{"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\/cs\/blog\/administrirovanie\/shema-razdeleniya-sekreta-shamira","title":{"rendered":"Schematika d\u011blen\u00ed tajemstv\u00ed Shamir\u016fv","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Zva\u017eme sc\u00e9n\u00e1\u0159, kdy je nutn\u00e9 zajistit bezpe\u010dnost bankovn\u00edho trezoru. Ten se pova\u017euje za zcela nepr\u016fst\u0159eln\u00fd bez kl\u00ed\u010de, kter\u00fd obdr\u017e\u00edte hned prvn\u00ed pracovn\u00ed den. Va\u0161\u00edm c\u00edlem je bezpe\u010dn\u011b uchovat kl\u00ed\u010d.<\/p>\n

P\u0159edpokl\u00e1dejme, \u017ee jste se rozhodli kl\u00ed\u010d neust\u00e1le nosit u sebe a poskytovat p\u0159\u00edstup k trezoru podle pot\u0159eby. Rychle v\u0161ak zjist\u00edte, \u017ee takov\u00e9 \u0159e\u0161en\u00ed se prakticky ne\u0161k\u00e1luje, proto\u017ee poka\u017ed\u00e9 je nutn\u00e1 va\u0161e fyzick\u00e1 p\u0159\u00edtomnost pro otev\u0159en\u00ed trezoru. A co dovolen\u00e9, kter\u00e9 v\u00e1m sl\u00edbili? Nav\u00edc je tu je\u0161t\u011b stra\u0161ideln\u011bj\u0161\u00ed ot\u00e1zka: co kdy\u017e ztrat\u00edte jedin\u00fd kl\u00ed\u010d?<\/p>\n

S my\u0161lenkou na dovolenou jste se rozhodli ud\u011blat kopii kl\u00ed\u010de a sv\u011b\u0159it ji jin\u00e9mu zam\u011bstnanci. Uv\u011bdomujete si v\u0161ak, \u017ee to tak\u00e9 nen\u00ed ide\u00e1ln\u00ed. Dvoj\u00ed po\u010det kl\u00ed\u010d\u016f znamen\u00e1 tak\u00e9 dvojn\u00e1sobn\u00e9 mo\u017enosti kr\u00e1de\u017ee kl\u00ed\u010de.<\/p>\n

Bezradn\u011b ni\u010d\u00edte dublikat a rozhodujete se rozd\u011blit p\u016fvodn\u00ed kl\u00ed\u010d na dv\u011b \u010d\u00e1sti. Te\u010f si \u0159\u00edk\u00e1te, \u017ee dva d\u016fv\u011bryhodn\u00ed lid\u00e9 s \u010d\u00e1stmi kl\u00ed\u010de mus\u00ed b\u00fdt fyzicky p\u0159\u00edtomni, aby kl\u00ed\u010d slo\u017eili a otev\u0159eli trezor. To znamen\u00e1, \u017ee zlod\u011bj mus\u00ed ukr\u00e1st dv\u011b \u010d\u00e1sti, co\u017e je dvakr\u00e1t t\u011b\u017e\u0161\u00ed ne\u017e ukr\u00e1st jeden kl\u00ed\u010d. Ale brzy si uv\u011bdom\u00edte, \u017ee tento pl\u00e1n nen\u00ed o moc lep\u0161\u00ed ne\u017e m\u00edt jen jeden kl\u00ed\u010d, proto\u017ee pokud n\u011bkdo ztrat\u00ed jednu \u010d\u00e1st kl\u00ed\u010de, cel\u00fd kl\u00ed\u010d nelze obnovit.
\n<\/a><\/noindex>
\nProbl\u00e9m lze vy\u0159e\u0161it s\u00e9ri\u00ed dal\u0161\u00edch kl\u00ed\u010d\u016f a z\u00e1mk\u016f, ale p\u0159i takov\u00e9m p\u0159\u00edstupu se rychle vy\u017eaduje mnoho<\/i> kl\u00ed\u010d\u016f a z\u00e1mk\u016f. Rozhodujete se, \u017ee v ide\u00e1ln\u00edm pl\u00e1nu by m\u011blo b\u00fdt rozd\u011bleno kl\u00ed\u010de, aby bezpe\u010dnost nespol\u00e9hala zcela na jednu osobu. Tak\u00e9 z\u00e1v\u011b\u0159\u00edte, \u017ee by m\u011blo existovat ur\u010dit\u00e9 prahov\u00e9 mno\u017estv\u00ed \u010d\u00e1st\u00ed, aby p\u0159i ztr\u00e1t\u011b jedn\u00e9 \u010d\u00e1sti (nebo pokud osoba odejde na dovolenou) z\u016fstal cel\u00fd kl\u00ed\u010d funk\u010dn\u00ed.<\/p>\n

