{"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":"Skema e ndarjes s\u00eb sekreteve t\u00eb Shamir","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Le t\u00eb shqyrtojm\u00eb nj\u00eb skenar kur \u00ebsht\u00eb e nevojshme t\u00eb sigurohet siguria e nj\u00eb depo bankare. Ajo konsiderohet krejt\u00ebsisht e paprekshme pa \u00e7el\u00ebsin, i cili ju jepet n\u00eb dit\u00ebn e par\u00eb t\u00eb pun\u00ebs. Q\u00ebllimi juaj \u00ebsht\u00eb t\u00eb ruani \u00e7el\u00ebsin n\u00eb m\u00ebnyr\u00eb t\u00eb sigurt.<\/p>\n

Supozoni se keni vendosur ta mbani gjithnj\u00eb \u00e7el\u00ebsin me vete, duke ofruar akses n\u00eb depo sipas nevoj\u00ebs. Por shpejt do t\u00eb kuptoni se ky zgjidhje n\u00eb praktik\u00eb nuk \u00ebsht\u00eb e q\u00ebndrueshme, sepse \u00e7do her\u00eb p\u00ebr t\u00eb hapur depon k\u00ebrkohet prania juaj fizike. \u00c7far\u00eb ndodh me pushimet q\u00eb ju jan\u00eb premtuar? P\u00ebr m\u00eb tep\u00ebr, mbetet nj\u00eb pyetje e frikshme: \u00e7far\u00eb ndodh n\u00ebse humbni \u00e7el\u00ebsin tuaj t\u00eb vet\u00ebm?<\/p>\n

Me mendimin p\u00ebr pushimet, vendos\u00ebt t\u00eb b\u00ebni nj\u00eb kopje t\u00eb \u00e7el\u00ebsit dhe t'ia besoni nj\u00eb kolegu tjet\u00ebr. Megjithat\u00eb, e kuptoni se kjo gjithashtu nuk \u00ebsht\u00eb ideale. Duke e dyfishuar numrin e \u00e7el\u00ebsave, gjithashtu keni dyfishuar mund\u00ebsit\u00eb p\u00ebr t\u00eb vjedhur \u00e7el\u00ebsin.<\/p>\n

Duke u d\u00ebshp\u00ebruar, ju shkaterroni nj\u00eb kopje dhe vendosni t\u00eb ndani \u00e7el\u00ebsin origjinal n\u00eb dy pjes\u00eb. Tani, mendoni se dy persona t\u00eb besuesh\u00ebm me fragmentet e \u00e7el\u00ebsit duhet t\u00eb jen\u00eb fizikisht t\u00eb pranish\u00ebm p\u00ebr t\u00eb mbledhur \u00e7el\u00ebsin dhe hapur ark\u00ebn. Kjo do t\u00eb thot\u00eb q\u00eb hajduti 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. Megjithat\u00eb, shpejt kuptoni se kjo skem\u00eb nuk \u00ebsht\u00eb shum\u00eb m\u00eb e mir\u00eb sesa thjesht nj\u00eb \u00e7el\u00ebs, sepse n\u00ebse dikush humbet nj\u00eb pjes\u00eb t\u00eb \u00e7el\u00ebsit, \u00e7el\u00ebsi i plot\u00eb nuk mund t\u00eb restaurohet.
\n<\/a><\/noindex>
\nProblemi mund t\u00eb zgjidhet me nj\u00eb seri \u00e7el\u00ebsash dhe \u00e7el\u00ebsash, por me k\u00ebt\u00eb qasje shpejt do t\u00eb nevojiten shum\u00eb<\/i> \u00e7el\u00ebsa dhe \u00e7el\u00ebsa. Ju vendosni q\u00eb 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 t\u00eb vet\u00ebm. Ju gjithashtu arrini n\u00eb p\u00ebrfundimin se duhet t\u00eb ekzistoj\u00eb nj\u00eb kufi i caktuar i numrit t\u00eb fragmenteve, k\u00ebshtu q\u00eb n\u00eb rast se humbet nj\u00eb fragment (ose n\u00ebse nj\u00eb person largohet p\u00ebr pushime) i gjith\u00eb \u00e7el\u00ebsi t\u00eb mbetet funksional.<\/p>\n

