The following question is frequently asked in Codeforces ([46390](https://codeforces.net/blog/entry/46390), [45157](https://codeforces.net/blog/entry/45157), [11324](https://codeforces.net/blog/entry/11324)): _Is there a 2D segment tree that supports range addition and range minimum?_ In this blog post I give evidence that such a data structure does not exist, or if it did exist it would not generalize to higher dimensions. In particular I show that if for all $d$ a $d$-dimensional data structure that performs such queries in $O(polylog(N))$ time did exist, then the [exponential time hypothesis](https://en.wikipedia.org/wiki/Exponential_time_hypothesis) would fail. Such a data structure exists for range addition and range sum, so this is a non-trivial claim separating the hardness of these problems.↵
↵
## Update:↵
I researched some literature on related topics and found that while lower bounds for this exact data structure have not been given before, there has been a lot of research on related topics. Some observations:↵
↵
1. Klee's measure problem is very related to Range-Add-Min structure.↵
2. The reduction from Clique to Klee's measure problem in [1] can be adapted to show that $d$-dimensional Range-Add-Min-structure with $O(polylog(N))$ queries could solve $d+1$-Clique in $O(n^2)$ time.↵
3. Therefore $d$-dimensional Range-Add-Min-structure for all $d$ would mean FPT=W[1]. Note that ETH implies FPT!=W[1].↵
4. Also, 2D Range-Add-Min structure could solve 3-clique in $O(n^2)$ time, which would be a breaktrough result.↵
5. Hardness of very related problems has been considered earlier this year in [2]. I'm not sure if their results directly imply hardness of 2D Range-Add-Min structure.↵
6. There is also a connection to MAX-2-CSP [3], which can be extended to connection to MAX-d-CSP for $d$-dimensional Range-Add-Min-structure. ↵
↵
↵
[1] T. M. Chan. A (slightly) faster algorithm for klee's measure problem. SCG 2008 [https://dl.acm.org/citation.cfm?id=1377693](https://dl.acm.org/citation.cfm?id=1377693)↵
↵
[2] L. Duraj, K. Kleiner, A. Polak, V. V. Williams. Equivalences between triangle and range query problems. arXiv 2019 [https://arxiv.org/abs/1908.11819](https://arxiv.org/abs/1908.11819)↵
↵
[3] R. Williams. A new algorithm for optimal 2-constraint satisfaction and its implications. TCS 2005 [https://www.sciencedirect.com/science/article/pii/S0304397505005438](https://www.sciencedirect.com/science/article/pii/S0304397505005438)↵
↵
↵
↵
A pdf version of this blog is in [http://laakeri.kapsi.fi/a/rmq.pdf](http://laakeri.kapsi.fi/a/rmq.pdf)↵
↵
## Formalization↵
Consider a $d$-dimensional integer array $A$ whose elements are indexed with $A[(i_1, \ldots, i_d)]$ for all $0 \le i_j < N$.↵
Furthermore assume that $A$ is initialized to zero.↵
The operation $add(v, l_1, r_1, \ldots, l_d, r_d)$ adds the value $v$ to all elements $A[(i_1, \ldots, i_d)]$ such that $l_j \le i_j < r_j$.↵
The operation $min(l_1, r_1, \ldots, l_d, r_d)$ returns the minimum value over all elements $A[(i_1, \ldots, i_d)]$ such that $l_j \le i_j < r_j$.↵
A data structure that supports these two operations is called _Range-Add-Min_-structure.↵
↵
In the Boolean satisfiability problem (SAT) the task is to decide if there is an assigment of $n$ binary variables to true/false so that it satisfies a given Boolean formula.↵
The formula consists of $m$ _clauses_, and an assigment of variables satisfies the formula if it satisfies every clause.↵
A clause is a logical or that consists of variables and negations of variables.↵
