{"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\/az\/blog\/administrirovanie\/shema-razdeleniya-sekreta-shamira","title":{"rendered":"\u015eamir'in gizli b\u00f6l\u00fc\u015f\u00fcm \u015femas\u0131","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Bir senaryoyu d\u00fc\u015f\u00fcnelim, bankac\u0131l\u0131k oturumunun g\u00fcvenli\u011fini sa\u011flamak gerekiyor. Bu, size i\u015fe ba\u015flad\u0131\u011f\u0131n\u0131z g\u00fcn verilen anahtar olmadan tamamen eri\u015filemez kabul edilir. Amac\u0131n\u0131z, anahtar\u0131 g\u00fcvenli bir \u015fekilde saklamakt\u0131r.<\/p>\n

Diyelim ki anahtar\u0131 her zaman yan\u0131n\u0131zda tutmaya karar verdiniz, gerekti\u011finde oturuma eri\u015fim sa\u011flayarak. Ancak, oturumu a\u00e7mak i\u00e7in her seferinde fiziksel varl\u0131\u011f\u0131n\u0131z\u0131n gerekti\u011fini fark edeceksiniz ki bu pratikte \u00f6l\u00e7eklenebilir bir \u00e7\u00f6z\u00fcm de\u011fil. Peki, size vaat edilen tatile ne olacak? Ayr\u0131ca, tek anahtar\u0131n\u0131z\u0131 kaybetmeniz durumunda daha korkutucu bir soru ortaya \u00e7\u0131k\u0131yor.<\/p>\n

Tatil d\u00fc\u015f\u00fcncesiyle, anahtar\u0131n bir kopyas\u0131n\u0131 yapmaya ve ba\u015fka bir \u00e7al\u0131\u015fana emanet etmeye karar verdiniz. Ancak bunun da ideal olmad\u0131\u011f\u0131n\u0131 anl\u0131yorsunuz. Anahtar say\u0131s\u0131n\u0131 iki kat\u0131na \u00e7\u0131kard\u0131\u011f\u0131n\u0131zda, anahtar\u0131n \u00e7al\u0131nma olas\u0131l\u0131\u011f\u0131n\u0131 da art\u0131rm\u0131\u015f oluyorsunuz.<\/p>\n

Umutsuzca, kopyay\u0131 yok ediyorsunuz ve as\u0131l anahtar\u0131 iki par\u00e7aya ay\u0131rmaya karar veriyorsunuz. Art\u0131k d\u00fc\u015f\u00fcn\u00fcyorsunuz ki, anahtar\u0131n par\u00e7alar\u0131n\u0131 ta\u015f\u0131yan iki g\u00fcvenilir ki\u015finin fiziksel olarak bulunmas\u0131 gerekiyor ki anahtar\u0131 birle\u015ftirip oturumu a\u00e7s\u0131nlar. Bu, birinin iki par\u00e7a \u00e7almas\u0131n\u0131 gerektiriyor ki bu da bir anahtar \u00e7almaktan iki kat daha zor. Ancak, birinin anahtar\u0131n yar\u0131s\u0131n\u0131 kaybetmesi durumunda, tam anahtar\u0131n geri kazan\u0131lamayaca\u011f\u0131n\u0131 anlamakta gecikmiyorsunuz.
\n<\/a><\/noindex>
\nBu sorunu bir dizi ek anahtar ve kilit ile \u00e7\u00f6zmek m\u00fcmk\u00fcnd\u00fcr, ancak bu yakla\u015f\u0131mda h\u0131zla bir\u00e7ok<\/i> anahtar ve kilide ihtiya\u00e7 duyulacakt\u0131r. \u0130deal \u015feman\u0131z\u0131n anahtar\u0131 b\u00f6lmek oldu\u011funu d\u00fc\u015f\u00fcn\u00fcyorsunuz, b\u00f6ylece g\u00fcvenlik tamamen tek bir ki\u015fiye ba\u011fl\u0131 kalmamal\u0131. Ayr\u0131ca, bir par\u00e7ay\u0131 kaybetti\u011finizde (ya da bir ki\u015fi tatile gitti\u011finde) t\u00fcm anahtar\u0131n i\u015flevsel kalabilmesi i\u00e7in belli bir par\u00e7a say\u0131s\u0131n\u0131n olmas\u0131 gerekti\u011fini d\u00fc\u015f\u00fcn\u00fcyorusunuz.<\/p>\n

