{"id":95846,"date":"2020-10-04T01:42:23","date_gmt":"2020-10-03T23:42:23","guid":{"rendered":"https:\/\/prohoster.info\/blog\/administrirovanie\/mozhno-li-generirovat-sluchajnye-chisla-esli-my-ne-doveryaem-drug-drugu-chast-2"},"modified":"2020-10-04T01:42:23","modified_gmt":"2020-10-03T23:42:23","slug":"mozhno-li-generirovat-sluchajnye-chisla-esli-my-ne-doveryaem-drug-drugu-chast-2","status":"publish","type":"post","link":"https:\/\/prohoster.info\/cs\/blog\/administrirovanie\/mozhno-li-generirovat-sluchajnye-chisla-esli-my-ne-doveryaem-drug-drugu-chast-2","title":{"rendered":"Je mo\u017en\u00e9 generovat n\u00e1hodn\u00e1 \u010d\u00edsla, pokud si navz\u00e1jem ned\u016fv\u011b\u0159ujeme? \u010c\u00e1st 2","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

\"Je<\/p>\n

Ahoj, Habr!<\/p>\n

V prvn\u00ed \u010d\u00e1sti<\/a><\/noindex> V \u010dl\u00e1nku jsme prodiskutovali, pro\u010d m\u016f\u017ee b\u00fdt nezbytn\u00e9 generovat n\u00e1hodn\u00e1 \u010d\u00edsla pro \u00fa\u010dastn\u00edky, kte\u0159\u00ed si vz\u00e1jemn\u011b ned\u016fv\u011b\u0159uj\u00ed, jak\u00e9 po\u017eadavky jsou kladeny na takov\u00e9 gener\u00e1tory n\u00e1hodn\u00fdch \u010d\u00edsel, a prozkoumali jsme dva p\u0159\u00edstupy k jejich realizaci.<\/p>\n

V t\u00e9to \u010d\u00e1sti \u010dl\u00e1nku podrobn\u011b prozkoum\u00e1me dal\u0161\u00ed p\u0159\u00edstup, kter\u00fd vyu\u017e\u00edv\u00e1 prahov\u00e9 podpisy.<\/p>\n

Trocha kryptografie<\/h3>\n

Abychom pochopili, jak funguj\u00ed prahov\u00e9 podpisy, mus\u00edme m\u00edt n\u011bjak\u00e9 z\u00e1kladn\u00ed znalosti o kryptografii. Pou\u017eijeme dv\u011b koncepce: scal\u00e1ry, nebo jednodu\u0161e \u010d\u00edsla, kter\u00e1 budeme ozna\u010dovat mal\u00fdmi p\u00edsmeny (x, y<\/em>) a body na eliptick\u00e9 k\u0159ivce, kter\u00e9 budeme ozna\u010dovat velk\u00fdmi p\u00edsmeny.<\/p>\n

Pro pochopen\u00ed z\u00e1klad\u016f prahov\u00fdch podpis\u016f nen\u00ed t\u0159eba rozum\u011bt tomu, jak funguj\u00ed eliptick\u00e9 k\u0159ivky, krom\u011b n\u011bkolika z\u00e1kladn\u00edch v\u011bc\u00ed:<\/p>\n

    \n
  1. \n

    Body na eliptick\u00e9 k\u0159ivce lze s\u010d\u00edtat a n\u00e1sobit scal\u00e1rem (n\u00e1soben\u00ed scal\u00e1rem budeme ozna\u010dovat jako xG<\/em>, i kdy\u017e notace Gx<\/em> se tak\u00e9 \u010dasto pou\u017e\u00edv\u00e1 v literatu\u0159e). V\u00fdsledek sou\u010dtu a n\u00e1soben\u00ed skal\u00e1rem \u2014 to je bod na elliptick\u00e9 k\u0159ivce.<\/p>\n<\/li>\n

  2. \n

    Zn\u00e1me-li pouze bod G<\/em> a jeho n\u00e1sobek s scal\u00e1rem xG<\/em> nelze vypo\u010d\u00edtat x<\/em>.<\/p>\n<\/li>\n<\/ol>\n

