# XORDETECTIVE - Editorial

Setter: Anton Trygub
Tester: Harris Leung
Editorialist: Trung Dang

2839

XOR

# PROBLEM:

There are some hidden non-negative numbers A and B, with 0 \leq A \lt B \lt 2^{29}. You were not able to determine them and were gonna cry. Luckily, XOR The Detective decided to save your day!

XOR The Detective is a great interrogator. In one query, he can ask any integer X such that 0 \le X \lt 2^{30}, and learn the value of (A + X) \oplus (B + X). Here \oplus denotes the bitwise XOR operation.

Help XOR The Detective to determine both A and B in at most 30 queries.

# EXPLANATION:

Let’s set up a goal on X: We want to find X such that B + X = 2^{29}. The reason being that since A < B, we know that A + X < 2^{29}, and therefore A + X and B + X don’t have any overlap bits, which makes it easier for us to recover A and B.

Let’s solve the easier case first: Suppose we know that the 28-th bit of B is 1 while the 28-th bit of A is 0 (we can easily check if this is the case using one question on X = 0). How can we proceed to find the suitable X (which I will now denote as \hat{X})?

We first see that \hat{X} \le 2^{28} (since B \ge 2^{28} due to our assumption). Observe that for any 0 \le X \le 2^{28}, A + X < 2^{29}. Therefore:

• If 0 \le X < \hat{X}, then (A + X) \oplus (B + X) < 2^{29} (because B + X < B + \hat{X} = 2^{29}).
• If \hat{X} \le X \le 2^{28}, then (A + X) \oplus (B + X) \ge 2^{29} (because B + X \ge B + \hat{X} = 2^{29}).

Would you look at that, we now have a condition to use in our binary search to find out \hat{X}! In this case though, we can implement this in an even easier way: simply loop through the 28-th to 0-th bit, turning it on one by one to see if \hat{X} exceeds this value or not. This takes us 30 queries exactly (one more from asking X = 0 at the beginning).

How do we generalize from our assumption? Suppose we query X = 0, and the first 1 bit in the returned result is at position P (where P = 28 would correspond to our assumption). We know that this means the P-th bit of B is 1, while the P-th bit of A is 0. Here’s an idea: we solve for these last P bits only. \hat{X} then has a slightly different semantic meaning: it is the smallest value such that B + \hat{X} has the last P bits being all 0. However, since the last P bits of B + \hat{X} are all 0's, we know that it will not affect the upper bits regardless of A, so we simply ignore these last P bits and solve the problem recursively with the upper bits, with our new A and B being A + \hat{X} and B + \hat{X}. We can prove that this recursive process of solving the last P bits and then ignoring them will ask one question for every bit (from binary lifting), which still takes us exactly 30 queries (with an initial query of X = 0 of course).

# TIME COMPLEXITY:

Time complexity is O(1).

# SOLUTION:

Setter's Solution
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

using namespace __gnu_pbds;
using namespace std;

using ll = long long;
using ld = double;

typedef tree<
pair<int, int>,
null_type,
less<pair<int, int>>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;

#define mp make_pair

int MOD = 998244353;

int mul(int a, int b) {
return (1LL * a * b) % MOD;
}

int add(int a, int b) {
int s = (a+b);
if (s>=MOD) s-=MOD;
return s;
}

int sub(int a, int b) {
int s = (a+MOD-b);
if (s>=MOD) s-=MOD;
return s;
}

int po(int a, ll deg)
{
if (deg==0) return 1;
if (deg%2==1) return mul(a, po(a, deg-1));
int t = po(a, deg/2);
return mul(t, t);
}

int inv(int n)
{
return po(n, MOD-2);
}

mt19937 rnd(time(0));

const int LIM = 1000005;

vector<int> facs(LIM), invfacs(LIM), invs(LIM);

void init()
{
facs[0] = 1;
for (int i = 1; i<LIM; i++) facs[i] = mul(facs[i-1], i);
invfacs[LIM-1] = inv(facs[LIM-1]);
for (int i = LIM-2; i>=0; i--) invfacs[i] = mul(invfacs[i+1], i+1);