Gizli bir s\u0131r nas\u0131l payla\u015f\u0131l\u0131r<\/h1>\n

\nBu t\u00fcr bir anahtar y\u00f6netim \u015femas\u0131 hakk\u0131nda Adi Shamir, 1979 y\u0131l\u0131nda yay\u0131nlad\u0131\u011f\u0131 \u00e7al\u0131\u015fmas\u0131nda d\u00fc\u015f\u00fcnm\u00fc\u015ft\u00fcr. \u00abGizli Payla\u015f\u0131m\u00bb<\/a><\/noindex>ba\u015fl\u0131kl\u0131 makalesinde anlat\u0131lm\u0131\u015ft\u0131r. Makalede, s\u00f6zde \"\u015eamir'in e\u015fik \u015femas\u0131 a\u00e7\u0131klanmakta ve gizli bir de\u011feri (\u00f6rne\u011fin, kriptografik anahtar) etkin bir \u015fekilde \"\u015eamir'in par\u00e7alara ay\u0131rma y\u00f6ntemini a\u00e7\u0131klamaktad\u0131r. Daha sonra, ancak en az \"\u015eamir'in biz \"\u015eamir'in par\u00e7a topland\u0131\u011f\u0131nda, s\u0131r kolayca geri kazan\u0131labilir. \"\u015eamir'in.<\/p>\n

G\u00fcvenlik a\u00e7\u0131s\u0131ndan, bu \u015feman\u0131n \u00f6nemli bir \u00f6zelli\u011fi, bir sald\u0131rgan\u0131n en az \"\u015eamir'in par\u00e7a yoksa tamamen hi\u00e7bir \u015fey \u00f6\u011frenmemesidir. Hatta \"\u015eamir'in par\u00e7a varl\u0131\u011f\u0131n\u0131n herhangi bir bilgi vermemesi gerekmektedir. Bu \u00f6zelli\u011fe anlamsal g\u00fcvenlik<\/i>.<\/p>\n

denir.<\/h1>\n

\nPolinom interpolasyonu \"\u015eamir'in \u015eamir\u2019in e\u015fik \u015femas\u0131, polinom interpolasyonu kavram\u0131 etraf\u0131nda in\u015fa edilmi\u015ftir.<\/b>Bu konseptl\u0259 tan\u0131\u015f deyilsinizs\u0259, \u0259slind\u0259 o, olduqca sad\u0259dir. \u00dcmumiyy\u0259tl\u0259, \u0259g\u0259r he\u00e7 vaxt qrafikd\u0259 n\u00f6qt\u0259l\u0259r \u00e7\u0259kib, sonra onlar\u0131 x\u0259t v\u0259 ya \u0259yril\u0259rl\u0259 birl\u0259\u015fdirmisinizs\u0259, art\u0131q onu istifad\u0259 etmisiniz!<\/p>\n

\"\u015eamir'in
\n\u0130ki n\u00f6qt\u0259d\u0259n sonsuz sayda 2-ci d\u0259r\u0259c\u0259li polinom ke\u00e7m\u0259k m\u00fcmk\u00fcnd\u00fcr. Onlardan yaln\u0131z birini se\u00e7m\u0259k \u00fc\u00e7\u00fcn \u00fc\u00e7\u00fcnc\u00fc bir n\u00f6qt\u0259 laz\u0131md\u0131r. \u0130ll\u00fcstrasiya: Vikipediya<\/a><\/noindex><\/i> <\/p>\n

1-ci d\u0259r\u0259c\u0259li bir polinomu n\u0259z\u0259rd\u0259n ke\u00e7ir\u0259k, \"\u015eamir'in. \u018fg\u0259r bu funksiyan\u0131 qrafikd\u0259 qurmaq ist\u0259yirsinizs\u0259, ne\u00e7\u0259 n\u00f6qt\u0259y\u0259 ehtiyac\u0131n\u0131z var? Bilirik ki, bu, x\u0259tti bir funksiyad\u0131r, bu da dem\u0259kdir ki, \u0259n az\u0131 iki n\u00f6qt\u0259 laz\u0131md\u0131r. \u0130ndi is\u0259 2-ci d\u0259r\u0259c\u0259li polinomiya baxaq, \"\u015eamir'in. Bu, kvadrat funksiyad\u0131r, buna g\u00f6r\u0259 d\u0259 qrafiki qurmaq \u00fc\u00e7\u00fcn \u0259n az\u0131 \u00fc\u00e7 n\u00f6qt\u0259 laz\u0131md\u0131r. \u00dc\u00e7\u00fcnc\u00fc d\u0259r\u0259c\u0259li polinom haqq\u0131nda nec\u0259? \u018fn az\u0131 d\u00f6rd n\u00f6qt\u0259. V\u0259 bel\u0259 davam edir.<\/p>\n

