[{"@context":"https:\/\/schema.org\/","@type":"Article","@id":"https:\/\/www.muronka.cz\/jak-silne-mate-heslo\/#Article","mainEntityOfPage":"https:\/\/www.muronka.cz\/jak-silne-mate-heslo\/","headline":"Jak siln\u00e9 m\u00e1te heslo?","name":"Jak siln\u00e9 m\u00e1te heslo?","description":"\u00dato\u010dn\u00edk se v\u017edy sna\u017e\u00ed takov\u00e9 heslo prolomit a pou\u017e\u00edv\u00e1 na to dva zp\u016fsoby. Bu\u010f se jedn\u00e1 o tzv. \u201ebrutte force attack\u201c, tedy \u00fatok za pomoci hrub\u00e9 s\u00edly, nebo vyu\u017e\u00edv\u00e1 datab\u00e1zi slov \u2013 slovn\u00edk. N\u011bkdy je mo\u017en\u00e9 heslo uhodnout, pokud zlo\u010dinec zn\u00e1 svou ob\u011b\u0165 a m\u016f\u017ee zkou\u0161et r\u016fzn\u00e9 kombinace jmen nap\u0159. v rodinn\u00e9m kruhu apod., nebo […]","datePublished":"2025-02-25","dateModified":"2023-04-29","author":{"@type":"Person","@id":"https:\/\/www.muronka.cz\/author\/#Person","name":"muronka.cz\n","url":"https:\/\/www.muronka.cz\/author\/","identifier":1,"image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/69bd9b72104e3ccfb6214cbd23cb226f96c6ecf33a2a351fa1d4682235118df3?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/69bd9b72104e3ccfb6214cbd23cb226f96c6ecf33a2a351fa1d4682235118df3?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"muronka.cz","logo":{"@type":"ImageObject","@id":"\/logo.png","url":"\/logo.png","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/www.muronka.cz\/wp-content\/uploads\/img_a311036_w1977_t1521985418.jpg","url":"https:\/\/www.muronka.cz\/wp-content\/uploads\/img_a311036_w1977_t1521985418.jpg","height":0,"width":0},"url":"https:\/\/www.muronka.cz\/jak-silne-mate-heslo\/","about":["Internet"],"wordCount":435,"articleBody":"\u00dato\u010dn\u00edk se v\u017edy sna\u017e\u00ed takov\u00e9 heslo prolomit a pou\u017e\u00edv\u00e1 na to dva zp\u016fsoby. Bu\u010f se jedn\u00e1 o tzv. \u201ebrutte force attack\u201c, tedy \u00fatok za pomoci hrub\u00e9 s\u00edly, nebo vyu\u017e\u00edv\u00e1 datab\u00e1zi slov \u2013 slovn\u00edk. N\u011bkdy je mo\u017en\u00e9 heslo uhodnout, pokud zlo\u010dinec zn\u00e1 svou ob\u011b\u0165 a m\u016f\u017ee zkou\u0161et r\u016fzn\u00e9 kombinace jmen nap\u0159. v rodinn\u00e9m kruhu apod., nebo t\u0159eba data narozen\u00ed. N\u011bkdy v\u0161ak m\u016f\u017ee b\u00fdt heslo velmi siln\u00e9 a p\u0159esto jej \u00fato\u010dn\u00edk z\u00edsk\u00e1 podvodn\u00fdm zp\u016fsobem, nap\u0159. phishingem, odchyt\u00e1v\u00e1n\u00edm hesla atd. K tomu slou\u017e\u00ed r\u016fzn\u00e9 typy \u0161kodliv\u00e9ho software. Pokud si zvol\u00edte dostate\u010dn\u011b siln\u00e9 heslo, je pro \u00fato\u010dn\u00edka prakticky v re\u00e1ln\u00e9m sv\u011bt\u011b a t\u00edm p\u00e1dem v \u010dasov\u00e9m omezen\u00ed lidsk\u00e9ho \u017eivota prakticky neprolomiteln\u00e9   