CPSC 490: Problem Solving in Computer Science 1 Convex sets A set S - - PowerPoint PPT Presentation

CPSC 490: Problem Solving in Computer Science

Lecture 16: Convex hull, binary/ternary search

Henry Xia, Brandon Zhang

based on CPSC 490 slides from 2014-2018

2019-03-14

University of British Columbia

Convex sets

A set S is convex ⇔ ∀x, y ∈ S, 0 ≤ λ ≤ 1, λx + (1 − λ)y ∈ S ⇔ the line segment x → y is inside the set

Figure 1: In a convex set, the line segment between any 2 points lies in a set

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Convex hull

A convex hull of a set is the smallest convex set containing the set.

Figure 2: Some sets and their convex hulls

Why do we care about convex hulls? They give us nice convex polygons, which we can

• ptimize over easier.

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Monotone chain algorithm

Step 1: scan from leħt to right, “wrap” around the top Step 2: scan from right to leħt, “wrap” around the bottom.

Figure 3: Upper and lower hulls built by monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

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Monotone chain algorithm

Algorithm summary

• Sort all points by x-coordinate
• Start from leħt most point
• Iteratively add points into the hull
• If adding a point causes CCW turn, pop points from hull until this is no longer the case

before adding!

• Repeat the above steps from right to leħt.

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Monotone chain algorithm

Edge cases

• Convex hull of 1, 2, 3 points
• Collinear points

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Monotone chain algorithm

Time complexity analysis

• Sort all points by x-coordinate: O(n log n)
• Each point is added and removed at most once: O(n)

⇒ Time complexity: O(n log n) Can we do better? Arrange n points on a circle equivalent to to sorting, n n . However, not all points on the hull, so actually O n h possible.

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Monotone chain algorithm

Time complexity analysis

• Sort all points by x-coordinate: O(n log n)
• Each point is added and removed at most once: O(n)

⇒ Time complexity: O(n log n) Can we do better? Arrange n points on a circle ⇒ equivalent to to sorting, Ω(n log n). However, not all points on the hull, so actually O(n log h) possible.

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Merging convex hulls

Consider the following alternate divide & conquer approach:

• 1. Split points into two halves
• 2. Compute convex hull of each half
• 3. Merge two convex hulls into one

How do we merge?

Figure 4: Two convex hulls

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Merging convex hulls

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Merging convex hulls

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Merging convex hulls

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Merging convex hulls

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Merging convex hulls

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Merging convex hulls

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Merging convex hulls

Basic idea

• Start with the nearest two points
• Alternate moving leħt side CW and right side CCW along hull
• Stop when cannot move anymore

⇒ Time complexity of merge: O(n) ⇒ Divide and conquer also solves convex hull in O(n log n).

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Merging convex hulls in 3D

Similar idea works in 3D!

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Merging convex hulls in 3D

Project to 2D, use 2D convex hulls to find “bottom edge”

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

“Giħt-wrap” to find faces, moving to neighbor vertex of current edge

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Merging convex hulls in 3D

Summary

• Compute hull of 2D projection and find “bottom edge”
• “Giħt-wrap” smartly to find each face around in a circle
• Sort neighbors of leħt vertex CCW, neighbors of right vertex CW
• For leħt & right vertices, find their “candidate neighbor”
• Candidate neighbor of leħt vertex = neighbor of leħt vertex such that leħt convex hull is entirely
• n one side of plane of new triangle
• There is only one possible candidate
• Giħt wrap: find which candidate neighbor (leħt or right) causes “least rotation” of the

plane and add that one

• Update new: for the new vertex sort neighbors and find candidate
• Update old: for vertex that did not move, new candidate neighbor must be further along

in the sorted list! ⇒ amortized O(degree)

⇒ Time complexity: O(n) merge ⇒ O(n log n) divide and conquer for 3D convex hull!

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Problem 1 – Binary search on a convex hull

Given N ≤ 100, 000 points on the plane, answer Q ≤ 100, 000 queries of the following type: If you “look” from point P, what can you “see”? P

Figure 5: Finding the leħtmost and rightmost point when viewed from P

Hint: to binary search on a circular list, first cut the list in half.

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Problem 1 – Solution

Observation 1: solution lies on convex hull Observation 2: pick an arbitrary point on hull, draw a line through, then one extremum on each side of the line. P

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Problem 1 – Solution

Observation 1: solution lies on convex hull Observation 2: pick an arbitrary point on hull, draw a line through, then one extremum on each side of the line. ⇒ binary search to find start/end vertices on hull for each half, then binary search on each half of the hull for when P → pi → pi+1 changes from CCW to CW (or vice versa). Time complexity: O(N log N + Q log N)

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Problem 2 – Dynamic programming on a convex hull

Given N ≤ 100 points on the plane, and K ≤ N, what’s the maximum area you can enclose by picking up to K out of N points to form a polygon?

Figure 6: Enclosing an area by picking a subset of points to form polygon.