For example if $x_1, x_2, \ldots, x_n$ are variables, the clause $(x_1 \vee \neg x_3 \vee x_6)$ is satisfied if $x_1$ is true or $x_3$ is false or $x_6$ is true.↵
↵
The problem $k$-SAT is Boolean satisfiability problem where each clause has at most $k$ variables or negations of variables.↵
$2$-SAT can be solved in polynomial time, but $3$-SAT is NP-hard.↵
Furthermore a widely believed _exponential time hypothesis_ conjectures that there exists $\varepsilon > 0$ such that $3$-SAT cannot be solved in time $O((1+\varepsilon)^n poly(n, m))$, where $poly(n, m)$ is polynomial function on $n$ and $m$.↵
↵
**Theorem 1:**↵
If there exists a function $f(d)$ such that $d$-dimensional _Range-Add-Min_-structure can process $Q$ operations in $O((\log_2(N) + \log_2(Q))^{f(d)} Q)$ time for all $d$, then exponential time hypothesis fails.↵
↵
## Proof↵
We prove Theorem 1 by assuming that such $d$-dimensional data structure exists, and constructing a $O(2^{3n/d} poly(n,m))$ time algorithm for 3-SAT, where $n$ is the number of variables and $m$ is the number of clauses.↵
↵
**Claim 1:**↵
For all $k$ and $d$, $k$-SAT can be solved with $O(2^{kn/d} m)$ operations with a $d$-dimensional _Range-Add-Min_-structure with $N=2^{n/d}$.↵
↵
**Proof:**↵
Suppose without loss of generality that $d$ divides $n$.↵
Let $N = 2^{n/d}$.↵
We use the $d$-dimensional array $A$ with indices $(i_1, \ldots, i_d)$, $0 \le i_j < N$.↵
$A$ has $2^n$ elements, because $N^d = (2^{n/d})^d = 2^n$.↵
↵
Let $x_0, \ldots, x_{n-1}$ be a bit string of length $n$.↵
The bijection $\phi$ with $$\phi(x_0, \ldots, x_{n-1}) = (1 x_0 + 2 x_1 + 4 x_2 + \ldots + x_{n/d-1} 2^{n/d - 1}, \ldots, 1 x_{n-n/d} + \ldots)$$↵
maps bit strings of length $n$ to the indices of $A$.↵
For example if $n=6$ and $d=2$, the bijection is $\phi(x_0, x_1, x_2, x_3, x_4, x_5) = (1 x_0 + 2 x_1 + 4 x_2, 1 x_3 + 2 x_4 + 4 x_5)$.↵
↵
An assignment of variables in SAT is a bit string of length $n$.↵
We use the array $A$ to represent if the assigment $x_0, x_1, \ldots, x_{n-1}$ is a satisfiable assignment of the SAT formula, using the bijection $\phi(x_0, \ldots, x_{n-1})$ to map between variable assignments and indices of $A$.↵
We iteratively add clauses to the formula, maintaining the invariant that $A[\phi(x_0, \ldots, x_{n-1})] = 0$ if and only if the assigment $x_0, \ldots, x_{n-1}$ satisfies all clauses added so far.↵
↵
**Claim 2:**↵
Given a clause that contains $k$ variables or negations of variables, we can add $1$ to all elements $A[\phi(x_0, \ldots, x_{n-1})]$ such that $x_0, \ldots, x_{n-1}$ does not satisfy the clause using at most $2^{kn/d}$ $add$-operations.↵
↵
**Proof:**↵
Let $v_1, \ldots, v_k$ be the variables that occur in the clause (possibly with negations).↵
Only the values $x_{v_1}, \ldots, x_{v_k}$ affect whether the clause is satisfied or not.↵
These values affect at most $k$ dimensions in the mapping $\phi(x_0, \ldots, x_{n-1})$.↵
Lets call these dimensions the non-trivial dimensions, and other dimensions the trivial dimensions.↵
If we choose the values of non-trivial dimensions, then by the inverse mapping $\phi^{-1}$ we choose the values of $x_{v_1}, \ldots, x_{v_k}$, and know if the clause is satisfied or not.↵
We brute force over all possible values of non-trivial dimensions.↵
For each value, if the clause is not be satisfied by $\phi^{-1}$, we use operation $add(1, l_1, r_1, \ldots, l_d, r_d)$, where $l_j = r_j - 1$ is the brute forced value of dimension $j$ if $j$ is a non-trivial dimension and otherwise $l_j = 0$ and $r_j = N$.↵
Each dimension has $N$ values, so brute forcing the non-trivial dimensions takes at most $N^k = (2^{n/d})^k = 2^{kn/d}$ $add$-operations. $\square$↵
