By DrSwad, history, 19 months ago,

## Inspiration

I'm very excited about this blog, as it took me quite a lot of effort and scavenging through the internet to completely grasp the concept of this technique(That's probably because I have almost zero knowledge in Linear Algebra or I'm just plain dumb). So I feel like I genuinely conquered a challenge, and I really want to share it with someone. But there's no way my CP friends circle will believe it, they'll think I'm just trying to show off :P

So here I am, sharing it on CF. I also created a personal blog, so that if I ever feel like sharing something again(not only about CP), I can write a blog there. I also added this same post there, you can read it there if you prefer dark theme. I'll be pleased to hear any thoughts on the blog or if I can improve it in some way ^_^

## Introduction

Since it concerns Linear Algebra, there needs to be a lot of formal stuff going on in the background. But, I'm too much inconfident on this subject to dare go much deep. So, whenever possible, I'll try to explain everything in intuitive and plain English words. Also, this blog might take a while to be read through completely, as there are quite a few observations to grasp, and the example problems aren't that easy either. So please be patient and try to go through it all, in several sits if needed. I believe it'll be worth the time. In any case, I've written the solutions, codes, and provided links to their editorials(if available). I'll provide more details in the solutions tomorrow and put more comments in the codes, since I'm really tired from writing this blog all day.

Now, the problems that can be solved using this technique are actually not much hard to identify. The most common scenario involves: you'll be given an array of numbers, and then the problem asks for an answer by considering all the xor-sums of the numbers in all possible subsets of the array. This technique can also be used in some online-query problems: the problem can provide queries of first type instructing you to insert numbers in the array(_without removal_, I don't know how to solve with deletion of elements) and in-between those queries, asking for answers in separate queries of second type.

The whole technique can be divided into two main parts, some problems can even be solved by using only the first part(Don't worry if you don't understand them completely now, I explain them in details right below): 1. Represent each given number in it's binary form and consider it as a vector in the $\mathbb{Z}_2^d$ vector space, where $d$ is the maximum possible number of bits. Then, xor of some of these numbers is equivalent to addition of the corresponding vectors in the vector space. 2. Somehow, relate the answer to the queries of second type with the basis of the vectors found in Part 1.

PS: Does anyone know any name for this technique? I'm feeling awkward referring to it as 'technique' this many times :P If it's not named yet, how about we name it something?

## Part 1: Relating XOR with Vector Addition in $\mathbb{Z}_2^d$

Let me explain the idea in plain English first, then we'll see what the $\mathbb{Z}_2^d$ and vector space means. I'm sure most of you have already made this observation by yourselves at some point.

Suppose, we're xor-ing the two numbers $2$ and $3.$ Let's do it below:

$\begin{equation*}\begin{array}{r} (10)_2\\ \underline{\oplus\;(11)_2}\\ (01)_2 \end{array}\end{equation*}$

Now, for each corresponding pair of bits in the two numbers, compare the result of their xor with the result of their sum taken modulo $2$:

Bit no. First number Second number $\oplus$ Sum Sum taken $\pmod 2$
$1$st bit $0$ $1$ $1$ $1$ $1$
$2$nd bit $1$ $1$ $0$ $2$ $0$

Notice the similarity between columns $4$ and $6$? So, we can see that taking xor between two numbers is essentially the same as, for each bit positions separately, taking the sum of the two corresponding bits in the two numbers modulo $2.$

Now, consider a cartesian plane with integer coordinates, where the coordinate values can only be $0$ or $1.$ If any of the coordinates, exceeds $1,$ or goes below $0,$ we simply take it's value modulo $2.$

This way, there can only be $4$ points in this plane: $(0, 0), (0, 1), (1, 0), (1, 1).$ Writing any other pair of coordinates will refer to one of them in the end, for example, point $(3, 2)$ is the same point as point $(1, 0)$ since $3 \equiv 1$ and $2 \equiv 0$ modulo $2.$

In view of this plane, we can represent the number $2 = (10)_2$ as the point $(0, 1),$ by setting the first bit of $2$ as the $x$ coordinate and the second bit as the $y$ coordinate in our plane. Refer to this point as $P(0, 1).$ Then, the position vector of $2$ will be $\overrightarrow{OP}$ where $O(0, 0)$ is the origin. Similarly, the position vector of $3$ will be $\overrightarrow{OQ}$ where $Q = (1, 1).$

An interesting thing happens here, if we add the two position vectors, the corresponding coordinates get added modulo $2,$ which actually gives us the position vector of the xor of these two position vectors. For example, adding vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ we get $\overrightarrow{OR}$ where $R(1, 0)$ turns out to be the point corresponding the xor of $2$ and $3.$

This is all there is to it. Transforming xor operations to bitwise addition modulo $2$ and, in some cases, vector addition in this way can be helpful in some problems. Let's see one such problem. Before that, let me explain in short what vector space and $\mathbb{Z}_2^b$ meant earlier. I apologize to any Linear Algebra fans, since I don't want to write formal definitions here to make things look harder than it is. I'll explain the idea of these terms the way I find them in my mind, kindly pardon me for any mistakes and correct me if I'm wrong.

$\underline{\text{Vector Space}}$: Just a collection of vectors.

