Statistics: Posted by Bisser — Fri Nov 01, 2019 8:14 am
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http://www.dcs.shef.ac.uk/research/ilas ... words.html
I may even be able to find my Qbasic program that reads it. The program was actually to do those competitions of the type: - how many words can you make from "petes qb site is best".
Statistics: Posted by Quibbler — Thu Aug 17, 2006 9:45 am
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Non native english person speaks good english if he knows 500 to 2K words.
Person who speaks english as hes/her home language, knows 2K to 10K words.
Here is first 850 http://www.langmaker.com/wordlist/basiclex.htm
Alltho, if i do remember right, ATM, english dictionary knows 200K words
Statistics: Posted by bungytheworm — Thu Aug 17, 2006 1:42 am
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Anagram solver? Well for a N letter word there are N! permutations - far too many. I have written a similar program and the best approach is to search a dictionary file.
First check is the length, then if the anagram and a dictionary word have the same length use INSTR to check the letters are all there.
Interesting approach, but where can you get such a dictionary file? To be used by a QB program, it would also have to be a plain ASCII text file.First check is the length, then if the anagram and a dictionary word have the same length use INSTR to check the letters are all there.
*****
Statistics: Posted by moneo — Wed Aug 16, 2006 6:36 pm
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Anagram solver? Well for a N letter word there are N! permutations - far too many. I have written a similar program and the best approach is to search a dictionary file.
First check is the length, then if the anagram and a dictionary word have the same length use INSTR to check the letters are all there.
Although, if you used INSTR to find the letters, you would come up with duplicates... (like moneo said) You could remove letters from each string and still open your dictionary file.First check is the length, then if the anagram and a dictionary word have the same length use INSTR to check the letters are all there.
Code:
EXAMPLE: NOT CODE!FREE and REEF are being tested.FREE - REEF (F is scanned and removed)_REE - REE_ (R is scanned and removed)__EE - _EE_ (One of the E's is scanned and removed)___E - __E_ (The other E is scanned and removed)____ - ____ (There are no more letters, the scan was a success)Statistics: Posted by Patz QuickBASIC Creations — Wed Aug 16, 2006 4:25 pm
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Statistics: Posted by Quibbler — Wed Aug 16, 2006 9:05 am
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I'm not an expert on this, but the numbers 123,132, etc. are a perfect set of the permutations of 3 numbers.
Since you didn't show us the PRINT statement, I don't know why you're printing the numbers of the letter positions instead of the letters themselves. My guess is that in order to print both the numbers and their corresponding letter, you probably need to keep 2 arrays.
The big problem that you will encounter later is when the input word contains duplicate letters, like HELLO which has 2 L's. The duplicate letters will generate duplicate permutations which you'll need to detect and eliminate. When you get to this part, let me know. I have an algorithm for this.
Good luck!
Regards..... Moneo
Statistics: Posted by moneo — Wed Aug 16, 2006 12:06 am
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[1-2-3] = [r-t-a]
[1-3-2] = [r-a-t]
[2-1-3] = [t-r-a]
[2-3-1] = [t-a-r]
[3-1-2] = [a-r-t]
[3-2-1] = [a-t-r]
grtz
Statistics: Posted by Seb McClouth — Tue Aug 15, 2006 1:14 am
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I just wake up so im not 100% sure what youre after.
Code:
DIM AS String Word, Letters()DIM AS Integer CounterLINE INPUT "Give in word: ";Word REDIM Letters(LEN(Word))PRINT "Wordlenght is "; LEN(word)FOR Counter = 1 to LEN(Word) Letters(Counter) = MID$(Word, Counter, 1)NEXTStatistics: Posted by bungytheworm — Tue Aug 15, 2006 1:10 am
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Currently I can only use up to 3 letters, anyone to step in and help me out?
Code:
DIM SHARED letters(9) AS STRINGLINE INPUT "Give in word: ", word$PRINT "Wordlenght is "; LEN(word$)FOR a = 1 to LEN(word$) letters(a) = MID$(word$, a, 1)NEXT[1-2-3]
[1-3-2]
[2-1-3]
[2-3-1]
[3-1-2]
[3-2-1]
grtz
Edited.
Statistics: Posted by Seb McClouth — Tue Aug 15, 2006 12:47 am
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