↵
By Claim 2 each clause can be processed with $2^{kn/d}$ $add$-operations.↵
Therefore when adding all $m$ clauses we use at most $m 2^{kn/d}$ $add$-operations.↵
After all clauses have been added, we use a single $min$-operation where $l_j = 0$ and $r_j = N$ for all dimensions $j$ to query if there are any elements in $A$ that have value $0$, and thus correspond to satisfiable assignments. $\square$↵
↵
**Claim 3:**↵
If the data structure of Theorem 1 exists, $k$-SAT can be solved in $O((1+\varepsilon)^n poly(n, m))$ for any $\varepsilon > 0$.↵
↵
**Proof:**↵
Given $k$ and $\varepsilon$, lets choose $d$ so that $2^{k/d} < 1 + \varepsilon$.↵
By using the algorithm of Claim 1 with the data structure of Theorem 1 we obtain a runtime of↵
$$O((\log_2(2^{n/d}) + \log_2(2^{kn/d} m))^{f(d)} 2^{kn/d} m)$$↵
$$= O((n/d + kn \log_2(m)/d)^{f(d)} 2^{kn/d} m)$$↵
$$= O((1 + \varepsilon)^n poly(n, m)).\square$$↵
↵
Claim 3 completes the proof, because the exponential time hypothesis states that there exists $\varepsilon > 0$ such that no $O((1 + \varepsilon)^n poly(n, m))$ algorithm exists for 3-SAT.↵
↵
## Conclusion↵
We proved that efficient $d$-dimensional _Range-Add-Min_-structure does not exist if the exponential time hypothesis holds.↵
Our proof allows the existance of such 2D data structure.↵
However we know that such 2D data structure must be developed with techniques that do not generalize to higher dimensions.↵
An 8-dimensional _Range-Add-Min_-structure would imply a state-of-the-art algorithm for 3-SAT trough our reduction.↵
Some open questions are:↵
↵
1. Is there a conditional hardness proof for 2D case?↵
↵
2. Likewise to $(min, +)$-semiring, our reduction also works for $(+, \cdot)$-semiring. Does it work for all commutative semirings?↵
↵
3. A $d$-dimensional data structure with $O(polylog(N))$ operations exists if the operations are addition and sum. Can we classify all pairs of operations so that the data structure either exists, or does not exist assuming ETH?↵
↵
4. Is it plausible that techniques for designing multidimensional data structures work for $d=2$ but not for some higher $d$?
↵
## Update:↵
I researched some literature on related topics and found that while lower bounds for this exact data structure have not been given before, there has been a lot of research on related topics. Some observations:↵
↵
1. Klee's measure problem is very related to Range-Add-Min structure.↵
2. The reduction from Clique to Klee's measure problem in [1] can be adapted to show that $d$-dimensional Range-Add-Min-structure with $O(polylog(N))$ queries could solve $d+1$-Clique in $O(n^2)$ time.↵
3. Therefore $d$-dimensional Range-Add-Min-structure for all $d$ would mean FPT=W[1]. Note that ETH implies FPT!=W[1].↵
4. Also, 2D Range-Add-Min structure could solve 3-clique in $O(n^2)$ time, which would be a breaktrough result.↵
5. Hardness of very related problems has been considered earlier this year in [2]. I'm not sure if their results directly imply hardness of 2D Range-Add-Min structure.↵
6. There is also a connection to MAX-2-CSP [3], which can be extended to connection to MAX-d-CSP for $d$-dimensional Range-Add-Min-structure. ↵
↵
↵
[1] T. M. Chan. A (slightly) faster algorithm for klee's measure problem. SCG 2008 [https://dl.acm.org/citation.cfm?id=1377693](https://dl.acm.org/citation.cfm?id=1377693)↵
↵
[2] L. Duraj, K. Kleiner, A. Polak, V. V. Williams. Equivalences between triangle and range query problems. arXiv 2019 [https://arxiv.org/abs/1908.11819](https://arxiv.org/abs/1908.11819)↵