    Budeme tak\u00e9 pou\u017e\u00edvat koncept polynomu p(x)<\/em> stupn\u011b k<\/em>-1. Konkr\u00e9tn\u011b budeme vyu\u017e\u00edvat n\u00e1sleduj\u00edc\u00ed vlastnost polynom\u016f: pokud zn\u00e1me hodnotu p(x) <\/em>pro jak\u00e9koli k <\/em>r\u016fzn\u00e9 x <\/em>(a nem\u00e1me \u017e\u00e1dn\u00e9 jin\u00e9 informace o p(x)<\/em>), m\u016f\u017eeme vypo\u010d\u00edtat p(x) <\/em>pro jak\u00e9koli jin\u00e9 x<\/em>.<\/p>\n

    Je zaj\u00edmav\u00e9, \u017ee pro jak\u00fdkoli polynom p(x)<\/em> a n\u011bjak\u00fd bod na k\u0159ivce G<\/em>, zn\u00e1me-li hodnotu p(x)G<\/em> pro jak\u00e9koli k<\/em> r\u016fzn\u00fdch hodnot x<\/em>, lze tak\u00e9 vypo\u010d\u00edtat p(x)G<\/em> pro jak\u00fdkoli x<\/em>.<\/p>\n

    Tyto informace sta\u010d\u00ed k tomu, abychom se pono\u0159ili do detail\u016f o tom, jak funguj\u00ed prahov\u00e9 podpisy a jak je pou\u017e\u00edt k generov\u00e1n\u00ed n\u00e1hodn\u00fdch \u010d\u00edsel.<\/p>\n

    Gener\u00e1tor n\u00e1hodn\u00fdch \u010d\u00edsel na prahov\u00fdch podpisech<\/h3>\n

    P\u0159edstavme si, \u017ee n<\/em> \u00fa\u010dastn\u00edk\u016f chce vygenerovat n\u00e1hodn\u00e9 \u010d\u00edslo, a chceme, aby \u00fa\u010dast jak\u00e9hokoli z nich byl dostate\u010dn\u00fd k vygenerov\u00e1n\u00ed \u010d\u00edsla, ale aby \u00fato\u010dn\u00edci, kte\u0159\u00ed kontroluj\u00ed k<\/em> -1 nebo m\u00e9n\u011b \u00fa\u010dastn\u00edk\u016f, nemohli p\u0159edpov\u011bd\u011bt nebo ovlivnit vygenerovan\u00e9 \u010d\u00edslo. k<\/em>P\u0159edstavme si, \u017ee existuje takov\u00fd polynom<\/p>\n

    \"Je<\/p>\n

    -1, \u017ee prvn\u00ed \u00fa\u010dastn\u00edk v\u00ed p(x)<\/em> stupn\u011b k<\/em>p(1) , druh\u00fd v\u00ed<\/em>p(2), a tak d\u00e1le ( <\/em>-\u00fd v\u00edn<\/em>p(n) ). Tak\u00e9 p\u0159edpokl\u00e1dejme, \u017ee pro n\u011bjak\u00fd p\u0159edem ur\u010den\u00fd bod<\/em>v\u0161ichni v\u011bd\u00ed G<\/em> pro v\u0161echny hodnoty p(x)G <\/em>. Budeme to naz\u00fdvat x<\/em>. Budeme to naz\u00fdvat p(i)<\/em> \"priv\u00e1tn\u00ed komponenta\u201c i<\/em>-\u00ed \u00fa\u010dastn\u00edka (proto\u017ee pouze i<\/em>-\u00fd \u00fa\u010dastn\u00edk to v\u00ed), a p(i)G<\/em> \"ve\u0159ejn\u00e1 komponenta\u201c i<\/em>-\u00ed \u00fa\u010dastn\u00edka (proto\u017ee v\u0161ichni \u00fa\u010dastn\u00edci to znaj\u00ed). Jak si pamatujete, znalost p(i)G <\/em>nen\u00ed dostate\u010dn\u00e1 pro obnoven\u00ed p(i).<\/em><\/p>\n

    Vytvo\u0159en\u00ed takov\u00e9ho polynomu tak, aby pouze i-<\/em>-\u00fd \u00fa\u010dastn\u00edk a nikdo jin\u00fd znal svou priv\u00e1tn\u00ed komponentu \u2013 to je nejt\u011b\u017e\u0161\u00ed a nejzaj\u00edmav\u011bj\u0161\u00ed \u010d\u00e1st protokolu, a to si podrobn\u011b rozebereme n\u00ed\u017ee. Zat\u00edm p\u0159edpokl\u00e1dejme, \u017ee takov\u00fd polynom m\u00e1me a v\u0161ichni \u00fa\u010dastn\u00edci znaj\u00ed sv\u00e9 priv\u00e1tn\u00ed komponenty.<\/p>\n