$\underline{\mathbb{Z_2}}$: $\mathbb{Z_m}$ is the set of remainders upon division by $m.$ So, $\mathbb{Z_2}$ is simply the set $\{0, 1\},$ since these are the only remainders possible when taken modulo $2.$

$\underline{\mathbb{Z_2^d}}$: A $d-$dimensional vector space consisting of all the different position vectors that consists of $d$ coordinates, all coordinates being elements of $\mathbb{Z_2}.$ For example, earlier our custom cartesian plane was a two-dimensional one. So, it was $\mathbb{Z_2^2}.$ $\mathbb{Z_2^3}$ would be a small $3d-$plane with only $2^3 = 8$ points, all coordinates taken modulo $2.$

So, what we've seen is that the xor-operation is equivalent to vector addition in a $\mathbb{Z}_2^d$ vector space. See how unnecessarily intimidating this simple idea sounds when written in formal math!

Anyways, the problem:

### Problem 1 (Division 2 — C)

Find the number of non-empty subsets, modulo $10^9 + 7,$ of a given set of size $1 \le n \le 10^5$ with range of elements $1 \le a_i \le 70,$ such that the product of it's elements is a square number.

If you'd like to solve the problem first, then kindly pause and try it before reading on further.

Solution

#### Solution

It's obvious that our solution will build on the constraint on $a_i,$ which is just $70.$

For a number to be square, each of it's prime divisors must have an even exponent in the prime factorization of the number. There are only $19$ primes upto $70.$ So, we can assign a mask of $19$ bits to each array element, denoting if the $i$'th prime occurs odd or even number of times in it by the $i$'th bit of the mask.

So, the problem just boils down to finding out the number of non-empty subsets of this array for which the xor-sum of it's elements' masks will be $0.$

I got stuck here for quite a while ;-; We can try to use dynamic programming, $\text{dp[at][msk]}$ states the number of subsets in $\{a_1, a_2, \ldots, a_{\text{at}}\}$ such that the xor-sum of it's elements' masks is $\text{msk}.$ Then,

$\text{dp[at][msk] = dp[at - 1][msk] + dp[at - 1][msk ^ mask[at]]}$

with the initial value $\text{dp[0][0] = 1}.$ The final answer would be, $\text{dp[n][0]}.$

But, the complexity is $O(n \cdot 2^{19}),$ which is way too high :(

The thing to notice here, is that, even if $n \le 10^5,$ the actual number of different possible $a_i$ is just $70.$ So, if we find the dp for these $70$ different masks, and if for each $1 \le \text{at} \le 70$ know the number of ways to select odd/even number of array elements with value $\text{at},$ then we can easily count the answer with the following dp:

$\text{dp[at][msk] = dp[at - 1][msk] * poss[at][0] + dp[at - 1][msk ^ mask[at]] * poss[at][1]}$

where, $\text{poss[at][0]}$ is the number of ways to select even number of array elements with value $\text{at},$ and similarly $\text{poss[at][1]}$ for odd number of elements.

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;
const int MAX_A = 70;
const int TOTAL_PRIMES = 19;
const int MOD = 1e9 + 7;

int n;
int poss[MAX_A + 1][2];
const int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67};
int dp[MAX_A + 1][1 << TOTAL_PRIMES];

int main() {
cin >> n;

for (int i = 1; i <= MAX_A; i++) poss[i][0] = 1;

for (int i = 1; i <= n; i++) {
int a;
scanf("%d", &a);

int tmp = poss[a][0];
poss[a][0] = (poss[a][0] + poss[a][1]) % MOD;
poss[a][1] = (poss[a][1] + tmp) % MOD;
}

for (int i = 1; i <= MAX_A; i++) {
for (int p = 0; p < TOTAL_PRIMES; p++) {
int cnt = 0;
int at = i;

while (at % primes[p] == 0) {
at /= primes[p];
cnt++;
}

if (cnt & 1) mask[i] |= (1 << p);
}
}

int max_mask = 1 << TOTAL_PRIMES;
dp[0][0] = 1;

for (int at = 1; at <= MAX_A; at++)
for (int msk = 0; msk < max_mask; msk++) {
dp[at][msk] = dp[at - 1][msk] * 1LL * poss[at][0] % MOD;

dp[at][msk] += dp[at - 1][msk ^ mask[at]] * 1LL * poss[at][1] % MOD;
dp[at][msk] %= MOD;
}

cout << (dp[MAX_A][0] + MOD - 1) % MOD << endl;

return 0;
}


Since the number of different possible masks were just $70$ in the previous problem, we had been able to use dynamic programming for checking all possible xors. But what if the constraint was much bigger, say $10^5.$ That is when we can use Part $2$ of this technique, which, in some cases, works even when the queries are online.

## Part 2: Bringing in Vector Basis

We need a couple of definitions now to move forward. All the vectors mentioned in what follows, exclude null vectors. I sincerely apologize for being so informal with these definitions.

$\underline{\text{Independent Vectors:}}$ A set of vectors $\vec{v_1}, \vec{v_2}, \ldots, \vec{v_n}$ is called independent, if none of them can be written as the sum of a linear combination of the rest.

$\underline{\text{Basis of a Vector Space:}}$ A set of vectors is called a basis of a vector space, if all of the element vectors of that space can be written uniquely as the sum of a linear combination of elements of that basis.

