{"id":53444,"date":"2019-12-02T00:00:00","date_gmt":"2019-12-01T21:00:00","guid":{"rendered":"https:\/\/prohoster.info\/blog\/blog_prohoster\/shema-razdeleniya-sekreta-shamira"},"modified":"2020-02-18T14:01:21","modified_gmt":"2020-02-18T11:01:21","slug":"shema-razdeleniya-sekreta-shamira","status":"publish","type":"post","link":"https:\/\/prohoster.info\/sq\/blog\/administrirovanie\/shema-razdeleniya-sekreta-shamira","title":{"rendered":"Schema e ndarjes s\u00eb sekretit t\u00eb Shamir","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Le t\u00eb shqyrtojm\u00eb nj\u00eb skenar ku \u00ebsht\u00eb e nevojshme t\u00eb sigurohet siguria e depozit\u00ebs bankare. Ajo konsiderohet plot\u00ebsisht e pap\u00ebrshkueshme pa \u00e7el\u00ebsin q\u00eb ju jepet n\u00eb dit\u00ebn e par\u00eb t\u00eb pun\u00ebs. Q\u00ebllimi juaj \u00ebsht\u00eb ta ruani \u00e7el\u00ebsin me besim.<\/p>\n

Supozoni se vendosni ta mbani gjithmon\u00eb \u00e7el\u00ebsin me vete, duke ofruar akses n\u00eb depozit\u00eb sipas nevoj\u00ebs. Por shpejt do t\u00eb kuptoni se ky zgjidhje nuk \u00ebsht\u00eb praktike pasi \u00e7do her\u00eb p\u00ebr t\u00eb hapur depozit\u00ebn k\u00ebrkohet prania juaj fizike. Dhe si ndodhet me pushimet q\u00eb ju jan\u00eb premtuar? P\u00ebr m\u00eb tep\u00ebr, pyetja e frikshme \u00ebsht\u00eb: \u00e7far\u00eb n\u00ebse e humbni \u00e7el\u00ebsin e vet\u00ebm?<\/p>\n

Me mendimin p\u00ebr pushimet, vendosni t\u00eb b\u00ebni nj\u00eb kopje t\u00eb \u00e7el\u00ebsit dhe t'ia besoni nj\u00eb kolegu tjet\u00ebr. Megjithat\u00eb, kuptoni se kjo gjithashtu nuk \u00ebsht\u00eb ideale. Duke dyfishuar numrin e \u00e7el\u00ebsave, ju gjithashtu keni dyfishuar mund\u00ebsit\u00eb p\u00ebr vjedhje t\u00eb \u00e7el\u00ebsit.<\/p>\n

T\u00eb d\u00ebshp\u00ebruar, e shkat\u00ebrroni kopjen dhe vendosni t\u00eb ndani \u00e7el\u00ebsin origjinal n\u00eb dy pjes\u00eb. Tani, mendoni se dy njer\u00ebz t\u00eb besuar me fragmente t\u00eb \u00e7el\u00ebsit duhet t\u00eb jen\u00eb fizikisht t\u00eb pranish\u00ebm p\u00ebr t\u00eb b\u00ebr\u00eb \u00e7el\u00ebsin dhe hapur depozit\u00ebn. Kjo do t\u00eb thot\u00eb se vjedh\u00ebsi duhet t\u00eb vjedh\u00eb dy fragmente, q\u00eb \u00ebsht\u00eb dyfish m\u00eb e v\u00ebshtir\u00eb se vjedhja e nj\u00eb \u00e7el\u00ebsi t\u00eb vet\u00ebm. Megjithat\u00eb, s\u00eb shpejti kuptoni se ky skem\u00eb nuk \u00ebsht\u00eb shum\u00eb m\u00eb e mir\u00eb se nj\u00eb \u00e7el\u00ebs i vet\u00ebm, pasi n\u00ebse dikush humb nj\u00eb nga pjes\u00ebt e \u00e7el\u00ebsit, \u00e7el\u00ebsi i plot\u00eb nuk mund t\u00eb rikuperohet.
\n<\/a><\/noindex>
\nProblemi mund t\u00eb zgjidhet me nj\u00eb seri \u00e7el\u00ebsash dhe \u00e7el\u00ebsh shtes\u00eb, por me k\u00ebt\u00eb qasje shpejt do t\u00eb nevojiten shum\u00eb<\/i> \u00e7el\u00ebsa dhe \u00e7el\u00ebsa. Ju vendosni se n\u00eb nj\u00eb skem\u00eb ideale duhet t\u00eb ndahet \u00e7el\u00ebsi, n\u00eb m\u00ebnyr\u00eb q\u00eb siguria t\u00eb mos mb\u00ebshtetet plot\u00ebsisht n\u00eb nj\u00eb person. Ju gjithashtu arrini n\u00eb p\u00ebrfundimin se duhet t\u00eb ekzistoj\u00eb nj\u00eb prag i numrit t\u00eb pjes\u00ebve, p\u00ebr t\u00eb b\u00ebr\u00eb q\u00eb humbja e nj\u00eb pjese (ose n\u00ebse dikush shkon n\u00eb pushime) t\u00eb mos \u00e7oj\u00eb n\u00eb mosfunksionimin e t\u00eb gjith\u00eb \u00e7el\u00ebsit.<\/p>\n