Bu x\u00fcsusiyy\u0259tin h\u0259qiq\u0259t\u0259n d\u0259 maraql\u0131 t\u0259r\u0259fi ondan ibar\u0259tdir ki, polinom funksiyas\u0131n\u0131n d\u0259r\u0259c\u0259sini v\u0259 \u0259n az\u0131 \"\u015eamir'in n\u00f6qt\u0259ni n\u0259z\u0259rd\u0259n ke\u00e7ir\u0259r\u0259k, bu polinom funksiyas\u0131 \u00fc\u00e7\u00fcn \u0259lav\u0259 n\u00f6qt\u0259l\u0259ri \u00e7\u0131xara bil\u0259rik. Bu \u0259lav\u0259 n\u00f6qt\u0259l\u0259rin ekstrapolyasiyas\u0131na polinomial interpolasiya<\/i>.<\/p>\n

Sirrinin t\u0259rtibi<\/h1>\n

\nB\u0259lk\u0259 d\u0259 burada \u015eamir ssenarisi i\u015f\u0259 d\u00fc\u015f\u00fcr. F\u0259rz ed\u0259k ki, bizim sirrimiz \"\u015eamir'in \u2014 bu \"\u015eamir'in. Biz bunu \"\u015eamir'in qrafikd\u0259 bir n\u00f6qt\u0259y\u0259 \u00e7evir\u0259 bil\u0259rik \"\u015eamir'in v\u0259 bu n\u00f6qt\u0259y\u0259 uy\u011fun olan \"\u015eamir'ind\u0259r\u0259c\u0259d\u0259n bir polinomiya funksiyas\u0131 icad ed\u0259 bil\u0259rik. Qeyd ed\u0259k ki, \"\u015eamir'in laz\u0131m olan par\u00e7a say\u0131n\u0131n e\u015fik n\u00f6qt\u0259sini t\u0259min ed\u0259c\u0259k, buna g\u00f6r\u0259 d\u0259 \u0259g\u0259r e\u015fi\u011fi \u00fc\u00e7 par\u00e7aya qoysak, iki d\u0259r\u0259c\u0259li bir polinomiya funksiyas\u0131 se\u00e7m\u0259liyik.<\/p>\n

Bizim polinomumuz a\u015fa\u011f\u0131dak\u0131 formada olacaq: \"\u015eamir'in, burada \"\u015eamir'in v\u0259 \"\u015eamir'in \u2014 t\u0259sad\u00fcfi se\u00e7ilmi\u015f m\u00fcsb\u0259t tam \u0259d\u0259dl\u0259r. Biz yaln\u0131z \"\u015eamir'ind\u0259r\u0259c\u0259li bir polinom qururuq, burada sabit \u0259msal \"\u015eamir'in\u00a0bizim sirrimizdir, \"\u015eamir'insonrak\u0131 h\u0259r bir \"\u015eamir'in \u00fczv\u00fcn t\u0259sad\u00fcfi se\u00e7ilmi\u015f m\u00fcsb\u0259t \u0259msal\u0131 var. \u018fvv\u0259lki misala qay\u0131daraq, \u0259g\u0259r \"\u015eamir'inolsalar, o zaman a\u015fa\u011f\u0131dak\u0131 funksiya \u0259ld\u0259 ed\u0259rik: \"\u015eamir'in.<\/p>\n

Bu m\u0259rh\u0259l\u0259d\u0259 biz par\u00e7alara yaratmaq \u00fc\u00e7\u00fcn \"\u015eamir'in unikal tam \u0259d\u0259dl\u0259ri birl\u0259\u015fdir\u0259r\u0259k \"\u015eamir'in, burada \"\u015eamir'in (\u00e7\u00fcnki bu, bizim sirrimizdir). Bu n\u00fcmun\u0259d\u0259 d\u00f6rd par\u00e7a payla\u015fmaq ist\u0259yirik, e\u015fik is\u0259 \u00fc\u00e7d\u00fcr, buna g\u00f6r\u0259 d\u0259 t\u0259sad\u00fcfi n\u00f6qt\u0259l\u0259r yaradaraq \"\u015eamir'in bu d\u00f6rd etibarl\u0131 insan, a\u00e7ar m\u00fchafiz\u0259\u00e7il\u0259rin\u0259 bir n\u00f6qt\u0259 g\u00f6nd\u0259ririk. H\u0259m\u00e7inin insanlara bildiririk ki, \"\u015eamir'in\u00e7\u00fcnki bu, ictimai m\u0259lumat say\u0131l\u0131r v\u0259 b\u0259rpa etm\u0259k \u00fc\u00e7\u00fcn laz\u0131md\u0131r. \"\u015eamir'in.<\/p>\n