    Jak m\u016f\u017eeme pou\u017e\u00edt takov\u00fd polynom k vygenerov\u00e1n\u00ed n\u00e1hodn\u00e9ho \u010d\u00edsla? Nejprve pot\u0159ebujeme n\u011bjak\u00fd \u0159et\u011bzec, kter\u00fd dosud nebyl pou\u017eit jako vstup pro gener\u00e1tor. V p\u0159\u00edpad\u011b blockchainu je hash posledn\u00edho bloku h<\/em> \u2014 dobr\u00fd kandid\u00e1t pro takov\u00fd \u0159et\u011bzec. Nech\u0165 \u00fa\u010dastn\u00edci cht\u011bj\u00ed vytvo\u0159it n\u00e1hodn\u00e9 \u010d\u00edslo, pou\u017e\u00edvaj\u00edc\u00ed h <\/em>jako seed. Nejprve \u00fa\u010dastn\u00edci p\u0159evedou h<\/em> na bod na k\u0159ivce pomoc\u00ed jak\u00e9koli p\u0159edem definovan\u00e9 funkce:<\/p>\n

    H = scalarToPoint(h)<\/em><\/p>\n

    Pot\u00e9 ka\u017ed\u00fd \u00fa\u010dastn\u00edk i<\/em> vypo\u010d\u00edt\u00e1 a zve\u0159ejn\u00ed Hi = p(i)H, <\/em>co\u017e mohou ud\u011blat, proto\u017ee znaj\u00ed p(i) a H. <\/em>Odhalen\u00ed H<\/em>i neumo\u017e\u0148uje ostatn\u00edm \u00fa\u010dastn\u00edk\u016fm obnovit priv\u00e1tn\u00ed komponentu i<\/em>-\u00ed \u00fa\u010dastn\u00edka, a proto jeden soubor priv\u00e1tn\u00edch komponent lze pou\u017e\u00edt od bloku k bloku. Tak\u017ee drah\u00fd algoritmus pro tvorbu polynomu, kter\u00fd je pops\u00e1n n\u00ed\u017ee, je t\u0159eba prov\u00e9st pouze jednou.<\/p>\n

    Kdy\u017e k<\/em> \u00fa\u010dastn\u00edci odhal\u00ed Hi = p(i)H, <\/em>v\u0161ichni mohou vypo\u010d\u00edtat H<\/em>x = p(x)H<\/em> pro v\u0161echny x<\/em> d\u00edky vlastnosti polynom\u016f, kter\u00e9 jsme prob\u00edrali v p\u0159edchoz\u00ed \u010d\u00e1sti. V tuto chv\u00edli v\u0161ichni \u00fa\u010dastn\u00edci vypo\u010d\u00edtaj\u00ed H0 = p(0)H, <\/em>a to je v\u00fdsledn\u00e9 n\u00e1hodn\u00e9 \u010d\u00edslo. V\u0161imn\u011bte si, \u017ee nikdo nev\u00ed p(0), <\/em>a tud\u00ed\u017e jedin\u00fd zp\u016fsob, jak spo\u010d\u00edtat p(0)H \u2013 <\/em>je interpolace p(x)H, <\/em>co\u017e je mo\u017en\u00e9 pouze tehdy, kdy\u017e k <\/em>hodnoty p(i)H <\/em>jsou zn\u00e1my. Odhalen\u00ed jak\u00e9hokoliv men\u0161\u00edho po\u010dtu p(i)H <\/em>ned\u00e1v\u00e1 \u017e\u00e1dn\u00e9 informace o p(0)H.<\/em><\/p>\n

