PROBLEM LINK:Author: Lalit Kundu DIFFICULTY:SIMPLE PREREQUISITES:DP, bits PROBLEM:You are given an array A of N integers and an integer K. Your task is to return the maximum possible value of F(P) xor K, where P is a subset of A and F(P) is defined as a xor of all values in P. If P is empty, then F(P) = 0. QUICK EXPLANATION:Since each element of A has a value of at most 1000, we can use dynamic programming dp[i][j] := 1 if and only if there exists a subset P of A[1..i] such that F(P) = j. In order to get the result, we return max 1 <= j <= 1023 { dp[n][j] * (j ^ k) } EXPLANATION:Let dp[i][j] := 1 if and only if there exists a subset P of A[1..i] such that F(P) = j, otherwise 0 The first observation is that F(P) can be at most 1023 since any input number is at most 1000. Initially we set all dp[i][j] := 0. Next, we set dp[0][0] := 1, since a xor of the empty set is 0. We iterate over all elements of A from left to right. For each A[i], we iterate over all possible values of the xor function i.e. a range from 0 to 1023 inclusive and update these values as follows:
The reason for this is that if there exists a subset P of A[1..i  1] such that F(P) = j then there exists also a subset of A[1..i] such that F(P) = j or if there exists a subset P of A[1..i  1] such that F(P) = j ^ a[i], then F(P) ^ a[i] = j, so there exists a subset P' of A[1..i] such that F(P') = j At the end we have dp[n][j] for all 0 <= j <= 1023, and we can return a maximum value of dp[n][j] * (j ^ k) for all j. Time Complexity: The time complexity per one testcase is O(N * 1023); AUTHOR'S AND TESTER'S SOLUTIONS:To be uploaded soon. RELATED PROBLEMS:To be uploaded soon.
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asked 15 Dec '14, 19:41

I've managed to solve the problem without DP. I've used C++ STL's set and maintained all possible xors in it. Since all numbers are in [0, 1000], any xor of them will lie in [0, 1023]. So, set can have at max of 1024 elements at any point of time. So the complexity should be little less than O(n * 1024 + 1024 * log(1024)) where second term is for the inserts in the worst case. Here's my solution: http://www.codechef.com/viewsolution/5592821 I've also solved this using Guassian Elimination, you can view my code here: http://www.codechef.com/viewsolution/5616381 Also, the problem does not even need a 2 Dimensional DP, it can be made simpler. Here's my DP solution: http://www.codechef.com/viewsolution/5616433 answered 15 Dec '14, 19:58
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This man is a genius.
(20 Dec '14, 18:04)
@chaitan94 About your Gaussian Elimination soulution: http://www.codechef.com/viewsolution/5616381 The logic is correct. But there is small flaw in implementation. The above code will give wrong output for following test case
This is because of how STL set works. For correct result, just swap these two statements
(22 Dec '14, 22:07)
Where else can we use gaussian elimination technique other than this example and the traditional example of solving linear equation?
(24 Dec '14, 08:05)
@bhavesh_munot Gaussian elimination technique can also be used to solve a system of congruences.
(29 Dec '14, 23:24)
Oh! So the xor basis reduction trick is actually called gaussian elimination , I never knew that. I did the same for my solution, but it's not too popular huh. I haven't seen tutorials on XOR G.E., so it might be good to cook up a good tutorial on it in the future.
(19 Jun '17, 06:22)