A few important properties of independent vectors and vector basis that we will need later on(I find these pretty intuitive, so I didn't bother with reading any formal proofs. Let me know in the comments if you need any help):

1. For a set of independent vectors, we can change any of these vectors by adding to it any linear combination of all of them, and the vectors will still stay independent. What's more fascinating is that, the set of vectors in the space representable by some linear combination of this independent set stays exactly the same after the change.

2. Notice that, in case of $\mathbb{Z}_2^d$ vector space, the coefficients in the linear combination of vectors must also lie in $\mathbb{Z}_2.$ Which means that, an element vector can either stay or not stay in a linear combination, there's no in-between.

3. The basis is actually the smallest sized set such that all other vectors in the vector space are representable by a linear combination of just the element vectors of that set.

4. The basis vectors are independent.

5. For any set with smaller number of independent vectors than the basis, not all of the vectors in the space will be representable.

6. And there cannot possibly be larger number of independent vectors than basis in a set. If $d$ is the size of the basis of a vector space, then the moment you have $d$ independent vectors in a set, it becomes a basis. You cannot add another vector into it, since that new vector is actually representable using the basis.

7. For a $d-$dimensional vector space, it's basis can have at most $d$ vector elements.

With just these few properties, we can experience some awesome solutions to a few hard problems. But first, we need to see how we can efficiently find the basis of a vector space of $n$ vectors, where each vector is an element of $\mathbb{Z}_2^d.$ The algorithm is quite awesome <3 And it works in $O(n \cdot d).$

### The Algorithm:

This algorithm extensively uses properties $1, 2, 3$ and $4,$ and also the rest in the background. All the vectors here belong to $\mathbb{Z}_2^d,$ so they are representable by a bitmask of length $d.$
Suppose at each step, we're taking an input vector $\vec{v_i}$ and we already have a basis of the previously taken vectors $\vec{v_1}, \vec{v_2}, \ldots, \vec{v_{i - 1}},$ and now we need to update the basis such that it can also represent the new vector $\vec{v_i}.$
In order to do that, we first need to check whether $\vec{v_i}$ is representable using our current basis or not.
If it is, then this basis is still enough and we don't need to do anything. But if it's not, then we just add this vector $vec{v_i}$ to the set of basis.
So the only difficuly that remains is, to efficiently check whether the new vector is representable by the basis or not. In order to facilitate this purpose, we use property $1$ to slightly modify any new vectors before inserting it in the basis, being careful not to break down the basis. This way, we can have more control over the form of our basis vectors. So here's the plan:
Let, $f(\vec{v})$ be the first position in the vector's binary representation, where the bit is set. We make sure that all the basis vectors each have a different $f$ value.
Here's how we do it. Initially, there are no vectors in the basis, so we're fine, there are no $f$ values to collide with each other. Now, suppose we're at the $i$'th step, and we're checking if vector $\vec{v_i}$ is representable by the basis or not. Since, all of our basis have a different $f$ value, take the one with the least $f$ value among them, let's call this basis vector $\vec{b_1}.$
If $f(\vec{v_i}) < f(\vec{b_1})$ then no matter how we take the linear combination, by property $2,$ no linear combination of the basis vectors' can have $1$ at position $f(\vec{v_i}).$ So, $\vec{v_i}$ will be a new basis vector, and since it's $f$ value is already different from the rest of the basis vectors, we can insert it into the set as it is and keep a record of it's $f$ value.
But, if $f(\vec{v_i}) == f(\vec{b_1}),$ then we must subtract $\vec{b_1}$ from $\vec{v_i}$ if we want to represent $\vec{v_i}$ as a linear combination of the basis vectors, since no other basis vector has bit $1$ at position $f(\vec{v_i}) = f(\vec{b_1}).$ So, we subtract $\vec{b_1}$ from $\vec{v_i}$ and move on to $\vec{b_2}.$
Note that, by changing the value of $\vec{v_i}$ we're not causing any problem according to property $1.$ $\vec{v_i}$ and $\vec{v_i} - \vec{b_1}$ is of same use to us. If in some later step we find out $\vec{v_i}$ is actually not representable by the current basis, we can still just insert it's changed value in the basis, since the set of vectors in the space representable by this new basis would've been the same if we inserted the original $\vec{v_i}$ instead.
If, after iterating through all the basis vector $\vec{b}$'s and subtracting them from $\vec{v_i}$ if needed, we still find out that $\vec{v_i}$ is not null vector, it means that the new changed $\vec{v_i}$ has a larger value of $f$ than all other basis vectors. So we have to insert it into the basis and keep a record of it's $f$ value.

Here's the implementation, the vectors being represented by bitmasks of length $d$:

int basis[d]; // basis[i] keeps the mask of the vector whose f value is i

int sz; // Current size of the basis

for (int i = 0; i < d; i++) {
if ((mask & 1 << i) == 0) continue; // continue if i != f(mask)

if (!basis[i]) { // If there is no basis vector with the i'th bit set, then insert this vector into the basis
++sz;

return;
}

mask ^= basis[i]; // Otherwise subtract the basis vector from this vector
}
}


Let's view some problems now:

### Problem 2a

Given a set $S$ of size $1 \le n \le 10^5$ with elements $0 \le a_i \lt 2^{20}.$ Find the number of distinct integers that can be represented using xor over the set of the given elements.

