euclidean distance geometry - École polytechniqueliberti/teaching/dgp/dgp-columbia.pdfeuclidean...

Euclidean Distance Geometry

Leo Liberti

CNRS LIX Ecole Polytechnique, France

IEOR, Columbia University, November 2015

[L., Lavor: Introduction to Euclidean Distance Geometry, in preparation]

Table of contents

1. Applications

2. Definition

3. Complexity primer

4. Complexity of the DGP

5. Number of solutions

6. Mathematical optimization formulations

7. Realizing complete graphs

8. The Branch-and-Prune algorithm

9. Symmetry in the KDMDGP

10. Tractability of protein instances

11. Finding vertex orders

12. Approximate realizations

2

Applications

1. Applications

2. Definition











3

Clock Synchronization

5

7

3

4

Atomic clock (S)16:27

ChuckAlice

Bob

↓

16:20 16:22 16:24 16:26 16:28 16:30

A C S B

16:21 16:23 16:25 16:27 16:29 16:31

[Singer, 2011]

4

Sensor network localization

0

29

0 / 2.50781

36

1 / 1.97906

51

2 / 1.13649

52

3 / 2.92955

804 / 2.5548

915 / 2.4284

175 / 2.49225177 / 1.4276

87

176 / 2.73147

211 / 0.989917214 / 2.31786

49

210 / 2.83291

84

213 / 3.02018

66 212 / 1.56417

268 / 1.76662

267 / 2.31042

271 / 1.75417

69269 / 2.0622

73

270 / 3.10479

85

272 / 1.15897

346 / 1.50056

1

18

6 / 1.90126

37

7 / 2.61352

41

8 / 1.6604

45

9 / 1.92913

57

10 / 2.66407

5811 / 0.658861

8312 / 1.32064

110 / 1.29792

111 / 0.840347

113 / 2.14071

114 / 2.03576

47

112 / 2.5425

215 / 2.12848

217 / 3.0822

218 / 1.98508

216 / 2.70003

229 / 2.97748

231 / 1.63275

232 / 2.32966

230 / 2.92252

249 / 1.49322

250 / 1.4467

251 / 3.11231

289 / 2.5056

81

290 / 2.54013

291 / 1.97931

2

5

13 / 1.62227

1414 / 2.01015

17

15 / 2.68422

20

16 / 3.05872

23

17 / 2.48332

4218 / 1.05756

64

19 / 2.21356

67

20 / 2.42912

8821 / 2.34146

93

22 / 1.0662

28 / 2.06104

29 / 1.80061

31 / 1.46243

33 / 1.3817835 / 1.01186

33

30 / 3.15392 6532 / 2.19726

70

34 / 3.17108

82 / 2.05285

83 / 1.9929

84 / 2.90088

85 / 0.965501

87 / 1.40579

88 / 2.51699

86 / 3.19003

98

89 / 2.50718

106 / 1.36517

107 / 2.2605

109 / 2.88336

55

108 / 2.81041

119 / 2.25758

123 / 0.666733

126 / 2.79445

122 / 1.26894

124 / 2.45029

71

125 / 2.61515

127 / 2.5523

56

120 / 2.46168

63

121 / 3.15779

143 / 2.62768

145 / 3.01294146 / 1.62871

89

147 / 3.16128

144 / 2.60777

233 / 1.59139

234 / 3.07459

235 / 1.62072

316 / 1.57213

318 / 2.54084

92

317 / 3.16387

326 / 2.30061

325 / 2.97621

327 / 2.62621

350 / 3.18927

323 / 0.844888

26 / 2.43216

24 / 1.43953

25 / 2.52302

27 / 1.2847

329 / 2.97101

331 / 1.47226

330 / 2.81533

79

339 / 3.17026

195 / 3.02256

196 / 2.72116

194 / 3.16431

53

193 / 0.88823

319 / 1.78484

322 / 3.15941

320 / 1.19371321 / 2.59083

332 / 2.88171

77

333 / 2.5427

6

36 / 2.6581637 / 2.84506

39 / 3.00913

39

38 / 0.977096

40 / 0.147583

90

41 / 2.54617 223 / 1.11811

225 / 1.74291

72

224 / 2.873

257 / 2.65041

7

10

42 / 1.67711

16

43 / 1.28092

3044 / 2.6039944

45 / 1.93798

46 / 0.90946

78

47 / 1.95204

86

48 / 3.1702

9749 / 2.91638

50 / 2.79803

59 / 2.64882

61 / 3.11897

63 / 2.57151

64 / 0.6334465 / 1.47059

15

58 / 2.53596 2160 / 2.72874

59

62 / 2.44092

98 / 2.45974102 / 2.89582

101 / 3.00823

100 / 2.72757

103 / 1.23805

104 / 2.65248

105 / 1.56149

24

99 / 2.76855

182 / 2.80251

183 / 2.17394

178 / 0.684621

184 / 2.04844

186 / 2.36739

179 / 1.75353

181 / 1.27559

62

180 / 2.82612

185 / 2.02673

244 / 2.90515

245 / 2.54461246 / 1.4908

248 / 2.19914

242 / 2.10075

243 / 1.81286

247 / 2.70375

334 / 2.85895

335 / 2.74921

345 / 0.96983896344 / 3.01517

852 / 1.59789

53 / 3.1109

13

51 / 2.96353

54 / 2.6062

94 55 / 2.44249

81 / 2.44372

31

80 / 3.09041

352 / 3.17363

95

353 / 1.69606

9

56 / 2.23711

57 / 2.87413

95 / 2.4309797 / 1.79475

90 / 0.575857

91 / 1.84882

60

92 / 3.1144161

93 / 0.745853

7494 / 1.08674

96 / 1.91462

132 / 2.48921134 / 1.67584

128 / 2.41661

129 / 2.54924

130 / 0.857403131 / 1.03428

133 / 1.33929

295 / 2.78515

296 / 3.03355

297 / 2.79379

292 / 2.23004

293 / 2.59059

76

294 / 2.25052

1122

66 / 2.54335

2567 / 1.59039

26

68 / 2.55644

3569 / 1.22183

4370 / 1.26323

46

71 / 0.65784172 / 1.4554873 / 1.45696

74 / 2.43597

99

75 / 0.710271

135 / 1.03939

137 / 2.64809

138 / 2.36917139 / 2.85603

141 / 1.99344

142 / 0.632129

28136 / 3.03397

140 / 1.63894

150 / 2.05658

151 / 2.68427

152 / 1.33314153 / 1.85365155 / 1.0403

156 / 1.22973

157 / 2.29037

154 / 2.67744

158 / 3.05556

159 / 2.34031

160 / 1.82595

162 / 2.70381

163 / 2.43938

161 / 1.90326

203 / 0.991465

204 / 1.81584

206 / 2.66247

207 / 2.45539

208 / 2.2551

209 / 1.2466

48205 / 3.06568

236 / 1.90515238 / 2.5944239 / 2.72018

240 / 3.16484

241 / 0.720351

237 / 2.87679

252 / 0.86256254 / 0.843728

255 / 2.43018

256 / 1.26966

253 / 2.83326

262 / 0.880856

263 / 3.06774

264 / 1.88647

261 / 1.9719

348 / 2.26962

349 / 2.1111

351 / 3.02766

12

78 / 2.35652