I used Gaussian elimination to transform the input array A into the lower echelon form which is an equivalent representation of the above array (in which the array of bits answered 16 Dec '14, 10:44
@adijo i tried with the same Gaussian Elimination technique but my code failed for subtask 2....any suggestions on where i went wrong...http://www.codechef.com/viewsolution/5549953
(16 Dec '14, 15:59)
So your code can solve it for A < 2^30 for example? I have to check that ;) I was not able to find the solution for that...
(16 Dec '14, 16:46)
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@betlista I think it should work! @skysarthak Can you explain your algorithm? I might be able to point out flaws there if any. If not just reimplement the algorithm more carefully
(16 Dec '14, 23:02)
Thanx @betlista & @adijo for help......I followed the first answer given here with slight modification that instead of starting with result = 0 i started with result = k where k is from F(p) xor k...and while traversing the echelon form i xor the row with result if and only if the corresponding bit in result is not set while the one in the array is. After each comparison i jump to the next row and next column. If the no. of columns > no. of rows(bit representation of number is long) then the last row is xor'ed with the result if need be.
(17 Dec '14, 01:10)
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gaussian elimination for me too ! :) got AC http://www.codechef.com/viewsolution/5505732
(17 Dec '14, 01:12)
@skysarthak Follow the algorithm in the second answer (with 4 upvotes) After you transform the array to the echelon form don't worry about which bit is set etc, just select the value if the XOR increases, like so: max = k for value in array: if max ^ value > max: max = max ^ value print max
(17 Dec '14, 10:50)
I am doing the same thing here: http://www.codechef.com/viewsolution/5608801 I just changed the last for loop of my XMAX solution but it gives WA
(17 Dec '14, 11:47)
Also, what is then wrong with the following approach http://www.codechef.com/viewsolution/5527772 , it's doing the same thing by just dividing the input set in two different sets as explained in my answer below.
(17 Dec '14, 12:00)
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@damn_me,upvote my comment!!! U are correct,Your logics are absolutly correct. Just scratch your mind on your codes once again :( :( You are pushing everything into vector.How can it be possible??!!! vec.clear()!!!!
(17 Dec '14, 13:05)
@rudra_sarraf Thanks a lot, seems u needed an upvote :D But what is then wrong with this: http://www.codechef.com/viewsolution/5609487
(17 Dec '14, 14:58)
@damn_me,What I said was Just scratch your mind on your codes once again :( :( But u didn't!!! Declaring a variable twice gives error.(ll row=0)(Always try to write clean codes.) This time no more bullshit neither I need more upvotes than I deserve!!so, http://www.codechef.com/viewsolution/5609023 If you find clear proof of analogy of gauss elimination method=>Then kindly comment the link (other than mathematics stack exchange=>maximization of xor operator!!). At last but not the least=>actually (2 upvotes>1upvote)!!!
(17 Dec '14, 16:32)
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@rudra_sarraf I have already got AC on this logic. First open the code or link I have mentioned in my comment carefully and then say!! I think discussion forums are for help people want to do themselves and no one is forcing you to see someone's problems or errors. So better not use any more harsh words here!!!! And also, I was just joking for that upvote thing.
(17 Dec '14, 16:43)
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Once again=>Declaring a variable twice gives error.( you have declared ll row=0 twice!!) Is that harsh??? OOPS!! And do you really think that I was taking your joke seriously??!!(just go on thinking blah blah...) Just skip it...(Otherwise you may complain to admin!!!)
(18 Dec '14, 01:41)
Where else can we use gaussian elimination technique other than this example and the traditional example of solving linear equation?
(24 Dec '14, 08:06)
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I made a recursive function for solving this. Considering that the biggest number we can make is 1023 and in any case, we can always use an empty set to get the answer as K^0 = K, the biggest number that can be attained is 1023, and the smallest (to be checked) is K. So the answer lies in the range from K to 1023. To take the initial input, I made an array with size 1001, using which I could directly set array[i]=1 if i was present in the input, else the value of array[i] would be 0. One more thing I did was to go through this array and make a new array which only holds all the elements in descending order. Now we basically need to check for each number from 1023 to K+1, whether that number can somehow be made with the other numbers present in the set. The most obvious(and ultimately required) case would be if the number is directly present in the set. The basic idea is that if some number r is required, and isn't directly present in the set, take a number say p from the set, and again call the function for finding r^p in the remaining set. This forms the basic recursion. But that would give a TLE, so I needed some constraints. I was using the array in decreasing order, so the first number would be the greatest, from this, one simple constraint I was able to make was that if the required number's Most significant bit(MSB) is greater than the currently largest number's MSB, then there is no way to make the required number. Using all this I was able to get AC with time 0.00 in all but 1 case, which was giving TLE. For further refinement I used the concept that, considering a certain subset of a larger set, if a certain required number could not be made using the larger set, then it can definitely not be made using this smaller subset. With this I was finally able to get AC that one case too, with a time 0.02 Here is my solution: http://www.codechef.com/viewsolution/5597629 answered 16 Dec '14, 21:05

I would like to explain the Gaussian Elimination method which I have used and got AC in 0.00 :) The link to my solution. (sorry for some dubugging statements in it).
About my solution: Bucket[i] contains all ibit no.s the M chosen is first element of bucket i.e. Bucket[i][0] x[] is modified_array[]
link
This answer is marked "community wiki".
answered 19 Dec '14, 00:28

I tried doing this in the problem. This is the link:: http://math.stackexchange.com/questions/48682/maximizationwithxoroperator This is same as what @adijo said, I believe. But, the main idea behind this approach is setting one bit of each number one and then xoring with the numbers below in the series if they have that specific bit set, just to ensure no 2 numbers have the same MSB (most significant bit) as 1. If this is to follow, then since we were given a But, it gave me a wrong answer. :/ answered 16 Dec '14, 13:18
Don't take anything about k into consideration and apply the algorithm as stated by llmari Karonen on stackexchange. The only difference to be made is in maximum finding algorithm is to initialize max to k instead of 0.
(18 Dec '14, 21:52)

I thought something similar to the XMAX solution as mentioned above by @h1ashdr@gon . What i did was: divided the whole array in two subarrays, the first that doesn't have any of the bits set as of k and the other that have atleast one bit set. In my maximum, xor of all elements in the first set has a fair chance of giving me the maximum. My task was reduced to finding the maximum of ((xor of set1^k), (xor of set2^k), (set1^set2^k). But then I got stuck in finding the maximum for set2 which is nothing but one of the subproblem of the original question(I tired many approaches though). So, it was just DP i should have thought about or any other way out in this approach?? answered 16 Dec '14, 01:15

I have tried this problem and 9 testcases are not passed . This is my solution link : http://www.codechef.com/viewsolution/5518114 Could anyone help me out . answered 17 Dec '14, 09:04