Solution

#### Solution

Think of each element as a vector of dimension $d = 20.$ Then the vector space is $\mathbb{Z}_2^{20}.$ We can find it's basis in $O(d \cdot n).$ For any linear combination of the basis vectors, we get a different possible xor of some subset. So, the answer would be $2^\text{size of basis}.$ It would fit in an integer type, since size of basis $\le d = 20$ by property $7.$

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10, LOG_A = 20;

int basis[LOG_A];

int sz;

for (int i = 0; i < LOG_A; i++) {
if ((mask & 1 << i) == 0) continue;

if (!basis[i]) {
++sz;

return;
}

}
}

int main() {
int n;
cin >> n;

while (n--) {
int a;
scanf("%d", &a);

insertVector(a);
}

cout << (1 << sz) << endl;

return 0;
}


### Problem 2b

We have a graph of $2^{k}$ nodes numbered from $0$ to $2^{k} - 1,$ $1 \le k \le 30.$ Also, we're given $1 \le M \le 10^5$ integers $x_1, x_2, \ldots, x_M$ within the range $0 \le x_i \le 2^{k} - 1.$ In the graph, two vertices $u$ and $v$ are connected with an edge iff $u \oplus v = x_i$ for some $i.$ Find the number of connected components in the graph.
Link to the editorial and reference code

### Problem 3

Given a set $S$ of size $1 \le n \le 10^5$ with elements $0 \le a_i \lt 2^{20}.$ What is the maximum possible xor of the elements of some subset of $S?$

Solution

#### Solution

In this problem, we need to slightly alter the definition of $f(\vec{b}).$ Instead of $f$ being the first position with a set bit, let it be the last position with a set bit.

Now, to get the maximum, we initialize our answer at 0 and we start iterating the basis vectors starting with the one that has the highest value of $f.$

Suppose, we're at basis vector $\vec{b}$ and we find that answer doesn't have the $f(\vec{b})$'th bit set, then we add $\vec{b}$ with answer. This greedy solution works because $f(\vec{b})$ is the most significant bit at the moment, and we must set it; doesn't matter if all the following bits turn to $0.$

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10, LOG_A = 20;

int basis[LOG_A];

for (int i = LOG_A - 1; i >= 0; i--) {
if ((mask & 1 << i) == 0) continue;

if (!basis[i]) {
return;
}

}
}

int main() {
int n;
cin >> n;

while (n--) {
int a;
scanf("%d", &a);

insertVector(a);
}

int ans = 0;

for (int i = LOG_A - 1; i >= 0; i--) {
if (!basis[i]) continue;

if (ans & 1 << i) continue;

ans ^= basis[i];
}

cout << ans << endl;

return 0;
}


### Problem 4 (1st Hunger Games — S)

We have an empty set $S$ and we are to do $1 \le n \le 10^6$ queries on it. Let, $X$ denote the set of all possible xor-sums of elements from a subset of $S.$ There are two types of queries.
Type $1$: Insert an element $1 \le k \le 10^9$ to the set(If it's already in the set, do nothing)
Type $2$: Given $k,$ print the $k$'th hightest number from $X.$ It's guaranteed that $k \le \mid X \mid.$ Link to the source

Solution

#### Solution

A bit like the previous one. For query type $2,$ again we'll iterate through the basis vectors according to their decreading order of $f$ values.
Suppose $\vec{b_h}$ is the one with the hightest $f$ value. Initially we know there are $2^\text{basis size}$ elements in $X.$ So, if $k <= \frac{2^\text{basis size}}{2},$ we set the $f(\vec{b_h})$'th bit of answer to $0.$ Otherwise we set it to $1$ and subtract $\frac{2^\text{basis size}}{2}$ from $k.$ Then we move on to the next basis vector and continue. In the end $k$ will be $1$ and we'll get our answer by setting $0$ in answer for all $f(\vec{b_i})$'th bits from that point forward.

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 10, LOG_K = 30;

int basis[LOG_K], sz;

for (int i = LOG_K - 1; i >= 0; i--) {
if ((mask & 1 << i) == 0) continue;

if (!basis[i]) {
sz++;

return;
}

}
}

int query(int k) {

int tot = 1 << sz;
for (int i = LOG_K - 1; i >= 0; i--)
if (basis[i]) {
int low = tot / 2;

if ((low < k && (mask & 1 << i) == 0) ||
(low >= k && (mask & 1 << i) > 0)) mask ^= basis[i];

if (low < k) k -= low;

tot /= 2;
}

}

int main() {
int n;
cin >> n;

while (n--) {
int t, k;
scanf("%d %d", &t, &k);

if (t == 1) insertVector(k);
else printf("%d\n", query(k));
}

return 0;
}


### Problem 5 (Division 2 — F)

You're given an array $0 \le a_i \lt 2^{20}$ of length $1 \le n \le 10^5.$ You have to answer $1 \le q \le 10^5$ queries.
In each query you'll be given two integers $1 \le l \le n$ and $0 \le x \lt 2^{20}.$ Find the number of subsequences of the first $l$ elements of this array, modulo $10^9 + 7,$ such that their bitwise-xor sum is $x.$