76 / 0.500095

77 / 3.13454

79 / 2.02458

173 / 2.85379

172 / 2.85207

174 / 2.12157

281 / 1.77315

280 / 2.13908

187 / 0.741716

188 / 1.28288

304 / 2.67349

298 / 3.19922

301 / 3.10155

303 / 1.2464

68

299 / 2.68553300 / 2.93307

82302 / 2.70564

306 / 3.166308 / 2.46248

305 / 0.378469

307 / 1.95331

341 / 2.70349340 / 1.87485

354 / 2.04535

149 / 1.71292

32148 / 2.63438

19

117 / 2.4561

118 / 1.55824

115 / 1.40904

50116 / 0.648045

259 / 2.59804

260 / 2.60371

258 / 2.04883

265 / 2.53475

266 / 1.20584

285 / 3.04711

284 / 1.28966

286 / 0.444549287 / 2.49879

283 / 0.739388

282 / 2.4463

288 / 2.40494

312 / 2.0258

313 / 1.03139314 / 2.65429

315 / 1.70638

34

189 / 2.92806

38

190 / 2.07339

40191 / 2.80145

192 / 2.35256

324 / 2.63064

323 / 2.58729

27

171 / 2.23482

167 / 2.38609

164 / 1.92141

165 / 2.6582

166 / 2.28031

168 / 2.41683 169 / 0.955716

170 / 0.967808199 / 2.18351

197 / 0.88776

198 / 1.51159

200 / 0.583974201 / 2.87535

202 / 1.31584

220 / 2.98526

219 / 1.26904

221 / 0.308754

222 / 1.8407

226 / 1.33417227 / 3.09905

228 / 1.31378

328 / 1.65874

338 / 1.28689

337 / 1.83675

75

336 / 3.11488

347 / 3.11676

310 / 2.37803

309 / 1.9722

311 / 1.13749

276 / 3.05065

277 / 2.40579

273 / 2.81885

275 / 3.02541

274 / 2.49444

54

278 / 3.02172

279 / 1.17834

343 / 1.31891

342 / 2.50522

−→

[Yemini, 1978]

5

Protein conformation from NMR data

1

2

3

4

5

9

6

8

10

7

11

−→

[Crippen & Havel 1988]

6

Definition

1. Applications

2. Definition











8

Distance Geometry Problem (DGP)

Given: Determine whether ∃:– a simple graph G = (V,E)

– an edge function d : E → R≥0– an integer K ∈ N

a realization x : V → RK s.t.

∀{u, v} ∈ E ‖xu − xv‖2 = duv

1

2

3

4

1

√3

1

1

1

1 1

2

3

4

Let n = |V |9

More applications

• Autonomous underwater vehicles [Bahr et al. 2009]

• Statics of rigid structures [Maxwell 1864]

• Matrix completion [Laurent 2009]

• Statistics [Boer 2013]

• Psychology [Kruskal 1964]

[Liberti et al., SIREV 2014]

10

Complexity primer

1. Applications

2. Definition











11

Definitions

• Decision problem: mathematical YES/NO-type question depending

on a parameter vector π

• Instance: same as above with π replaced by given values v

• Certificate: proof that a given answer is true

• P: all decision problems solvable in at most p(|π|) steps

where p is a polynomial

• NP: all decision problems with |YES certificate| ≤ p(|π|)where p is a polynomial

12

Reductions

• P,Q: decision problems

• If ∃ algorithm A which:

1. reformulates instances P of P into instances Q of Q

2. has answer(P ) = YES iff answer(A(P )) = YES

3. is polytime in the instance size |P |

then A is a reduction of P to Q

13

NP-hardness

• Q is NP-hard if every problem in NP reduces to Q

• Q is NP-complete if it is NP-hard and is in NP

Why does it work?

any P in NPpolytime reduction−−−−−−−−−−−−−−−−−−→ Q: how hard?

• Suppose Q easier than P

• Solve P by reducing to Q in polytime and then solve Q

• Then P as easy as Q, against assumption

• ⇒ Q at least as hard as P

So if Q is NP-hard it is as hard as any problem in NP

⇒ Q is as hard as the hardest problem in NP

14

NP-hardness proofs

Given a new problem Q, take any known

NP-hard problem P and reduce it to Q

Why does it work?

P : NP-hardpolytime reduction−−−−−−−−−−−−−−−−−−→ Q: how hard?

• As before: Suppose . . . (etc.) ⇒ Q at least as hard as P

• Since P is NP-hard, it is hardest in NP, and so is Q

⇒ Q is NP-hard

15

Complexity of the DGP

1. Applications

2. Definition











16

DGP ∈ NP?

• NP: YES/NO problems with polytime-checkable proofs for YES

• DGP is a YES/NO problem

• DGP1 ∈ NP, since duv = |xu − xv| ⇒ (d ∈ Q→ x ∈ Q)

• Solutions might involve irrational numbers when K > 1

• Some empirical evidence that DGP 6∈ NP [Beeker et al. 2013]

17

The DGP is NP-hard

PartitionGiven a = (a1, . . . , an) ∈ Nn, ∃ I ⊆ {1, . . . , n} s.t.

∑

i∈Iai =

∑

i 6∈Iai ?

• Reduce (NP-hard) Partition to DGP1

• a −→ cycle C with V (C) = {1, . . . , n}, E(C) = {{1,2}, . . . , {n,1}}• For i < n let di,i+1 = ai, and dn,n+1 = dn1 = an

• E.g. for a = (1,4,1,3,3), get cycle graph:

1

2 3

4

5

1

4

1

33

[Saxe, 1979]

18

Partition is YES ⇒ DGP1 is YES

• Given: I ⊂ {1, . . . , n} s.t.∑

i∈Iai =

∑

i 6∈Iai

• Construct: realization x of C in R

1. x1 = 0 // start

2. induction step: suppose xi known

if i ∈ I

let xi+1 = xi + di,i+1 // go right

else

xi+1 = xi − di,i+1 // go left

• Correctness proof: by the same induction

but careful when i = n: have to show xn+1 = x1

19

I = {1,2,3}

0 1 2 3 4 5 6

x1

0 1 2 3 4 5 6

x1 x2

1 ∈ I

0 1 2 3 4 5 6

x1 x2 x3

1 ∈ I2 ∈ I

0 1 2 3 4 5 6

x1 x2 x3 x4

1 ∈ I2 ∈ I

3 ∈ I

0 1 2 3 4 5 6

x1 x2 x3 x4x5

1 ∈ I2 ∈ I

3 ∈ I

4 6∈ I

0 1 2 3 4 5 6

x6 = x1? x2 x3 x4x5

1 ∈ I2 ∈ I

3 ∈ I

4 6∈ I5 6∈ I

Partition is YES ⇒ DGP1 is YES

(1) =∑

i∈I(xi+1 − xi) =

∑

i∈Idi,i+1 =

=∑

i∈Iai =

∑

i 6∈Iai =

=∑

i 6∈Idi,i+1 =

∑

i 6∈I(xi − xi+1) = (2)