Sirrinin b\u0259rpas\u0131<\/h1>\n

\nBiz art\u0131q polinomial interpolasiya anlay\u0131\u015f\u0131n\u0131 v\u0259 onun \u015eamir infrastrukturunun \u0259sas\u0131n\u0131 m\u00fczakir\u0259 etmi\u015fik. \"\u015eamir'in\u0130st\u0259nil\u0259n \u00fc\u00e7 d\u00f6rd etibarl\u0131 \u015f\u0259xsl\u0259r \"\u015eamir'insirrini b\u0259rpa etm\u0259k ist\u0259dikd\u0259, yaln\u0131z \"\u015eamir'in \u00f6z unikal n\u00f6qt\u0259l\u0259ri il\u0259 interpolasiya etm\u0259l\u0259ri laz\u0131md\u0131r. Bunun \u00fc\u00e7\u00fcn onlar \u00f6z n\u00f6qt\u0259l\u0259rini m\u00fc\u0259yy\u0259n ed\u0259 bil\u0259rl\u0259r. \"\u015eamir'in v\u0259 Lagrange interpolasiya polinomunu a\u015fa\u011f\u0131dak\u0131 formuldan istifad\u0259 ed\u0259r\u0259k hesablamaq. \u018fg\u0259r proqramla\u015fd\u0131rma sizin \u00fc\u00e7\u00fcn riyaziyyatdan daha ayd\u0131nd\u0131rsa, onda pi \u0259slind\u0259 operatoru t\u0259svir edir. for<\/code>, b\u00fct\u00fcn n\u0259tic\u0259l\u0259ri \u00e7arpan, sigma is\u0259 for<\/code>, h\u0259r \u015feyi toplayand\u0131r.<\/p>\n

\"\u015eamir'in<\/p>\n

<\/p>\n

\"\u015eamir'in<\/p>\n

\n\u0130stifad\u0259 \"\u015eamir'in bunu a\u015fa\u011f\u0131dak\u0131 kimi h\u0259ll ed\u0259 bil\u0259rik v\u0259 ilkin polinom funksiyam\u0131z\u0131 qaytara bil\u0259rik:<\/p>\n

\"\u015eamir'in<\/p>\n

\n\u00c7\u00fcnki bilirik ki, \"\u015eamir'in, b\u0259rpa \"\u015eamir'in sad\u0259c\u0259 h\u0259yata ke\u00e7irilir:<\/p>\n

\"\u015eamir'in<\/p>\n

<\/p>\n

T\u0259hl\u00fck\u0259siz olmayan tam \u0259d\u0259d arifmetikas\u0131ndan istifad\u0259<\/h1>\n

\nBizi \u015eamirinin \u0259sas ideyas\u0131n\u0131 u\u011furla t\u0259tbiq etmi\u015f olsaq da, \"\u015eamir'in, indiy\u0259 q\u0259d\u0259r yaddaqalan bir problemin qald\u0131\u011f\u0131n\u0131 g\u00f6r\u00fcr\u00fck. Bizim polinom funksiyam\u0131z t\u0259hl\u00fck\u0259siz olmayan tam \u0259d\u0259d arifmetikas\u0131n\u0131 istifad\u0259 edir. H\u0259r \u0259lav\u0259 n\u00f6qt\u0259 \u00fc\u00e7\u00fcn, h\u00fccum ed\u0259nl\u0259r bizim funksiyan\u0131n qrafikind\u0259 \u0259ld\u0259 etdiyi zamana, dig\u0259r n\u00f6qt\u0259l\u0259r \u00fc\u00e7\u00fcn az\u0131 imkanlar qal\u0131r. Bunu tam \u0259d\u0259di arifmetika istifad\u0259 ed\u0259r\u0259k polinom funksiyas\u0131n\u0131n n\u00f6qt\u0259 say\u0131n\u0131 art\u0131rd\u0131qca qrafik \u00e7iz\u0259nd\u0259 \u00f6z g\u00f6zl\u0259rinizl\u0259 g\u00f6r\u0259 bil\u0259rsiniz. Bu, biz\u0259 t\u0259hl\u00fck\u0259sizliyi n\u0259z\u0259rd\u0259 tutdu\u011fumuz m\u0259qs\u0259d\u0259 ziddir, \u00e7\u00fcnki h\u00fccum ed\u0259n he\u00e7 bir \u015fey bilm\u0259m\u0259lidir, n\u0259 zamana q\u0259d\u0259r \u0259n az\u0131 \"\u015eamir'in fragments.<\/p>\n