    \"Je<\/p>\n

    V\u00fd\u0161e uveden\u00fd gener\u00e1tor m\u00e1 v\u0161echny vlastnosti, kter\u00e9 chceme: \u00fato\u010dn\u00edci, kte\u0159\u00ed kontroluj\u00ed pouze k-<\/em>1 \u00fa\u010dastn\u00edk\u016f, nebo m\u00e9n\u011b, nemaj\u00ed \u017e\u00e1dn\u00e9 informace a vliv na v\u00fdstup, zat\u00edmco jak\u00fdkoli k<\/em> \u00fa\u010dastn\u00edk\u016f m\u016f\u017ee vypo\u010d\u00edtat v\u00fdsledn\u00e9 \u010d\u00edslo, a jak\u00fdkoli podmno\u017eina z k<\/em> \u00fa\u010dastn\u00edk\u016f v\u017edy dos\u00e1hne stejn\u00e9ho v\u00fdsledku pro stejn\u00fd seed.<\/p>\n

    Existuje jeden probl\u00e9m, kter\u00e9mu jsme se shodou okolnost\u00ed vyhnuli d\u0159\u00edve. Aby interpolace fungovala, je d\u016fle\u017eit\u00e9, aby hodnota H<\/em>i, kterou zve\u0159ejnil ka\u017ed\u00fd \u00fa\u010dastn\u00edk, i<\/em> skute\u010dn\u011b byla rovna p(i)H.<\/em> Jeliko\u017e nikdo krom\u011b i<\/em>-\u00e9ho \u00fa\u010dastn\u00edka nev\u00ed p(i), <\/em>nikdo krom\u011b i-<\/em>\u00e9ho \u00fa\u010dastn\u00edka nem\u016f\u017ee ov\u011b\u0159it, \u017ee Ahoj, <\/em>skute\u010dn\u011b bylo spr\u00e1vn\u011b spo\u010d\u00edt\u00e1no, a bez n\u011bjak\u00e9ho kryptografick\u00e9ho d\u016fkazu spr\u00e1vnosti, H<\/em>i \u00fato\u010dn\u00edk m\u016f\u017ee zve\u0159ejnit jakoukoli hodnotu jako Ahoj, <\/em>a libovoln\u011b ovlivnit v\u00fdstup gener\u00e1toru n\u00e1hodn\u00fdch \u010d\u00edsel.:<\/em><\/p>\n

    \"JeR\u016fzn\u00e9 hodnoty H_1, zaslan\u00e9 prvn\u00edm \u00fa\u010dastn\u00edkem, vedou k r\u016fzn\u00fdm v\u00fdsledn\u00fdm H_0.<\/p>\n

    Existuj\u00ed alespo\u0148 dva zp\u016fsoby, jak prok\u00e1zat spr\u00e1vnost H<\/em>i, kter\u00e9 probereme pot\u00e9, co se budeme v\u011bnovat generaci polynomu.<\/p>\n

    Generace polynomu<\/h3>\n

    V p\u0159edchoz\u00edm odd\u00edle jsme p\u0159edpokl\u00e1dali, \u017ee m\u00e1me takov\u00fd polynom p(x)<\/em> stupn\u011b k<\/em>-1, kter\u00fd \u00fa\u010dastn\u00edk i<\/em> zn\u00e1, p(i)<\/em>a nikdo jin\u00fd nem\u00e1 \u017e\u00e1dn\u00e9 informace o t\u00e9to hodnot\u011b. V n\u00e1sleduj\u00edc\u00edm odd\u00edle bude tak\u00e9 nutn\u00e9, aby pro n\u011bjak\u00fd p\u0159edem stanoven\u00fd bod G<\/em> v\u0161ichni v\u011bd\u011bli p(x)G <\/em>pro v\u0161echny x<\/em>.<\/p>\n

    V tomto odd\u00edle budeme p\u0159edpokl\u00e1dat, \u017ee ka\u017ed\u00fd \u00fa\u010dastn\u00edk m\u00e1 lok\u00e1ln\u011b n\u011bjak\u00fd soukrom\u00fd kl\u00ed\u010d xi, <\/em>tak, \u017ee v\u0161em je zn\u00e1m odpov\u00eddaj\u00edc\u00ed ve\u0159ejn\u00fd kl\u00ed\u010d X<\/em>i.<\/p>\n

    Jeden mo\u017en\u00fd protokol generace polynomu je n\u00e1sleduj\u00edc\u00ed:<\/p>\n

