Algorithms and Computational Complexity: an Overview
Winter 2011
Larry Ruzzo
Thanks to Paul Beame, James Lee, Kevin Wayne for some slides
1
goals
Design of Algorithms – a taste
design methods
common or important types of problems
analysis of algorithms - efficiency
2
goals
Complexity & intractability – a taste
solving problems in principle is not enough
algorithms must be efficient
some problems have no efficient solution
NP-complete problems
important & useful class of problems whose solutions (seemingly) cannot be found efficiently
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complexity example Cryptography (e.g. RSA, SSL in browsers)
Secret: p,q prime, say 512 bits each
Public: n which equals p x q, 1024 bits
In principle
there is an algorithm that given n will find p and q: try all 2512 possible p’s, but an astronomical number
In practice
no fast algorithm known for this problem (on non-quantum computers)
security of RSA depends on this fact
(and research in “quantum computing” is strongly driven by the possibility of changing this)
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algorithms versus machines
Moore’s Law and the exponential improvements in hardware...
Ex: sparse linear equations over 25 years
10 orders of magnitude improvement!
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algorithms or hardware?
Seconds
25 years progress solving sparse linear systems
Hardware alone: 4 orders of magnitude
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G.E. / CDC 3600
G.E. = Gaussian Elimination
CDC 6600
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CDC 7600
Cray 1
105
Cray 2
104
Cray 3 (Est.)
103
102
101
100
1960
!
Source: Sandia, via M. Schultz 1970
1980
1990
6
2000
algorithms or hardware?
Seconds
25 years progress solving sparse linear systems
Hardware alone: 4 orders of magnitude
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G.E. / CDC 3600
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CDC 7600
Cray 1
105
Cray 2
104
103
Cray 3 (Est.)
Sparse G.E.
Gauss-Seidel
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Software alone: 6 orders of magnitude
G.E. = Gaussian Elimination
SOR = Successive OverRelaxation
CG = Conjugate Gradient
CDC 6600
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100
1960
SOR
CG
!
Source: Sandia, via M. Schultz 1970
1980
1990
7
2000
algorithms or hardware? The N-Body Problem:
in 30 years 107 hardware 1010 software
!
Source: T.Quinn
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algorithms: a definition
Procedure to accomplish a task or solve a well-specified problem
Well-specified: know what all possible inputs look like and what output looks like given them
“accomplish” via simple, well-defined steps
Ex: sorting names (via comparison)
Ex: checking for primality (via +, -, *, /, ≤)
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algorithms: a sample problem
Printed circuit-board company has a robot arm that solders components to the board
Time: proportional to total distance the arm must move from initial rest position around the board and back to the initial position
For each board design, find best order to do the soldering
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printed circuit board
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printed circuit board
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more precise problem definition
Input: Given a set S of n points in the plane
Output: The shortest cycle tour that visits each point in the set S.
Better known as “TSP”
How might you solve it?
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nearest neighbor heuristic Start at some point p0
Walk first to its nearest neighbor p1
Walk to the nearest unvisited neighbor p2, then nearest unvisited p3, … until all points have been visited
Then walk back to p0
heuristic: A rule of thumb, simplification, or educated guess that reduces or limits the search for solutions in domains that are difficult and poorly understood. May be good, but usually not guaranteed to give the best or fastest solution.
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nearest neighbor heuristic
p 0!
p 1!
p 6!
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an input where nn works badly
length ~ 84
16!
4!
1! .9! 2! p 0!
8! 16
an input where nn works badly
optimal soln for this example length ~ 64
16!
4!
1! .9! 2! p 0!
8! 17
revised heuristic – closest pairs first
Repeatedly join the closest pair of points
(such that result can still be part of a single loop in the end. I.e., join endpoints, but not points in middle, of path segments already created.)
?
How does this work on our bad example?
16!
4!
1! .9! 2! p 0!
8! 18
a bad example for closest pair
1! 1.5!
1.5!
19
a bad example for closest pair
1! 1.5!
1.5!
6+√10 = 9.16 !
vs !
8!
20
something that works
“Brute Force Search”:
For each of the n! = n(n-1)(n-2)…1 orderings of the points, check the length of the cycle;
Keep the best one
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two notes The two incorrect algorithms were greedy
Often very natural & tempting ideas
They make choices that look great “locally” (and never reconsider them)
When greed works, the algorithms are typically efficient
BUT: often does not work - you get boxed in
Our correct alg avoids this, but is incredibly slow
20! is so large that checking one billion per second would take 2.4 billion seconds (around 70 years!)