Si t\u00eb ndani nj\u00eb sekret<\/h1>\n

\nP\u00ebr k\u00ebt\u00eb lloj skeme t\u00eb menaxhimit t\u00eb \u00e7el\u00ebsave mendohej nga Adi Shamir n\u00eb vitin 1979, kur publikoi punimin e tij \"Si t\u00eb ndahet nj\u00eb sekret\"<\/a><\/noindex>. Artikulli shpjegon shkurtimisht q\u00eb \u00ebsht\u00eb e ashtuquajtura \"Skema skema pragmatike p\u00ebr ndarjen efektive t\u00eb nj\u00eb vler\u00eb sekrete (p.sh., \u00e7el\u00ebsi kriptografik) n\u00eb \"Skema pjes\u00eb. Pastaj, kur dhe vet\u00ebm kur t\u00eb pakt\u00ebn \"Skema nga \"Skema pjes\u00ebt jan\u00eb mbledhur, mund t\u00eb rikuperohet leht\u00ebsisht sekreti \"Skema.<\/p>\n

Nga pik\u00ebpamja e siguris\u00eb, nj\u00eb nga atributet e r\u00ebnd\u00ebsishme t\u00eb k\u00ebsaj skeme \u00ebsht\u00eb se nj\u00eb sulmues nuk duhet t\u00eb dij\u00eb absolutisht asgj\u00eb, n\u00ebse ai nuk ka t\u00eb pakt\u00ebn \"Skema pjes\u00eb. Edhe posedimi i \"Skema pjes\u00ebve nuk duhet t\u00eb jap\u00eb asnj\u00eb informacion. Ne e quajm\u00eb k\u00ebt\u00eb atribut siguria semantike<\/i>.<\/p>\n

Interpolimi polinomial<\/h1>\n

\nSkema pragmatike e Shamir \"Skema nd\u00ebrtohet rreth konceptit t\u00eb interpolimit polinomial<\/b>. N\u00ebse nuk jeni njohur me k\u00ebt\u00eb koncept, ai n\u00eb fakt \u00ebsht\u00eb mjaft i thjesht\u00eb. N\u00eb p\u00ebrgjith\u00ebsi, n\u00ebse ndonj\u00ebher\u00eb keni vizatuar pika n\u00eb nj\u00eb grafik dhe m\u00eb pas i keni lidhur ato me linja ose kurba, at\u00ebher\u00eb keni p\u00ebrdorur tashm\u00eb at\u00eb!<\/p>\n

\"Skema
\nP\u00ebrmes dy pikave, mund t\u00eb t\u00ebrheqni nj\u00eb num\u00ebr t\u00eb pakufizuar polinom\u00ebsh t\u00eb grad\u00ebs 2. P\u00ebr t\u00eb zgjedhur vet\u00ebm nj\u00eb t\u00eb till\u00eb - nevojitet nj\u00eb pik\u00eb e tret\u00eb. Ilustrimi: Wikipedia<\/a><\/noindex><\/i> <\/p>\n

Le t\u00eb shqyrtojm\u00eb nj\u00eb polinom me grad\u00eb nj\u00eb, \"Skema. N\u00ebse d\u00ebshironi t\u00eb nd\u00ebrtoni k\u00ebt\u00eb funksion n\u00eb grafik, sa pika ju nevojiten? E dim\u00eb se ky \u00ebsht\u00eb nj\u00eb funksion linear, i cili formon nj\u00eb vij\u00eb dhe p\u00ebr k\u00ebt\u00eb arsye nevojiten t\u00eb pakt\u00ebn dy pika. M\u00eb pas, le t\u00eb shqyrtojm\u00eb nj\u00eb funksion polinomial me grad\u00eb dy, \"Skema. Ky \u00ebsht\u00eb nj\u00eb funksion katror, prandaj p\u00ebr nd\u00ebrtimin e grafik\u00ebs k\u00ebrkohen t\u00eb pakt\u00ebn tre pika. Si p\u00ebr nj\u00eb polinom me grad\u00eb tre? T\u00eb pakt\u00ebn kat\u00ebr pika. Dhe k\u00ebshtu me radh\u00eb.<\/p>\n

Ajo q\u00eb \u00ebsht\u00eb v\u00ebrtet e mrekullueshme n\u00eb k\u00ebt\u00eb pron\u00eb \u00ebsht\u00eb se, duke marr\u00eb parasysh grad\u00ebn e funksionit polinom, dhe, t\u00eb pakt\u00ebn, \"Skema pikave, ne mund t\u00eb nxjerrim pika t\u00eb tjera p\u00ebr k\u00ebt\u00eb funksion polinomial. Ekstrapolimi i k\u00ebtyre pikave t\u00eb tjera ne e quajm\u00eb interpolimi polinomial<\/i>.<\/p>\n

