CS280 HW1 Solution Set Spring2002. now, we need to get rid of the n term. We know that:

Transcription

1 CS80 HW Solutio Set Sprig00 Solutios by: Shddi Doghmi -) -) 4 * ) 4 b-) ) ) ) * ) ) ) 0 ) c-) ) ) ) ) ) ow we eed to get rid of the term. We ow tht: ) ) ) ) substitute ito the recursive epressio: ) so the solutio is: 6 *3 * d-) Two possible solutios I c thi of: solutio : ) solutio :

2 by similr resoigctully its ectly the sme!!) to c-) we get: 4 -) We wt to prove iductively tht: The ple is divided by lies) the regios c be colored usig blue d red such tht o two djcet regios hve the sme color) Note tht two regios re djcet if d oly if they shre edge I will use the word edge isted of side here to void mbiguity) Bsis: The ple is divided by 0 lies >we c color the oly regio blue> we c color it usig blue d red such tht o two djcet regios hve the sme color Iductive Hypothesis: The ple is divided by lies) the regios c be colored usig blue d red such tht o two djcet regios hve the sme color) Iductio: Give y lies dividig the ple we c remove lie i the ple to me totl of lies i the ple. Therefore the rrgemet of lies c be formed s such from rrgemet of lies by ddig the desired lie to me rrgemet of lies i the ple. Let us strt with our iitil lies. By the Iductio Hypothesis we c color the regios such tht o two djcet regios hve the sme color. Color them to chieve this. No two djcet regios hve the sme color if d oly if ll edges obey predicte P: Ps): At y poit o edge s the color to tht poit s right is differet from the color to tht poit s left Now dd the desired lie let s cll it lie λ. Ivert ll the colors o the right of tht lie blue becomes red d red becomes blue). For ll edges ot o λ we ow tht prior to the iversio ll such edges obeyed predicte P. After the iversio ech edge to the right of λ hd the colors o both its right d its left iverted which implies tht the colors remied differet!

3 Blue-Red becme Red-Blue). Edges to the left of λ were ot ffected d still obey predicte P. Therefore: Eve fter ddig lie λ d ivertig colors to λ s right ll edges NOT o λ still obey predicte P. Now wht bout edges ON lie λ? Whe we dded lie λ d before ivertig colors ech edge o λ pssed through wht ws sigle regio before λ cme ito beig by the defiitio of edge). Therefore ech edge o λ hd the sme colors to both it s left d it s right. The cme our iversio which iverted ll colors to the right of λ cusig ech edge o λ to hve differet colors to the left d right. Therefore fter ddig λ d ivertig the right side of λ: All edges o λ hve differet colors to the left d right Thus we coclude tht fter ddig λ d ivertig colors to λ s right: All edges be they o λ or ot hve differet colors to their right d left Hece give the iductio hypothesis we hve proved tht rbitrry rrgemet of lies c be colored such tht o two djcet regios hve the sme color: The ple is divided by lies) the regios c be colored usig blue d red such tht o two djcet regios hve the sme color) Coclusio: Give the bsis cse d the iductio we coclude tht: The ple is divided by lies) the regios c be colored usig blue d red such tht o two djcet regios hve the sme color) 3-) Bsis: 0 lies i the ple ple is divided ito regio. ple is divided ito 00)/ regios Iductive Hypothesis: lies i the ple ple is divided ito )/ regios Iductio: I rrgemet of lies we c remove lie to get rrgemet of lies. Therefore the rrgemet of lies c be formed from rrgemet of lies by ddig the desired lie.

4 Accordig to the iductive hypothesis before ddig the etr lie the ple hs )/ regios. Now we dd the desired lie λ. Sice the lie is ifiite ot segmet) d is ot prllel to y other lie ccordig to the questio specifictio) it must itersect ech of the lies lredy i the ple. Sice the questio lso specifies tht o 3 lies c itersect t poit we ow tht our ew lie λ will itersect ech of the lies i the ple seprtely. Sice λ will itersect lies seprtely these lies will chop λ up ito segmets this is esy to see ituitively but c be proved iductively). Ech of the segmets psses through wht ws oe regio before λ ws dded. Hece regios re divided ito ech geertig more regios. So fter ddig λ we ed up with this my regios: )/ ))/ Therefore if our iductive hypothesis is true the: lies i the ple ple is divided ito ))/ regios Or more precisely: IH ) IH ) Where IH is the iductive hypothesis predicte pplied to vrible. Coclusio: lies i the ple ple is divided ito )/ regios) 4-) The sttemet is flse. Proof by couter-emple: Te ) 4* 43 4*43 is obviously divisible by 4 which is ot itself or. This mes tht it is ot prime. Therefore it is ot true tht 4 is prime for ll turl umbers. 5-) We wt to prove iductively tht:

5 ... Bsis: For 0 the bsolute vlue of the sum of 0 umbers 0. The sum of the bsolute vlues of 0 umbers lso0. They re equl. Therefore the iductive hypothesis below is true for 0. Iductive Hypothesis:... Iductio: For y... : ow if d hve the sme sigve or -ve) the the bove epressio is equl to: otherwise if they hve differet sigs the the epressio is equl to the differece betwee their bsolute vlues: Therefore the gretest vlue c hve is: d it follows tht: IEQ: Usig our iductio hypothesis: IH: Combie IH d IEQ to get:

6 So give our IH ) predicte we proved tht:... Which is IH ) Coclusio:... 6-) The iductive by itself step is flwless. The bsis step by itself is lso flwless. However it is ot correct to combie them to form iductive proof. The reso is tht the iductive step uses d to prove the iductive hypothesis for. Therefore y vlid Bsis step MUST defie the first two elemets of the sequece. Oly by defiig the first two elemets 0 d ) d provig the iductive hypothesis for these elemets c we begi to costruct vlid iductive proof. If we modify the Bsis step d defie 0 d we would relize tht the Iductive Hypothesis ctully DOES NOT hold for the Bsis cse hece destroyig the iductive rgumet before it eve begis.