Si t\u00eb ndahen sekretet<\/h1>\n

\nP\u00ebr nj\u00eb lloj t\u00eb till\u00eb skeme menaxhimi t\u00eb \u00e7el\u00ebsave mendonte Adi Shamir n\u00eb vitin 1979, kur publikoi punimin e tij \u00abSi t\u00eb ndahen sekretet\u00bb<\/a><\/noindex>. N\u00eb artikull shpjegohet shkurtimisht ajo q\u00eb quhet \"Schema skema e pragut p\u00ebr ndarjen efektive t\u00eb nj\u00eb vlere sekrete (p\u00ebr shembull, nj\u00eb \u00e7el\u00ebs kriptografik) n\u00eb \"Schema pjes\u00eb. M\u00eb pas, kur dhe vet\u00ebm kur t\u00eb pakt\u00ebn \"Schema nga \"Schema pjes\u00eb t\u00eb mbledhura, mund t\u00eb rikuperohet leht\u00eb sekreti \"Schema.<\/p>\n

Nga k\u00ebndv\u00ebshtrimi i siguris\u00eb, nj\u00eb ve\u00e7ori e r\u00ebnd\u00ebsishme e k\u00ebsaj skeme \u00ebsht\u00eb se sulmuesi nuk duhet t\u00eb dij\u00eb asgj\u00eb n\u00ebse nuk ka t\u00eb pakt\u00ebn \"Schema pjes\u00eb. Edhe prania e \"Schema pjes\u00ebve nuk duhet t'u jap\u00eb asnj\u00eb informacion. Ne e quajm\u00eb k\u00ebt\u00eb ve\u00e7ori siguri semantike<\/i>.<\/p>\n

Interpolimi polinominal<\/h1>\n

\nSkema e Shamir \"Schema nd\u00ebrtohet rreth konceptit interpolimit polinominal<\/b>. N\u00ebse nuk jeni t\u00eb njohur me k\u00ebt\u00eb koncept, n\u00eb t\u00eb v\u00ebrtet\u00eb \u00ebsht\u00eb mjaft i thjesht\u00eb. N\u00eb p\u00ebrgjith\u00ebsi, n\u00ebse keni \u00e7izmuar ndonj\u00ebher\u00eb pik\u00eb n\u00eb nj\u00eb grafik dhe pastaj i keni lidhur ato me linja ose kurba, at\u00ebher\u00eb keni p\u00ebrdorur at\u00eb!<\/p>\n

\"Schema
\nP\u00ebrmes dy pikave mund t\u00eb kalojn\u00eb nj\u00eb num\u00ebr t\u00eb pakufizuar polinomesh t\u00eb grad\u00ebs 2. P\u00ebr t\u00eb zgjedhur nj\u00eb t\u00eb vetme - ne kemi nevoj\u00eb p\u00ebr nj\u00eb pik\u00eb t\u00eb tret\u00eb. Ilustrimi: Wikipedia<\/a><\/noindex><\/i> <\/p>\n