Solution

#### Solution

We can answer the queries online. Iterate through the prefix of the array, and for each prefix remember the basis vectors of that prefix.
Then, iterate through the queries. To answer a query, we check if $x$ is actually representable by the prefix of $l$ elements or not, with slight modification to the insertVector function(We don't need to add $x,$ just check if it's representable or not).
If it's not representable, then the answer to the query is $0.$ If it is representable, then the answer will be $2^(l - b),$ where $b$ is the basis size for the first $l$ elements. It is so, because for each subset of the $(l - b)$ non-basis vectors in the prefix, we find a unique linear combination to yield xor-sum $x.$

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> ii;
#define x first
#define y second

const int N = 1e5 + 10;
const int LOG_A = 20;
const int MOD = 1e9 + 7;

int n;
int a[N];

int q;
ii q_data[N];
vector<int> q_at[N];
int powers[N];
int ans[N];

int base[LOG_A], sz;

for (int i = 0; i < LOG_A; i++) {
if ((mask & 1 << i) == 0) continue;

if (!base[i]) return false;

}

return true;
}

for (int i = 0; i < LOG_A; i++) {
if ((mask & 1 << i) == 0) continue;

if (!base[i]) {
sz++;

return;
}

}
}

int main() {
cin >> n >> q;

for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

for (int i = 1; i <= q; i++) {
scanf("%d %d", &q_data[i].x, &q_data[i].y);
q_at[q_data[i].x].push_back(i);
}

powers[0] = 1;
for (int i = 1; i < N; i++)
powers[i] = powers[i - 1] * 2LL % MOD;

for (int at = 1; at <= n; at++) {
insertVector(a[at]);

for (int at_q : q_at[at])
if (checkXor(q_data[at_q].y)) {
ans[at_q] = powers[at - sz];
}
}

for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);

return 0;
}


### Problem 6 (Education Round — G)

You are given an array $0 \le a_i \le 10^9$ of $1 \le n \le 2 \cdot 10^5$ integers. You have to find the maximum number of segments this array can be partitioned into, such that -
1. Each element is contained in exactly one segment
2. Each segment contains at least one element
3. There doesn't exist a non-empty subset of segments such that bitwise-xor of the numbers from them is equal to $0$
Print $-1$ if no suitable partition exists.

Solution

#### Solution

Notice that, saying all subsets of a set yeild non-zero xor is equivalent to saying all subsets of that set yeild different xor-sum. The the xor-sums of segments in the answer partition need to be independent vectors. This is the first of the two main observations.

The second one is that, suppose we picked some segments $[l_1 = 1, r_1], [l_2 = r_1 + 1, r_2], \ldots, [l_k = r_{k - 1} + 1, r_k].$ Let, $p_i$ be the xor of the xor-sums of the first $i$ segments. Then, observe that, every possible xor of the numbers from some non-empty subset of these segments can also be obtained by xor-ing some subset from the set $\{p_1, p_2, ldots, p_k\}$ and vice versa. Which means that the set of xor-sums of these segments and the set of prefix xors of these segments produces the exact same set of vectors in $\mathbb{Z}_2^{31}.$ So, if the xor-sums of these segments has to be independent, then so does the prefix xors of these segments. Thus, the answer simply equals the basis size of the n prefix xors of the array. The only exception when the answer equals $-1$ happens, when the xor-sum of all the elements in the array is $0.$

I'll write this solution in more detail tomorrow. I'm half asleep right now.

#### Reference Code

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10, LOG_PREF = 31;

int n;
int basis[LOG_PREF];

for (int i = 0; i < LOG_PREF; i++) {
if ((mask & 1 << i) == 0) continue;

if (!basis[i]) {
return;
}

}
}

int main() {
cin >> n;

int pref = 0;

for (int i = 1; i <= n; i++) {
int a;
scanf("%d", &a);

pref ^= a;
insertVector(pref);
}

if (pref == 0) {
cout << -1 << endl;
return 0;
}

int ans = 0;

for (int i = 0; i < LOG_PREF; i++) {
ans += (basis[i] > 0);
}

cout << ans << endl;

return 0;
}


## Conclusion

This is my first take on writing tutorial blogs on CF. I hope it'll be of use to the community.

I apologize for my terrible Linear Algebra knowledge. I would write this blog without using any of it if I could. I don't want to spread any misinformation. So please let me know in comments if you find any mistakes/wrong usage of notations.

I plan to write on Hungarian Algorithm next. There's just so many prerequisites to this algorithm. It'll be an enjoyable challenge to write about. I'd be glad if you can provide me some resource links in the comments to learn it from, though I already have quite a few.