(1) = (2)⇒∑

i∈I(xi+1 − xi) =

∑

i 6∈I(xi − xi+1)⇒

∑

i≤n(xi+1 − xi) = 0

⇒ (xn+1 − xn) + (xn − xn−1) + · · ·+ (x3 − x2) + (x2 − x1) = 0

⇒ xn+1 = x1

26

Partition is NO ⇒ DGP1 is NO

• By contradiction: suppose DGP1 is YES, x realization of C

• F = {{u, v} ∈ E(C) | xu ≤ xv}, E(C)rF = {{u, v} ∈ E(C) | xu > xv}• Trace x1, . . . , xn: follow edges in F (→) and in E(C) r F (←)

0 1 2 3-1-2-3

x1 x2x3x4 x5

∑

{u,v}∈F(xv − xu) =

∑

{u,v}6∈F(xu − xv)

∑

{u,v}∈F|xu − xv| =

∑

{u,v}6∈F|xu − xv|

∑

{u,v}∈Fduv =

∑

{u,v}6∈Fduv

• Let J = {i < n | {i, i+1} ∈ F} ∪ {n | {n,1} ∈ F}

⇒∑

i∈Jai =

∑

i 6∈Jai

• So J solves Partition instance, contradiction

• ⇒ DGP is NP-hard, DGP1 is NP-complete

27

Number of solutions

1. Applications

2. Definition











28

With congruences

• (G,K): DGP instance

• X ⊆ RKn: set of solutions

• Congruence: composition of translations, rotations, reflections

• C = set of congruences in RK

• x ∼ y means ∃ρ ∈ C (y = ρx):

distances in x are preserved in y through ρ

• ⇒ if |X| > 0, |X| = 2ℵ0

29

Modulo congruences

• Congruence is an equivalence relation ∼ on X

(reflexive, symmetric, transitive)

∼

• Partitions X into equivalence classes

• X = X/∼: sets of representatives of equivalence classes

• Focus on |X| rather than |X|

30

Cardinality of X

• infeasible ⇔ |X| = 0

• rigid graph ⇔ |X| < ℵ0

• globally rigid graph ⇔ |X| = 1

• flexible graph ⇔ |X| = 2ℵ0

• |X| = ℵ0: impossible by Milnor’s theorem

31

Milnor’s theorem implies |X| 6= ℵ0

• System S of polynomial equations of degree 2

∀i ≤ m pi(x1, . . . , xnK) = 0

• Let X be the set of x ∈ RnK satisfying S

• Number of connected components of X is O(3nK)

[Milnor 1964]

• If |X| is countable then G cannot be flexible

⇒ incongruent elements of X are separate connected components

⇒ by Milnor’s theorem, there’s finitely many of them

32

Examples

V 1 = {1,2,3}E1 = {{u, v} | u < v}d1 = 1

x1 x2

x3

ρ congruence in R2

⇒ ρx valid realization

|X| = 1

V 2 = V 1 ∪ {4}E2 = E1 ∪ {{1,4}, {2,4}}d2 = 1 ∧ d14 =

√2

x1 x2

x3

x4

ρ reflects x4 wrt x1, x2


|X| = 2 ( , )

V 3 = V 2

E3 = {{u, u+1}|u ≤ 3}∪ {1,4}d1 = 1

x1 x2

x3x4ρ rotates x2x3, x1x4 by θ


|X| is uncountable

( , , , , . . .)

33

Mathematical optimization formulations

1. Applications

2. Definition











34

System of quadratic constraints

∀{u, v} ∈ E ‖xu − xv‖2 = d2uv

• Around 10 vertices

• Computationally useless

35

Quadratic objective

minx∈RnK

∑

{u,v}∈E(‖xu − xv‖2 − d2uv)

2

• Globally optimal value zero iff x is a realization of G

• sBB: 10-100 vertices, exact solutions

• heuristics: 100-1000 vertices, poor quality

[Lavor et al., 2006]

36

Convexity and concavity

maxx∈RnK

∑

{u,v}∈E‖xu − xv‖2

∀{u, v} ∈ E ‖xu − xv‖2 ≤ d2uv

• Convex constraints, concave objective

• Computationally no better than “quadratic objective”

37

Pointwise reformulation

maxx∈RnK

∑

{u,v}∈E,k≤Kθuvk(xuk − xvk)

∀{u, v} ∈ E ‖xu − xv‖2 ≤ d2uv

• Convex subproblem in stochastic iterative heuristics

“guess θ and solve”

• 100-1000 vertices, good quality

[L. IOS14/MAGO14(slides)]

38

SDP formulation

minX�0

∑

{u,v}∈E(Xuu +Xvv − 2Xuv)

∀{u, v} ∈ E Xuu +Xvv − 2Xuv ≥ d2uv

• Similar to those of Ye, Wolkowicz — works better for proteins

• 100 vertices, good quality

[D’Ambrosio et al., in progress]

39

Realizing complete graphs

1. Applications

2. Definition











40

Cliques

2-clique

12

3-clique

1

2

3

4-clique

1

2

3

4

(K +1)-clique = K-clique ⊕ a vertex

Given a realization of the K-clique, find the position of the vertex

41

Trilateration

1

2

3

11

2

−→1 2 3

0 1 2

Example: realize triangle on a line

• From ‖x3 − x1‖ = 2 and ‖x3 − x2‖ = 1 get

x23 − 2x1x3 + x21 = 4 (1)

x23 − 2x2x3 + x22 = 1. (2)

• (2)− (1) yields

2x3(x1 − x2) = x21 − x22 − 3

⇒ 2x3 = 4,

• Hence x3 = 2

42

Realizing a (K +1)-clique in RK−1

• Apply trilateration inductively on Kassume x1, . . . , xK ∈ RK−1 known, compute y = xK+1

• K quadratic eqns (∀j ≤ K ‖y − xj‖2 = d2j,K+1) in K − 1 vars

‖y‖2 − 2x1 · y + ‖x1‖2 = d21,K+1 [1]...

...‖y‖2 − 2xK · y + ‖xK‖2 = d2K,K+1 [K]

• Form system ∀j ≤ K − 1 ([j]− [K])

2(x1 − xK) · y = ‖x1‖2 − ‖xK‖2 − d21,K+1 + d2K,K+1 [1]− [K]...

...2(xK−1 − xK) · y = ‖xK−1‖2 − ‖xK‖2 − d2K−1,K+1 + d2K,K+1 [K − 1]− [K]

• This is a (K − 1)× (K − 1) linear system Ay = b

Solve to find y

[Dong, Wu 2002]

43

“Solve”?