for (int i = 1; i<LIM; i++) invs[i] = mul(invfacs[i], facs[i-1]);
}

int C(int n, int k)
{
if (n<k) return 0;
if (n<0 || k<0) return 0;
return mul(facs[n], mul(invfacs[k], invfacs[n-k]));
}

struct DSU
{
vector<int> sz;
vector<int> parent;
void make_set(int v) {
parent[v] = v;
sz[v] = 1;
}

int find_set(int v) {
if (v == parent[v])
return v;
return find_set(parent[v]);
}

void union_sets(int a, int b) {
a = find_set(a);
b = find_set(b);

if (a != b) {
if (sz[a] < sz[b])
swap(a, b);
parent[b] = a;
sz[a] += sz[b];
}
}

DSU (int n)
{
parent.resize(n);
sz.resize(n);
for (int i = 0; i<n; i++) make_set(i);
}
};

void print(vector<int> a)
{
for (auto it: a) cout<<it<<' ';
cout<<endl;
}

void print(vector<bool> a)
{
for (auto it: a) cout<<it<<' ';
cout<<endl;
}
/*
void print(vector<pair<ll, ll>> a)
{
for (auto it: a) cout<<it.first<<' '<<it.second<<"| ";
cout<<endl;
}*/

void print(vector<pair<int, int>> a)
{
for (auto it: a) cout<<it.first<<' '<<it.second<<"| ";
cout<<endl;
}

/*const int mod = 998244353;

template<int mod>
struct NTT {
static constexpr int max_lev = __builtin_ctz(mod - 1);

int prod[2][max_lev - 1];

NTT() {
int root = find_root();//(mod == 998244353) ? 31 : find_root();
int rroot = power(root, mod - 2);
vector<vector<int>> roots(2, vector<int>(max_lev - 1));
roots[0][max_lev - 2] = root;
roots[1][max_lev - 2] = rroot;
for (int tp = 0; tp < 2; ++tp) {
for (int i = max_lev - 3; i >= 0; --i) {
roots[tp][i] = mul(roots[tp][i + 1], roots[tp][i + 1]);
}
}
for (int tp = 0; tp < 2; ++tp) {
int cur = 1;
for (int i = 0; i < max_lev - 1; ++i) {
prod[tp][i] = mul(cur, roots[tp][i]);
cur = mul(cur, roots[tp ^ 1][i]);
}
}
}

template<bool inv>
void fft(int *a, int lg) const {
const int n = 1 << lg;
int pos = max_lev - 1;
for (int it = 0; it < lg; ++it) {
const int h = inv ? lg - 1 - it : it;
const int shift = (1 << (lg - h - 1));
int coef = 1;
for (int start = 0; start < (1 << h); ++start) {
for (int i = start << (lg - h); i < (start << (lg - h)) + shift; ++i) {
if (!inv) {
const int y = mul(a[i + shift], coef);
a[i + shift] = a[i];
inc(a[i], y);
dec(a[i + shift], y);
} else {
const int y = mul(a[i] + mod - a[i + shift], coef);
inc(a[i], a[i + shift]);
a[i + shift] = y;
}
}
coef = mul(coef, prod[inv][__builtin_ctz(~start)]);
}
}
}

vector<int> product(vector<int> a, vector<int> b) const {
if (a.empty() || b.empty()) {
return {};
}
const int sz = a.size() + b.size() - 1;
const int lg = 32 - __builtin_clz(sz - 1), n = 1 << lg;
a.resize(n);
b.resize(n);
fft<false>(a.data(), lg);
fft<false>(b.data(), lg);
for (int i = 0; i < n; ++i) {
a[i] = mul(a[i], b[i]);
}
fft<true>(a.data(), lg);
a.resize(sz);
const int rn = power(n, mod - 2);
for (int &x : a) {
x = mul(x, rn);
}
return a;
}

private:
static inline void inc(int &x, int y) {
x += y;
if (x >= mod) {
x -= mod;
}
}