Krijimi i nj\u00eb sekreti<\/h1>\n

\nMund t\u00eb keni kuptuar tashm\u00eb se k\u00ebtu hyn n\u00eb loj\u00eb skema e men\u00e7ur e Shamir. Le t\u00eb supozojm\u00eb se sekreti yn\u00eb \"Skema \u2014 kjo \u00ebsht\u00eb \"Skema. Ne mund ta kthejm\u00eb \"Skema n\u00eb nj\u00eb pik\u00eb n\u00eb grafik \"Skema dhe t\u00eb krijojm\u00eb nj\u00eb funksion polinomial me grad\u00eb \"Skema, i cili p\u00ebrmbush k\u00ebt\u00eb pik\u00eb. Le t\u00eb kujtojm\u00eb se \"Skema do t\u00eb jet\u00eb prag i k\u00ebrkuar p\u00ebr fragmentet, prandaj n\u00ebse e vendosim pragun n\u00eb tre fragmente, duhet t\u00eb zgjedhim nj\u00eb funksion polinom me grad\u00eb dy.<\/p>\n

Polinomi yn\u00eb do t\u00eb ket\u00eb form\u00ebn \"Skema, ku \"Skema dhe \"Skema \u2014 numra t\u00eb rast\u00ebsish\u00ebm pozitiv\u00eb t\u00eb plot\u00eb. Ne thjesht po nd\u00ebrrmarim nj\u00eb polinom me grad\u00eb \"Skema, ku koeficienti i lir\u00eb \"Skema\u00a0\u2014 \u00ebsht\u00eb sekreti yn\u00eb \"Skema, nd\u00ebrsa secili nga an\u00ebtar\u00ebt e mbetur \"Skema ka nj\u00eb koeficient pozitiv t\u00eb zgjedhur rast\u00ebsisht. N\u00ebse kthehemi n\u00eb shembullin fillestar dhe supozojm\u00eb se \"Skema, at\u00ebher\u00eb do t\u00eb marrim funksionin \"Skema.<\/p>\n

N\u00eb k\u00ebt\u00eb faz\u00eb, mund t\u00eb gjenerojm\u00eb fragmentet, duke lidhur \"Skema numra t\u00eb plot\u00eb unik\u00eb n\u00eb \"Skema, ku \"Skema (sepse ky \u00ebsht\u00eb sekreti yn\u00eb). N\u00eb k\u00ebt\u00eb rast, ne duam t\u00eb japim kat\u00ebr fragmente me prag tre, prandaj gjenerojm\u00eb rast\u00ebsisht pik\u00ebt \"Skema dhe d\u00ebrgojm\u00eb nga nj\u00eb pik\u00eb p\u00ebr secilin nga kat\u00ebr personat e besuar, ruajt\u00ebsit e \u00e7el\u00ebsave. Ne gjithashtu i njoftojm\u00eb njer\u00ebzit se \"Skema, pasi kjo konsiderohet informacion publik dhe \u00ebsht\u00eb e nevojshme p\u00ebr rikuperimin \"Skema.<\/p>\n

Rikuperimi i sekretit<\/h1>\n

\nKemi diskutuar tashm\u00eb konceptin e interpolimit polinom dhe at\u00eb q\u00eb q\u00ebndron n\u00eb baz\u00eb t\u00eb skem\u00ebs s\u00eb pragut t\u00eb Shamir \"Skema. Kur tre personat nga kat\u00ebr t\u00eb besuesh\u00ebm duan t\u00eb rikuperojn\u00eb \"Skema, ata duhet vet\u00ebm t\u00eb nd\u00ebrtojn\u00eb \"Skema me pikat e tyre unike. P\u00ebr k\u00ebt\u00eb ata mund t\u00eb p\u00ebrcaktojn\u00eb pikat e tyre \"Skema dhe t\u00eb llogarisin polinomin e interpolimit t\u00eb Lagrange-it, duke p\u00ebrdorur formul\u00ebn n\u00eb vijim. N\u00ebse programimi ju kuptohet m\u00eb mir\u00eb se matematika, pi \u00ebsht\u00eb n\u00eb thelb nj\u00eb operator p\u00ebr<\/code>, i cili shum\u00ebzon t\u00eb gjitha rezultatet, dhe sigma \u00ebsht\u00eb p\u00ebr<\/code>, i cili gjith\u00e7ka i shton.<\/p>\n