Merrni nj\u00eb polinom me grad\u00eb nj\u00eb, \"Schema. N\u00ebse d\u00ebshironi ta nd\u00ebrtoni k\u00ebt\u00eb funksion n\u00eb grafik, sa pika ju nevojiten? E dim\u00eb se ky \u00ebsht\u00eb nj\u00eb funksion linear q\u00eb formon nj\u00eb vij\u00eb dhe prandaj nevojiten t\u00eb pakt\u00ebn dy pika. Tani, le t\u00eb shqyrtojm\u00eb nj\u00eb funksion polinomial me grad\u00eb dy, \"Schema. Ky \u00ebsht\u00eb nj\u00eb funksion kuadratik, prandaj p\u00ebr t\u00eb nd\u00ebrtuar grafik nevojiten t\u00eb pakt\u00ebn tre pika. Si q\u00ebndron me nj\u00eb polinom me grad\u00eb tre? T\u00eb pakt\u00ebn kat\u00ebr pika. Dhe k\u00ebshtu me radh\u00eb.<\/p>\n

Gj\u00ebja v\u00ebrtet e shk\u00eblqyer lidhur me k\u00ebt\u00eb ve\u00e7ori \u00ebsht\u00eb se, duke marr\u00eb parasysh grad\u00ebn e funksionit polinomial dhe, t\u00eb pakt\u00ebn \"Schema pik\u00eb, ne mund t\u00eb nxjerrim pika shtes\u00eb p\u00ebr k\u00ebt\u00eb funksion polinomial. Ekstrapolimin e k\u00ebtyre pikave shtes\u00eb e quajm\u00eb interpolim polinomial<\/i>.<\/p>\n

Krijimi i sekreteve<\/h1>\n

\nMund t\u00eb keni kuptuar tashm\u00eb se k\u00ebtu hyjn\u00eb n\u00eb loje skemat e men\u00e7ura t\u00eb Shamir. Le t\u00eb supozojm\u00eb se sekreti yn\u00eb \"Schema \u2014 \u00ebsht\u00eb \"Schema. Ne mund ta kthejm\u00eb \"Schema n\u00eb nj\u00eb pik\u00eb n\u00eb grafik \"Schema dhe t\u00eb krijojm\u00eb nj\u00eb funksion polinomial me grad\u00eb \"Schema, q\u00eb plot\u00ebson k\u00ebt\u00eb pik\u00eb. Kujtojm\u00eb se \"Schema do t\u00eb jet\u00eb prag i fragmenteve t\u00eb k\u00ebrkuara, prandaj n\u00ebse e vendosim pragun n\u00eb tre fragmente, duhet t\u00eb zgjedhim nj\u00eb funksion polinomial me grad\u00eb dy.<\/p>\n

Polinomi yn\u00eb do t\u00eb ket\u00eb form\u00ebn \"Schemaindex \"Schema dhe \"Schema \u2014 numra t\u00eb plot\u00eb pozitiv\u00eb t\u00eb zgjedhur rast\u00ebsisht. Ne thjesht po nd\u00ebrtojm\u00eb nj\u00eb polinom me grad\u00eb \"Schema, ku koeficenti i lir\u00eb \"Schema\u00a0\u00ebsht\u00eb sekreti yn\u00eb \"Schema, dhe \u00e7do nj\u00eb nga t\u00eb tjer\u00ebt \"Schema an\u00ebtar\u00ebt kan\u00eb nj\u00eb koeficient pozitiv t\u00eb zgjedhur rast\u00ebsisht. N\u00ebse kthehemi te shembulli fillestar dhe supozoni se \"Schema, at\u00ebher\u00eb do t\u00eb marrim funksionin \"Schema.<\/p>\n