static inline void dec(int &x, int y) {
x -= y;
if (x < 0) {
x += mod;
}
}

static inline int mul(int x, int y) {
return (1LL * x * y) % mod;
}

static int power(int x, int y) {
if (y == 0) {
return 1;
}
if (y % 2 == 0) {
return power(mul(x, x), y / 2);
}
return mul(x, power(x, y - 1));
}

static int find_root() {
for (int root = 2; ; ++root) {
if (power(root, (1 << max_lev)) == 1 && power(root, (1 << (max_lev - 1))) != 1) {
return root;
}
}
}
};

NTT<mod> ntt;
*/

{
cout<<"? "<<x<<endl;
int val; cin>>val; return val;
}

const int M = (1<<29);

void solve()
{
int q; cin>>q;

int a = 0; int b = 0;
int bit = 0;
for (int i = 0; i<29; i++) if (X&(1<<i)) bit = i;

b+=(1<<bit);

for (int i = bit+1; i<29; i++)
{
int x = (1<<i) - b;
if ((res^X)&(1<<(i+1))) b+=(1<<i);
}

for (int i = bit-1; i>=0; i--)
{
int x = M - b - (1<<i);
if (res&M) b+=(1<<i);
}
a = b^X;
cout<<"! "<<a<<' '<<b<<endl;
}

int main()
{
ios_base::sync_with_stdio(0);
cin.tie(nullptr);

int t; cin>>t;
while (t--) solve();

}

Tester's Solution
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
const int N=2e5+1;
const int iu=29;
cout << "? " << x << endl;
ll res;cin >> res;
return res;
}
void solve(){
int antonbaby;cin >> antonbaby;
int z=0;
while(1<<(z+1)<=king) z++;
ll a=0,b=(1<<z);
ll love=(1<<z);
for(int i=z+1; i<iu ;i++){
int flag=((res^king)>>(i+1))&1;
if(flag) a|=(1<<i),b|=(1<<i);
else love|=(1<<i);
}
love=0;
int last=z;
for(int i=z-1; i>=0 ;i--){
if((king>>i)&1){
int flag=(res^king)>>(last+1);
if(!flag) swap(a,b);
b|=(1<<i);
last=i;
}
else{
int flag=(res!=king);
if(flag) a|=(1<<i),b|=(1<<i);
else love|=(1<<i);
}
}
if(a>b) swap(a,b);
cout << "! " << a << ' ' << b << endl;
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);
int t;cin >> t;while(t--) solve();
}

Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;

cout << "? " << u << endl;
int ans; cin >> ans; return ans;
}

int main() {
int t; cin >> t;
while (t--) {
int q; cin >> q;
int top = __lg(sum);
for (int i = top; i > lst; i--) {
if (ret < (1 << (top + 1))) {
} else {
sum = ret;
}
}
add += (1 << (lst + 1)); lst = top;
}
int b = 1 << (__lg(sum)), a = sum - b;
cout << "! " << a << " " << b << endl;
}
}


Is there a typo here, if X <=2^{29} how can A+X <2^{29} ?

Yea it’s X \le 2^{28} … thank you for spotting

My code gives WA for subtask 1. Am I missing something?

#include<bits/stdc++.h>
using namespace std;

int main() {
cin.tie(0)->sync_with_stdio(0);

const int MX = 128;
map<vector<int>, pair<int, int>> mp;
vector<int> q(MX);
for (int a = 0; a < MX; a++) {
for (int b = a + 1; b < MX; b++) {
for (int x = 0; x < MX; x++) {
q[x] = (a + x) ^ (b + x);
}
mp[q] = {a, b};
}
}
int t;
for (cin >> t; t; t--) {
for (int i = 0; i < MX; i++) {
cout << "? " << i << endl << flush;
cin >> q[i];
}
assert(mp.count(q));
auto [a, b] = mp[q];
cout << "! " << a << ' ' << b << endl << flush;
}
return 0;
}

can anyone explain the approach in little bit easy way or other method