Tam \u0259d\u0259d arifmetikas\u0131n\u0131n z\u0259ifliyi il\u0259 ba\u011fl\u0131 bir ssenari g\u00f6st\u0259r\u0259 bil\u0259rik. H\u00fccum ed\u0259n iki n\u00f6qt\u0259 \u0259ld\u0259 ed\u0259rs\u0259 \"\u015eamir'in v\u0259 ictimai m\u0259lumatlar\u0131n oldu\u011funu bilir ki, \"\u015eamir'in. Bu m\u0259lumatdan o, \"\u015eamir'in, iki b\u0259rab\u0259r olmaq \u00fczr\u0259 v\u0259 m\u0259lum d\u0259y\u0259rl\u0259ri formulaya qoymaq \u00fc\u00e7\u00fcn istifad\u0259 ed\u0259 bil\u0259r. \"\u015eamir'in v\u0259 \"\u015eamir'in.<\/p>\n

\"\u015eamir'in<\/p>\n

\n Sonra, h\u00fccum ed\u0259n \"\u015eamir'intapmaq \u00fc\u00e7\u00fcn \"\u015eamir'in:<\/p>\n

\"\u015eamir'in<\/p>\n

\nhesablaya bil\u0259r. \"\u015eamir'in \u00c7\u00fcnki \"\u015eamir'int\u0259sad\u00fcfi se\u00e7ilmi\u015f tam \u0259d\u0259d m\u00fcsb\u0259t \u0259d\u0259dl\u0259r kimi m\u00fc\u0259yy\u0259n etdiyimiz \u00fc\u00e7\u00fcn, m\u00fcmk\u00fcn olan say\u0131 m\u0259hduddur. \"\u015eamir'inBu m\u0259lumatla, h\u00fccum ed\u0259n \"\u015eamir'in tapmaq imkan\u0131na malikdir, \u00e7\u00fcnki be\u015fd\u0259n b\u00f6y\u00fck olan h\u0259r \u015fey \"\u015eamir'in<\/p>\n

m\u0259nfi edir. Bu do\u011frudur, \u00e7\u00fcnki biz \"\u015eamir'inSonra, h\u00fccum ed\u0259n m\u00fcmk\u00fcn d\u0259y\u0259rl\u0259ri hesablaya bil\u0259r \"\u015eamir'in daxilind\u0259dir. \"\u015eamir'in:<\/p>\n

\"\u015eamir'in<\/p>\n

\n, \"\u015eamir'in \u00fc\u00e7\u00fcn m\u0259hdud se\u00e7im t\u0259qdim edil\u0259rk\u0259n \"\u015eamir'innec\u0259 asanl\u0131qla uy\u011fun d\u0259y\u0259rl\u0259ri tapmaq v\u0259 yoxlamaq m\u00fcmk\u00fcn oldu\u011funu g\u00f6st\u0259rir.<\/p>\n

Burada yaln\u0131z be\u015f variant var.<\/h1>\n

\nT\u0259hl\u00fck\u0259siz olmayan tam \u0259d\u0259d arifmetikas\u0131n\u0131n problemini h\u0259ll etm\u0259k \"\u015eamir'in buna \"\u015eamir'in, burada \"\u015eamir'in v\u0259 \"\u015eamir'in Bu z\u0259ifliyi aradan qald\u0131rmaq \u00fc\u00e7\u00fcn, \u015eamir modulyar arifmetikadan istifad\u0259 etm\u0259yi t\u0259klif edir,<\/p>\n