    \"Je<\/p>\n

      \n
    1. \n

      Ka\u017ed\u00fd \u00fa\u010dastn\u00edk i<\/em> lok\u00e1ln\u011b vytvo\u0159\u00ed libovoln\u00fd polynom pi(x) stupn\u011b k-1. <\/em>Pot\u00e9 ka\u017ed\u00fd z nich za\u0161le ka\u017ed\u00e9mu \u00fa\u010dastn\u00edkovi j <\/em>hodnotu p<\/em>i(j), za\u0161ifrovanou ve\u0159ejn\u00fdm kl\u00ed\u010dem Xj. <\/em>T\u00edmto zp\u016fsobem pouze i-<\/em>\u00fd <\/em>a j-\u00fd \u00fa\u010dastn\u00edk v\u00ed<\/em>\u00fd <\/em>i(j). \u00da\u010dastn\u00edk p<\/em>tak\u00e9 ve\u0159ejn\u011b ozn\u00e1m\u00ed i<\/em> pi(j)G v\u010detn\u011b. <\/em>pro v\u0161echny j <\/em>od 1 <\/em>do k <\/em>V\u0161ichni \u00fa\u010dastn\u00edci pou\u017e\u00edvaj\u00ed n\u011bjak\u00fd konsensus, aby vybrali<\/p>\n<\/li>\n

    2. \n

      \u00fa\u010dastn\u00edky, jejich\u017e polynomy budou pou\u017eity. Jeliko\u017e n\u011bkte\u0159\u00ed \u00fa\u010dastn\u00edci mohou b\u00fdt offline, nem\u016f\u017eeme \u010dekat, a\u017e v\u0161ichni k <\/em>\u00fa\u010dastn\u00edci zve\u0159ejn\u00ed polynomy. V\u00fdsledek tohoto kroku je mno\u017eina n <\/em>skl\u00e1daj\u00edc\u00ed se alespo\u0148 <\/em>Z<\/em><\/strong> <\/em>polynom\u016f vytvo\u0159en\u00fdch v kroku (1) k <\/em>\u00da\u010dastn\u00edci se ujist\u00ed, \u017ee jimi zn\u00e1m\u00e9 hodnoty.<\/em><\/p>\n<\/li>\n

    3. \n

      i(j) odpov\u00eddaj\u00ed ve\u0159ejn\u011b ozn\u00e1men\u00fdm p<\/em>pi(j)G. Po tomto kroku by m\u011bly z\u016fstat pouze polynomy, pro kter\u00e9 soukrom\u011b p\u0159edan\u00e9 <\/em>vypo\u010d\u00edt\u00e1v\u00e1 svou soukromou komponentu <\/em>Z <\/em><\/strong>p(j) p<\/em>pi(j)G. Po tomto kroku by m\u011bly z\u016fstat pouze polynomy, pro kter\u00e9 soukrom\u011b p\u0159edan\u00e9<\/em><\/p>\n<\/li>\n

    4. \n

      Ka\u017ed\u00fd \u00fa\u010dastn\u00edk j <\/em>jako sou\u010det i(j) pro v\u0161echny <\/em>. Ka\u017ed\u00fd \u00fa\u010dastn\u00edk tak\u00e9 vypo\u010d\u00edt\u00e1 v\u0161echny hodnoty p<\/em>pi(x)G pro v\u0161echny i, i<\/em> v Z<\/em><\/strong>p(x) \u2013 p(x)G <\/em>. Ka\u017ed\u00fd \u00fa\u010dastn\u00edk tak\u00e9 vypo\u010d\u00edt\u00e1 v\u0161echny hodnoty to je skute\u010dn\u011b polynom stupn\u011b <\/em>v <\/em>Z<\/em><\/strong>.<\/em><\/p>\n<\/li>\n<\/ol>\n

      \"Je<\/p>\n

      V\u0161imn\u011bte si, \u017ee k-1, <\/em>proto\u017ee je to sou\u010det jednotliv\u00fdch i(x), p\u0159i\u010dem\u017e ka\u017ed\u00fd z nich je polynom stupn\u011b. <\/em>proto\u017ee to je sou\u010det jednotliv\u00fdch p<\/em>i(x), p\u0159i\u010dem\u017e ka\u017ed\u00fd z nich je polynom stupn\u011b k<\/em>-1. Pot\u00e9 si uv\u011bdomte, \u017ee zat\u00edmco ka\u017ed\u00fd \u00fa\u010dastn\u00edk j<\/em> zn\u00e1, p(j), <\/em>nemaj\u00ed \u017e\u00e1dn\u00e9 informace o p(x)<\/em> pro x \u2260 j<\/em>. Skute\u010dn\u011b, aby tento hodnotu spo\u010d\u00edtali, mus\u00ed v\u011bd\u011bt v\u0161echny pi(x), <\/em>a dokud \u00fa\u010dastn\u00edk j <\/em>nezn\u00e1 alespo\u0148 jeden z vybran\u00fdch polynom\u016f, nemaj\u00ed dostatek informac\u00ed o p(x).<\/em><\/p>\n

