algorithm - Proof: Check if two integer arrays are permutations of each other using linear time and constant space -

i interested in creating simple array problem running time , space constraints. seems have found solution problem. please read initial description comment of problem in following java code:

 /*  * problem: given 2 integer arrays, , b, return whether array permutation of array b.  * running time complexity: o(n)  * space complexity: o(1)  * example 1:  * = [1, 2, -3]  * b = [2, 3, 1]  * false  * example 2:  * = [1, 2, 3]  * b = [0, 1]  * false  * example 3:  * = [2, 7, 3, 5]  * b = [5, 7, 2, 3]  * true  * example 4:  * = [1, -2, 10000]  * b = [10000, 1, -2]  * true  * example 5:  * = [1, 2, 2, 3]  * b = [3, 2, 1, 3]  * false  * example 6:  * = [2, 2, 4, 4]  * b = [3, 3, 3, 3]  * false  * example 7:  * = [4, 4, 2, 2, 4, 4]  * b = [4, 3, 3, 3, 3, 4]  * false  * ----------------------  * input 2 space separated lines of integers  * output true or false  * terminal example:  * 1 4 9 25  * 4 25 9 2  * false  * ----------------------  * solution:  * 1. average displacement (delta) between elements of array , array b equals 0  * ,  * 2. xor-ing of values between elements of array , array b equals 0  * ,  * 3. mins same , maxs same  * @author (david brewster)  * @version (27.01.2016) (requires java 1.8)  */  import java.util.scanner; import java.util.arrays; public class arrayprob {     public static int xorcomparison(int[] a, int[] b, int i, int xortotal)     {         return == a.length ?                 xortotal : xorcomparison(a, b, + 1, xortotal ^ a[i] ^ b[i]);     }     public static int deltacomparison(int[] a, int[] b, int i, int deltatotal)     {         return == a.length ?                 deltatotal : deltacomparison(a, b, + 1, deltatotal + a[i] - b[i]);     }     public static int mincomparison(int[] a, int[] b, int i, int amin, int bmin)     {         return == a.length ?                 amin-bmin : mincomparison(a, b, + 1,                 a[i] < amin ? a[i] : amin,                 b[i] < bmin ? b[i] : bmin);     }     public static int maxcomparison(int[] a, int[] b, int i, int amax, int bmax)     {         return == a.length ?                 amax-bmax : maxcomparison(a, b, i+1,                 a[i] > amax ? a[i] : amax,                 b[i] > bmax ? b[i] : bmax);     }     public static boolean arepermutations(int[] a, int[] b)     {         if (a.length == b.length)         {             boolean d = xorcomparison(a, b, 0, 0) == 0;             boolean e = deltacomparison(a, b, 0, 0) == 0;             boolean f = maxcomparison(a, b, 0, integer.min_value, integer.min_value) == 0;             boolean g = mincomparison(a, b, 0, integer.max_value, integer.max_value) == 0;             return d && e && f && g;         }         else         {             return false;         }     }     public static void main(string[] args)     {         scanner input = new scanner(system.in);         int[] = arrays.stream(input.nextline().split(" ")).maptoint(integer::parseint).toarray();         int[] b = arrays.stream(input.nextline().split(" ")).maptoint(integer::parseint).toarray();         system.out.println(arepermutations(a, b));     } } 

even thought algorithm seems work cases (at least have tested far), how go proving solution correct 100% of time?

what trying computing fixed size fingerprint of each array (representing multiset) , determining equality of multisets comparing fingerprints.

this not possible because there infinite number of multisets if space constant, there limited number of fingerprints. inevitably find examples 2 different multisets map same fingerprint.