\u2014 b\u00fct\u00fcn sad\u0259 \u0259d\u0259dl\u0259rin c\u0259midir. \"\u015eamir'inModulyar arifmetikan\u0131n nec\u0259 i\u015fl\u0259diyini tez yad\u0131ma salaq. Saatlar \u2014 art\u0131q tan\u0131\u015f konsepsiyad\u0131r. Bu, saatlar\u0131 istifad\u0259 edir, hans\u0131 ki, olar ki, onlardir. Saat g\u00f6st\u0259ricisi on iki ke\u00e7\u0259nd\u0259 bir\u0259 qay\u0131d\u0131r. Bu sistemin maraql\u0131 x\u00fcsusiyy\u0259ti odur ki, sad\u0259c\u0259 saatlara baxaraq, saat g\u00f6st\u0259ricisi ne\u00e7\u0259 d\u00f6vr ke\u00e7diyini bilm\u0259k olmur. Ancaq \u0259g\u0259r biz bils\u0259k ki, saat g\u00f6st\u0259ricisi 12-d\u0259n d\u00f6rd d\u0259f\u0259 ke\u00e7ib, sad\u0259 bir formulla ke\u00e7\u0259n saatlar\u0131n say\u0131n\u0131 tam m\u00fc\u0259yy\u0259n etm\u0259k m\u00fcmk\u00fcnd\u00fcr. \"\u015eamir'in, burada \"\u015eamir'in\u00a0\u2014 bu bizim b\u00f6l\u00fcc\u00fcm\u00fczd\u00fcr (burada \"\u015eamir'in), \"\u015eamir'in\u00a0\u2014 bu katsay\u0131d\u0131r (b\u00f6l\u00fcc\u00fc, tam olarak ka\u00e7 kez ana say\u0131ya ge\u00e7er, burada \"\u015eamir'in), v\u0259 \"\u015eamir'in\u00a0\u2014 bu genellikle mod operat\u00f6r\u00fc \u00e7a\u011fr\u0131s\u0131n\u0131n d\u00f6nd\u00fcrd\u00fc\u011f\u00fc kaland\u0131r (burada \"\u015eamir'in). Bu de\u011ferlerin hepsini bilmek, denklemi \u00e7\u00f6zmemizi sa\u011flar, \"\u015eamir'in, ancak katsay\u0131y\u0131 atlarsak, ba\u015flang\u0131\u00e7 de\u011ferini asla geri kazanamay\u0131z.<\/p>\n

Bunun, \u00f6nceki \u00f6rne\u011fimize uygulayarak ve \"\u015eamir'ing\u00fcvenli\u011fimizi nas\u0131l art\u0131rd\u0131\u011f\u0131n\u0131 g\u00f6sterebiliriz. Yeni polinom fonksiyonumuz \"\u015eamir'in, yeni noktalar ise \"\u015eamir'in. Art\u0131k anahtar sahipleri, toplama ve \u00e7arpma i\u015flemlerinin mod alarak ger\u00e7ekle\u015ftirilmesi gerekti\u011fi bu sefer polinom interpolasyonunu yeniden kullanabilirler \"\u015eamir'in (\u00f6rne\u011fin, \"\u015eamir'in).<\/p>\n

Bu yeni \u00f6rne\u011fi kullanarak, varsayal\u0131m ki bir sald\u0131rgan bu yeni noktalardan ikisini \u00f6\u011frendi, \"\u015eamir'in, ve kamuya a\u00e7\u0131k bilgi \"\u015eamir'in. Bu sefer sald\u0131rgan, elindeki t\u00fcm bilgilere dayanarak \u015fu i\u015flevleri \u00e7\u0131kar\u0131yor, burada \"\u015eamir'in\u00a0\u2014 t\u00fcm pozitif tam say\u0131lar\u0131n k\u00fcmesi, ve \"\u015eamir'in mod\u00fcl\u00fcn katsay\u0131s\u0131n\u0131 temsil eder. \"\u015eamir'in.<\/p>\n

\"\u015eamir'in<\/p>\n

\n\u015eimdi sald\u0131rgan\u0131m\u0131z tekrar \u015funlar\u0131 buluyor, \"\u015eamir'in, hesaplayarak \"\u015eamir'in:<\/p>\n

\"\u015eamir'in<\/p>\n

\nSonra tekrar \u00e7\u0131karmaya \u00e7al\u0131\u015f\u0131yor, \"\u015eamir'inSonra, h\u00fccum ed\u0259n m\u00fcmk\u00fcn d\u0259y\u0259rl\u0259ri hesablaya bil\u0259r \"\u015eamir'in daxilind\u0259dir. \"\u015eamir'in:<\/p>\n