And growing: n! ~ √2 π n • (n/e)n ~ 2O(n log n)
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the morals of the story
Algorithms are important
Many performance gains outstrip Moore’s law
Simple problems can be hard
Factoring, TSP, many others
Simple ideas don’t always work
Nearest neighbor, closest pair heuristics
Simple algorithms can be very slow
Brute-force factoring, TSP
A point we hope to make: for some problems, even the best algorithms are slow
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my plan
A brief overview of the theory of algorithms
Efficiency & asymptotic analysis
Some scattered examples of simple problems where clever algorithms help
A brief overview of the theory of intractability
Especially NP-complete problems
“Basics every educated CSE student should know”
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computational complexity
The complexity of an algorithm associates a number T(n), the worst-case time the algorithm takes, with each problem size n.
Mathematically,
T: N+ → R+
i.e.,T is a function mapping positive integers (problem sizes) to positive real numbers (number of steps).
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computational complexity
Time!
T(n)!
Problem size ! 27
computational complexity: general goals Characterize growth rate of worst-case run time as a function of problem size, up to a constant factor
Why not try to be more precise?
Average-case, e.g., is hard to define, analyze
Technological variations (computer, compiler, OS, …) easily 10x or more
Being more precise is a ton of work
A key question is “scale up”: if I can afford this today, how much longer will it take when my business is 2x larger? (E.g. today: cn2, next year: c(2n)2 = 4cn2 : 4 x longer.)
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computational complexity
T(n)!
Time!
2n log2n!
n log2n! Problem size ! 29
asymptotic analysis & big-O Given two functions f and g: N→R, f(n) is O(g(n)) iff
∃ constant c > 0 so that f(n) is eventually always ≤ c g(n)
Example:
10n2-16n+100 is O(n2)
(and also O(n3)…)
why?:
10n2-16n+100 ≤ 11n2 for all n ≥ 10
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polynomial vs exponential
For all r > 1 (no matter how small) and all d > 0, (no matter how large) nd = O(rn).
1.01n
n100
In short, every exponential grows faster than every polynomial!
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the complexity class P: polynomial time
P: Running time O(nd) for some constant d
(d is independent of the input size n)
Nice scaling property: there is a constant c s.t. doubling n, time increases only by a factor of c.
(E.g., c ~ 2d)
Contrast with exponential: For any constant c, there is a d such that n → n+d increases time by a factor of more than c.
(E.g., 2n vs 2n+1)
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polynomial vs exponential growth
22n! 22n
2n/10!
2n/10
1000n2
1000n2!
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why it matters
not only get very big, but do so abruptly, which likely yields erratic performance on small instances
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another view of poly vs exp Next year's computer will be 2x faster. If I can solve problem of size n0 today, how large a problem can I solve in the same time next year?
Complexity
Increase
O(n) O(n2) O(n3) 2n /10 2n
n0 → 2n0 n0 → √2 n0 n0 → 3√2 n0 n0 → n0+10 n0 → n0 +1
E.g. T=1012 1012 106 104 400 40
→ → → → →
2 x 1012 1.4 x 106 1.25 x 104 410 41
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complexity summary
Typical initial goal for algorithm analysis is to find an
asymptotic
upper bound on
worst case running time
as a function of problem size
This is rarely the last word, but often helps separate good algorithms from blatantly poor ones - concentrate on the good ones!
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why “polynomial”? Point is not that n2000 is a nice time bound, or that the differences among n and 2n and n2 are negligible.
Rather, simple theoretical tools may not easily capture such differences, whereas exponentials are qualitatively different from polynomials, so more amenable to theoretical analysis.
“My problem is in P” is a starting point for a more detailed analysis
“My problem is not in P” may suggest that you need to shift to a more tractable variant, or otherwise readjust expectations
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algorithm design techniques
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algorithm design techniques
We will survey two:
Later: Dynamic programming
Orderly solution of many smaller sub-problems, typically non-disjoint
Can give exponential speedups compared to more bruteforce approaches
Today: Divide & Conquer
Reduce problem to one or more sub-problems of the same type, typically disjoint
Often gives significant, usually polynomial, speedup
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algorithm design techniques
Divide & Conquer
Reduce problem to one or more sub-problems of the same type
Each sub-problem’s size a fraction of the original
Subproblem’s typically disjoint
Often gives significant, usually polynomial, speedup
Examples:
Mergesort, Binary Search, Strassen’s Algorithm, Quicksort (roughly)
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divide & conquer – the key idea
Suppose we've already invented DumbSort, taking time n2
Try Just One Level of divide & conquer:
DumbSort(first n/2 elements)
DumbSort(last n/2 elements)
Merge results
Time: 2 (n/2)2 + n = n2/2 + n << n2
D&C in a
nutshell
Almost twice as fast!