N\u00eb k\u00ebt\u00eb faz\u00eb ne mund t\u00eb gjenerojm\u00eb pjes\u00eb duke lidhur \"Schema numra t\u00eb ve\u00e7ant\u00eb t\u00eb plot\u00eb n\u00eb \"Schemaindex \"Schema (sepse \u00ebsht\u00eb sekreti yn\u00eb). N\u00eb k\u00ebt\u00eb shembull ne duam t\u00eb shp\u00ebrndajm\u00eb kat\u00ebr pjes\u00eb me prag tre, prandaj gjenerojm\u00eb rast\u00ebsisht pik\u00ebt \"Schema dhe d\u00ebrgojm\u00eb nj\u00eb pik\u00eb p\u00ebr secilin nga kat\u00ebr individ\u00ebt e besuar, ruajt\u00ebsit e \u00e7el\u00ebsit. Ne gjithashtu i njoftojm\u00eb ata se \"Schema, pasi kjo konsiderohet informacion publik dhe \u00ebsht\u00eb e nevojshme p\u00ebr rikuperimin \"Schema.<\/p>\n

Rikuperimi i sekrecionit<\/h1>\n

\nNe kemi diskutuar tashm\u00eb konceptin e interpolimit polinomik dhe se \u00e7far\u00eb q\u00ebndron n\u00eb baz\u00eb t\u00eb skem\u00ebs s\u00eb pragut t\u00eb Shamir \"Schema. Kur \u00e7do tre nga kat\u00ebr individ\u00ebt e besuar duan t\u00eb rikuperojn\u00eb \"Schema, ata vet\u00ebm duhet t\u00eb interpozojn\u00eb \"Schema me pik\u00ebt e tyre unike. P\u00ebr k\u00ebt\u00eb ata mund t\u00eb p\u00ebrcaktojn\u00eb pik\u00ebt e tyre \"Schema dhe t\u00eb llogaritin polinomin e interpolimit t\u00eb Lagranges, duke p\u00ebrdorur formul\u00ebn e m\u00ebposhtme. N\u00ebse programimi \u00ebsht\u00eb m\u00eb i kuptuesh\u00ebm p\u00ebr ju se matematika, at\u00ebher\u00eb pi \u00ebsht\u00eb n\u00eb thelb operatori p\u00ebr<\/code>, i cili shumon t\u00eb gjith\u00eb rezultatet, nd\u00ebrsa sigma \u00ebsht\u00eb p\u00ebr<\/code>, i cili i shton t\u00eb gjitha.<\/p>\n

\"Schema<\/p>\n

<\/p>\n

\"Schema<\/p>\n

\nKur \"Schema ne mund ta zgjidhim k\u00ebt\u00eb n\u00eb k\u00ebt\u00eb m\u00ebnyr\u00eb dhe t\u00eb kthejm\u00eb funksionin ton\u00eb polinomial origjinal:<\/p>\n

\"Schema<\/p>\n

\nPasi e dim\u00eb q\u00eb \"Schema, rikuperimi \"Schema ndodh thjesht:<\/p>\n

\"Schema<\/p>\n

<\/p>\n

P\u00ebrdorimi i aritmetik\u00ebs konsideruese t\u00eb plot\u00eb t\u00eb pasigurt<\/h1>\n

\nMegjith\u00ebse kemi aplikuar me sukses iden\u00eb themelore t\u00eb Shamir \"Schema, na mbetet nj\u00eb problem q\u00eb e kemi injoruar deri tani. Funksioni yn\u00eb polinomial p\u00ebrdor aritmetik\u00ebn konsideruese t\u00eb plot\u00eb t\u00eb pasigurt. Kini parasysh se p\u00ebr secil\u00ebn pik\u00eb shtes\u00eb q\u00eb sulmuesi merr n\u00eb grafik\u00ebn e funksionit ton\u00eb, mbetet m\u00eb pak mund\u00ebsi p\u00ebr pik\u00ebt e tjera. Ju mund ta shihni k\u00ebt\u00eb me sy tuaj kur p\u00ebrb\u00ebni nj\u00eb grafik me rritjen e numrit t\u00eb pik\u00ebve p\u00ebr funksionin polinomial duke p\u00ebrdorur aritmetik\u00ebn e plot\u00eb. Kjo \u00ebsht\u00eb kontraproduktive p\u00ebr q\u00ebllimin ton\u00eb t\u00eb deklaruar t\u00eb siguris\u00eb, sepse sulmuesi nuk duhet t\u00eb dij\u00eb asgj\u00eb derisa t\u00eb ket\u00eb t\u00eb pakt\u00ebn \"Schema pjes\u00eb.<\/p>\n