\"\u015eamir'in<\/p>\n

\nBu sefer ciddi bir problemi var. Form\u00fclde de\u011ferler eksik, \"\u015eamir'in, \"\u015eamir'in v\u0259 \"\u015eamir'in. Bu de\u011fi\u015fkenlerin sonsuz say\u0131da kombinasyonu oldu\u011fundan, ek bir bilgi elde edemez.<\/p>\n

G\u00fcvenlik d\u00fc\u015f\u00fcnceleri<\/h1>\n

\nShamir'in gizli payla\u015f\u0131m \u015femas\u0131, bilgi teorisi a\u00e7\u0131s\u0131ndan g\u00fcvenlik sunar.<\/i>. Bu, matemati\u011fin s\u0131n\u0131rs\u0131z hesaplama g\u00fcc\u00fcne sahip bir sald\u0131rgana kar\u015f\u0131 bile dayan\u0131kl\u0131 oldu\u011fu anlam\u0131na gelir. Ancak \u015feman\u0131n hala bilinen baz\u0131 sorunlar\u0131 vard\u0131r.<\/p>\n

\u00d6rne\u011fin, Shamir'in \u015femas\u0131, do\u011frulanabilir par\u00e7alar<\/i>olu\u015fturmaz, yani insanlar sahte par\u00e7alar\u0131 serbest\u00e7e sunabilir ve do\u011fru s\u0131rr\u0131n kurtar\u0131lmas\u0131n\u0131 engelleyebilir. Yeterli bilgiye sahip olan d\u00fc\u015fman bir par\u00e7a bile \u00fcretebilir, \"\u015eamir'in istedi\u011fi \u015fekilde de\u011fi\u015ftirerek. Bu sorun, do\u011frulanabilir gizli payla\u015f\u0131m \u015femalar\u0131<\/i>, Feldman \u015femas\u0131 gibi, kullan\u0131larak \u00e7\u00f6z\u00fcl\u00fcr.<\/p>\n

Ba\u015fka bir sorun ise, herhangi bir par\u00e7an\u0131n uzunlu\u011funun ilgili s\u0131ran\u0131n uzunlu\u011funa e\u015fit olmas\u0131d\u0131r; b\u00f6ylece s\u0131ran\u0131n uzunlu\u011fu kolayca belirlenebilir. Bu sorun, s\u0131ray\u0131 sabit bir uzunlu\u011fa kadar rastgele say\u0131larla doldurarak basit\u00e7e \u00e7\u00f6z\u00fcl\u00fcr.<\/i> \u015f\u0259kil uzunlu\u011funa q\u0259d\u0259r t\u0259sad\u00fcfi \u0259d\u0259dl\u0259rin sirri.<\/p>\n

N\u0259hay\u0259t, bizim t\u0259hl\u00fck\u0259sizlikl\u0259 ba\u011fl\u0131 narahatl\u0131qlar\u0131m\u0131z yaln\u0131z \u0259sas sxeml\u0259 kifay\u0259tl\u0259nmir. H\u0259qiqi kriptoqrafiya t\u0259tbiql\u0259rind\u0259 \u00e7ox vaxt xarici kanallar vasit\u0259sil\u0259 h\u00fccumlar t\u0259hl\u00fck\u0259si var, burada h\u00fccum\u00e7u proqram\u0131n icra m\u00fcdd\u0259tind\u0259n, ke\u015fl\u0259m\u0259, q\u0259za v\u0259 s. m\u0259lumatlar\u0131 \u0259ld\u0259 etm\u0259y\u0259 \u00e7al\u0131\u015f\u0131r. \u018fg\u0259r bu narahatl\u0131q do\u011furursa, inki\u015faf zaman\u0131 m\u00fchafiz\u0259 t\u0259dbirl\u0259rinin istifad\u0259sini diqq\u0259tl\u0259 n\u0259z\u0259rd\u0259n ke\u00e7irm\u0259k, m\u0259s\u0259l\u0259n, sabit icra m\u00fcdd\u0259tin\u0259 \u0259saslanan funksiyalar\u0131 n\u0259z\u0259r\u0259 almaq, yadda\u015f\u0131 diske saxlama\u011f\u0131 qar\u015f\u0131s\u0131n\u0131 almaq v\u0259 bu m\u0259qal\u0259d\u0259n k\u0259narda bir s\u0131ra dig\u0259r m\u0259s\u0259l\u0259l\u0259ri d\u00fc\u015f\u00fcnm\u0259k laz\u0131md\u0131r.<\/p>\n