\"Skema<\/p>\n

<\/p>\n

\"Skema<\/p>\n

\nN\u00eb \"Skema ne mund ta zgjidhim k\u00ebt\u00eb duke e b\u00ebr\u00eb n\u00eb k\u00ebt\u00eb m\u00ebnyr\u00eb dhe t\u00eb kthejm\u00eb funksionin ton\u00eb polinomial origjinal:<\/p>\n

\"Skema<\/p>\n

\nDuke ditur se \"Skema, rikuperimi \"Skema realizohet thjesht:<\/p>\n

\"Skema<\/p>\n

<\/p>\n

P\u00ebrdorimi i aritmetik\u00ebs s\u00eb sigurt\u00eb t\u00eb numrave t\u00eb plot\u00eb<\/h1>\n

\nMegjith\u00ebse kemi aplikuar me sukses iden\u00eb kryesore t\u00eb Shamirit \"Skema, ne na mbetet nj\u00eb problem, t\u00eb cilin e kemi injoruar deri m\u00eb tani. Funksioni yn\u00eb polinomial p\u00ebrdor aritmetik\u00eb t\u00eb pa sigurt t\u00eb numrave t\u00eb plot\u00eb. Mbani parasysh se p\u00ebr \u00e7do pik\u00eb shtes\u00eb q\u00eb nj\u00eb sulmues merr n\u00eb grafik\u00ebn e funksionit ton\u00eb, mbetet nj\u00eb num\u00ebr m\u00eb i vog\u00ebl mund\u00ebsish p\u00ebr pika t\u00eb tjera. Mund ta shihni k\u00ebt\u00eb me sy kur nd\u00ebrtoni grafik me rritjen e numrit t\u00eb pikave p\u00ebr funksionin polinomial duke p\u00ebrdorur aritmetik\u00eb t\u00eb numrave t\u00eb plot\u00eb. Kjo \u00ebsht\u00eb kontraproduktive p\u00ebr q\u00ebllimin ton\u00eb t\u00eb pretenduar t\u00eb siguris\u00eb, sepse nj\u00eb sulmues nuk duhet t\u00eb dij\u00eb asgj\u00eb derisa t\u00eb ket\u00eb t\u00eb pakt\u00ebn \"Skema fragmente.<\/p>\n

P\u00ebr t\u00eb demonstruar sa e dob\u00ebt \u00ebsht\u00eb skema e aritmetik\u00ebs s\u00eb numrave t\u00eb plot\u00eb, le t\u00eb shqyrtojm\u00eb nj\u00eb skenar n\u00eb t\u00eb cilin nj\u00eb sulmues ka marr\u00eb dy pika \"Skema dhe di informacionin publik q\u00eb \"Skema. Nga ky informacion ai mund t\u00eb nxjerr\u00eb \"Skema, i cili \u00ebsht\u00eb i barabart\u00eb me dy, dhe t\u00eb lidh\u00eb n\u00eb formul\u00eb vlerat e njohura \"Skema dhe \"Skema.<\/p>\n

\"Skema<\/p>\n

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

\"Skema<\/p>\n

\nDuke marr\u00eb parasysh se ne e kemi p\u00ebrcaktuar \"Skema si numra t\u00eb plot\u00eb pozitiv\u00eb t\u00eb zgjedhur rast\u00ebsisht, ka nj\u00eb num\u00ebr t\u00eb kufizuar mund\u00ebsish \"Skema. Me k\u00ebt\u00eb informacion, nj\u00eb sulmues mund t\u00eb nxjerr\u00eb \"Skema, pasi \u00e7do gj\u00eb m\u00eb shum\u00eb se 5 do ta b\u00ebj\u00eb \"Skema negative. Kjo rezulton t\u00eb jet\u00eb e v\u00ebrtet\u00eb, pasi ne e kemi p\u00ebrcaktuar \"Skema<\/p>\n

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

\"Skema<\/p>\n

\nMe nj\u00eb grup t\u00eb kufizuar mund\u00ebsish p\u00ebr \"Skema b\u00ebhet e qart\u00eb se sa leht\u00eb \u00ebsht\u00eb t\u00eb gjenden dhe t\u00eb verifikohen vlerat \"Skema. K\u00ebtu ka vet\u00ebm pes\u00eb mund\u00ebsi.<\/p>\n