1. What if A is singular?

2. Or: A nonsingular but instance is NO

44

Singularity: rkA = K − 2

One row xj − xK of A depends on the others

K = 2 triangle in R1 x1 − x2 = 0x1=x2 x3?x3?

K = 3 4-clique in R2 x1, x2, x3 on a linex1 x2

x3

x4?

x4?

K = 4 5-clique in R3 x1, . . . , x4 in a plane x1x2 x3 x4

x5?

x5?

Trend continues: rk A = K − 2⇒ |X| = 2 (see later)

45

Singularity: rkA = K − 3

Two rows xj − xk depend on the others

K = 3 4-clique in R2 x1 = x2 = x3 x1=x2=x3

x4?

K = 4 5-clique in R3 x1, . . . , x4 on a line

x1

x2

x3

x4

x5?

Trend continues: [Hendrickson, 1992]

Thm. 5.8. If a graph G is connected, flexible and has more than K vertices, X

contains almost always a submanifold diffeomorphic to a circle

46

Hendrickson’s theorem also applies to non-cliques

47

Nonsingular matrix A with NO instance

• Infeasible quadratic system ∀j ≤ K ‖xK+1 − xj‖2 = d2j,K+1

• Take differences, get nonsingular A and value for xK+1

• . . . but it’s wrong!

Shit happens!

Every time you solve the linear system Ay = b

check feasibility with quadratic system

48

Algorithm for realizing complete graphs in RK

• Assume:

(i) G = (V,E) complete

(ii) |V | = n ≥ K +2

(iii) we know x1, . . . , xK+1

• Increase K: we know how to realize xK+2 in RK

• Use this inductively for each i ∈ {K +2, . . . , n}

49

Algorithm for realizing complete graphs in RK

// realize next vertex iteratively

for i ∈ {K +2, . . . , n} do// use (K +1) immediate adjacent predecessors to compute xi

if rkA = K thenxi = A−1b // A, b defined as above

elsexi =∞ // A singular, mark ∞ and exit

breakend if// check that xi is feasible w.r.t. other distances

for {j ∈ N(i) | j < i} doif ‖xi − xj‖ 6= dij then

// if not, mark infeasible and exit loop

xi = ∅break

end ifend forif xi = ∅ then

breakend if

end forreturn x

50

Complexity of Alg. 1

• Outer loop: O(n)

• Rank and inverse of A: O(K3)

• Inner loop: O(n)

• Get O(n2K3)

• But in most applications K is fixed

• Get O(n2)

But how do we find the realization of the first K +1 vertices?

51

Realizing (K +1)-cliques in RK

• Realizing (K +1)-cliques in RK−1 yields “flat simplices”

(e.g. triangles on lines)

• Use “natural” embedding dimension RK

• Same reasoning as above:

get system Ay = b where y = xK+1 and Aj = 2(xj − xK)

• But now A is (K − 1)×K

• Same as previous case with A singular

52

Almost square

How can you solve the following system Ay = b:

a11 a12 . . . a1K... ... . . . ...

aK−1,1 aK−1,2 . . . aK−1,K

y1...

yK

=

b1...

bK−1

where A has one more column than rows and rank K − 1?

53

Basics and nonbasics

• Since rkA = K − 1, ∃ K − 1 linearly independent columns

• B: set of their indices

• N : index of remaining column

• B: (K − 1)× (K − 1) square matrix of columns in B

• ⇒ B is nonsingular

• Can partition columns as A = (B|N)

Column j corresponds to variable yj

• Variables yB are called basic variables

• Variable yN is called nonbasic variable

54

The dictionary

(B|N)y = b

⇒ ByB+NyN = b

⇒ yB = B−1b−B−1NyN

Basics expressed in function of nonbasic

55

One quadratic equation

• From value of yN , can use dictionary to get yB

• Use one quadratic equation

1. Pick any h ∈ {1, . . . ,K − 1}, equation is ‖xh − y‖22 = d2hK

2. y = (yB|yN )⊤

3. Replace yB with B−1b−B−1NyN in equation

4. Solve resulting quadratic equation in one variable yN

5. Get 0,1 or 2 values for yN

6. ⇒ Get 0,1 or 2 positions for xK+1

56

What if B−1N is zero?

• yB = B−1b−B−1NyN reduces to yB = B−1b

• Use one quadratic equation

1. Pick any h ∈ {1, . . . ,K − 1}, equation is ‖xh − y‖22 = d2hK

2. y = (yB|yN )⊤

3. Replace yB with B−1b in equation

4. Solve resulting quadratic equation in one variable yN

5. Get 0,1 or 2 values for yN

6. ⇒ Get 0,1 or 2 positions for xK+1

57

The difference

• B−1N 6= 0: yNdictionary−−−−−−−→ yB

• Different values y+N 6= y−N −→ y+, y− with different components

• B−1N = 0: yBquadratic eqn.−−−−−−−−−−→ yN

• Even if y+N 6= y−N , K − 1 components of y+, y− are equal

aff(x1, . . . , xK−1) = {y ∈ RK | yN = 0}

58

The case of no solutions

• No realizations exist for this (K +1)-clique in RK

• DGP instance is NO

59

The case of one solution

• Assume for simplicity: N = K, h = 1, B−1N 6= 0Then ‖xh − y‖2 = d2h,K+1 becomes:

λy2K − 2µyK + ν = 0, where

λ = 1+∑

ℓ,j<K

β2ℓja

2jK

µ = x1K +∑

ℓ,j<K

βℓjajK(βℓjbℓ − x1ℓ)

ν =∑

ℓ,j<K

βℓjbℓ(βℓjbℓ − 2x1ℓ) + ‖x1‖2 − d21,K+1

• (Exactly one solution for yK) ⇔ µ2 = λν, not a tautology

• The set of all (K + 1)-clique DGP instances in RK s.t. µ2 = λν

has Lebesgue measure 0

• Ignore them, they happen with probability∗ 0!

∗ Assuming continuous distributions over the reals. For floating point number, who knows? . . .

but we’ll ignore these instances anyhow

60

Discriminant > 0, = 0

4

4

4

1

3

1

2

3

1

2

32

61

The case of two solutions

• K spheres SK−11 , . . . , SK−1K in RK

centered at x1, . . . , xK

with radii d1,K+1, . . . , dK,K+1

• xK+1 must be at the intersection of SK−11 , . . . , SK−1K

• If⋂

j SK−1j 6= ∅, then |⋂j S

K−1j | = 2 in general

• will not mention “probability 0” or “in general” anymore

[Coope 2000]

62

Mirror images

• Let x+ = {x1, . . . , xK, x+K+1}, x− = {x1, . . . , xK, x−K+1}assume dim aff(x1, . . . , xK) = K − 1 (†)

• Theorem

x+, x− are reflections w.r.t. hyperplane defined by x1, . . . , xK

• Proof

1. x+, x− congruent by construction

2. ∀i ≤ K xi ∈ x+ ∩ x− → x+, x− not translations

3. |x+ ∩ x−| = K < |x+| = |x−| → x+, x− not rotations by (†)4. ⇒ must be reflections