Jak rozd\u011blit tajemstv\u00ed<\/h1>\n

\nO takov\u00e9m typu sch\u00e9matu spr\u00e1vy kl\u00ed\u010d\u016f p\u0159em\u00fd\u0161lel Adi Shamir v roce 1979, kdy\u017e publikoval svou pr\u00e1ci \u201eJak rozd\u011blit tajemstv\u00ed\u201c<\/a><\/noindex>. V \u010dl\u00e1nku se stru\u010dn\u011b vysv\u011btluje tzv. \"Schematika prahov\u00e9 sch\u00e9ma pro efektivn\u00ed rozd\u011blen\u00ed tajn\u00e9 hodnoty (nap\u0159\u00edklad kryptografick\u00e9ho kl\u00ed\u010de) na \"Schematika \u010d\u00e1sti. A teprve kdy\u017e jsou shrom\u00e1\u017ed\u011bny alespo\u0148 \"Schematika z \"Schematika \u010d\u00e1sti, lze tajemstv\u00ed snadno obnovit. \"Schematika.<\/p>\n

Z hlediska bezpe\u010dnosti je d\u016fle\u017eitou vlastnost\u00ed tohoto sch\u00e9matu to, \u017ee \u00fato\u010dn\u00edk by nem\u011bl zjistit absolutn\u011b nic, pokud nem\u00e1 alespo\u0148 \"Schematika \u010d\u00e1st\u00ed. I p\u0159\u00edtomnost \"Schematika \u010d\u00e1st\u00ed by nem\u011bla poskytovat \u017e\u00e1dn\u00e9 informace. Tento vlastnost naz\u00fdv\u00e1me semastrickou bezpe\u010dnost\u00ed<\/i>.<\/p>\n

Polynomi\u00e1ln\u00ed interpolace<\/h1>\n

\nPrahov\u00e9 sch\u00e9ma Shamir \"Schematika je postaveno kolem konceptu polynomi\u00e1ln\u00ed interpolace.<\/b>Pokud nejste s t\u00edmto konceptem obezn\u00e1meni, je to vlastn\u011b docela jednoduch\u00e9. V podstat\u011b, pokud jste n\u011bkdy kreslili body na grafu a pak je spojovali \u010darami nebo k\u0159ivkami, ji\u017e jste ho pou\u017eili!<\/p>\n

\"Schematika
\nMezi dv\u011bma body lze nakreslit neomezen\u00e9 mno\u017estv\u00ed polynom\u016f druh\u00e9ho stupn\u011b. Abychom vybrali jedin\u00fd, pot\u0159ebujeme t\u0159et\u00ed bod. Ilustrace: Wikipedia<\/a><\/noindex><\/i> <\/p>\n

Zva\u017eme polynom prvn\u00edho stupn\u011b, \"Schematika. Kolik bod\u016f pot\u0159ebujete, pokud chcete nakreslit tuto funkci na grafu? No, v\u00edme, \u017ee je to line\u00e1rn\u00ed funkce, kter\u00e1 tvo\u0159\u00ed \u010d\u00e1ru, a proto pot\u0159ebujeme alespo\u0148 dva body. D\u00e1le zva\u017eme polynomi\u00e1ln\u00ed funkci druh\u00e9ho stupn\u011b, \"Schematika. Je to kvadratick\u00e1 funkce, a proto k vykreslen\u00ed grafu pot\u0159ebujeme alespo\u0148 t\u0159i body. Co \u0159\u00edk\u00e1te na polynom t\u0159et\u00edho stupn\u011b? Alespo\u0148 \u010dty\u0159i body. A tak d\u00e1le.<\/p>\n