## References:

• +425

 » 19 months ago, # |   +31 Wow
 » 19 months ago, # |   0 I saw some coder using this technique, what is the name of this technique?
•  » » 19 months ago, # ^ |   0 I think some people refer to it as Gaussian Elimination modulo $2$ or bitwise Gaussian Elimination, since we use a similar idea to efficiently find a basis of $\mathbb{Z}_2^d$ vector space. But, personally, I don't like it much. Because, Gaussian Elimination is not the main theme of this technique, it's the use of xor->vector space transition and then finding a basis of that vector space.If it really doesn't have a common name yet, how does xor-basis sound? Simple and easy to remember. Also perfectly resembles the idea of this technique.
•  » » » 19 months ago, # ^ |   0 I usually consider it as either — Addition of vectors over GF(2)/Z(2) Or just system of linear equations mod 2.
•  » » » 19 months ago, # ^ |   +2 To be very particular about the term, Gaussian elimination is the very concept of having different pivots ( leftmost $1$ ) for each row in a matrix. The main highlight here is that itself, "Let, $f(v)$ be the first position in the vector's binary representation, where the bit is set. We make sure that all the basis vectors each have a different $f$ value."But this is all just language, apart from that, thanks a lot. I never tried to learn Gaussian elim because it looked tough ( saw long codes everywhere, and couldn't understand without the correlation with Linear Algebra ). But now I feel it's so easy.
•  » » » 11 months ago, # ^ |   0 Isn't this really just a weird variant of the Gram-Schmidt procedure to find an orthogonal basis of a subspace?
•  » » » » 9 months ago, # ^ | ← Rev. 3 →   0 There is a $\mathbb{Z}_2$ analogue of the Gram-Schmidt orthogonalization procedure (using the least significant bit of popcnt(x & y) as a $\mathbb{Z}_2$ analogue of the inner product) to find a basis, but that isn't described here. Over the real numbers this variant of the Gaussian elimination process can be represented as a limit of Gram-Schmidt processes with (non-standard) inner products increasingly skewed to weigh the first coordinate more and more heavily, but I'm not aware of any way to (even "formally") take such a limit over $\mathbb{Z}_2$. (Of course, one could still use a relative of $\mathbb{Z}_2$ dual numbers to justify the analogy, but I think that seems rather artificial.)EDIT: Oops! I clearly misremembered something. This pseudo-inner-product is unsuitable for use with the Gram-Schmidt process over $\mathbb{Z}_2^n$, as it is degenerate, lacking positive-definiteness for $n>1$. That type of construction can still be useful in discrete linear algebra, but not in this context.
•  » » 19 months ago, # ^ |   +37 I'd just say "I used some linear algebra".
 » 19 months ago, # |   0 Link to the source of the problem 4 is not working :/
•  » » 19 months ago, # ^ |   0 You have to join the group first to access the problem. The group name is probably The 1st Hunger Games.
•  » » » 19 months ago, # ^ |   0 Done.
•  » » 19 months ago, # ^ |   0 Sorry, I'll edit the blog to mention about it. This problem is from a CF group. You need to join the group as a spectator first to view it.Group Link.More information.
 » 19 months ago, # | ← Rev. 2 →   0 Can you please explain why the answer for problem 5 is 2 ^ (l — b). I couldn't understand properly.
•  » » 19 months ago, # ^ | ← Rev. 2 →   0 There are $2 ^ {(l - b)}$ subsets of the set containing only the non-basis element vectors. I claimed that, for each of these subsets, if we take the vectors from this subset, then there will a unique way to select some more vectors from the basis, so that addition of all these vectors equals $x.$To see this, pick some subset from the set of non-basis element vectors and let the xor-sum of it's elements be $y.$ Then, we need to pick some elements from the basis so that their xor-sum is $y \oplus x,$ which can be done in a unique way by the definition of basis.
•  » » » 19 months ago, # ^ |   +5 Thanks, it is clear now.
 » 19 months ago, # | ← Rev. 3 →   +5 For Problem 3, can't we just bruteforce on the basis vector to find the maximum value of XOR in a subset ? The maximum size of basis vector would be 20, and we can do exponential brute force of 2^20 for elements in basis vector and get the maximum xor ? I think proof can be something like the maximum XOR value (say) X is linear combination of (say) n vectors a1, a2, ...., an, i.e, X = a1^a2^....^an. Then obviously this X can be represented as linear combination of basis vectors too. Is this approach correct ?
•  » » 19 months ago, # ^ |   0 Yes, it works :) Nicely utilized the constraints.
 » 19 months ago, # |   +2 The following problem uses the concept of Problem 2 of in this blog: https://www.codechef.com/COOK106A/problems/XORCMPNTSeriously speaking, I am finding this topic hard to comprehend(I mean solving XOR/bits related problems with Gaussian Elimination modulo 2). Do I need a degree in Mathematics for that? XD
•  » » 19 months ago, # ^ |   0 Thank you :) Added the problem. Nice problem solaimanope vai ^_^
 » 19 months ago, # | ← Rev. 2 →   +10 Thank u very much. Almost everything in this post was completely clear for me. Very good.
•  » » 19 months ago, # ^ |   +13 Glad it helped! ^_^ It feels very nice to receive appreciative words for my work :)Btw I wonder, how did you come across this post once it went out of the "recent actions" list?
•  » » » 19 months ago, # ^ |   +5 My friend shared a link.