↵
[3] R. Williams. A new algorithm for optimal 2-constraint satisfaction and its implications. TCS 2005 [https://www.sciencedirect.com/science/article/pii/S0304397505005438](https://www.sciencedirect.com/science/article/pii/S0304397505005438)↵
↵
↵
↵
A pdf version of this blog is in [http://laakeri.kapsi.fi/a/rmq.pdf](http://laakeri.kapsi.fi/a/rmq.pdf)↵
↵
## Formalization↵
Consider a $d$-dimensional integer array $A$ whose elements are indexed with $A[(i_1, \ldots, i_d)]$ for all $0 \le i_j < N$.↵
Furthermore assume that $A$ is initialized to zero.↵
The operation $add(v, l_1, r_1, \ldots, l_d, r_d)$ adds the value $v$ to all elements $A[(i_1, \ldots, i_d)]$ such that $l_j \le i_j < r_j$.↵
The operation $min(l_1, r_1, \ldots, l_d, r_d)$ returns the minimum value over all elements $A[(i_1, \ldots, i_d)]$ such that $l_j \le i_j < r_j$.↵
A data structure that supports these two operations is called _Range-Add-Min_-structure.↵
↵
In the Boolean satisfiability problem (SAT) the task is to decide if there is an assigment of $n$ binary variables to true/false so that it satisfies a given Boolean formula.↵
The formula consists of $m$ _clauses_, and an assigment of variables satisfies the formula if it satisfies every clause.↵
A clause is a logical or that consists of variables and negations of variables.↵
For example if $x_1, x_2, \ldots, x_n$ are variables, the clause $(x_1 \vee \neg x_3 \vee x_6)$ is satisfied if $x_1$ is true or $x_3$ is false or $x_6$ is true.↵
↵
The problem $k$-SAT is Boolean satisfiability problem where each clause has at most $k$ variables or negations of variables.↵
$2$-SAT can be solved in polynomial time, but $3$-SAT is NP-hard.↵
Furthermore a widely believed _exponential time hypothesis_ conjectures that there exists $\varepsilon > 0$ such that $3$-SAT cannot be solved in time $O((1+\varepsilon)^n poly(n, m))$, where $poly(n, m)$ is polynomial function on $n$ and $m$.↵
↵
**Theorem 1:**↵
If there exists a function $f(d)$ such that $d$-dimensional _Range-Add-Min_-structure can process $Q$ operations in $O((\log_2(N) + \log_2(Q))^{f(d)} Q)$ time for all $d$, then exponential time hypothesis fails.↵
↵
## Proof↵
We prove Theorem 1 by assuming that such $d$-dimensional data structure exists, and constructing a $O(2^{3n/d} poly(n,m))$ time algorithm for 3-SAT, where $n$ is the number of variables and $m$ is the number of clauses.↵
↵
**Claim 1:**↵
For all $k$ and $d$, $k$-SAT can be solved with $O(2^{kn/d} m)$ operations with a $d$-dimensional _Range-Add-Min_-structure with $N=2^{n/d}$.↵
↵
**Proof:**↵
Suppose without loss of generality that $d$ divides $n$.↵
Let $N = 2^{n/d}$.↵
We use the $d$-dimensional array $A$ with indices $(i_1, \ldots, i_d)$, $0 \le i_j < N$.↵
$A$ has $2^n$ elements, because $N^d = (2^{n/d})^d = 2^n$.↵
↵
Let $x_0, \ldots, x_{n-1}$ be a bit string of length $n$.↵
The bijection $\phi$ with $$\phi(x_0, \ldots, x_{n-1}) = (1 x_0 + 2 x_1 + 4 x_2 + \ldots + x_{n/d-1} 2^{n/d - 1}, \ldots, 1 x_{n-n/d} + \ldots)$$↵
maps bit strings of length $n$ to the indices of $A$.↵
For example if $n=6$ and $d=2$, the bijection is $\phi(x_0, x_1, x_2, x_3, x_4, x_5) = (1 x_0 + 2 x_1 + 4 x_2, 1 x_3 + 2 x_4 + 4 x_5)$.↵
↵
An assignment of variables in SAT is a bit string of length $n$.↵
We use the array $A$ to represent if the assigment $x_0, x_1, \ldots, x_{n-1}$ is a satisfiable assignment of the SAT formula, using the bijection $\phi(x_0, \ldots, x_{n-1})$ to map between variable assignments and indices of $A$.↵