Opravdu skv\u011bl\u00e1 v\u011bc na t\u00e9to vlastnosti je, \u017ee, vezmeme-li v \u00favahu stupe\u0148 polynomi\u00e1ln\u00ed funkce a alespo\u0148 \"Schematika bod\u016f, m\u016f\u017eeme odvodit dal\u0161\u00ed body pro tuto polynomi\u00e1ln\u00ed funkci. Extrapolaci t\u011bchto dal\u0161\u00edch bod\u016f naz\u00fdv\u00e1me polynomi\u00e1ln\u00ed interpolace.<\/i>.<\/p>\n

Sestavov\u00e1n\u00ed tajemstv\u00ed<\/h1>\n

\nMo\u017en\u00e1 u\u017e ch\u00e1pete, \u017ee zde vstupuje do hry chytr\u00fd Shamir\u016fv pl\u00e1n. P\u0159edpokl\u00e1dejme, \u017ee na\u0161e tajemstv\u00ed je \"Schematika \u2014 to je \"Schematika. M\u016f\u017eeme p\u0159ev\u00e9st \"Schematika na bod na grafu \"Schematika a vymyslet polynomi\u00e1ln\u00ed funkci stupn\u011b \"Schematika, kter\u00e1 tento bod spl\u0148uje. P\u0159ipome\u0148me, \u017ee \"Schematika bude na\u0161\u00edm prahem po\u017eadovan\u00fdch \u010d\u00e1st\u00ed, tak\u017ee pokud nastav\u00edme pr\u00e1h na t\u0159i \u010d\u00e1sti, mus\u00edme zvolit polynomi\u00e1ln\u00ed funkci stupn\u011b dva.<\/p>\n

N\u00e1\u0161 polynom bude m\u00edt tvar \"Schematika, kde \"Schematika a \"Schematika \u2014 n\u00e1hodn\u011b vybran\u00e1 kladn\u00e1 cel\u00e1 \u010d\u00edsla. Jen stav\u00edme polynom stupn\u011b \"Schematika, kde voln\u00fd koeficient \"Schematika\u00a0je na\u0161e tajemstv\u00ed, \"Schematikaa ka\u017ed\u00fd z n\u00e1sleduj\u00edc\u00edch \"Schematika \u010dlen\u016f je n\u00e1hodn\u011b vybran\u00fd kladn\u00fd koeficient. Pokud se vr\u00e1t\u00edme k p\u016fvodn\u00edmu p\u0159\u00edkladu a p\u0159edpokl\u00e1d\u00e1me, \u017ee \"Schematika, pak z\u00edsk\u00e1me funkci \"Schematika.<\/p>\n

V t\u00e9to f\u00e1zi m\u016f\u017eeme generovat \u00faseky p\u0159ipojen\u00edm \"Schematika unik\u00e1tn\u00edch cel\u00fdch \u010d\u00edsel v \"Schematika, kde \"Schematika (proto\u017ee to je na\u0161e tajemstv\u00ed). V tomto p\u0159\u00edkladu chceme rozdat \u010dty\u0159i \u00faseky s prahem t\u0159i, tak\u017ee n\u00e1hodn\u011b generujeme body \"Schematika a pos\u00edl\u00e1me jeden bod ka\u017ed\u00e9mu ze \u010dty\u0159 d\u016fv\u011bryhodn\u00fdch osob, str\u00e1\u017ec\u016f kl\u00ed\u010de. Tak\u00e9 informujeme lidi, \u017ee \"Schematika, proto\u017ee se pova\u017euje za ve\u0159ejnou informaci a je pot\u0159ebn\u00e1 pro obnovu \"Schematika.<\/p>\n