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d&c approach, cont. Moral 1: “two halves are better than a whole”
Two problems of half size are better than one full-size problem, even given O(n) overhead of recombining, since the base algorithm has super-linear complexity.
Moral 2: “If a little's good, then more's better”
Two levels of D&C would be almost 4 times faster, 3 levels almost 8, etc., even though overhead is growing.
In the limit: you’ve just rediscovered mergesort.
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mergesort (review)
T(n) = 2T(n/2)+cn, n≥2
T(1) = 0
Solution: O(n log n)
Log n levels!
Mergesort: (recursively) sort 2 half-lists, then merge results.
O(n)
work
per
level!
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A Divide & Conquer Example: Closest Pair of Points
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closest pair of points: non-geometric version Given n points and arbitrary distances between them, find the closest pair. (E.g., think of distance as airfare – definitely not Euclidean distance!)
n) edges…)
(… and all the rest of the (2
Must look at all n choose 2 pairwise distances, else any one you didn’t check might be the shortest.
Also true for Euclidean distance in 1-2 dimensions?
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closest pair of points: 1 dimensional version Given n points on the real line, find the closest pair
Closest pair is adjacent in ordered list
Time O(n log n) to sort, if needed
Plus O(n) to scan adjacent pairs
Key point: do not need to calc distances between all pairs: exploit geometry + ordering
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closest pair of points. 2d, Euclidean distance: 1st try
Divide. Sub-divide region into 4 quadrants.
L
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closest pair of points: 1st try
Divide. Sub-divide region into 4 quadrants.
Obstacle. Impossible to ensure n/4 points in each piece.
L
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closest pair of points
Algorithm.
Divide: draw vertical line L with ≈ n/2 points on each side.
L
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closest pair of points
Algorithm.
Divide: draw vertical line L with ≈ n/2 points on each side.
Conquer: find closest pair on each side, recursively.
L
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12
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closest pair of points
Algorithm.
Divide: draw vertical line L with ≈ n/2 points on each side.
Conquer: find closest pair on each side, recursively.
Combine: find closest pair with one point in each side.
seems Return best of 3 solutions.
like Θ(n2) ?
L
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21
12
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closest pair of points
Find closest pair with one point in each side, assuming that distance < δ.
L
21
12
δ = min(12, 21)
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closest pair of points
Find closest pair with one point in each side, assuming that distance < δ.
Observation: suffices to consider points within δ of line L.
L
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δ = min(12, 21)
12
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δ
closest pair of points
Find closest pair with one point in each side, assuming that distance < δ.
Observation: suffices to consider points within δ of line L.
Almost the one-D problem again: Sort points in 2δ-strip by their y coordinate.
L
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5
4
6
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δ = min(12, 21)
12
3
2
1
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δ
closest pair of points
Find closest pair with one point in each side, assuming that distance < δ.
Observation: suffices to consider points within δ of line L.
Almost the one-D problem again: Sort points in 2δ-strip by their y coordinate. Only check pts within 8 in sorted list!
L
7
5
4
6
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δ = min(12, 21)
12
3
2
1
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δ
closest pair of points
Def. Let si be the point in the 2δ-strip, with the ith smallest ycoordinate.
j
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Claim. If |i – j| > 8, then the distance between si and sj is > δ.
i Pf: No two points lie in same ½δ-by-½δ box; only 8 boxes within δ
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½δ
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δ
δ
½δ
closest pair of points: analysis
Number of pairwise distance calculations:
⎧ 0 D(n) ≤ ⎨2D n /2 + 7n ) ⎩ (
n = 1⎫ ⎬ n > 1⎭ ⇒ D(n) = O(n log n)
€ (A
mostly superfluous detail: straightforward implementation gives a running time that is a factor of log n larger, due to sorting in the various subproblems. Run time can be reduced to O(n log n) also, roughly by the trick of sorting by x at the top level, and returning/merging y-sorted lists from the subcalls.