      To je cel\u00fd proces generov\u00e1n\u00ed polynomu, kter\u00fd byl pot\u0159ebn\u00fd v p\u0159edchoz\u00ed \u010d\u00e1sti. Krok 1, 2 a 4 v\u00fd\u0161e maj\u00ed dostate\u010dn\u011b zjevnou realizaci. Krok 3 v\u0161ak nen\u00ed tak trivi\u00e1ln\u00ed.<\/p>\n

      Konkr\u00e9tn\u011b, pot\u0159ebujeme b\u00fdt schopni dok\u00e1zat, \u017ee za\u0161ifrovan\u00e9 p<\/em>i(j) skute\u010dn\u011b odpov\u00eddaj\u00ed publikovan\u00fdm Po tomto kroku by m\u011bly z\u016fstat pouze polynomy, pro kter\u00e9 soukrom\u011b p\u0159edan\u00e9 <\/em>Pokud to nedok\u00e1\u017eeme, \u00fato\u010dn\u00edk i <\/em>m\u016f\u017ee poslat odpad m\u00edsto p<\/em>i(j) \u00fa\u010dastn\u00edkovi j<\/em>, a \u00fa\u010dastn\u00edk j <\/em>nebude schopen z\u00edskat skute\u010dnou hodnotu pi(j), <\/em>a nebude schopen vypo\u010d\u00edtat svou soukromou komponentu.<\/em><\/p>\n

      Existuje kryptografick\u00fd protokol, kter\u00fd umo\u017e\u0148uje vytvo\u0159it dal\u0161\u00ed zpr\u00e1vu proof<\/em>i(j), tak\u017ee ka\u017ed\u00fd \u00fa\u010dastn\u00edk, maj\u00edc\u00ed ur\u010ditou hodnotu e, <\/em>a tak\u00e9 proofi(j) <\/em>a p<\/em>i(j)G, m\u016f\u017ee lok\u00e1ln\u011b ov\u011b\u0159it, \u017ee e<\/em> \u2013 je skute\u010dn\u011b pi(j), <\/em>za\u0161ifrovan\u00e9 kl\u00ed\u010dem \u00fa\u010dastn\u00edka j. <\/em>Bohu\u017eel, velikost takov\u00e9ho d\u016fkazu je neuv\u011b\u0159iteln\u011b velk\u00e1, a vzhledem k tomu, \u017ee je t\u0159eba publikovat O(nk) <\/em>takov\u00fdch d\u016fkaz\u016f, nelze je pro tento \u00fa\u010del pou\u017e\u00edt.<\/p>\n

      M\u00edsto toho, abychom dok\u00e1zali, \u017ee pi(j) <\/em>odpov\u00edd\u00e1 p<\/em>i(j)G m\u016f\u017eeme v protokolu generov\u00e1n\u00ed polynomu vy\u010dlenit velmi dlouhou dobu, b\u011bhem n\u00ed\u017e v\u0161ichni \u00fa\u010dastn\u00edci kontroluj\u00ed obdr\u017een\u00e9 za\u0161ifrovan\u00e9 pi(j), <\/em>a pokud de\u0161ifrovan\u00e1 zpr\u00e1va neodpov\u00edd\u00e1 ve\u0159ejn\u00e9mu p<\/em>i(j)G, publikuj\u00ed kryptografick\u00fd d\u016fkaz, \u017ee obdr\u017een\u00e1 za\u0161ifrovan\u00e1 zpr\u00e1va je chybn\u00e1. Dok\u00e1zat, \u017ee zpr\u00e1va ne <\/em>odpov\u00edd\u00e1 pi(G)<\/em> je mnohem jednodu\u0161\u0161\u00ed, ne\u017e dok\u00e1zat, \u017ee odpov\u00edd\u00e1. Je t\u0159eba poznamenat, \u017ee to vy\u017eaduje, aby se ka\u017ed\u00fd \u00fa\u010dastn\u00edk objevil v s\u00edti alespo\u0148 jednou b\u011bhem \u010dasu p\u0159id\u011blen\u00e9ho k vytv\u00e1\u0159en\u00ed takov\u00fdch d\u016fkaz\u016f, a spol\u00e9hat se na p\u0159edpoklad, \u017ee pokud takov\u00fd d\u016fkaz publikovali, doraz\u00ed ke v\u0161em ostatn\u00edm \u00fa\u010dastn\u00edk\u016fm do stejn\u00e9ho p\u0159id\u011blen\u00e9ho \u010dasu.<\/p>\n