63

Algorithm for realizing (K +1)-cliques in RK

// realize 1 at the origin

x1 = (0, . . . ,0)

// realize next vertex iteratively

for ℓ ∈ {2, . . . ,K +1} do

// at most two positions in Rℓ−1 for vertex ℓ

S =⋂

i<ℓSℓ−2i

if S = ∅ then

// warn: infeasible

return ∅end if

// arbitrarily choose one of the two points

choose any xℓ ∈ S

end for

// return feasible realization

return x

64

Complexity of Alg. 2

• Outer loop: O(K)

• Gaussian elimination on A: O(K3)

• Some messing about to obtain x+K+1, x−K+1: +O(K2)

• Get O(K4)

• But in most applications K is fixed

• Get O(1)

65

Back to complete graphs

• Alg. 2: realize 1, . . . ,K +1 in RK: O(1)

• Alg. 1: Realize K +2, . . . , n: O(n2)

• ⇒ O(n2)

• What about |X|?– Alg. 1 is deterministic: one solution from x1, . . . , xK+1– Alg. 2 is stochastic: pick one of two values K times

⇒ |X| = 2K

66

Let’s look at sparser graphs

67

K-laterative graphs

• In Alg. 1 we only need each v > K + 1 to have K + 1 adjacent

predecessors in order to find a unique solution for xv

• Determination of xv from K+1 adjacent predecessors: K-lateration

• K-laterative graph:

(i) has a vertex order ensuring this property

(ii) the initial K +1 vertices induce a (K +1)-clique

the order is called K-lateration order

• Alg. 1 realizes all K-laterative graphs

The DGP restricted to K-laterative graphs in RK is tractable

[Eren et al. 2004]

68

The story so far

• Lots of nice applications

• DGP is NP-hard

• May have 0,1, finitely many or 2ℵ0 solutions modulo congruences

• Continuous optimization techniques don’t scale well

• Using K +1 adjacent predecessors, realize K-laterative graphs in

RK in polytime

• Do we need K +1 adjacent predecessors, or can we do with

less?

69

The Branch-and-Prune algorithm

1. Applications

2. Definition











70

Fewer adjacent predecessors

• Alg. 2 only needs K adjacent predecessor

• Extend to n vertices: (K − 1)-laterative graphs

• Can we realize (K − 1)-laterative graphs in RK?

• A small case: graph consisting of two K +1 cliques11

1

5

5

5

222

33

3 44

4

71

Take a closer look. . .

5

5

22

33

44

• Realization of a K +1 clique in RK knowing x1, . . . , xK

• We know how to do that!

• Consistent with 2 solutions for x5, reflected across plane through

x2, x3, x4

72

Discretization and pruning edges

• (K − 1)-laterative graph G = (V,E):

∀v > K ∃Uv ⊂ V (|Uv| = K ∧ ∀u ∈ Uv(u < v) ∧ {u, v} ∈ E)

• Discretization edges:

ED = {{u, v} ∈ E | u, v ≤ K}︸︷︷︸

initial clique

∪ {{u, v} ∈ E | v > K ∧ u ∈ Uv}︸︷︷︸

vertex order

• Pruning edges EP = E r ED

10

1

2

3

4

5

6

7

8

9

11

12

13

14

15

16

17

1819

73

Role of discretization edges

Missing discretization edge

⇒ non-rigid structure

⇒ X uncountable

Else: X finite

74

Role of pruning edges

No pruning edges: 8 incongruent realizations in R2

1

2

3

4

5 1

2

3

4

5 1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

75

Role of pruning edges

Pruning edge {1,4}: only 4 realizations remain valid

1

2

3

4

5 1

2

3

4

5 1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

76

Motivation

Protein backbones

• Total order < on V

• Covalent bond distances: {u− 1, u} ∈ E

• Covalent bond angles: {u− 2, u} ∈ E

• NMR experiments: {u− 3, u} ∈ E

(and other edges {u, v} with v − u > 3)

Generalize “3” to K

[Lavor et al., COAP 2012]

77

KDMDGP graphs

K = 2 K = 3

1

2

3

4

5

Generalization of protein backbone order:

v > K is adjacent to K immediate predecessors v − 1, . . . , v −K

KDMDGP: Discretizable Molecular Distance Geometry Problem

78

The Branch-and-Prune (BP) algorithm

BP(v, x, X):

1. Given v > K, realization x = (x1, . . . , xv−1)

2. Compute S =⋂

u∈Uv

SK−1u

3. For each xv ∈ S s.t. ∀{u, v} ∈ EP (u < v → ‖xu − xv‖ = duv)

(a) let x = (x, xv)

(b) if v = n add x to X, else call BP(v +1, x, X)

• Recursive: starts with BP(K + 1, (x1, . . . , xK), ∅)

• All realizations in X are incongruent∗

• Can be easily modified to find only p solutions for given p

• Applies to all (K − 1)-laterative graphs in RK

• Specialize to KDMDGP graph by setting Uv = {v − 1, . . . , v −K}

∗ with probability 1, and aside from one reflection at v = K +1

[L. et al. ITOR 2008]

79

Complexity of BP

• Most operations are O(Kh) for some fixed h ⇒ O(1)

• Distance check at Step 3: O(n)

• Recursion on at most 2 branches at each call: binary tree

• Only recurse when v > K, v < n: 2n−K nodes

• Overall O(n2n−K) = O(2n)

Worst-case exponential behaviour

80

Hardness of KDMDGP

• The KDMDGP is NP-hard for each K

– every DGP instance is also DMDGP if K = 1

– reduction from Partition can be extended to any K

• (K − 1)-lateration graphs are NP-hard by inclusion

• No polytime algorithm unless P=NP

Trilaterative graphs in RK are complexitywise borderline at K

81

Correctness

Thm.

When BP terminates, X contains every incongruent realization of G

Proof.

• Let y be any realization of G

• Since G has an initial K-clique, can rotate/translate/reflect y to

y[K] = x[K] for all x ∈ X

• BP exhaustively constructs every extension of x[K] which is feasible

with all distances, so y ∈ X

for a realization y, y[h] = (y1, . . . , yh) is the initial segment of y

82

Two examples

83

Empirical observations

• Fast: up to 10k vertices in a few seconds on 2010 hardware

• Precise: errors in range O(10−9)-O(10−12)

• Number of solutions always a power of 2:

obvious if EP = ∅, but otherwise mysterious

• Linear-time behaviour on proteins:

this really shouldn’t happen

84

Symmetry in the KDMDGP

1. Applications

2. Definition











[L. et al. DAM 2014]

85

Partial reflections

• For each v > K, let

gv(x) = (x1, . . . , xv−1, Rvx(xv), . . . , R

vx(xn))

be the partial reflection of x w.r.t. v

xv−3

xv−2

xv−1

• Note: the gv’s are idempotent operators

• GD = (V,ED): subgraph of G given by discretization edges

• ∀v > K reflection Rvx gives a binary choice in general∗

• XD ⊂ RnK contains 2n−K incongruent realizations of GD

∗ subsequent results hold “with probability 1”