 » 18 months ago, # |   +5 Thank you so much! It helps me a lot. But a small question in the code about how to get the basis in n vectors.You said "basis[i] keeps the index of the vector whose f value is i" , and then you just made "basis[i] = mask;" .I find it okey, but in my opion the "mask" means the vector value, instead of the index of that mask.Perhaps I just get a wrong understanding QAQ. At last thank you!
•  » » 18 months ago, # ^ |   0 Sorry, you probably read it from my blog. I fixed it here, but forgot to do so in the blog. Thanks for noting it!
•  » » 9 months ago, # ^ |   0 DrSwad advgr1997 Why cant we store indices for basis[i] instead of mask ? I got WA when i submitted the solution 2b) considering basis[i]=(1<
 » 15 months ago, # |   0 How can i find a and b, if i have a*b and a^b?Thank you.
 » 15 months ago, # |   0 This is pretty well known...
 » 15 months ago, # |   0 Thank you a lot! It's the best blog on xor-basis i've seen so far
 » 15 months ago, # |   0 If we have to count the number of distinct integers that can be possible using OR over the set of the given elements. then can we solve this problem using gaussian elimination technique? How can we solve this problem?
 » 12 months ago, # | ← Rev. 2 →   +11 I recently came up with a super easy solution with xor basis for Problem 5: 73340139
 » 11 months ago, # |   +11 Please don't stop writing!
•  » » 11 months ago, # ^ | ← Rev. 2 →   +13 Yes Please. Especially problem 5 :P <3
•  » » » 11 months ago, # ^ |   +13 ROFL! Even I found that one particularly interesting
 » 11 months ago, # | ← Rev. 3 →   0 I liked how you baited us to your blog by dark theme, also a great article learned a lot of things.
 » 11 months ago, # |   +3 Lovely explanation. Never thought, I'd be using vectors and vector spaces for finding if product is square. Honestly, it still won't come to me at first sight. LOLWill read the blog again.
 » 11 months ago, # | ← Rev. 2 →   0 For Problem 1, the second DP represents something different than the first DP. dp[at][mask] = # of non-empty subsets using values <= at (from the array) with this mask. Just thought that was unclear.Also, the poss[at] array can be computed as 2^(count[at] — 1) directly, by the binomial theorem (irrespective of even/odd).
 » 10 months ago, # |   -8 Pardon me, but for arbitrary large dimensions of $d$, this algorithm actually works in $O(nd^2)$, correct? Here you assume that XOR can be performed in $O(1)$, but that is not generally the case, if say d=10^4. Would using bitsets help?
•  » » 6 months ago, # ^ | ← Rev. 2 →   +21 You are right and the actual complexity is just $O(n \cdot d^2 / 32)$. It's just that $d$ usually is just binary length of numbers from the input so it's around $32$ in 90% of cases.
 » 10 months ago, # |   0 how many distinct numbers can be generated by doing bitwise OR of one or more integers between A and B (inclusive)? where 1<=a<=b<=2^60as it is problem for ongoing contest can u plz help me with this after 1 day..the contest will be over till then..thanks in advance i am very much curious to know its solution that solves it in an optimal time.
 » 9 months ago, # |   0 very nice tutorial, thank you for your efforts
 » 9 months ago, # |   +3 Thanks for the nice tutorial. A nice problem that can be solved using this technique.
 » 9 months ago, # |   +5 Thank you for you explanation !
 » 9 months ago, # |   +8 Another problem using this technique: Problem A of atcoder grand contest 045.
•  » » 9 months ago, # ^ |   +1 Thanks man, I landed on this page searching for explanation for XOR Battle because of your comment.
 » 9 months ago, # |   0 In the problem 5 you said we can answer the queries online. But you have solved it offline right? Or did I misunderstood the definition of offline queries?
 » 9 months ago, # |   0 In code of part 2, it says mask ^= basis[i]; // Otherwise subtract the basis vector from this vector Can someone explain how subtracting is synonymous to XOR here ? Is this an error ?
•  » » 6 months ago, # ^ |   +3 It is explained earlier in the blog that XOR related to addition/subtraction under $\mathbb{Z}_2^d$. Here, you are trying to represent $mask$ as sum of some $basis[i]$. If the representation contains $basis[i]$, then xor-ing it again with $mask$ will exclude it from $mask$.
 » 9 months ago, # |   0 A much-needed blog for today's contest! :P Thanks for sharing!
 » 8 months ago, # |   +3 Your work is really appreciable. Thanks a lot and keep bringing more such content!!
 » 8 months ago, # |   -8 what u are actually storing in basis[i]?
 » 8 months ago, # |   -8 If i insert 7,3 then i can represent vector space as [1,1,1], [1,0,0] as basis. Am i right or not someone please clarify?
•  » » 6 months ago, # ^ |   +6 Yes. The first number is 111 and it will mark the least significant bit as taken. Then 011 is changed into 100.
 » 8 months ago, # | ← Rev. 4 →   -16 DrSwad , -is-this-fft- in problem 5, Is the idea is like this, addition in a vector space is closure and x should be present in the given vector space. So to form a subsequence with xor sum equals to x, from non-basis vectors we have xor sum of the form (K xor x), and as addition is closure so it is possible to form vector of the form K from basis vectors?
•  » » 8 months ago, # ^ |   +5 Downvote for tags.Also I didn't understand anything from what you said, can you please clarify?
•  » » » 8 months ago, # ^ |   +5 I am asking about the basic idea behind the number of subsequences whose xor sum is equal to X.