      \"Je<\/p>\n

      Pokud se \u00fa\u010dastn\u00edk v tomto obdob\u00ed v s\u00edti neobjevil a opravdu m\u011bl alespo\u0148 jednu nespr\u00e1vnou komponentu, pak tento konkr\u00e9tn\u00ed \u00fa\u010dastn\u00edk nebude moci \u00fa\u010dastnit dal\u0161\u00ed generace \u010d\u00edsel. Protokol v\u0161ak bude st\u00e1le fungovat, pokud existuje alespo\u0148 k<\/em> \u00fa\u010dastn\u00edk\u016f, kte\u0159\u00ed bu\u010f pr\u00e1v\u011b obdr\u017eeli spr\u00e1vn\u00e9 komponenty, nebo stihli poskytnout d\u016fkaz nespr\u00e1vnosti v p\u0159id\u011blen\u00e9m \u010dase.<\/p>\n

      D\u016fkazy spr\u00e1vnosti H_i<\/h3>\n

      Posledn\u00ed \u010d\u00e1st, kterou je t\u0159eba prodiskutovat, je, jak prok\u00e1zat spr\u00e1vnost publikovan\u00fdch H<\/em>i, konkr\u00e9tn\u011b \u017ee Hi = p(i)H, <\/em>bez odhalen\u00ed p(i).<\/em><\/p>\n

      Vzpome\u0148me, \u017ee hodnoty H, G, p(i)G <\/em>jsou ve\u0159ejn\u00e9 a zn\u00e1m\u00e9 v\u0161em. <\/em>Operace z\u00edsk\u00e1v\u00e1n\u00ed p(i) <\/em>znaje p(i)G <\/em>a G <\/em>se naz\u00fdv\u00e1 diskr\u00e9tn\u00ed logarithmus nebo dlog, <\/em>a chceme prok\u00e1zat, \u017ee:<\/p>\n

      dlog(p(i)G, G) = dlog(H<\/em>i, H<\/em>)<\/p>\n

      S takovou konstrukc\u00ed ka\u017ed\u00fd \u00fa\u010dastn\u00edk spole\u010dn\u011b s Ahoj, <\/em>odes\u00edl\u00e1 d\u016fkaz spr\u00e1vnosti podle konstrukce.<\/p>\n

      Kdy\u017e je n\u00e1hodn\u00e9 \u010d\u00edslo generov\u00e1no, \u010dasto je t\u0159eba pou\u017e\u00edt ho \u00fa\u010dastn\u00edk\u016fm, kte\u0159\u00ed se li\u0161\u00ed od t\u011bch, kdo ho vygenerovali. T\u011bmto \u00fa\u010dastn\u00edk\u016fm je t\u0159eba spolu s \u010d\u00edslem pos\u00edlat v\u0161echny Ahoj,<\/em> a souvisej\u00edc\u00ed d\u016fkazy.<\/p>\n

      Zv\u00eddav\u00fd \u010dten\u00e1\u0159 se m\u016f\u017ee zeptat: proto\u017ee fin\u00e1ln\u00ed n\u00e1hodn\u00e9 \u010d\u00edslo je H<\/em>0, a p(0)G \u2013 <\/em>je to ve\u0159ejn\u00e1 informace, pro\u010d je pot\u0159eba d\u016fkaz pro ka\u017ed\u00e9 jednotliv\u00e9 H<\/em>i, pro\u010d m\u00edsto toho neodeslat d\u016fkaz, \u017ee<\/p>\n

