Word unscrambling

**pukebucket** · 01-07-2006

Ok. So, is the ctrie part the letter count thing?

Also, do you have msn? It would be easy to talk that way

And thanks again for the help

**treenef** · 01-07-2006

ok let's go back to basics:

Do you know how to loop through an array of characters?

So let's say I have an array

char array[100] = "treenef";

Code:

+--+--+--+--+--+--+--+
| t|r |e |e |n |e | f|
+--+--+--+--+--+--+--+
 
 0   1  2  3  4  5  6

to loop through each character I would do this:

Code:

for (int i=0; i<7; i++)
{
   cout<<array[i]<<endl;
}

**dwks** · 01-07-2006

Or better yet

Code:

#include <cstring>

size_t len = strlen(array);
for(size_t i = 0; i < len; i ++) {
    cout << array[i] << endl;
}

**dwks** · 01-07-2006

And then, instead of couting the characters, you could count them . . .

Code:

#include <climits>
#include <cstring>
#include <cctype>

unsigned charcount[UCHAR_MAX] = {0};
size_t len = strlen(array);

for(size_t i = 0; i < len; i ++) {
    charcount[array[i]] ++;
}

for(int x = 0; x < UCHAR_MAX; x ++) {
    if(charcount[x]) {
        cout << '\'' << isprint(x) ? x : '-' << "': " << charcount[x] << endl;
    }
}

**pukebucket** · 01-07-2006

Yes I understand that, It would print all 7 charactors stored in the array.

**treenef** · 01-07-2006

OK, that's good.

So the word treenef...

has three 'e's
one 'f'
one 't'

etc.

Would you know how to write a program to do that?

**dwks** · 01-07-2006

My first post would [edit]"print all 7 charectors stored in the array"[/edit], and treenef's one would too, but not my most recent one. My previous post would print something like

Code:

e: 3
f: 1
n: 1
r: 1
t: 1

**dwks** · 01-07-2006

Would you know how to write a program to do that?

Too late.

**pukebucket** · 01-07-2006

I see. No I don't think I could. My tutor is looking pretty bad right now. He taught us nothing like that.

To dwk: I'm not sure that I understand your code.

Sorry to be a pain people

**treenef** · 01-07-2006

I see...

Erm in regards to dwks code, it's probably a bit too complicated.

You need something more intuitive...

Have a look at this and see if you understand the concept behind it:

Code:

#include <iostream>
#include <string>

using namespace std;

int main()
{
    char array[100]="treenef";
    
    int size = strlen(array);
    
    char alphabet[50]="abcdefghijklmnopqrstuvwxyz";
    
    int letter_count[26];
    for(int i=0; i<26; i++)
    {
        letter_count[i]=0;
    }
        
    
    for(int i=0; i<size; i++)
    {
        for(int j=0; j<=26; j++)
        {
            if(array[i]==alphabet[j])
            {
                letter_count[j]++;
            }
        }
    }            
        
       for(int j=0; j<26; j++)
        {
            
          cout<<alphabet[j]<<" "<<letter_count[j]++<<endl;
            
        }
    
    cin.get();
    return 0;
}

**pukebucket** · 01-07-2006

Ok I'll try to explain your code to show how much I understand.

Code:

char array[100]="treenef"; //Declares a charactor array called array,
 of 100 bytes and saves treenaf into it
    
    int size = strlen(array); //Declares an integer called size,
 Then it assigns its value as the length of the array called array
    
    char alphabet[50]="abcdefghijklmnopqrstuvwxyz"; // Declares
 a charactor array called alphabet, of 50 bytes. Then it saves the aphabet into it.
    
    int letter_count[26]; // Declares an integer called letter_count, consisting of 26 bytes.

    for(int i=0; i<26; i++) // A loop. It creates an integer called i 
and assigns it a value of 0. It then says that as long as i is under 
26 to run the next code. The i++ makes the integer i increase by 1 each loop.

    {
        letter_count[i]=0; // Im not sure. It looks like it assigns each

 letter of the alphabet to 0?
    }
        
    
    for(int i=0; i<size; i++) //Another loop. Assigns integer i to 0, 
loops if i is under the variable size, and i increases by 1 each loop.


    {
        for(int j=0; j<=26; j++) //Another loop. Sets j to 0, loops if j 
is bigger or equal to 26, j increases by 1 each loop

        {
            if(array[i]==alphabet[j]) //Hrm. Another loop, if value i 
array is eqaul  to alphabet value j then do the next code.

            {
                letter_count[j]++; //Not sure
            }
        }
    }            
        
       for(int j=0; j<26; j++) //Same as the one above
        {
            
          cout<<alphabet[j]<<" "<<letter_count[j]++<<endl; // Print 
the alphabet value j, a space, then the letter_count j. Not sure 
what this means.
            
        }
    
    cin.get();
    return 0;
}

Hrm, I understand quite a bit (I think, and hope) but theres still a
bit I don't quite get. Sorry about the way its layed out :S

**treenef** · 01-07-2006

Yeah some of what u got is right some wrong.

Draw out a flow chart to and follow each bit with your finger.

Think about it for a day or so and then come back. I'll explain any stuff you don't understand

I've got to go now.

However, you should be able to see the reason why a letter frequency count is useful for your problem.

Code:

treenef,  feneert,  eertnef,

All have the same letter frequency count.

**pukebucket** · 01-07-2006

Ok thanks. I understand why I need it now. Thanks for all the help. If I need more I'll come and ask, as for now, I got some learning to do!

Thanks and bye people.
Thanks alot treenef, your help has been invaluable, so was everyone elses.

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