Computational Geometry

Transcription

1 Computatioal Geometry Chapter 4 Liear programmig Duality Smallest eclosig disk O the Ageda Liear Programmig Slides courtesy of Craig Gotsma Liear Programmig - Example Defie: (amout amout cosumed per day) i types of foods ( i d). j types of vitamis ( j ). x i the amout of food of type I cosumed per day). a ji the amout of vitami j i oe uit of food i. c i the umber of calories i oe uit of food i. b j miimal required amout of vitami j. Costraits (we eed to cosume some miimal amout of each vitami): ax + a x + L + ad xd b T Miimize : c x M Subject to : Ax b a x + a x + L + a x b d d Miimize: the total umber of calories cosumed: C( x) = c x + c x + L+ cd xd 4. Liear Programmig The Geometry Each costrait defies defies a half-space regio i d-dimesioal space. The feasible regio is the (covex) itersectio of these half-spaces. We will treat the case d =, where each costrait defies a half-plae. 44. More Geometry Degeerate Cases The feasible regio may be: The solutio to the liear program is a poit i the feasible regio that is extreme i the directio of the target fuctio. Theorem: Ay bouded liear program that is feasible has a uique solutio, whics a vertex of the feasible regio. Proof: Covexity c Empty Ubouded The solutio may be: Not uique

2 The Simplex Algorithm Assume WLOG that the cost fuctio poits dowwards. Costruct (some of) the vertices of the feasible regio. Walk edge by edge dowwards util reachig a local miimum (whics also a global miimum). I R d, the umber of vertices might be Θ ( d/ ). c LP History Mid 0 th cetury: Simplex algorithm, time complexity Θ ( d/ ) i the worst case. 980 s (Khachiya) ellipsoid algorithm with time complexity poly(,d). 980 s (Karmakar) iterior-poit algorithm with time complexity poly(,d). 984 (Megiddo) parametric search algorithm with time complexity O(C d ) where C d is a costat depedet oly o d. E.g. C d = d^. The holy grail: A algorithm with complexity idepedet of d. I practice the simplex algorithm is used because of its liear expected rutime O( log ) ) D Liear Programmig Iput: half plaes. Cost fuctio that WLOG poits dow. Algorithm:. Partitio the half-plaes ito two groups.. Compute, recursively, the feasible regio for each group.. Compute the itersectio of the two feasible regios. 4. Check the cost fuctio o the regio vertices. Divide ad Coquer Complexity Aalysis Stage : Itersectio of two covex polygos plae sweep algorithm. No more tha four segmets are ever i the SLS ad o more tha eight evets i the EQ O(). Stage 4: Fid the miimal cost vertex - O(). T() = T(/)+O() T() = O( log ) O( ) Icremetal Algorithm Icremetal Algorithm - Symbols The idea: Start by itersectig two halfplaes. Add halfplaes oe by oe ad update optimal vertex by solvig oe-dimesioal LP problem o ew lie if eeded. l i the i th half plae the lie that defies h C l l v h l v C i the feasible regio after i costraits C h the optimal vertex of C i 4. 4.

3 Icremetal Algorithm Basic Theorem Basic Theorem - Cot. Theorem:. if -, the = -. // O() check, othig to do. if -, the either C i = // termiate or C i = C i- ad lies o l i // ru D LP Proof:. Trivial. Otherwise would ot have bee optimum before Assume that is ot o l i. must be i C i- By covexity, also the lie - is i C i-. Cosider poit v j - the itersectio of - with l i. v j is i both C i- ad C i, ad is better tha. Cotradictio. v j l i Fidig give l i (oe-dimesioal LP) Complexity Aalysis Itersect each h j (j<i) with l i, geeratig i- rays represetig (ubouded) itervals. Itersect the i- itervals i O(i) time. If the itersectio is empty the report o solutio, else report the lowest poit. T ( ) = i= O( i) = O( ) Icremetal Algorithm O() Radomized Versio Exactly like the determiistic versio, oly the order of the lies is radom. Theorem: The expected rutime of the radom icremetal algorithm (over all! permutatios of the iput costraits) is O(). Complexity Aalysis The expected rutime is: [ O()( E( xi )) + O( i) E( xi )] O( ) + [ O( i) E( xi )] i= i= where x i is a radom variable: vi vi // ru D LP xi = 0 vi = vi // do othig

4 Backward aalysis Probability Aalysis Questio: Whe give a solutio after i halfplaes, what is the probability that the last half-plae affected the solutio? Aswer: Exactly /i, because a chage ca occur oly if the last halfplae iserted is oe of the two halfplaes thru. (ote that depeds o the i half-plaes, but ot o their order) Complexity Aalysis E( xi ) = Pr( vi vi ) i O( ) + O( i) E( xi ) = O( ) + O i = O( ) i= i= i Just to Make Sure False Claim: The probabilistic aalysis is for the average iput. Hece there exist bad sets of costraits for which the algorithm s expected rutime is more tha O(), ad there exist good sets of costraits for which the algorithm s expected rutime is less tha O(). True Claim: The probabilistic aalysis is valid for all iputs. The expected complexity is over all permutatios of this iput. Smallest Eclosig Disk Iput: poits. Output: Disk with miimal radius that cotais all the poits. Theorem: For ay fiite set of poits i geeral positio, the smallest eclosig disk either has at least three poits o its boudary, or two poits which form a diameter. If there are three poits, they subdivide the circle ito three arcs of legth o more tha π each. Prove! This immediately implies a O( 4 ) algorithm (why?) Basic Theorem Theorem: Usig a icremetal algorithm, where D i is the updated disk after seeig the first i poits p,.., p i : Proof: If p i D i- the p i is o the boudary of D i. Observatio: If r <r the a<π. p i D i- r <r => a<π p i D i- q,q,q D i- Arc(q,q )>π. Cotradictio. D i- q q a q r r D i p i Icremetal O() ) Expected Time Algorithm Costruct the procedures: MiDisk(P) fid a smallest eclosig disk for a set of poits P. MiDisk(P,q) fid a eclosig disk for a set of poits P which touches poit q. MiDisk(P,q,q ) fid a eclosig disk for a set of poits P, which touches poits q ad q. Disk(q,q,q ) fid a disk thru poits q q ad q (easy)

5 MiDisk(P) Icremetal Algorithm MiDisk(P,q) Icremetal Algorithm D = the miimal disk through p ad p. For each poit p i i radom order ( i ): If p i D i- the D i = D i- // do othig Else D i = MiDisk(P i-,p i ). // look for other two poits o disk D = the miimal disk through q ad p. For each poit p i ( i ): If p i D i- the D i = D i- // do othig Else D i = MiDisk(P i-,q,p i ). // look for other oe poit o disk Retur D Retur D MiDisk(P,q,q ) Icremetal Algorithm D 0 = the miimal disk through q ad q. For each poit p i ( i ): If p i D i- the D i = D i- // do othig Else D i = Disk(q,q,p i ). // form disk Retur D Complexity Aalysis Use backward aalysis o poit orderig. Total time complexity: i= O( i) = O( ) i Liear expected rutime. Worst case: O( )