Regardless of this nuance, the big picture is the same: divideand-conquer allows sharp speed gain over a naive n2 method.)
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Integer Multiplication
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integer arithmetic
Add. Given two n-digit integers a and b, compute a + b.
Add
O(n) bit operations.
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Multiply. Given two n-digit integers a and b, compute a × b. The “grade school” method: Θ(n2) bit operations.
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Multiply
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integer arithmetic
Add. Given two n-digit integers a and b, compute a + b.
Add
O(n) bit operations.
1
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Multiply. Given two n-digit integers a and b, compute a × b. The “grade school” method: Θ(n2) bit operations.
*
0
1
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1
1
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0
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0
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Multiply
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divide-and-conquer multiplication: warmup
To multiply two 2-digit integers:
Multiply four 1-digit integers.
Add, shift some 2-digit integers to obtain result.
x y xy
= 10⋅ x1 + x 0 = 10⋅ y1 + y 0 = (10⋅ x1 + x 0 ) (10⋅ y1 + y 0 ) = 100 ⋅ x1 y1 + 10⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0
Same idea works for long integers –
can split them into 4 half-sized ints
4
5
y1 y0
3
2
x1 x0
1
0
x0⋅y0
0
8
x0⋅y1
1
5
x1⋅y0
1
2
x1⋅y1
1
4
4
0
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divide-and-conquer multiplication: warmup
To multiply two n-digit integers:
Multiply four n/2-digit integers.
Add, shift some n/2-digit integers to obtain result.
x = 2 n / 2 ⋅ x1 + y = 2 n / 2 ⋅ y1 + xy = (2 n / 2 ⋅ x1 + = 2 n ⋅ x1 y1 +
x0 y0 x 0 ) (2 n / 2 ⋅ y1 + y 0 ) 2 n / 2 ⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0
T(n) = 4T (n /2 ) + Θ(n) ⇒ T(n) = Θ(n 2 ) recursive calls
€
add, shift
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1
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y1 y0
*
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x1 x0
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x0⋅y0
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x0⋅y1
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x1⋅y1
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1
assumes n is a power of 2
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key trick: 2 multiplies for the price of 1 x y xy
= 2 n / 2 ⋅ x1 + x 0 = 2 n / 2 ⋅ y1 + y 0 = (2 n / 2 ⋅ x1 + x 0 ) (2 n / 2 ⋅ y1 + y 0 )
Well, ok, 4 for 3 is more accurate…
= 2 n ⋅ x1 y1 + 2 n / 2 ⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0
€
α β αβ
( x1 y 0 + x 0 y1 )
= = = = =
x1 + x 0 y1 + y 0 ( x1 + x 0 ) ( y1 + y 0 ) x1 y1 + ( x1 y 0 + x 0 y1 ) + x 0 y 0 αβ − x1 y1 − x 0 y 0 68
€
Karatsuba multiplication To multiply two n-digit integers:
Add two ½n digit integers.
Multiply three ½n-digit integers.
Add, subtract, and shift ½n-digit integers to obtain result.
x = 2 n / 2 ⋅ x1 y = 2 n / 2 ⋅ y1 xy = 2 n ⋅ x1 y1 = 2 n ⋅ x1 y1
+ + + +
x0 y0 2 n / 2 ⋅ ( x1 y0 + x0 y1 ) + x0 y0 2 n / 2 ⋅ ( (x1 + x0 ) (y1 + y0 ) − x1 y1 − x0 y0 ) + x0 y0
A
B
A
C
C
Theorem. [Karatsuba-Ofman, 1962] Can multiply two n€ digit integers in O(n1.585) bit ops.
T(n) ≤ 3 T ( n /2) + O(n) recursive calls
⇒ T(n) = O(n
add, subtract, shift
log 2 3
) = O(n1.585 )
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multiplication – the bottom line
Naïve:
Θ(n2)
Karatsuba:
Θ(n1.59…)
Amusing exercise: generalize Karatsuba to do 5 size n/3 subproblems → Θ(n1.46…)
Best known:
Θ(n log n loglog n)
"Fast Fourier Transform"
but mostly unused in practice (unless you need really big numbers - a billion digits of π, say)
High precision arithmetic IS important for crypto
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d & c summary
Idea:
“Two halves are better than a whole”
if the base algorithm has super-linear complexity.
“If a little's good, then more's better”
repeat above, recursively
Applications: Many.
Binary Search, Merge Sort, (Quicksort), Closest points, Integer multiply,…
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