We iteratively add clauses to the formula, maintaining the invariant that $A[\phi(x_0, \ldots, x_{n-1})] = 0$ if and only if the assigment $x_0, \ldots, x_{n-1}$ satisfies all clauses added so far.↵
↵
**Claim 2:**↵
Given a clause that contains $k$ variables or negations of variables, we can add $1$ to all elements $A[\phi(x_0, \ldots, x_{n-1})]$ such that $x_0, \ldots, x_{n-1}$ does not satisfy the clause using at most $2^{kn/d}$ $add$-operations.↵
↵
**Proof:**↵
Let $v_1, \ldots, v_k$ be the variables that occur in the clause (possibly with negations).↵
Only the values $x_{v_1}, \ldots, x_{v_k}$ affect whether the clause is satisfied or not.↵
These values affect at most $k$ dimensions in the mapping $\phi(x_0, \ldots, x_{n-1})$.↵
Lets call these dimensions the non-trivial dimensions, and other dimensions the trivial dimensions.↵
If we choose the values of non-trivial dimensions, then by the inverse mapping $\phi^{-1}$ we choose the values of $x_{v_1}, \ldots, x_{v_k}$, and know if the clause is satisfied or not.↵
We brute force over all possible values of non-trivial dimensions.↵
For each value, if the clause is not be satisfied by $\phi^{-1}$, we use operation $add(1, l_1, r_1, \ldots, l_d, r_d)$, where $l_j = r_j - 1$ is the brute forced value of dimension $j$ if $j$ is a non-trivial dimension and otherwise $l_j = 0$ and $r_j = N$.↵
Each dimension has $N$ values, so brute forcing the non-trivial dimensions takes at most $N^k = (2^{n/d})^k = 2^{kn/d}$ $add$-operations. $\square$↵
↵
By Claim 2 each clause can be processed with $2^{kn/d}$ $add$-operations.↵
Therefore when adding all $m$ clauses we use at most $m 2^{kn/d}$ $add$-operations.↵
After all clauses have been added, we use a single $min$-operation where $l_j = 0$ and $r_j = N$ for all dimensions $j$ to query if there are any elements in $A$ that have value $0$, and thus correspond to satisfiable assignments. $\square$↵
↵
**Claim 3:**↵
If the data structure of Theorem 1 exists, $k$-SAT can be solved in $O((1+\varepsilon)^n poly(n, m))$ for any $\varepsilon > 0$.↵
↵
**Proof:**↵
Given $k$ and $\varepsilon$, lets choose $d$ so that $2^{k/d} < 1 + \varepsilon$.↵
By using the algorithm of Claim 1 with the data structure of Theorem 1 we obtain a runtime of↵
$$O((\log_2(2^{n/d}) + \log_2(2^{kn/d} m))^{f(d)} 2^{kn/d} m)$$↵
$$= O((n/d + kn \log_2(m)/d)^{f(d)} 2^{kn/d} m)$$↵
$$= O((1 + \varepsilon)^n poly(n, m)).\square$$↵
↵
Claim 3 completes the proof, because the exponential time hypothesis states that there exists $\varepsilon > 0$ such that no $O((1 + \varepsilon)^n poly(n, m))$ algorithm exists for 3-SAT.↵
↵
## Conclusion↵
We proved that efficient $d$-dimensional _Range-Add-Min_-structure does not exist if the exponential time hypothesis holds.↵
Our proof allows the existance of such 2D data structure.↵
However we know that such 2D data structure must be developed with techniques that do not generalize to higher dimensions.↵
An 8-dimensional _Range-Add-Min_-structure would imply a state-of-the-art algorithm for 3-SAT trough our reduction.↵
Some open questions are:↵
↵
1. Is there a conditional hardness proof for 2D case?↵
↵
2. Likewise to $(min, +)$-semiring, our reduction also works for $(+, \cdot)$-semiring. Does it work for all commutative semirings?↵
↵
3. A $d$-dimensional data structure with $O(polylog(N))$ operations exists if the operations are addition and sum. Can we classify all pairs of operations so that the data structure either exists, or does not exist assuming ETH?↵
↵
4. Is it plausible that techniques for designing multidimensional data structures work for $d=2$ but not for some higher $d$?