P\u00ebr t\u00eb demonstruar se sa e dob\u00ebt \u00ebsht\u00eb skema me aritmetik\u00eb t\u00eb plot\u00eb, shqyrtoni nj\u00eb skenar ku sulmuesi ka marr\u00eb dy pika \"Schema dhe di informacionin publik q\u00eb \"Schema. Nga k\u00ebto informacione ai mund t\u00eb nxjerr\u00eb \"Schema, i barabart\u00eb me dy, dhe t\u00eb lidhet n\u00eb formul\u00eb me vlerat e njohura \"Schema dhe \"Schema.<\/p>\n

\"Schema<\/p>\n

\n Pastaj sulmuesi mund t\u00eb gjej\u00eb \"Schema, duke llogaritur \"Schema:<\/p>\n

\"Schema<\/p>\n

\nPasi e kemi p\u00ebrcaktuar \"Schema si numra t\u00eb rinj pozitiv, ka nj\u00eb num\u00ebr t\u00eb kufizuar mund\u00ebsish \"Schema. Me k\u00ebt\u00eb informacion, sulmuesi mund t\u00eb nxjerr\u00eb \"Schema, pasi gjith\u00e7ka m\u00eb shum\u00eb se 5 do t\u00eb b\u00ebj\u00eb \"Schema negative. Kjo del t\u00eb jet\u00eb e v\u00ebrtet\u00eb, pasi e kemi p\u00ebrcaktuar \"Schema<\/p>\n

Pastaj, sulmuesi mund t\u00eb llogaris\u00eb vlerat e mundshme \"Schema, duke z\u00ebvend\u00ebsuar \"Schema n\u00eb \"Schema:<\/p>\n

\"Schema<\/p>\n

\nMe nj\u00eb grup t\u00eb kufizuar mund\u00ebsish p\u00ebr \"Schema \u00ebsht\u00eb e qart\u00eb se sa leht\u00eb \u00ebsht\u00eb t\u00eb p\u00ebrshtaten dhe verifikohen vlerat \"Schema. Ka gjithsej pes\u00eb mund\u00ebsi.<\/p>\n

Zgjidhja e problemit me aritmetik\u00ebn e sigurt t\u00eb numrave t\u00eb plot\u00eb<\/h1>\n

\nP\u00ebr t\u00eb eliminuar k\u00ebt\u00eb dob\u00ebsi, Shamir sugjeron p\u00ebrdorimin e aritmetik\u00ebs modulare, duke z\u00ebvend\u00ebsuar \"Schema n\u00eb \"Schemaindex \"Schema dhe \"Schema \u2014 grupi i t\u00eb gjitha numrave t\u00eb thjesht\u00eb.<\/p>\n

Le t\u00eb p\u00ebrkujtojm\u00eb shpejt se si funksionon aritmetika modulare. Orar\u00ebt me dritare jan\u00eb nj\u00eb koncept i njohur. Ajo p\u00ebrdor oraret q\u00eb jan\u00eb \"Schema. Sa her\u00eb q\u00eb shigjeta e or\u00ebs kalon nga dymb\u00ebdhjeta, ajo kthehet tek nj\u00eb. Nj\u00eb ve\u00e7ori interesante e k\u00ebsaj sistemi \u00ebsht\u00eb se thjesht duke par\u00eb or\u00ebn, ne nuk mund ta nxjerrim se sa her\u00eb ka kaluar shigjeta e or\u00ebs. Megjithat\u00eb, n\u00ebse e dim\u00eb se shigjeta e or\u00ebs ka kaluar 12 kat\u00ebr her\u00eb, mund t\u00eb p\u00ebrcaktojm\u00eb plot\u00ebsisht numrin e or\u00ebve t\u00eb kaluara me nj\u00eb formul\u00eb t\u00eb thjesht\u00eb \"Schemaindex \"Schema\u00a0\u2014 kjo \u00ebsht\u00eb ndarja jon\u00eb (k\u00ebtu \"Schema), \"Schema\u00a0\u2014 kjo \u00ebsht\u00eb koeficienti (sa her\u00eb ndarja kalon n\u00eb numrin origjinal pa mbetje, k\u00ebtu \"Schema\u2014 emri i funksionit ( \"Schema\u00a0\u2014 kjo \u00ebsht\u00eb mbetja, e cila zakonisht kthehet nga operatori i modulit (k\u00ebtu \"Schema). T\u00eb diturit t\u00eb gjitha k\u00ebto vlera na lejon t\u00eb zgjidhim ekuacionin p\u00ebr \"Schema, por n\u00ebse humbasim koeficientin, kurr\u00eb nuk do t\u00eb mund t\u00eb rikthejm\u00eb vler\u00ebn origjinale.<\/p>\n