Obnova tajemstv\u00ed<\/h1>\n

\nJi\u017e jsme se zab\u00fdvali konceptem polynomick\u00e9 interpolace a t\u00edm, \u017ee le\u017e\u00ed v z\u00e1kladech Shamirova sch\u00e9matu \"Schematika. Kdy\u017e jak\u00e9koli t\u0159i z \u010dty\u0159 d\u016fv\u011bryhodn\u00fdch osob cht\u011bj\u00ed obnovit \"Schematika, pot\u0159ebuj\u00ed pouze interpolovat \"Schematika s jejich unik\u00e1tn\u00edmi body. K tomu mohou ur\u010dit sv\u00e9 body \"Schematika a vypo\u010d\u00edtat interpoluj\u00edc\u00ed Lagrange\u016fv polynom pomoc\u00ed n\u00e1sleduj\u00edc\u00edho vzorce. Pokud v\u00e1m programov\u00e1n\u00ed d\u00e1v\u00e1 v\u011bt\u0161\u00ed smysl ne\u017e matematika, pak pi je v podstat\u011b oper\u00e1tor for<\/code>, kter\u00fd n\u00e1sob\u00ed v\u0161echny v\u00fdsledky, a sigma je for<\/code>, kter\u00e1 v\u0161echno s\u010d\u00edt\u00e1.<\/p>\n

\"Schematika<\/p>\n

<\/p>\n

\"Schematika<\/p>\n

\nP\u0159i \"Schematika m\u016f\u017eeme to vy\u0159e\u0161it n\u00e1sleduj\u00edc\u00edm zp\u016fsobem a vr\u00e1tit na\u0161i p\u016fvodn\u00ed polynomickou funkci:<\/p>\n

\"Schematika<\/p>\n

\nJeliko\u017e v\u00edme, \u017ee \"Schematika, obnova \"Schematika se prov\u00e1d\u00ed jednodu\u0161e:<\/p>\n

\"Schematika<\/p>\n

<\/p>\n

Pou\u017eit\u00ed nezabezpe\u010den\u00e9 celo\u010d\u00edseln\u00e9 aritmetiky<\/h1>\n

\nI kdy\u017e jsme \u00fasp\u011b\u0161n\u011b aplikovali z\u00e1kladn\u00ed my\u0161lenku Shamira \"Schematika, zb\u00fdv\u00e1 n\u00e1m probl\u00e9m, kter\u00fd jsme dosud ignorovali. Na\u0161e polynomick\u00e1 funkce pou\u017e\u00edv\u00e1 nezabezpe\u010denou celo\u010d\u00edselnou aritmetiku. M\u011bjte na pam\u011bti, \u017ee s ka\u017ed\u00fdm dal\u0161\u00edm bodem, kter\u00fd \u00fato\u010dn\u00edk z\u00edsk\u00e1v\u00e1 na grafu na\u0161\u00ed funkce, z\u016fst\u00e1v\u00e1 m\u00e9n\u011b mo\u017enost\u00ed pro ostatn\u00ed body. To m\u016f\u017eete vid\u011bt na vlastn\u00ed o\u010di, kdy\u017e kresl\u00edte graf s rostouc\u00edm po\u010dtem bod\u016f pro polynomickou funkci s pou\u017eit\u00edm celo\u010d\u00edseln\u00e9 aritmetiky. To je kontraproduktivn\u00ed pro n\u00e1\u0161 deklarovan\u00fd c\u00edl bezpe\u010dnosti, proto\u017ee \u00fato\u010dn\u00edk by nem\u011bl zjistit absolutn\u011b nic, dokud nem\u00e1 alespo\u0148 \"Schematika fragments.<\/p>\n

