 pack per day

7.       Male, 21-34, More than 1 pack per day

8.       Male, 35-64, More than 1 pack per day

9.       Male, 65-99, More than 1 pack per day

10.   Female, 21-34, Non-Smoker

11.   Female, 35-64, Non-Smoker

12.   Female, 65-99, Non-Smoker

13.   Female, 21-34, Less than 1 pack per day

14.   Female, 35-64, Less than 1 pack per day

15.   Female, 65-99, Less than 1 pack per day

16.   Female, 21-34, More than 1 pack per day

17.   Female, 35-64, More than 1 pack per day

18.   Female, 65-99, More than 1 pack per day

Now, each of these factors can have any number of options (except for Sex, which can only be male or female.)

I have wasted 2 days on this problem. If anybody can help me, you'd be saving the rest of my hair. Thanks.
0
Question by:henningero
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LVL 48

Expert Comment

ID: 9912883
>> generate every possible permutation of these three factors individual options.

This sounds too abstract. What exactly you need? For example, this is pseudo-code for printing all permutations:

for each 3 years from 0 to 1000
{
for each sex
{
for each smoking state
{
print line
}
}
}

0

Author Comment

ID: 9913069
I basically need every possible combination of each arrays elements. So, if I have 3 arrays, each with 2 elements, i need the combinations of each element of each array.

string[] gender = new string{"Male", "Female"}

string[] smoking = new string{"Non-Smoker", "Smoker"}

string[] age = new string{"21-34", "34-99"}

I need the combinations like so:

Male, Non-Smoker, 21-34
Male, Non-Smoker, 34-99
Male, Smoker, 21-34
Male, Smoker, 35-99
Female, Non-Smoker, 21-34
Female, Non-Smoker, 34-99
Female, Smoker, 21-34
Female, Smoker, 35-99

The problem is, I don't know how many elements each array will have (except for gender, which will only have Male or Female). Smoking can be Non-Smoker, Smoker, Less than 1 pack per day, More than 1 pack per day, Not in last 5 years, or any user input. Age can be any number of user defined age bands, like 21-34, 35-64, 65-99 in any amounts and combination.

I do know that there will be at least one array (either age, sex, or smoking) and no more than three arrays (age, sex AND smoking). The arrays sized can all be different, except gender(only Male or Female). I hope this is clear enough.
0

LVL 22

Accepted Solution

ID: 9913323

very clear.

To make it a little more flexible, I am going to use ArrayLists

ArrayList sex = new ArrayList();
ArrayList smoke= new ArrayList();
ArrayList age= new ArrayList();

etc...

ArrayList myData = new ArrayList();

for(int i=0;i<sex.count;i++)
for(int j=0<j<smoke.count;j++)
for(int k=0;k<age.count;k++)
myData.Add(sex[i] + "," + smoke[j] + "," + age[k]);

for(int x=0;x<myData.Count;x++)
Console.Writeline(myData[x].ToString());

That's it.

If you want to know how many possible combinations there are it is simply multiplying each factor by the next criteria. That is: age.count * smoke.count * sex.count.

0

LVL 7

Expert Comment

ID: 9913689
this is in javascript solution but you can adopt it to your needs
<script language='javascript'>

var Gender = new Array();
Gender[0] = "Male";
Gender[1] = "Female";

var Smoke = new Array();
Smoke[0] = "Non-Smoker";
Smoke[1] = "Smoker";

var Age = new Array();
Age[0] = "21-34";
Age[1] = "34-99";
Age[2] = "35-99";
Age[3] = "45-99";

function fnPermutations()
{
for(a=0; a<Gender.length; a++)
{
for(b=0; b<Smoke.length; b++)
{
for(c=0; c<Age.length; c++)
{

document.write(Gender[a] + " , " + Smoke[b] + " , " + Age[c]);
document.write("<br>");
}
}
}
}

</script>
0

LVL 7

Expert Comment

ID: 9913690
oups I didn't know the solution was already posted
0

Author Comment

ID: 9913691
Excellent. Thank you very much _TAD_! You are truly a man among boys!
0

LVL 7

Expert Comment

ID: 9913697
< ---- slow
0

LVL 11

Expert Comment

ID: 9913706
i suppose the solution _TAD_ gave you works just fine, however with large sets it can be time consuming.

so just as an idea: if you put  the three(or more) kinds of atributes in 3(or more) database tables you will only have to do a 'cross join' because what you need to do is campute the cartesian product of the three sets of values.

A.
0

Author Comment

ID: 9913707
Thanks, TransBind. I think you just missed it. But an excellent answer nonetheless! Thanks to you both!
0

LVL 22

Expert Comment

ID: 9913852

Agarici>

That did actually cross my mind (using a cartesian join on database tables).  However, in my experience it is really just as quick to load several arrays as long as the data is not too complex and the amount of data is relatively trivial.

if henningero were to ultimately have more than... let's say 10,000 distinct permutaions then the solution i provided would work, but it would be extremely slow (perhaps 6 to 12 seconds of load time).

In a realistic setting, I don't foresee henningero's final data payload to be more than 50 or 100 distinct items (200 items on the high side).  Hence the quick and dirty, brute force method I provided.

but I like the way you think... efficiency is right near the top of the list when it comes to writing code

0

LVL 22

Expert Comment

ID: 9913858

Thanks for the points!
0

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