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

\nP\u00ebr t\u00eb eliminuar k\u00ebt\u00eb dob\u00ebsi, Shamir sugjeron t\u00eb p\u00ebrdoren aritmetika modulo, duke z\u00ebvend\u00ebsuar \"Skema n\u00eb \"Skema, ku \"Skema dhe \"Skema \u2014 grupi i t\u00eb gjith\u00eb numrave t\u00eb thjesht\u00eb.<\/p>\n

Le t\u00eb rikujtojm\u00eb shpejt si funksionon aritmetika modulo. Ora me shifra \u00ebsht\u00eb nj\u00eb koncept i njohur. Ajo p\u00ebrdor or\u00ebt q\u00eb jan\u00eb \"Skema. Sa her\u00eb q\u00eb akrepi i or\u00ebs kalon 12, ai kthehet n\u00eb nj\u00eb. Nj\u00eb karakteristik\u00eb interesante e k\u00ebsaj sistemi \u00ebsht\u00eb se vet\u00ebm duke e par\u00eb or\u00ebn, nuk mund t\u00eb nxjerrim se sa her\u00eb ka kaluar akrepi i or\u00ebs. Megjithat\u00eb, n\u00ebse e dim\u00eb q\u00eb akrepi i 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 \"Skema, ku \"Skema\u00a0\u2014 ky \u00ebsht\u00eb ndar\u00ebsi yn\u00eb (k\u00ebtu \"Skema), \"Skema\u00a0\u2014 \u00ebsht\u00eb koeficienti (sa her\u00eb ndahen pa mbetje ndar\u00ebsi n\u00eb numrin fillestar, k\u00ebtu \"Skema), nd\u00ebrsa \"Skema\u00a0\u2014 \u00ebsht\u00eb mbetja, q\u00eb zakonisht e kthen thirrja e operatorit modulo (k\u00ebtu \"Skema). Njohja e t\u00eb gjitha k\u00ebtyre vlerave na lejon t\u00eb zgjidhim ekuacionin p\u00ebr \"Skema, por n\u00ebse e anashkalojm\u00eb koeficientin, kurr\u00eb nuk do t\u00eb jemi n\u00eb gjendje t\u00eb rikuperojm\u00eb vler\u00ebn fillestare.<\/p>\n

Mund t\u00eb demonstrojm\u00eb se si kjo p\u00ebrmir\u00ebson sigurin\u00eb e skem\u00ebs ton\u00eb, duke aplikuar skem\u00ebn n\u00eb shembullin ton\u00eb t\u00eb m\u00ebparsh\u00ebm dhe duke p\u00ebrdorur \"Skema. Funksioni yn\u00eb polinom i ri \"Skema, dhe pik\u00ebt e reja \"Skema. Tani ruajt\u00ebsit e \u00e7el\u00ebsit mund t\u00eb p\u00ebrdorin p\u00ebrs\u00ebri interpolimin polinom p\u00ebr t\u00eb rikuperuar funksionin ton\u00eb, por k\u00ebt\u00eb her\u00eb operacionet e mbledhjes dhe shum\u00ebzimit duhet t\u00eb shoq\u00ebrohen me reduktimin modulo \"Skema (p.sh. \"Skema).<\/p>\n

Duke p\u00ebrdorur k\u00ebt\u00eb shembull t\u00eb ri, supozoni se nj\u00eb sulmues ka m\u00ebsuar dy nga k\u00ebto pika t\u00eb reja, \"Skema, dhe informacioni publik \"Skema. K\u00ebt\u00eb her\u00eb, sulmuesi bazuar n\u00eb gjith\u00eb informacionin q\u00eb ka nxjerr\u00eb funksionet n\u00eb vijim, ku \"Skema\u00a0\u2014 \u00ebsht\u00eb grupi i t\u00eb gjith\u00eb numrave t\u00eb plot\u00eb pozitiv\u00eb, dhe \"Skema p\u00ebrfaq\u00ebson koeficientin e modulit \"Skema.<\/p>\n

\"Skema<\/p>\n

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

\"Skema<\/p>\n

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

\"Skema<\/p>\n

\nK\u00ebt\u00eb her\u00eb ai ka nj\u00eb problem serioz. N\u00eb formul\u00eb mungojn\u00eb vlerat \"Skema, \"Skema dhe \"Skema. Duke qen\u00eb se ekzistojn\u00eb nj\u00eb num\u00ebr t\u00eb pafund t\u00eb kombinimeve t\u00eb k\u00ebtyre variablave, ai nuk mund t\u00eb marr\u00eb ndonj\u00eb informacion t\u00eb m\u00ebtejsh\u00ebm.<\/p>\n