Demo<\/h1>\n

\n\u00dc\u00e7\u00fcn bu s\u0259hif\u0259d\u0259<\/a><\/noindex> \u015eamirin gizli b\u00f6l\u00fc\u015fd\u00fcrm\u0259 sxeminin interaktiv n\u00fcmayi\u015fi var. N\u00fcmayi\u015f ssss-js<\/a><\/noindex>, populyar proqram\u0131n JavaScript portu olan ssss<\/a><\/noindex>d\u0131r. B\u00f6y\u00fck d\u0259y\u0259rl\u0259ri hesablamaq \u00fc\u00e7\u00fcn \"\u015eamir'in, \"\u015eamir'in v\u0259 \"\u015eamir'in biraz vaxt laz\u0131m ola bil\u0259r.
\n
M\u0259nb\u0259: habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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O, ilk i\u015f g\u00fcn\u00fcn\u00fczd\u0259 siz\u0259 veril\u0259n a\u00e7ar olmadan tamamil\u0259 qorunmaz hesab olunur.","canonical_url":"https:\/\/prohoster.info\/az\/blog\/administrirovanie\/shema-razdeleniya-sekreta-shamira","robots":"max-image-preview:large","keywords":"","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"az_AZ","og:site_name":"ProHoster | \u041a\u0443\u043f\u0438\u0442\u044c \u043d\u0430\u0434\u0435\u0436\u043d\u044b\u0439 \u0445\u043e\u0441\u0442\u0438\u043d\u0433 \u0434\u043b\u044f \u0441\u0430\u0439\u0442\u043e\u0432 \u0441 \u0437\u0430\u0449\u0438\u0442\u043e\u0439 \u043e\u0442 DDoS, VPS VDS \u0441\u0435\u0440\u0432\u0435\u0440\u044b","og:type":"article","og:title":"\ud83e\udd47\u0421\u0445\u0435\u043c\u0430 \u0440\u0430\u0437\u0434\u0435\u043b\u0435\u043d\u0438\u044f \u0441\u0435\u043a\u0440\u0435\u0442\u0430 \u0428\u0430\u043c\u0438\u0440\u0430 | ProHoster","og:description":"\u0420\u0430\u0441\u0441\u043c\u043e\u0442\u0440\u0438\u043c 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\u0440\u0430\u0431\u043e\u0442\u044b.","og:url":"https:\/\/prohoster.info\/az\/blog\/administrirovanie\/shema-razdeleniya-sekreta-shamira","og:image":"https:\/\/prohoster.info\/wp-content\/uploads\/2021\/11\/logo-350.jpg","og:image:secure_url":"https:\/\/prohoster.info\/wp-content\/uploads\/2021\/11\/logo-350.jpg","og:image:width":350,"og:image:height":350,"article:published_time":"2019-12-01T21:00:00+00:00","article:modified_time":"2020-02-18T11:01:21+00:00","article:publisher":"https:\/\/www.facebook.com\/prohoster","article:author":"https:\/\/www.facebook.com\/prohoster"},"aioseo_meta_data":{"post_id":"53444","title":null,"description":null,"keywords":null,"keyphrases":null,"primary_term":null,"canonical_url":null,"og_title":null,"og_description":null,"og_object_type":"default","og_image_type":"default","og_image_url":null,"og_image_width":null,"og_image_height":null,"og_image_custom_url":null,"og_image_custom_fields":null,"og_video":null,"og_custom_url":null,"og_article_section":null,"og_article_tags":null,"twitter_use_og":false,"twitter_card":"default","twitter_image_type":"default","twitter_image_url":null,"twitter_image_custom_url":null,"twitter_image_custom_fields":null,"twitter_title":null,"twitter_description":null,"schema":{"blockGraphs":[],"customGraphs":[],"default":{"data":{"Article":[],"Course":[],"Dataset":[],"FAQPage":[],"Movie":[],"Person":[],"Product":[],"ProductReview":[],"Car":[],"Recipe":[],"Service":[],"SoftwareApplication":[],"WebPage":[]},"graphName":"","isEnabled":true},"graphs":[]},"schema_type":null,"schema_type_options":null,"pillar_content":false,"robots_default":true,"robots_noindex":false,"robots_noarchive":false,"robots_nosnippet":false,"robots_nofollow":false,"robots_noimageindex":false,"robots_noodp":false,"robots_notranslate":false,"robots_max_snippet":null,"robots_max_videopreview":null,"robots_max_imagepreview":"large","priority":null,"frequency":null,"local_seo":null,"seo_analyzer_scan_date":"2026-01-24 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