Abychom demonstrovali, jak slab\u00e1 je sch\u00e9ma s celo\u010d\u00edselnou aritmetikou, zva\u017eme sc\u00e9n\u00e1\u0159, ve kter\u00e9m \u00fato\u010dn\u00edk z\u00edskal dva body \"Schematika a v\u00ed ve\u0159ejnou informaci, \u017ee \"Schematika. Z t\u011bchto informac\u00ed m\u016f\u017ee odvodit \"Schematika, rovn\u00fd dv\u011bma, a dosadit do vzorce zn\u00e1m\u00e9 hodnoty \"Schematika a \"Schematika.<\/p>\n

\"Schematika<\/p>\n

\n Pak m\u016f\u017ee \u00fato\u010dn\u00edk zjistit \"Schematika, spo\u010d\u00edt\u00e1n\u00edm \"Schematika:<\/p>\n

\"Schematika<\/p>\n

\nJeliko\u017e jsme ur\u010dili \"Schematika jako n\u00e1hodn\u011b vybran\u00e1 cel\u00e1 kladn\u00e1 \u010d\u00edsla, existuje omezen\u00fd po\u010det mo\u017en\u00fdch \"Schematika. Pomoc\u00ed t\u011bchto informac\u00ed m\u016f\u017ee \u00fato\u010dn\u00edk odvodit \"Schematika, jeliko\u017e v\u0161e, co je v\u011bt\u0161\u00ed ne\u017e 5, ud\u011bl\u00e1 \"Schematika negativn\u00edm. To se ukazuje jako pravdiv\u00e9, proto\u017ee jsme ur\u010dili \"Schematika<\/p>\n

Pak m\u016f\u017ee \u00fato\u010dn\u00edk spo\u010d\u00edtat mo\u017en\u00e9 hodnoty \"Schematika, nahrazen\u00edm \"Schematika v \"Schematika:<\/p>\n

\"Schematika<\/p>\n

\nS omezenou sadou variant pro \"Schematika se st\u00e1v\u00e1 jasn\u00e9, jak snadn\u00e9 je odhadnout a ov\u011b\u0159it hodnoty \"Schematika. Tady je celkem p\u011bt variant.<\/p>\n

\u0158e\u0161en\u00ed probl\u00e9mu s nebezpe\u010dnou celo\u010d\u00edselnou aritmetikou<\/h1>\n

\nAby se tato zranitelnost odstranila, navrhuje Shamir pou\u017e\u00edt modul\u00e1rn\u00ed aritmetiku, nahrazen\u00edm \"Schematika na \"Schematika, kde \"Schematika a \"Schematika \u2014 mno\u017estv\u00edm v\u0161ech prvo\u010d\u00edsel.<\/p>\n

Rychle si p\u0159ipome\u0148me, jak funguje modul\u00e1rn\u00ed aritmetika. Hodiny s ru\u010di\u010dkami jsou ji\u017e zn\u00e1m\u00fd koncept. Pou\u017e\u00edvaj\u00ed hodiny, kter\u00e9 jsou \"Schematika. Jakmile ru\u010di\u010dka hodin projde kolem dvan\u00e1cti, vrac\u00ed se k jedn\u00e9. Zaj\u00edmavou vlastnost\u00ed tohoto syst\u00e9mu je to, \u017ee pouh\u00fdm pohledem na hodiny nem\u016f\u017eeme odvodit, kolikr\u00e1t se ru\u010di\u010dka oto\u010dila. Nicm\u00e9n\u011b pokud v\u00edme, \u017ee ru\u010di\u010dka hodin pro\u0161la 12 \u010dty\u0159ikr\u00e1t, m\u016f\u017eeme pomoc\u00ed jednoduch\u00e9ho vzorce zcela ur\u010dit po\u010det uplynul\u00fdch hodin \"Schematika, kde \"Schematika\u00a0\u2014 je n\u00e1\u0161 d\u011blitel (zde \"Schematika), \"Schematika\u00a0\u2014 je koeficient (kolikr\u00e1t d\u011blitel bez zbytku p\u0159ech\u00e1z\u00ed do p\u016fvodn\u00edho \u010d\u00edsla, zde \"Schematika), a \"Schematika\u00a0\u2014 je zbytek, kter\u00fd obvykle vrac\u00ed vol\u00e1n\u00ed oper\u00e1toru podle modulu (zde \"Schematika). Znalost v\u0161ech t\u011bchto hodnot n\u00e1m umo\u017e\u0148uje vy\u0159e\u0161it rovnici pro \"Schematika, ale pokud koeficient vynech\u00e1me, nikdy nebudeme moci obnovit p\u016fvodn\u00ed hodnotu.<\/p>\n