86

Discretization group

• GD = 〈gv | v > K〉: the discretization group of G w.r.t. K

subgroup of a Cartesian product of reflection groups

• An element g ∈ GD has the form⊗

v>Kgavv , where av ∈ {0,1}

• Action of GD on XD: g(x) =(

gaK+1K+1 ◦ · · · ◦ gann

)

(x)

87

Commutativity of partial reflections

Lemma A GD is Abelian

Proof Assume K < u < v. Then

gugv(x) = gu(x1, . . . , xv−1, Rvx(xv), . . . , R

vx(xn))

= (x1 . . . , xu−1, Rugv(x)

(xu), . . . , Rugv(x)

Rvx(xv), . . . , R

ugv(x)

Rvx(xn))

= (x1 . . . , xu−1, Rux(xu), . . . , R

vgu(x)

Rux(xv), . . . , R

vgu(x)

Rux(xn))

= gv(x1, . . . , xu−1, Rux(xu), . . . , R

ux(xn))

= gvgu(x)

where equality of these terms holds by a Technical Lemma

(next slide)

[L. et al. 2013]

88

Commutativity of partial reflections

Technical Lemma

(Proof sketch for K = 2) Let y⊥Aff(xv−1, . . . , xv−K) and ρy = Rvx

ρz

ρy

t

O

z

y

ρzy

ρyt

ρzt

ρρzy

ρzρyt = ρρzyρzt

89

One realization generates all others

Lemma B The action of GD on XD is transitive

K = 2 x

y

g

gg

54

3

∃g ∈ GD (y = g(x)): namely, y = g5(g4(g3(x)))

Proof By induction on v: assume result holds to v− 1 with g′, then either

it holds for v and g = g′, else flip and let g = gvg′

[L. et al. 2013]

90

Structure and invariance

• GD is Abelian and generated by n−K idempotent elements

⇒ GD∼= Cn−K

2

• GD ≤ Aut(XD) by construction

91

Solution sets

• X: set of incongruent realizations of G

• GD defined on same vertices but fewer edges

⇒ fewer distance constraints on realizations

⇒ more realizations

• All realizations of G are also realizations of GD

⇒ X ⊆ XD

92

Losing invariance on pruning edges

Lemma C Let Wuv = {u+K +1, . . . , v} be the range of {u, v}∀x ∈ X, u,w, v ∈ V (w ∈Wuv ↔ ‖xu − xv‖ 6= ‖gw(x)u − gw(x)v‖)

Proof sketch for K = 2

u

w v

Corollary If {u, v} ∈ EP and w ∈Wuv, gw(x) 6∈ X

[L. et al. 2013]

93

Pruning group

Define:

Γ = {gw ∈ GD | w > K ∧ ∀{u, v} ∈ EP (w 6∈Wuv)}GP = 〈Γ〉

Lemma D X is invariant w.r.t. GP

Proof

Follows by corollary, invariance of XD w.r.t. GD and X ⊆ XD

94

Transitivity of the pruning group

Lemma E The action of GP on X is transitive

• Given x, y ∈ X, aim to show ∃g ∈ GP (y = g(x))

• Lemma B ⇒ ∃g ∈ GD with y = g(x) ∈ XD

• Suppose g 6∈ GP and aim for a contradiction

• ⇒ ∃{u, v} ∈ EP and w ∈W uv s.t. gw is a component of g

• Lemma C ⇒ ‖gw(x)u − gw(x)v‖ 6= duv

• If w is the only such vertex, ⇒ y = g(x) 6∈ X, contradiction

• Suppose ∃ another z ∈W uv s.t. gz is a component of g

• Set of cases s.t. ‖xu − xv‖ = ‖gzgw(x)u − gzgw(x)v‖ given ‖gw(x)u − gw(x)v‖ 6=‖xu − xv‖ 6= ‖gz(x)u − gz(x)v‖ has Lebesgue measure 0 in all DGP inputs

• By induction, holds for any number of components gz of g with z ∈W uv

• ⇒ y = g(x) 6= x against hypothesis, done

[L. et al. 2013]

95

The main result

Theorem |X| = 2|Γ|

• Lemma A ⇒ GD∼= Cn−K

2 ⇒ |GD| = 2n−K

• GP ≤ GD ⇒ ∃ℓ ∈ N (GP∼= Cℓ

2) , with ℓ = |Γ|

• Lemma E ⇒ ∀x ∈ X GPx = X

• Idempotency ⇒ ∀g ∈ GP g−1 = g

⇒ ∀g, h ∈ GP , x ∈ X (gx = hx→ h−1gx = x→ hgx = x→ hg = e→ h = g−1= g)

⇒ the mapping GPx→ GP given by gx→ g is injective

• ∀g, h ∈ GP , x ∈ X (g 6= h→ gx 6= hx)

⇒ the mapping gx→ g is surjective

• ⇒ the mapping gx→ g is a bijection

• ⇒ |GPx| = |GP |• ⇒ ∀x ∈ X |X| = |GPx| = |GP | = 2|Γ|

[L. et al. 2013]

96

Symmetry-aware BP

• Don’t need to explore all branches of BP tree

• Build Γ as a pre-processing step

• Run BP, terminating as soon as |X| = 1

• For each g ∈ GP , compute gx

[Mucherino et al. JBCB 2012]

97

Complexity

• Computing Γ: O(mn)

1. initialize indicator vector ι = (ιK+1, . . . , ιn) for gv ∈ Γ

2. initialize ι = 1

3. for each {u, v} ∈ EP and w ∈Wuv let ιw = 0

• BP: O(2n)

• Compute gx for each g ∈ GP : O(2|Γ|)

• Overall: O(2n)

• Gains depend on the instance

98

Tractability of protein instances

1. Applications

2. Definition











[L. et al. 2013]

99

Let’s handle the BP tree

1

2

3

4 29

5 17

6 13

7 12

8

9

10

11

14 15

16

18 22

19 21

20

23 24

25

26

27

28

30 42

31 38

32 37

33

34

35

36

39 40

41

43 47

44 46

45

48 49

50

51

52

53

Max depth: n, looks good! Aim to prove width is bounded

100

Number of solutions at each BP tree levelDepends on range of longer pruning edge incident to level v

2 4 8 16 32 64 128 256 512

256

128

64

32

16

8

2

1 2 3 4 5 6 7 8 9

2 4 8 16 32 64 128

2 4 8 16 32 64

2 4 8 16 32

2 4 8 16

2 4 8

2 4

2 4

K−trilaterative

K+

Green path

8 nodes at level K +4

{3,K+5} ∈ EP ⇒ gK+4, gK+5 6∈ Γ

⇒ no symmetry at levels K + 4

and K +5

⇒ only 4 nodes at level K +5

shortest pruning edge {6,K +7}

longer {5,K +7}

longest {1,K +7}

no pruning edges

101

Periodic pruning edges

1

16

2

3

4

5

6

7

8

9

10

11

12

13

14

1517

115 6 7 8 9

2 4 8 16 32 64 128 256 512

256

128

64

32

16

8

2

2 4 8 16 32 64 128

2 4 8 16 32 64

2 4 8 16 32

2 4 8 16

2 4 8

2 4

2 4

1210 ...4

• 2ℓ growth up to level ℓ, then constant: O(2ℓn) nodes in BP tree

• BP is Fixed-Parameter Tractable (FPT) in a bunch of cases

• For all tested protein backbones, ℓ ≤ 5⇒ BP linear on proteins!

