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Thread: Program Question
 01092011, 06:09 AM #1Member
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Program Question
I do not want to seem as if I just want an answer to a homework problem. But I just want to know what exactly I could do to write code to accomplish the task.
I am in a Java with Data Structures course and my professor gave me this:
Solve A/BC + D/EF + G/HI = 1 where A, B,C,... I are {1,2,3,...9} and each digit is used exactly once. Here BC, EF, HI are concatenations rather than multiplications. e.g. If B = 3 and C = 5, BC would be the number 35 (not the product 3x5 = 15). (The solution is unique up to ordering of the fractions.)
I would really appreciate some help.
Thanks,
Jim
 01092011, 07:01 AM #2Member
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this is what i have so far:
Java Code:public class StringProblem { public static void main(String[] args) { double a = 1; double d = 4; double g = 7; String b = "2"; String c = "3"; String e = "5"; String f = "6"; String h = "8"; String i = "9"; String first = b+c; String second = e+f; String third = h+i; double bc = Integer.parseInt(first); double ef = Integer.parseInt(second); double hi = Integer.parseInt(third); double answer = (a/bc) + (d/ef) + (g/hi); System.out.println(answer); } }
 01092011, 09:39 AM #3
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First let's get rid of those ugly divisions: your problem is equvalent to the problem: A*EF*HI+D*BC*HI+G*BC*EF == BC*EF*HI. Variables A .. I are a permutation of the values 1 ... 9. We have to find a permutation such that the equality holds. The equation makes sense in mathematical notation as:
A*(10*E+F)*(10*H+I)+D*(10*B+C)*(10*H+I)+G*(10*B+C) *(10*E+F) ==
(10*B+C)*(10*E+F)*(10*H+I)
(I hope I got that right ;)
Finding permutations isn't a trivial task but very well doable. The next method finds a next permutation (if there is one) given a current permutation in lexicographical order. Let a current permutation be P0 P1 P2 ... Pn;
the steps are:
Java Code:1) find a largest i such that Pi < P(i+1) 2) if no such i exists there is no next permutation 3) find a largest j such that j > i and Pj > Pi (such a value j always exists) 4) swap values Pi and Pj 5) reverse the subsequence P(i+1) .. Pn
You have to check all permutations starting at permutation 1, 2, 3 ... 9 and evaluate the equality above. If it holds you have found values for A ... I.
kind regards,
JosLast edited by JosAH; 01092011 at 10:54 AM.
cenosillicaphobia: the fear for an empty beer glass
 01122011, 11:16 PM #4Member
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I'm also having problems with this. I understand doing the A*EF*HI+D*BC*HI+G*BC*EF == BC*EF*HI to get ride of the divisions. Could you explain how A*(10*E+F)*(10*H+I)+D*(10*B+C)*(10*H+I)+G*(10*B+C) *(10*E+F) ==
(10*B+C)*(10*E+F)*(10*H+I) is equal to the other equation.
Also, the first step you said to do is find a largest i such that Pi < P(i+1). Does that mean I must increase all the variables by 1? How am I supposed to set up the first permutation? Do I just set A to 1, B to 2, and so on?
Sorry I'm very confused. I'd appreciate any help. Thank You.
 01132011, 08:02 AM #5
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If e.g. BC == 12 then B == 1 and C == 2; so mathemaTically 10*B+C == 10*1+2 == 12. The first notation is a symbolic notation, while the second is a methematical notation.
If you have a permutation, say, 14532, finding a largest i such that Pi < P(i+1) results in i == 1 because 4 < 5 and 4 is located at position 1. Next you have to find a largest j such that Pj > Pi. Pi == 4 and if j == 2 then Pj == 5 and 5 > 4. Swap Pi and Pj, so we get 15432 and next we have to reverse the subsequence starting at position i+1: 432, reversed we get 15234. Got it?
You do have to set up the first permutation yourself, which is easy: 123456789
kind regards,
Jos
edit: After rereading my own pseudo code I noticed that it was quite terse, so here is a Java implementation. This class permutes a char array with five elements:
Java Code:import java.util.Arrays; public class T { private static void swap(char[] s, int i, int j) { char t = s[i]; s[i] = s[j]; s[j] = t; } public static char[] permute(char[] s) { int i, j; for (i = s.length  1; i >= 0;) if (s[i] < s[i + 1]) break; if (i < 0) return null; for (j = s.length; j > i;) if (s[i] < s[j]) break; swap(s, i, j); for (j = s.length; ++i < j;) swap(s, i, j); return s; } public static void main(String[] args) { char[] names = "ABCDE".toCharArray(); do { System.out.println(Arrays.toString(names)); } while (permute(names) != null); } }
Last edited by JosAH; 01132011 at 08:57 AM.
cenosillicaphobia: the fear for an empty beer glass
 01132011, 06:09 PM #6Member
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