•  » » » » 8 months ago, # ^ |   +5 Assume you have these $\ell$ vectors and suppose $b$ of them form a basis $B$. Then all other $\ell - b$ vectors can be represented as the XOR of some basis vectors. I will write "$\operatorname{span} B$" to denote the set of all vectors that can be represented as the XOR of some basis vectors, notice that if $x, y \in \operatorname{span} B$, then also $x \oplus y \in \operatorname{span} B$.Suppose we take a vector $x \in \operatorname{span} B$. Pick any subset of the "other $\ell - b$ vectors", let their XOR be $K$, then also $K \in \operatorname{span} B$. If we wanted to pick some vectors such that the XOR is $x$, and the XOR of the non-basis vectors we chose is $K$, then we need to pick such basis vectors that their XOR is $K \oplus x$, and there is exactly one way to do that (because $B$ is a basis).If we want to form $x$, we have $2^{\ell - b}$ ways to choose from the "non-basis vectors" and 1 way to choose from the basis vectors, so the answer is $2^{\ell - b}$.
•  » » » » » 8 months ago, # ^ |   +5 Got it, thanks.
•  » » » 8 months ago, # ^ |   +5 Sorry its closure not commutative.
•  » » » » 8 months ago, # ^ |   +5 OK, it makes a bit more sense now.
 » 8 months ago, # |   0 Goes to my favorite blogs without doubt
 » 7 months ago, # | ← Rev. 2 →   -8 Deleted.
 » 5 months ago, # |   +8 Why is the maximum size of basis for a d-dimensional vector atmost d?
•  » » 5 months ago, # ^ |   +1 Each vector in your basis is responisble for one of the dimensions from the $d$-dimensions ( responsible refers to the fact that only that vector has a $1$ in that dimension and other vectors have a $0$ ). Since your vectors are $d$-dimensional, there can be atmost $d$ vectors that each take care of different dimensions.
•  » » » 5 months ago, # ^ |   0 Consider a case where all numbers (in space)of a given dimension have 1. Then we cannot find a basis (with size>1) for which a value could be responsible for that same dimension ( since all values of basis would have 1) so how to use atmost d numbers in basis?
•  » » » 5 months ago, # ^ |   0 Okay got it! We can always perform linear combination to get the same form which you described !
 » 4 months ago, # |   +5 I got goosebumps after reading your solution of problem 2a,just amazing!!
 » 6 weeks ago, # |   0 Problem 2a is giving wrong output at set s={3,4,5} it should give 4 ans but it is giving 8 can anyone look
•  » » 6 weeks ago, # ^ |   0 You're probably taking XOR on pairs only. The problem requires you to take XORs on all the subsets. Here are the results I'm getting: ∅ = 0 3 = 3 4 = 4 5 = 5 3 ^ 3 = 0 3 ^ 4 = 7 3 ^ 5 = 6 4 ^ 4 = 0 4 ^ 5 = 1 5 ^ 5 = 0 3 ^ 4 ^ 5 = 2 
•  » » » 6 weeks ago, # ^ |   0 okay.. thanks for explanation
 » 4 weeks ago, # | ← Rev. 3 →   0 Thanks for the awesome blog
 » 7 days ago, # |   0 How do we solve the problem 1 with bigger constraints (But what if the constraint was much bigger, say 10^5)? Can someone please help with the solution?
•  » » 7 days ago, # ^ | ← Rev. 2 →   +13 I assume you already know how to quickly factorize, and that you understand the solution for the small version.Let's learn how to solve "how many subsets are there with xor sum $0$" first. Compute prime masks as in the small version, and the answer is simply $2^{no.\ of\ non-basis\ elements}$. Indeed, for any non-basis subsets, the basis can form exactly the same xor sum. So the number of $0$-xor subsets is just the number of non-basis subsets. The blog already described how to build the basis.Understand up to this part first before proceeding. I am not sure whether there's an OJ problem of "counting $0$ xor sum subsets" aside from that problem 1.Now, you may be thinking "there are too many primes to consider". You're right, there are exactly $9592$ primes, which is a little too much. The catch here is that there can be at most one "large prime" larger than $\sqrt{10^5} \approx 316$, and there are only $65$ small primes (smaller than $316$). Let's see how we can speed up the algorithm with this observation.We will build the basis from the most significant bit down (like problem 3). The algorithm for adding a mask into the basis can be seen as follow: Find the most significant bit position of our mask (or terminate if the mask is $0$). If the respective basis position is unoccupied (i.e. there is no mask for this row yet), add our mask for this row, and terminate. Else, this position is occupied, xor our mask with the basis mask in this position. Now, since there is at most one large prime, there is also at most one bit out of the $9527$ larger bit positions, and this is also the most significant bit. Since there is at most one bit out of these, either it immediately gets added to said basis position, or it immediately gets xor-ed with the mask at said basis position, and there will be no more large bit that are set. So you can continue doing normal basis building for the $65$ small bits. In terms of implementation it should look something like this: function add(x, msb): // x has 9592 bits // msb stores the most significant "large" bit position if any if (basis[msb] > 0): basis[msb] = x return else: x ^= basis[msb] // now build the basis normally by for-looping bit positions from 64 to 0 And the answer is once again $2^{no.\ of\ non-basis\ elements}$. You need a bitset or something like that to store basis this large.Do ask if there are some things unclear or I made any mistakes.