102

The story so far

• Nice applications, problem is hard, could have many solutions

• Continuous methods don’t scale

• If certain vertex orders are present, use mixed-combinatorial methods

• Realize K-laterative in polytime but (K − 1)-laterative are hard

• If adjacent predecessors are immediate, theory of symmetries

• Number of solutions is a power of two

• For proteins, BP is linear time

• How do we find these vertex orders?

103

Finding vertex orders

1. Applications

2. Definition











[Cassioli et al., DAM]

104

. . . wasn’t the backbone providing them?

• NMR data not as clean as I pretended

• Have to mess around with side chains

• What about other applications, anyhow?

Methods for finding trilaterative orders automatically

105

Mostly bad news

• Finding K-laterative orders is NP-complete :-(

• But also FPT :-)

• Finding KDMDGP orders is NP-complete for all K :-(

• It’s also really hard in practice, and methods don’t scale well

106

Definitions

• Trilateration Ordering Problem (TOP)

Given a connected graph G = (V,E) and a positive integer

K, does G have a K-lateration order?

• Contiguous Trilateration Ordering Problem (CTOP)

Given a connected graph G = (V,E) and a positive integer

K, does G have a (K − 1)-lateration order such that Uv =

{v − 1, . . . , v −K} for each v > K?

Both problems are in NP

107

Hardness of TOP

• Essentially due to finding the initial clique

– brute force: test all(nK

)

subsets of V

–(nK

)

is O(nK), polytime if K fixed

• Reduction from K-Clique problem:

Given a graph, does it have a K-clique?

[Mucherino et al., OPTL 2012]

108

Reduction from K-Clique

reduction

GG′ = G⊕ U

K = 3 K − 1 dummy nodes

• If K-Clique instance is YES– start with α = (initial clique of G,U)

– induction: if αv−1 defined, pick αv at shortest path distance 1 from⋃

α

• If K-Clique instance is NO– By contradiction: suppose ∃ trilateration order α in G′

– Initial clique α[K] = (α1, . . . , αK) must have K − 1 vertices in G, 1 in U– αK+1 must be in G, hence ∃ K-clique in G


109

sh. path

K

U

n+ 1 n+ 2

ℓ = 1ℓ = 2

ℓ− 1 ℓ

110

Once the initial clique is known

Greedily grow a trilateration order α

• Initialize α with initial K-clique K

• Let W = V rK

• ∀v > K av = |vertices in K adjacent to v|// at termination, av will be the number of adjacent predecessors of v

• While W 6= ∅:

1. choose v ∈W with largest av2. if av < K, no trilateration order from this K, terminate

3. α← (α, v)

4. for all u ∈W adjacent to v, increase au5. W ←W r {v}

• Instance is YES

[Mucherino et al., OPTL 2012]

111

Greedy algorithm is correct

• Assume TOP instance is YES, proceed by induction

– start: by maximality, aK+1 > K

– assume α is a valid TOP up to v − 1, suppose av < K

– but instance is YES so there is another z ∈W with az ≥ K

– contradicts maximality of av

• Assume TOP instance is NO

– “YES” termination when W = ∅ contradicts the NO

– hence it must terminate with W 6= ∅ and “NO” answer

112

Complexity

• Outer while loop: O(n)

• Choice of largest av: O(n)

• Inner loop on W : O(n)

• Overall: O(n2)

• If we add brute force initial clique: O(2Kn2)

• Polytime if K fixed, FPT otherwise

113

CTOP is hard

• Reduction from Hamiltonian Path (HP)

Given a graph G, does it have a path passing through each

vertex exactly once?

• α a H. path in G ⇒ ∀v 6= 1, n αv is adjacent to αv−1, αv+1

• Apart from initial 1-clique α1

every αv is adjacent to its immediate predecessor

• ⇒ α is a KDMDGP order in G with K = 1

• HP is the same as CTOP with K = 1

• ⇒ By inclusion, CTOP is NP-hard


114

CTOP is hard for all K

• Reduction from HP

reduction

1

2

3C1

C2

C3

C12

C23

• Technical proof


115

How do we find KDMDGP orders?

Mathematical optimization & CPLEX

• xvi = 1 iff vertex v has rank i in the order

• Each vertex has a unique order rank:

∀v ∈ V∑

i∈nxvi = 1;

• Each rank value is assigned a unique vertex:

∀i ∈ n∑

v∈Vxvi = 1;

• There must be an initial K-clique:

∀v ∈ V, i ∈ {2, . . . ,K}∑

u∈N(v)

∑

j<i

xuj ≥ (i− 1)xvi;

• Each vertex with rank > K must have at least K contiguous adjacent predecessors

∀v ∈ V, i > K∑

u∈N(v)

∑

i−K≤j<i

xuj ≥ Kxvi.

• Do not expect too much; scales up to 100 vertices

116

How about those 10k-atom backbones?

We have Carlile for those

• Note the repetitions — they serve a purpose!

• Repetition orders are also hard to find for any K

• . . . but Carlile knows how to handcraft them!

[Lavor et al. JOGO 2013]

117

And what about the side-chains?The Carlile+Antonio tool!

[Costa et al. JOGO, 2014]

118

Approximate realizations

1. Applications

2. Definition











119

Data errors

Distance values are never precise

• Covalent bonds are fairly precise

• NMR data is a mess [Berger, J. ACM 1999]

– experimental errors yield intervals [dLuv, dUuv]

– NMR outputs frequencies of (atom type pair,distance value)

weighted graph reconstruction yields systematic error

– some atom type pairs yield more error (“only trust H—H”)

• Properties of specific molecules give rise to other constraints

• The protein graph may not be (K − 1)-laterative based on

the backbone

120

The Lavorder comes to the rescue!

• Carlile’s handcrafted repetition orders properties:

– repetitions allow a “virtual backbone” of H atoms only

– discretization edges: {v, v − i} covalent bonds for i ∈ {1,2},{v, v − 3} sometimes covalent sometimes from NMR

– most NMR data restricted to pruning edges

• When dv,v−3 is an interval: intersect two spheres with sph. shelldL

dU

• Discretize circular segments and run BP with modified S

Algorithm no longer exhaustive

121

Die Symmetriktheoriedammerung

• Intervals and discretization break the theory of symmetries��

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duv

• Only some bounds for the number b of BP solutions:

∃ℓ, k 2ℓqk ≤ b ≤ 2n−3qM

q = |discretization points|, M = |NMR discretization edges|

122

But at least it’s producing results

Joint work with Institut Pasteur

[Cassioli et al., BMC Bioinf., 2015]

123

General approximate methods

• All these methods are specialized to

protein distance data from NMR

• What about general approximate methods?

