Teen 3sum

Teen 3sum
From Wikipedia, the free encyclopedia
"3sum" redirects here. For the malt beverage, see Comparison of alcopops § Beer-based .
Unsolved problem in computer science :

^ Grønlund & Pettie 2014 .

^ Freund 2017 .

^ Gold & Sharir 2017 .

^ Jump up to: a b Chan 2018 .

^ Kane, Lovett & Moran 2018 .

^ Jump up to: a b Kopelowitz, Tsvi; Pettie, Seth; Porat, Ely (2014). "3SUM Hardness in (Dynamic) Data Structures". arXiv : 1407.6756 [ cs.DS ].

^ Cormen, Thomas H. ; Leiserson, Charles E. ; Rivest, Ronald L. ; Stein, Clifford (2009) [1990]. Introduction to Algorithms (3rd ed.). MIT Press and McGraw-Hill. ISBN 0-262-03384-4 . Ex. 30.1–7, p. 906.

^ Visibility Graphs and 3-Sum by Michael Hoffmann

^ For a reduction in the other direction, see Variants of the 3-sum problem .

^ Jump up to: a b c Patrascu, M. (2010). Towards polynomial lower bounds for dynamic problems . Proceedings of the 42nd ACM symposium on Theory of computing - STOC '10. p. 603. doi : 10.1145/1806689.1806772 . ISBN 9781450300506 .

^ Demaine, Erik ; Erickson, Jeff; O'Rourke, Joseph (20 August 2006). "Problem 41: Sorting X + Y (Pairwise Sums)" . The Open Problems Project . Retrieved 23 September 2014 .

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Is there an algorithm to solve the 3SUM problem in time

O
(

n

2
−
ϵ

)

{\displaystyle O(n^{2-\epsilon })}

, for some

ϵ
>
0

{\displaystyle \epsilon >0}

?
In computational complexity theory , the 3SUM problem asks if a given set of

n

{\displaystyle n}

real numbers contains three elements that sum to zero. A generalized version, k -SUM, asks the same question on k numbers. 3SUM can be easily solved in

O
(

n

2

)

{\displaystyle O(n^{2})}

time, and matching

Ω
(

n

⌈
k

/

2
⌉

)

{\displaystyle \Omega (n^{\lceil k/2\rceil })}

lower bounds are known in some specialized models of computation ( Erickson 1999 ).

It was conjectured that any deterministic algorithm for the 3SUM requires

Ω
(

n

2

)

{\displaystyle \Omega (n^{2})}

time.
In 2014, the original 3SUM conjecture was refuted by Allan Grønlund and Seth Pettie who gave a deterministic algorithm that solves 3SUM in

O
(

n

2

/

(

log
⁡
n

/

log
⁡
log
⁡
n

)

2

/

3

)

{\displaystyle O(n^{2}/({\log n}/{\log \log n})^{2/3})}

time. [1]
Additionally, Grønlund and Pettie showed that the 4- linear decision tree complexity of 3SUM is

O
(

n

3

/

2

log
⁡
n

)

{\displaystyle O(n^{3/2}{\sqrt {\log n}})}

.
These bounds were subsequently improved. [2] [3] [4]
The current best known algorithm for 3SUM runs in

O
(

n

2

(
log
⁡
log
⁡
n

)

O
(
1
)

/

log

2

⁡
n

)

{\displaystyle O(n^{2}(\log \log n)^{O(1)}/{\log ^{2}n})}

time. [4]
Kane, Lovett, and Moran showed that the 6- linear decision tree complexity of 3SUM is

O
(
n

log

2

⁡
n

)

{\displaystyle O(n{\log ^{2}n})}

. [5] The latter bound is tight (up to a logarithmic factor).
It is still conjectured that 3SUM is unsolvable in

O
(

n

2
−
Ω
(
1
)

)

{\displaystyle O(n^{2-\Omega (1)})}

expected time. [6]

When the elements are integers in the range

[
−
N
,
…
,
N
]

{\displaystyle [-N,\dots ,N]}

, 3SUM can be solved in

O
(
n
+
N
log
⁡
N
)

{\displaystyle O(n+N\log N)}

time by representing the input set

S

{\displaystyle S}

as a bit vector , computing the set

S
+
S

{\displaystyle S+S}

of all pairwise sums as a discrete convolution using the fast Fourier transform , and finally comparing this set to

S

{\displaystyle S}

. [7]

Suppose the input array is

S
[
0..
n
−
1
]

{\displaystyle S[0..n-1]}

. In integer ( word RAM ) models of computing, 3SUM can be solved in

O
(

n

2

)

{\displaystyle O(n^{2})}

time on average by inserting each number

S
[
i
]

{\displaystyle S[i]}

into a hash table , and then, for each index

i

{\displaystyle i}

and

j

{\displaystyle j}

, checking whether the hash table contains the integer

−
(
S
[
i
]
+
S
[
j
]
)

{\displaystyle -(S[i]+S[j])}

.

It is also possible to solve the problem in the same time in a comparison -based model of computing or real RAM , for which hashing is not allowed. The algorithm below first sorts the input array and then tests all possible pairs in a careful order that avoids the need to binary search for the pairs in the sorted list, achieving worst-case

O
(

n

2

)

{\displaystyle O(n^{2})}

time, as follows. [8]

The following example shows this algorithm's execution on a small sorted array. Current values of a are shown in red, values of b and c are shown in magenta.

The correctness of the algorithm can be seen as follows. Suppose we have a solution a + b + c = 0. Since the pointers only move in one direction, we can run the algorithm until the leftmost pointer points to a. Run the algorithm until either one of the remaining pointers points to b or c, whichever occurs first. Then the algorithm will run until the last pointer points to the remaining term, giving the affirmative solution.