Mund t\u00eb demonstrohet se si kjo p\u00ebrmir\u00ebson sigurin\u00eb e skem\u00ebs son\u00eb duke zbatuar skem\u00ebn n\u00eb shembullin ton\u00eb t\u00eb m\u00ebparsh\u00ebm dhe duke p\u00ebrdorur \"Schema. Funksioni yn\u00eb polinomial i ri \"Schema, dhe pik\u00ebt e reja \"Schema. Tani ruajt\u00ebsit e \u00e7el\u00ebsit mund t\u00eb p\u00ebrdorin p\u00ebrs\u00ebri interpolimin polinomial p\u00ebr t\u00eb rikthyer funksionin ton\u00eb, megjithat\u00eb, k\u00ebt\u00eb her\u00eb operacionet e shtimit dhe shum\u00ebzimit duhet t\u00eb shoq\u00ebrohen me zvog\u00eblimin modulas \"Schema (p.sh. \"Schema).<\/p>\n

Duke p\u00ebrdorur k\u00ebt\u00eb shembull t\u00eb ri, le t\u00eb supozojm\u00eb se sulmuesi ka m\u00ebsuar dy nga k\u00ebto pika t\u00eb reja, \"Schema, dhe informacioni publik \"Schema. N\u00eb k\u00ebt\u00eb rast, sulmuesi, duke u mb\u00ebshtetur n\u00eb t\u00eb gjitha informacionet n\u00eb dispozicion, nxjerr funksionet e m\u00ebposhtme, ku \"Schema\u00a0\u00ebsht\u00eb\u96c6\u5408t\u00eb gjith\u00eb numrat e plot\u00eb pozitiv\u00eb, dhe \"Schema p\u00ebrdor koeficientin e modulit \"Schema.<\/p>\n

\"Schema<\/p>\n

\nTani, sulmuesi yn\u00eb p\u00ebrs\u00ebri gjen \"Schema, duke llogaritur \"Schema:<\/p>\n

\"Schema<\/p>\n

\nPastaj ai p\u00ebrpiqet s\u00ebrish t\u00eb nxjerr\u00eb \"Schema, duke z\u00ebvend\u00ebsuar \"Schema n\u00eb \"Schema:<\/p>\n

\"Schema<\/p>\n

\nN\u00eb k\u00ebt\u00eb rast, ai ka nj\u00eb problem serioz. N\u00eb formul\u00eb mungojn\u00eb vlerat \"Schema, \"Schema dhe \"Schema. Duke marr\u00eb parasysh se ekziston nj\u00eb num\u00ebr t\u00eb pafund kombinimesh k\u00ebthese, ai nuk mund t\u00eb marr\u00eb asnj\u00eb informacion shtes\u00eb.<\/p>\n

Konsideratat e siguris\u00eb<\/h1>\n

\nSkema e ndarjes s\u00eb sekreteve t\u00eb Shamir ofron siguri nga k\u00ebndv\u00ebshtrimi i teoris\u00eb s\u00eb informacionit<\/i>. Kjo do t\u00eb thot\u00eb se matematika \u00ebsht\u00eb e fort\u00eb, madje edhe p\u00ebrball\u00eb nj\u00eb sulmuesi me kapacitete t\u00eb pakufizuara llogarit\u00ebse. Megjithat\u00eb, skema t\u00ebrheq disa probleme t\u00eb njohura.<\/p>\n