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

\nSkema e ndarjes s\u00eb sekretit t\u00eb Shamires ofron siguri nga perspektiva e teoris\u00eb s\u00eb informacionit<\/i>. Kjo do t\u00eb thot\u00eb se matematika \u00ebsht\u00eb e q\u00ebndrueshme edhe p\u00ebrball\u00eb nj\u00eb sulmuesi me fuqin\u00eb e pakufizuar t\u00eb llogaritjes. Sidoqoft\u00eb, skema ende p\u00ebrmban disa probleme t\u00eb njohura.<\/p>\n

P\u00ebr shembull, skema e Shamires nuk krijon fragmente verifikues<\/i>, dometh\u00ebn\u00eb njer\u00ebzit mund t\u00eb paraqesin lirisht fragmente fals dhe t\u00eb pengojn\u00eb rikonsolidimin e sekretit t\u00eb sakt\u00eb. Nj\u00eb ruajt\u00ebs armiq\u00ebsor i fragmenteve me informacion t\u00eb mjaftuesh\u00ebm mund t\u00eb prodhoj\u00eb madje nj\u00eb fragment tjet\u00ebr, duke ndryshuar \"Skema sip\u00ebrfaq\u00ebsisht. Ky problem zgjidhet me skema verifikues t\u00eb ndarjes s\u00eb sekretit<\/i>, si\u00e7 \u00ebsht\u00eb skema e Feldmanit.<\/p>\n

Nj\u00eb tjet\u00ebr problem \u00ebsht\u00eb se gjat\u00eb e \u00e7do fragmeni \u00ebsht\u00eb e barabart\u00eb me gjat\u00ebsi e sekretit p\u00ebrkat\u00ebs, k\u00ebshtu q\u00eb gjat\u00ebsi e sekretit p\u00ebrcaktohet leht\u00ebsisht. Ky problem zgjidhet me nj\u00eb mbushje triviale<\/i> sekret me numra t\u00eb rastit deri n\u00eb gjat\u00ebsi t\u00eb caktuar.<\/p>\n

N\u00eb fund, \u00ebsht\u00eb e r\u00ebnd\u00ebsishme t\u00eb theksohet se shqet\u00ebsimet tona p\u00ebr sigurin\u00eb mund t\u00eb kalojn\u00eb p\u00ebrtej vet\u00eb skem\u00ebs. P\u00ebr aplikacionet reale t\u00eb kriptografis\u00eb, shpesh ka rreziqe t\u00eb sulmeve nga kanale t\u00eb jashtme, kur nj\u00eb sulmues p\u00ebrpiqet t\u00eb nxjerr\u00eb informacion t\u00eb dobish\u00ebm nga koha e ekzekutimit t\u00eb aplikacionit, ndihm\u00ebsve, d\u00ebshtimeve, etj. N\u00ebse ky \u00ebsht\u00eb nj\u00eb shqet\u00ebsim, gjat\u00eb zhvillimit duhet t\u00eb shqyrtohet me kujdes p\u00ebrdorimi i masave mbrojt\u00ebse, si funksionet dhe k\u00ebrkimi me koh\u00eb t\u00eb q\u00ebndruesh\u00ebm t\u00eb ekzekutimit, t\u00eb parandalohet ruajtja e memorjes n\u00eb disk dhe t\u00eb merren parasysh nj\u00eb s\u00ebr\u00eb gj\u00ebrash q\u00eb kalojn\u00eb p\u00ebrtej k\u00ebtij artikulli.<\/p>\n

Demo<\/h1>\n

\nN\u00eb k\u00ebt\u00eb faqe<\/a><\/noindex> ka nj\u00eb demo interaktive t\u00eb skem\u00ebs s\u00eb ndarjes s\u00eb sekretit t\u00eb Shamir. Demonstra \u00ebsht\u00eb b\u00ebr\u00eb n\u00eb baz\u00eb t\u00eb bibliotek\u00ebs ssss-js<\/a><\/noindex>, e cila vet\u00eb \u00ebsht\u00eb nj\u00eb port JavaScript i programit t\u00eb njohur ssss<\/a><\/noindex>. Vini re se llogaritja e vlerave t\u00eb m\u00ebdha \"Skema, \"Skema dhe \"Skema mund t\u00eb marr\u00eb disa koh\u00eb.
\n
Burimi: habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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