Instead of looking for numbers whose sum is 0, it is possible to look for numbers whose sum is any constant C . The simplest way would be to modify the original algorithm to search the hash table for the integer

(
C
−
(
S
[
i
]
+
S
[
j
]
)
)

{\displaystyle (C-(S[i]+S[j]))}

.

For example, if A=[1,2,3,4] and if you are asked to find 3SUM for C =4, then subtract 4/3 from all the elements of A, and solve it in the usual 3sum way, i.e.,

(
a
−
C

/

3
)
+
(
b
−
C

/

3
)
+
(
c
−
C

/

3
)
=
0

{\displaystyle (a-C/3)+(b-C/3)+(c-C/3)=0}

.

Instead of searching for the 3 numbers in a single array, we can search for them in 3 different arrays. I.e., given three arrays X, Y and Z, find three numbers a ∈ X , b ∈ Y , c ∈ Z , such that

a
+
b
+
c
=
0

{\displaystyle a+b+c=0}

. Call the 1-array variant 3SUM×1 and the 3-array variant 3SUM×3.

Given a solver for 3SUM×1, the 3SUM×3 problem can be solved in the following way (assuming all elements are integers):

By the way we transformed the arrays, it is guaranteed that a ∈ X , b ∈ Y , c ∈ Z . [9]

Instead of looking for arbitrary elements of the array such that:

the convolution 3sum problem (Conv3SUM) looks for elements in specific locations: [10]

Given a solver for 3SUM, the Conv3SUM problem can be solved in the following way. [10]

Given a solver for Conv3SUM, the 3SUM problem can be solved in the following way. [6] [10]

The reduction uses a hash function . As a first approximation, assume that we have a linear hash function, i.e. a function h such that:

Suppose that all elements are integers in the range: 0... N -1, and that the function h maps each element to an element in the smaller range of indices: 0... n -1. Create a new array T and send each element of S to its hash value in T , i.e., for every x in S (

∀
x
∈
S

{\displaystyle \forall x\in S}

):

Initially, suppose that the mappings are unique (i.e. each cell in T accepts only a single element from S ). Solve Conv3SUM on T . Now:

This idealized solution doesn't work, because any hash function might map several distinct elements of S to the same cell of T . The trick is to create an array

T

∗

{\displaystyle T^{*}}

by selecting a single random element from each cell of T , and run Conv3SUM on

T

∗

{\displaystyle T^{*}}

. If a solution is found, then it is a correct solution for 3SUM on S . If no solution is found, then create a different random

T

∗

{\displaystyle T^{*}}

and try again. Suppose there are at most R elements in each cell of T . Then the probability of finding a solution (if a solution exists) is the probability that the random selection will select the correct element from each cell, which is

(
1

/

R

)

3

{\displaystyle (1/R)^{3}}

. By running Conv3SUM

R

3

{\displaystyle R^{3}}

times, the solution will be found with a high probability.

Unfortunately, we do not have linear perfect hashing, so we have to use an almost linear hash function , i.e. a function h such that:

This requires to duplicate the elements of S when copying them into T , i.e., put every element

x
∈
S

{\displaystyle x\in S}

both in

T
[
h
(
x
)
]

{\displaystyle T[h(x)]}

(as before) and in

T
[
h
(
x
)
]
−
1

{\displaystyle T[h(x)]-1}

. So each cell will have 2 R elements, and we will have to run Conv3SUM

(
2
R

)

3

{\displaystyle (2R)^{3}}

times.

A problem is called 3SUM-hard if solving it in subquadratic time implies a subquadratic-time algorithm for 3SUM. The concept of 3SUM-hardness was introduced by Gajentaan & Overmars (1995) . They proved that a large class of problems in computational geometry are 3SUM-hard, including the following ones. (The authors acknowledge that many of these problems are contributed by other researchers.)

By now there are a multitude of other problems that fall into this category. An example is the decision version of X + Y sorting : given sets of numbers X and Y of n elements each, are there n ² distinct x + y for x ∈ X , y ∈ Y ? [11]

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Published: 06:02 BST, 3 November 2016 | Updated: 06:09 BST, 3 November 2016
A young couple allegedly had a threesome with a 14-year-old girl after she was lured to their home from a shopping centre.
The 23-year-old woman allegedly met the teenage girl at a Hornsby shopping centre on Sydney's north shore on Monday.
The pair later went to a Denison Street property in Hornsby where the girl was allegedly plied with alcohol.
She was then allegedly sexually assaulted by the woman and her 31-year-old male partner during a suspected threesome.
A couple allegedly had a threesome with a 14-year-old girl after she was lured to their home from a Sydney shopping centre on Monday (stock image)
The girl left the home around 9.30pm that night and reported the matter to police.
Police were told the teen was with another 14-year-old girl when she met the woman at the shopping centre. The girl also attended the home but left before the alleged sexual assault took place.
Following extensive investigations by the State Crime Command's Child Abuse Squad, police executed a crime scene warrant at the property on Wednesday.
While carrying out the warrant, the man and woman returned to the home where they were arrested, before taken to Hornsby Police Station.
Following extensive investigations, police executed a crime scene warrant at the property
The woman has been charged with commit act of indecency with person under 16 years, aggravated indecent assault in company, and five counts of aggravated sexual intercourse with person under 16 years.
The man was charged with aggravated indecent assault in company and five counts of aggravated sexual intercourse with person under 16 years.
They were both refused bail and are due to face Hornsby Local Court on Thursday.
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Part of the Daily Mail, The Mail on Sunday & Metro Media Group

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