M\u016f\u017ee b\u00fdt demonstrov\u00e1no, jak to zlep\u0161uje bezpe\u010dnost na\u0161eho sch\u00e9matu, pou\u017eit\u00edm sch\u00e9matu na na\u0161em p\u0159edchoz\u00edm p\u0159\u00edkladu a pou\u017eit\u00edm \"Schematika. Na\u0161e nov\u00e1 polynomick\u00e1 funkce \"Schematika, a nov\u00e9 body \"Schematika. Nyn\u00ed mohou spr\u00e1vci kl\u00ed\u010de znovu pou\u017e\u00edt polynomickou interpolaci pro obnovu na\u0161\u00ed funkce, ale tentokr\u00e1t mus\u00ed b\u00fdt operace s\u010d\u00edt\u00e1n\u00ed a n\u00e1soben\u00ed doprov\u00e1zeny zjednodu\u0161en\u00edm podle modulu \"Schematika (nap\u0159. \"Schematika).<\/p>\n

Pou\u017eit\u00edm tohoto nov\u00e9ho p\u0159\u00edkladu p\u0159edpokl\u00e1dejme, \u017ee \u00fato\u010dn\u00edk zjistil dv\u011b z t\u011bchto nov\u00fdch bod\u016f, \"Schematika, a ve\u0159ejn\u00e9 informace \"Schematika. Tentokr\u00e1t \u00fato\u010dn\u00edk na z\u00e1klad\u011b v\u0161ech dostupn\u00fdch informac\u00ed odvozuje n\u00e1sleduj\u00edc\u00ed funkce, kde \"Schematika\u00a0\u2014 mno\u017eina v\u0161ech kladn\u00fdch cel\u00fdch \u010d\u00edsel, a \"Schematika p\u0159edstavuje koeficient modulu \"Schematika.<\/p>\n

\"Schematika<\/p>\n

\nNyn\u00ed n\u00e1\u0161 \u00fato\u010dn\u00edk op\u011bt nach\u00e1z\u00ed \"Schematika, vypo\u010d\u00edtav\u00e1 \"Schematika:<\/p>\n

\"Schematika<\/p>\n

\nPot\u00e9 se znovu pokou\u0161\u00ed odvodit \"Schematika, nahrazen\u00edm \"Schematika v \"Schematika:<\/p>\n

\"Schematika<\/p>\n

\nTentokr\u00e1t m\u00e1 v\u00e1\u017en\u00fd probl\u00e9m. Ve vzorci chyb\u00ed hodnoty \"Schematika, \"Schematika a \"Schematika. Jeliko\u017e existuje nekone\u010dn\u00e9 mno\u017estv\u00ed kombinac\u00ed t\u011bchto prom\u011bnn\u00fdch, nem\u016f\u017ee z\u00edskat \u017e\u00e1dn\u00e9 dal\u0161\u00ed informace.<\/p>\n

Zabezpe\u010den\u00ed<\/h1>\n

\nSch\u00e9ma d\u011blen\u00ed tajemstv\u00ed Shamirovy nab\u00edz\u00ed bezpe\u010dnost z hlediska teorie informace<\/i>. To znamen\u00e1, \u017ee matematika je odoln\u00e1 i v\u016f\u010di \u00fato\u010dn\u00edkovi s neomezen\u00fdm v\u00fdpo\u010detn\u00edm v\u00fdkonem. P\u0159esto sch\u00e9ma st\u00e1le obsahuje n\u011bkolik zn\u00e1m\u00fdch probl\u00e9m\u016f.<\/p>\n