• Assume large-sized input data with errors

• No assumptions on graph structure

124

The Isomap method: Ingredients

• PDM = Partial Distance Matrix (a representation of G)

• EDM = Euclidean Distance Matrix

1. Complete the given PDM d to a symmetric matrix D

2. Find a realization x (in some dimension K)

s.t. the EDM (‖xu − xv‖) is “close” to D

3. Project x from dimension K to dimension K,

keeping pairwise distances approximately equal

125

Completing the distance matrix

• ∀{u, v} 6∈ E let Duv = length of the shortest path u→ v

• Use Floyd-Warshall’s algorithm O(n3)

1: // n× n array Dij to store distances

2: D = 0

3: for {i, j} ∈ E do

4: Dij = dij5: end for

6: for k ∈ V do

7: for j ∈ V do

8: for i ∈ V do

9: if Dik +Dkj < Dij then

10: // Dij fails to satisfy triangle inequality, update

11: Dij = Dik +Dkj

12: end if

13: end for

14: end for

15: end for

126

Finding a realization

• Let’s give ourselves many dimensions, say K = n

• Attempt to find x : V → Rn with (‖xu − xv‖2) ≈ (Duv)

• If we had the Gram matrix B of x, then:

1. find eigen(value/vector) matrices Λ, Y of B

2. since B is PSD, Λ ≥ 0⇒√Λ exists

3. ⇒ B = Y ΛY ⊤ = (Y√Λ)(Y

√Λ)⊤

4. x⊤ = Y√Λ is such that x⊤x = B

• Can we compute B from D?

127

Schoenberg’s theorem

• Standard method for computing B from D2

• Also known as classic MultiDimensional Scaling (MDS)

• Apply many algebraic manipulations to

d2uv = ‖xu − xv‖2 = xu⊤xu + xv

⊤xv − 2xu⊤xv

where the centroid∑

k≤nxuk = 0 for all u ≤ n

• Get B = −12(In −

1n1n)D

2(In − 1n1n), i.e.

xu · xv =1

2n

∑

k≤n(d2uk + d2kv)− d2uv −

1

2n2

∑

h≤nk≤n

d2hk

• D “approximately” EDM ⇒ B “approximately” Gram

[Schoenberg, Annals of Mathematics, 1935]

128

Project to RK for a given K

• Only use the K largest eigenvalues of Λ

• Y [K] = K columns of Y corresp. to K largest eigenvalues

• Λ[K] = K largest eigenvalues of Λ on diagonal

• x⊤ = Y [K]√

Λ[K] is a K × n matrix

• Y [K] span the subspace where x “fills more space”, i.e. neglecting other

dimensions causes smaller errors w.r.t. the realization in Rn

This method is called Principal Component Analysis (PCA)

129

Isomap

Given K and PDM d:

1. D = FloydWarshall(d)

2. B = MDS(D)

3. x = PCA(B,K)

−→

[Tenenbaum et al. Science 2000]

130

Bonus material

Other topics

• DGP on a sphere

• DGP with different norms

ℓ1? ℓ∞? geodesics?

• Unassigned DGP

132

Realizing rigid graphs on a unit sphere

trivial extension: we get one more constraint ‖x‖2 = 1 for free!

Everything’s simpler on SK

main building block: K-lateration

Realize y knowing x1, . . . , xK+1and distances di from xi to y (i ≤ K +1)

‖xi − y‖2 = d2i

‖xi‖2 + ‖y‖2 − 2xiy = d2i

1 + 1− 2xiy = d2i

xiy = 1− d2i /2

Ay = b

where A =

x11 x12 · · · x1,K+1... ... . . . ...

xK+1,1 xK+1,2 · · · xK+1,K+1

b = (1− d2i /2 | i ≤ K +1)⊤

134

(K − 1)-lateration on SK

As in the Euclidean case, replace dictionary in ‖y‖ = 1, get

‖y‖2 = 1

‖yB‖2 + y2N = 1

‖B−1b− yNB−1N‖2 + y2N = 1

yN =−(B−1b)(B−1N)±

√

(B−1bB−1N)2 − ‖B−1N‖4 +1

‖B−1N‖2 +1

yB = B−1b− yNB−1N

Again, ≤ 2 possible positions for y

135

When distances are geodesic

Godel’s Distance Geometry theorem

Thm.

Any weighted 4-clique which can be realized in R3 but not R2 can

also be realized on rS2 (for some r > 0) with geodesic distances.

137

Proof 1/3: geodesic and chord lengths

• a1, . . . , a6: given distance lengths on 4-clique edges

aim to realize them on a sphere as geodesics

• cx(α): length of chord of geodesic α having radius r = 1/xα

cx(α)

1/x

• τ(x): tetrahedron in R3 having side lengths cx(ai) for i ≤ 6

• φ(x): radius inverse of circumscribed sphere about τ(x)

138

Proof 2/3: straighten the geodesics

• T : realization a1, . . . , a6 in R3

• r: radius of the sphere circumscribed around T

• limx→0

cx(α) = α

as 1/x→∞ implies chord lengths = geodesic lengths

• ⇒ limx→0

τ(x) = T and limx→0

φ(x) = 1/r

139

Proof 3/3: the fixed point

• By existence of T , φ(0) = 1/r makes φ continuous at 0

• Technical claim: φ has a fixed point in the interval I = (0, π/a′),where a′ = max(a1, . . . , a6)

• Let y by the fixed point of φ, i.e. φ(y) = y

• τ(y) circumscribed by a sphere σ having radius 1/y

• By defn. τ(y): tetrahedron s.t. side lengths = chords subtending

geodesics a1, . . . , a6 on circumscribed sphere

• ⇒ rlz. τ(y) on σ has geodesic lengths equal to a1, . . . , a6

140

Other norms?

Open research question

143

Unassigned DGP

Given two integers K > 0, n > 0 and a sequence d = (d1, . . . , dh) for

some h, find a configuration x = (x1, . . . , xn) ⊆ RK such that:

∀i ≤ h ∃u < v ≤ n ‖xu − xv‖2 = di. (3)

Another open research question

But look for the works of Phil Duxbury

144

Some references

• L. Liberti, C. Lavor, N. Maculan, A. Mucherino, Euclidean distance geometry

and applications, SIAM Review, 56(1):3-69, 2014

• L. Liberti, B. Masson, J. Lee, C. Lavor, A. Mucherino, On the number

of realizations of certain Henneberg graphs arising in protein conformation,Discrete Applied Mathematics, 165:213-232, 2014

• L. Liberti, C. Lavor, A. Mucherino, The discretizable molecular distance geometry

problem seems easier on proteins, in [see below], 47-60

• A. Mucherino, C. Lavor, L. Liberti, N. Maculan (eds.), Distance Geometry:

Theory, Methods and Applications, Springer, New York, 2013

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