      dlog(p(0)G, G) = dlog(H<\/em>0, H<\/em>)<\/p>\n

      Probl\u00e9mem je, \u017ee s pomoc\u00ed Schnorr Protocol nelze vytvo\u0159it takov\u00fd d\u016fkaz, proto\u017ee nikdo nezn\u00e1 hodnotu p(0)<\/em>, pot\u0159ebnou k vytvo\u0159en\u00ed d\u016fkazu, a nav\u00edc cel\u00fd gener\u00e1tor n\u00e1hodn\u00fdch \u010d\u00edsel je zalo\u017een na tom, \u017ee nikdo nezn\u00e1 tuto hodnotu. Proto je nutn\u00e9 m\u00edt v\u0161echny hodnoty Ahoj, <\/em>a jejich individu\u00e1ln\u00ed d\u016fkazy, aby bylo mo\u017en\u00e9 prok\u00e1zat spr\u00e1vnost. H<\/em>0.<\/p>\n

      Pokud by v\u0161ak na bodech eliptick\u00fdch k\u0159ivin byla n\u011bjak\u00e1 operace, kter\u00e1 je s\u00e9manticky podobn\u00e1 n\u00e1soben\u00ed, prok\u00e1z\u00e1n\u00ed spr\u00e1vnosti H0 <\/em>by bylo trivi\u00e1ln\u00ed, prost\u011b bychom se ujistili, \u017ee<\/p>\n

      H<\/em>0 \u00d7 G<\/em> = p(0)G \u00d7 H<\/em><\/p>\n

      Pokud vybran\u00e1 k\u0159ivka podporuje elliptic curve pairings<\/u><\/a><\/noindex>, takov\u00fd d\u016fkaz funguje. V takov\u00e9m p\u0159\u00edpad\u011b H<\/em>0 \u2013 to nen\u00ed jen v\u00fdstup gener\u00e1toru n\u00e1hodn\u00fdch \u010d\u00edsel, kter\u00fd m\u016f\u017ee zkontrolovat ka\u017ed\u00fd \u00fa\u010dastn\u00edk, jen\u017e zn\u00e1 G, H<\/em> a p(0)G. H<\/em>0 \u2013 to je tak\u00e9 podpis na zpr\u00e1v\u011b, kter\u00e1 byla pou\u017eita jako seed, potvrzuj\u00edc\u00ed, \u017ee k<\/em> a n <\/em>\u00fa\u010dastn\u00edci podepsali tuto zpr\u00e1vu. T\u00edm p\u00e1dem, pokud seed \u2013 <\/em>je to hash bloku v protokolu blockchainu, pak H0<\/em> \u2013 to je z\u00e1rove\u0148 multi-podpis na bloku a velmi dobr\u00e9 n\u00e1hodn\u00e9 \u010d\u00edslo.<\/p>\n

      Na z\u00e1v\u011br<\/h4>\n

      Tento \u010dl\u00e1nek je sou\u010d\u00e1st\u00ed s\u00e9rie technick\u00fdch \u010dl\u00e1nk\u016f na blogu. NEAR<\/a><\/noindex>. NEAR je blokchain protokol a platforma pro v\u00fdvoj decentralizovan\u00fdch aplikac\u00ed se zam\u011b\u0159en\u00edm na snadnost v\u00fdvoje a u\u017eivatelskou p\u0159\u00edv\u011btivost pro koncov\u00e9 u\u017eivatele.<\/p>\n

      K\u00f3d protokolu je otev\u0159en\u00fd, na\u0161e implementace je naps\u00e1na v jazyce Rust, lze ji nal\u00e9zt. tady<\/a><\/noindex>.<\/p>\n

      Pod\u00edvat se, jak vypad\u00e1 v\u00fdvoj na NEAR a vyzkou\u0161et si to v online IDE, lze. zde<\/a><\/noindex>.<\/p>\n

      Sledovat v\u0161echny novinky v ru\u0161tin\u011b lze v skupin\u011b na Telegramu<\/a><\/noindex> a v skupin\u011b na VKontakte<\/a><\/noindex>, a anglicky na ofici\u00e1ln\u00edm Twitteru<\/a><\/noindex>.<\/p>\n

      Na shledanou!<\/p>\n

      Zdroj: habr.com<\/a> <\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

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      bez zve\u0159ejn\u011bn\u00ed p(i)<\/em>. Konstrukce pro takov\u00e9 d\u016fkazy existuj\u00ed, nap\u0159\u00edklad Schnorr Protocol<\/u><\/a><\/noindex>.<\/p>\n