Nap\u0159\u00edklad Shamirovo sch\u00e9ma nevytv\u00e1\u0159\u00ed ov\u011b\u0159iteln\u00e9 \u00fatr\u017eky<\/i>, co\u017e znamen\u00e1, \u017ee lid\u00e9 mohou svobodn\u011b p\u0159edkl\u00e1dat fale\u0161n\u00e9 \u00fatr\u017eky a komplikovat obnovu spr\u00e1vn\u00e9ho tajemstv\u00ed. Nep\u0159\u00e1telsk\u00fd spr\u00e1vce \u00fatr\u017ek\u016f s dostate\u010dn\u00fdmi informacemi m\u016f\u017ee dokonce vyprodukovat jin\u00fd \u00fatr\u017eek, kdy\u017e zm\u011bn\u00ed \"Schematika podle sv\u00e9ho uv\u00e1\u017een\u00ed. Tento probl\u00e9m se \u0159e\u0161\u00ed pomoc\u00ed ov\u011b\u0159iteln\u00fdch sch\u00e9mat d\u011blen\u00ed tajemstv\u00ed<\/i>, jako je Feldmanovo sch\u00e9ma.<\/p>\n

Dal\u0161\u00edm probl\u00e9mem je to, \u017ee d\u00e9lka jak\u00e9hokoli \u00fatr\u017eku je rovna d\u00e9lce odpov\u00eddaj\u00edc\u00edho tajemstv\u00ed, tak\u017ee d\u00e9lku tajemstv\u00ed lze snadno ur\u010dit. Tento probl\u00e9m se \u0159e\u0161\u00ed trivi\u00e1ln\u00edm napln\u011bn\u00edm<\/i> tajemstv\u00ed libovoln\u00fdmi \u010d\u00edsly do pevn\u00e9 d\u00e9lky.<\/p>\n

Nakonec je d\u016fle\u017eit\u00e9 poznamenat, \u017ee na\u0161e obavy o zabezpe\u010den\u00ed mohou p\u0159esahovat samo sch\u00e9ma. Pro skute\u010dn\u00e9 kryptografick\u00e9 aplikace \u010dasto existuje hrozba \u00fatok\u016f prost\u0159ednictv\u00edm bo\u010dn\u00edch kan\u00e1l\u016f, kdy se \u00fato\u010dn\u00edk pokou\u0161\u00ed z\u00edskat u\u017eite\u010dn\u00e9 informace z doby b\u011bhu aplikace, vyrovn\u00e1vac\u00ed pam\u011bti, selh\u00e1n\u00ed atd. Pokud to vyvol\u00e1v\u00e1 obavy, je vhodn\u00e9 p\u0159i v\u00fdvoji pe\u010dliv\u011b zv\u00e1\u017eit pou\u017eit\u00ed ochrann\u00fdch opat\u0159en\u00ed, jako jsou funkce a vyhled\u00e1v\u00e1n\u00ed s konstantn\u00ed dobou b\u011bhu, zamezit ukl\u00e1d\u00e1n\u00ed pam\u011bti na disk a promyslet \u0159adu dal\u0161\u00edch v\u011bc\u00ed, kter\u00e9 p\u0159esahuj\u00ed tento \u010dl\u00e1nek.<\/p>\n

Demo<\/h1>\n

\nNa tato str\u00e1nka<\/a><\/noindex> existuje interaktivn\u00ed demonstrace Shamirova sch\u00e9ma d\u011blen\u00ed tajemstv\u00ed. Demonstrace byla vytvo\u0159ena na z\u00e1klad\u011b knihovny ssss-js<\/a><\/noindex>, kter\u00e1 je sama o sob\u011b JavaScriptov\u00fdm portem popul\u00e1rn\u00edho programu ssss<\/a><\/noindex>Vezm\u011bte pros\u00edm na v\u011bdom\u00ed, \u017ee v\u00fdpo\u010det velk\u00fdch hodnot \"Schematika, \"Schematika a \"Schematika m\u016f\u017ee trvat n\u011bjakou dobu.
\n
Zdroj: habr.com<\/a><\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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