Jak na to? Prvn\u00ed z v\u011bc\u00ed, na kterou se zam\u011b\u0159te, je d\u00e9lka hesla v po\u010dtu znak\u016f. Uv\u011bdomte si, \u017eehackersk\u00fd algoritmus pro prolamov\u00e1n\u00ed hesla hrubou silou (pokud tedy pomineme jednoduch\u00e1 slovn\u00edkov\u00e1 a uhodnuteln\u00e1 hesla typu Tonda 1, m\u00e1ma, 12345) mus\u00ed ka\u017ed\u00fd dal\u0161\u00ed znak lu\u0161tit n\u00e1r\u016fstem kombinac\u00ed exponenci\u00e1ln\u00ed \u0159adou. Nap\u0159. u 4 znak\u016f mal\u00fdch p\u00edsmen je to jen necel\u00fdch p\u016fl milionu kombinac\u00ed, zat\u00edmco u p\u011bti znak\u016f u\u017e t\u00e9m\u011b\u0159 12 milion\u016f a u osmi znak\u016f u\u017e je to 209 miliard kombinac\u00ed. Dvan\u00e1ctim\u00edstn\u00e9 heslo vy\u017eaduje t\u00e9m\u011b\u0159 kvadriliony kombinac\u00ed. Z matematick\u00e9ho hlediska se jedn\u00e1 o v\u00fdpo\u010det tzv. variac\u00ed s opakov\u00e1n\u00edm.  Velkou v\u00e1hu m\u00e1 vz\u00e1jemn\u00e1 kombinace \u010d\u00edsel a p\u00edsmen. Budeme-li uva\u017eovat nap\u0159. heslo slo\u017een\u00e9 ze 6 znak\u016f, mno\u017estv\u00ed kombinac\u00ed bude velmi odli\u0161n\u00e9 u samotn\u00e9ho hesla z \u010d\u00edslic, u hesla z p\u00edsmen, a u kombinace p\u00edsmen a \u010d\u00edslic. Heslo z \u010d\u00edslic (\u0161estim\u00edstn\u00e9) d\u00e1v\u00e1 pouh\u00fd milion kombinac\u00ed, heslo z p\u00edsmen abecedy (mal\u00e1 p\u00edsmena) celkem 309 milion\u016f kombinac\u00ed, stejn\u011b dlouh\u00e9 heslo slo\u017een\u00e9 z p\u00edsmen a \u010d\u00edslic ji\u017e p\u0159es dv\u011b miliardy kombinac\u00ed.  Nejsiln\u011bj\u0161\u00ed jsou hesla slo\u017een\u00e1 z mal\u00fdch a velk\u00fdch p\u00edsmen a \u010d\u00edslic, ta vy\u017eaduj\u00ed celkem v\u00edce ne\u017e 55 miliard kombinac\u00ed a pokud je je\u0161t\u011b prokl\u00e1d\u00e1te ob\u010dasn\u00fdm speci\u00e1ln\u00edm znakem, nap\u0159. paragrafem, m\u0159\u00ed\u017eemi, podtr\u017e\u00edtkem, zavin\u00e1\u010dem, vzroste po\u010det kombinac\u00ed na t\u00e9m\u011b\u0159 jeden bilion. Pokud takov\u00e9 heslo m\u00e1 v\u00edce ne\u017e \u0161est znak\u016f, kombinace jdou do kvintilion\u016f a pro jeho prolomen\u00ed by bylo zapot\u0159eb\u00ed zhruba n\u011bkolik milion\u016f let.                                                                                                                                                                                                                                                                                                                                                                                         4.5\/5 - (6 votes)        "},{"@context":"https:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Jak siln\u00e9 m\u00e1te heslo?","item":"https:\/\/www.muronka.cz\/jak-silne-mate-heslo\/#breadcrumbitem"}]}]