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Problem 2: Solution

Observation 1: must pick points on the convex hull Observation 2: convex hull is convex polygon ⇒ run DP on intervals of vertices on the convex hull Main idea: build the polygon by adding triangles one by one dp(l, r, k) = maximum enclosed area using k vertices out of l, l + 1, . . . , r given that l → r is an edge on the boundary = max

l<x<r Area(l, x, r) + dp(x, r, k − 1)

Time complexity: O(N3K)

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Ternary search

We can use binary search to find points of interest on a monotonically increasing function. What about minimizing a unimodal function (one that decreases, then increases)? Evaluate it at two points in the middle, if leħt < right, then already increasing in middle section, so minimum not in right section!

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Ternary search

What if two probe points are equal? Must be strictly unimodal (i.e. strictly decrease to minimum, then possibly stay flat at minimum, then strictly increase), to avoid failure!

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1 2

and 1 2 ?

• This is just binary search on derivative
• Make sure

not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios
• ne probe reused next iteration
• Let d

R L, 5 1 2, then probe at L d, R d What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle 1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1/2 − ϵ and 1/2 + ϵ?
• This is just binary search on derivative
• Make sure ϵ not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios
• ne probe reused next iteration
• Let d

R L, 5 1 2, then probe at L d, R d What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle 1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1/2 − ϵ and 1/2 + ϵ?
• This is just binary search on derivative
• Make sure ϵ not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios
• ne probe reused next iteration
• Let d

R L, 5 1 2, then probe at L d, R d What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle 1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1/2 − ϵ and 1/2 + ϵ?
• This is just binary search on derivative
• Make sure ϵ not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios

⇒ one probe reused next iteration

• Let d = R − L, ϕ = (

√ 5 − 1)/2, then probe at L + ϕd, R − ϕd What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle 1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1/2 − ϵ and 1/2 + ϵ?
• This is just binary search on derivative
• Make sure ϵ not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios

⇒ one probe reused next iteration

• Let d = R − L, ϕ = (

√ 5 − 1)/2, then probe at L + ϕd, R − ϕd What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle+1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search

What if we put probe points closer to middle?

• Converges faster! (by a constant factor)
• So choose 1/2 − ϵ and 1/2 + ϵ?
• This is just binary search on derivative
• Make sure ϵ not too small, otherwise you’ll run into numerical errors.

Can we use less probes per iteration?

• Golden section search: probe at golden ratios

⇒ one probe reused next iteration

• Let d = R − L, ϕ = (

√ 5 − 1)/2, then probe at L + ϕd, R − ϕd What about on an integer domain?

• This is easier then usual! Just use two probes at middle and middle+1.

Does it work in 2D/3D? Yes if f is convex.

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Ternary search (continuous)

Find minimum of strictly unimodal function f : R → R

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def ternary_search(l, r, f):

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while r - l > EPS:

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mid1 = l + (r-l)/3

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mid2 = l + 2*(r-l)/3

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if f(mid1) < f(mid2): rb = mid2

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else: lb = mid1

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return (l+r)/2

Time complexity: O(log r−l

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Ternary search (discrete)

Find minimum of strictly unimodal function f : Z → R

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def ternary_search(l, r, f):

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while l < r:

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set mid1 = (l+r)/2

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set mid2 = mid1+1

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if f(mid1) < f(mid2): rb = mid2

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else: lb = mid1

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return lb

Time complexity: O(log(r − l))

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Problem 3 – Convex optimization

Find the point that minimizes the total distance to all the other points. The points could be in d dimensions (d ≤ 5).

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Problem 3 – Solution

Observation 1: distance function is convex. Observation 2: sum of convex functions is convex ⇒ do d-dimensional ternary search

• Ternary search on dimension 1
• Given a probe point in dimension 1, do (d − 1)-dimensional ternary search to find

minimum with dimension 1 fixed, take this as the “value” of the function Time complexity: O(n logd ε−1)

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Problem 4 – Rectangle through L-Shaped Corridor

Consider if we had to move a long piece of furniture that had dimensions ℓ × w with ℓ ≤ w through an L-shaped corrider. If we know the dimensions of the corridor a and b, as well as the length ℓ of the furniture, how wide can the furniture be?

Source: Codeforces 98C

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Problem 4 – Solution

Consider the following cases when a ≤ b.

• ℓ ≤ a ≤ b then even if ℓ = w this can fit through the corridor
• a < ℓ ≤ b then a is the maximum. Clearly this fits through the corridor, and no larger

w can, since both dimensions can’t be larger than ℓ

• a ≤ b < ℓ then we need to rotate the furniture through the corridor. Consider if we

rotate with two corners of the furniture touching the outer sides of the corridor, then

• ur answer is the minimum distance of this line to the opposite corner.

This distance is unimodal, so we can ternary search for the answer as we rotate one side!

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References

• http://www.cs.uu.nl/docs/vakken/ga/slides1.pdf
• http://www.loria.fr/~pougetma/enseignement/webimpa/

2-3D-enveloppe-convexe-od.pdf

• http://www3.jouy.inra.fr/miaj/public/vigneron/talks/diam.pdf
• http://www.cs.umd.edu/~mount/754/Lects/754lects.pdf

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