P\u00ebr shembull, skema e Shamir-it nuk krijon fragmente t\u00eb verifikueshme<\/i>, dometh\u00ebn\u00eb, njer\u00ebzit mund t\u00eb paraqesin lirsh\u00ebm fragmente fallco dhe ta pengojn\u00eb rikuperimin e sekreteve t\u00eb sakta. Nj\u00eb ruajt\u00ebs armiqesor i fragmenteve me informacion t\u00eb mjaftuesh\u00ebm mund t\u00eb krijoj\u00eb madje nj\u00eb fragment tjet\u00ebr, duke modifikuar \"Schema sip\u00ebr d\u00ebshir\u00ebs s\u00eb tij. Ky problem zgjidhet p\u00ebrmes schema t\u00eb verifikueshme t\u00eb ndarjes s\u00eb sekreteve<\/i>, si\u00e7 \u00ebsht\u00eb skema e Feldman.<\/p>\n

Nj\u00eb problem tjet\u00ebr \u00ebsht\u00eb se gjat\u00ebsia e \u00e7do fragmenti \u00ebsht\u00eb e barabart\u00eb me gjat\u00ebsin\u00eb e sekreteve p\u00ebrkat\u00ebse, k\u00ebshtu q\u00eb gjat\u00ebsia e sekreteve \u00ebsht\u00eb e leht\u00eb p\u00ebr t'u p\u00ebrcaktuar. Ky problem zgjidhet p\u00ebrmes nj\u00eb mbushjeje<\/i> t\u00eb sekreteve me numra t\u00eb rast\u00ebsish\u00ebm deri n\u00eb nj\u00eb gjat\u00ebsi fikse.<\/p>\n

S\u00eb fundi, \u00ebsht\u00eb e r\u00ebnd\u00ebsishme t\u00eb theksohet se shqet\u00ebsimet tona p\u00ebr sigurin\u00eb mund t\u00eb tejkalojn\u00eb veten e skem\u00ebs. P\u00ebr aplikacione reale kriptografike shpesh ka k\u00ebrc\u00ebnim nga sulmet p\u00ebrmes kanaleve an\u00ebsore, kur sulmuesi p\u00ebrpiqet t\u00eb nxjerr\u00eb informacion t\u00eb dobish\u00ebm nga koha e ekzekutimit t\u00eb aplikacionit, caches, d\u00ebshtime, etj. N\u00ebse kjo ngre shqet\u00ebsime, duhet t\u00eb shqyrtohen me kujdes masa mbrojt\u00ebse si funksionet dhe k\u00ebrkimi me koh\u00eb t\u00eb q\u00ebndrueshme t\u00eb ekzekutimit, p\u00ebr t\u00eb parandaluar ruajtjen e t\u00eb dh\u00ebnave n\u00eb disk dhe t\u00eb mendohet p\u00ebr nj\u00eb s\u00ebr\u00eb gj\u00ebrash q\u00eb kalojn\u00eb jasht\u00eb k\u00ebtij artikulli.<\/p>\n

Demo<\/h1>\n

\nN\u00eb n\u00eb k\u00ebt\u00eb faqe<\/a><\/noindex> eksiston nj\u00eb demonstrim interaktiv t\u00eb skem\u00ebs s\u00eb ndarjes s\u00eb sekreteve t\u00eb Shamir. Demonstrimi \u00ebsht\u00eb i bazuar n\u00eb bibliotek\u00ebn ssss-js<\/a><\/noindex>, e cila vet\u00eb \u00ebsht\u00eb nj\u00eb port t\u00eb njohur t\u00eb programit JavaScript ssss<\/a><\/noindex>. Ju kujtojm\u00eb se llogaritja e vlerave t\u00eb m\u00ebdha \"Schema, \"Schema dhe \"Schema mund t\u00eb marr\u00eb pak koh\u00eb.
\n
Burimi: habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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