2a:null 2b:["$","title",null,{"children":"RSA | Attila Bagossy | Observable"}] 30:[] 2e:{"id":"cc57630a906297bb","slug":"rsa","title":"RSA","thumbnail":"b7b35f94963b53c55ab4d919bce2020b8b0f46f785cd90b4e53c62901ada0255","access_level":"public","create_time":"$D2019-12-05T16:16:04.187Z","update_time":"$D2021-04-13T15:34:27.448Z","view_time":null,"first_public_time":"$D2019-12-06T10:12:56.482Z","fork":false,"stars":0,"starred":false,"owner":{"type":"team","id":"ef739838e5a1d607","login":"battila7","name":"Attila Bagossy","avatar_url":"https://avatars-next.observableusercontent.com/90b1460199ec3e040db6a772368bce18487f1a014e7f564bfbc73d51132a4e07","tier":"starter_2024"},"version":1035,"publish_version":1035,"role":null,"authored":false,"authors":[{"type":"individual","id":"6afecd2e2d5dcc78","login":"battila7","name":"Attila Bagossy","avatar_url":"https://avatars-next.observableusercontent.com/90b1460199ec3e040db6a772368bce18487f1a014e7f564bfbc73d51132a4e07","tier":null}],"members":[{"type":"individual","id":"6afecd2e2d5dcc78","login":"battila7","name":"Attila Bagossy","avatar_url":"https://avatars-next.observableusercontent.com/90b1460199ec3e040db6a772368bce18487f1a014e7f564bfbc73d51132a4e07","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":2,"collection":{"id":"f61262f1bb38bd9f","slug":"az-informatikai-biztonsag-alapjai","title":"Az informatikai biztonság alapjai","owner":{"id":"ef739838e5a1d607","login":"battila7","name":"Attila Bagossy"}},"type":"notebook","body":{"title":"RSA","cells":[{"id":0,"value":"md`# RSA`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1033,"value":"md`([English translation by Ádám Vécsi](https://observablehq.com/@beardofdoom/rsa))`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":358,"value":"md` ## Előzetes ismeretek`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":362,"value":"md`### Aszimmetrikus titkosítás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":366,"value":"md`\nAmikor a titkosítás szóba kerül, akkor leggyakrabban az úgynevezett szimmetrikus titkosítási módszerek jutnak eszünkbe, hiszen ezek évezredek óta alkalmazásban állnak. Szimmetrikus rejtjelezés esetén ugyanazt a kulcsot használjuk a titkosításhoz és később a visszafejtéshez is*. Ez azt jelenti, hogy a titkosított kommunikációban résztvevő mindkét félnek, azaz a küldőnek és a fogadónak is rendelkeznie kell ezzel a kulccsal. Ebből következik, hogy a titkosított üzenetváltást megelőzően a feleknek meg kell egyezniük a használt kulcsban.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":781,"value":"md`\n> **Léteznek olyan szimmetrikus módszerek is, melyek két kulcsot alkalmaznak. Ekkor azonban az egyik kulcsból könnyen meghatározható a másik.*\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":372,"value":"md`\nSzámos esetben azonban erre nem adódik mód. Erre kínál megoldást az aszimmetrikus titkosítás, mely egy publikus és egy privát kulcsból álló kulcspárt alkalmaz. A titkosításhoz a nyilvánosan közzétett publikus kulcsot, míg a visszafejtéshez a kizárólag a fogadó fél által ismert privát kulcsot használjuk.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":377,"value":"md`### Euler-függvény`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":380,"value":"md`\n**Definíció:** Legyen ${tex`n`} egy pozitív egész. Ekkor ${tex`\\varphi(n)`} jelöli az ${tex`1, 2,\\ldots, n`} értékek közül azok számát, melyek ${tex`n`}-hez relatív prímek.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":424,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":412,"value":"md`\nHa ${tex`p`} prím, ${tex`k`} pedig egy pozitív egész, akkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":414,"value":"tex.block`\\varphi(p^{k}) = p^{k} - p^{k - 1}`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":421,"value":"md`\nEz azt jelenti, hogy ha ${tex`p`} prím, akkor ${tex`\\varphi(p) = p - 1`}.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":428,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":431,"value":"md`\nHa ${tex`n`} és ${tex`m`} relatív prímek, akkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":435,"value":"tex.block`\\varphi(nm) = \\varphi(n)\\varphi(m)`\n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":388,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":441,"value":"md`\nHa ${tex`p`} és ${tex`q`} prímek, akkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":443,"value":"tex.block`\\varphi(pq) = (p - 1)(q - 1)`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":447,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":457,"value":"md`\n**Euler-Femat tétel**: Ha ${tex`a`} és ${tex`n`} relatív prímek, akkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":459,"value":"tex.block`\na^{\\varphi(n)} \\equiv 1 \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":465,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":450,"value":"md`\n${tex`\\varphi(n)`} értékét egy tetszőleges ${tex`n`} pozitív egészre akkor tudjuk gyorsan meghatározni, ha ismerjük ${tex`n`} prímfaktorizációját. Ebből következik, hogy amennyiben ${tex`n`} nagyon nagy, akkor ${tex`\\varphi(n)`} nem számolható hatékonyan.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":468,"value":"md`## Az ötlet`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":471,"value":"md`\nSzeretnénk készíteni egy aszimmetrikus titkosítási eljárást. Ekkor követelmény, hogy a visszafejtés csak akkor legyen hatékony, ha valaki ismeri a privát kulcsot. Tehát, a publikus kulcsot felhasználva ne legyünk képesek meghatározni a visszafejtéshez használható privát kulcsot.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":481,"value":"md`\nIdézzük fel az Euler-Fermat tételt! Ez azt állítja, hogy ha ${tex`a`} és ${tex`n`} relatív prímek, akkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":484,"value":"tex.block`\na^{\\varphi(n)} \\equiv 1 \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":488,"value":"md`\nNyilvánvaló, hogy bármely ${tex`k`} esetén\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":491,"value":"tex.block`\na^{\\varphi(n) \\cdot k} \\equiv 1^{k} \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":503,"value":"md`\nTovábbá, a kongruencia tulajdonságait kihasználva\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":506,"value":"tex.block`\na^{\\varphi(n) \\cdot k} \\cdot a \\equiv 1^{k} \\cdot a \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":512,"value":"md`azaz`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":516,"value":"tex.block`\na^{\\varphi(n) \\cdot k + 1}\\equiv a \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":519,"value":"md`\nTehát van arra mód, hogy modulo ${tex`n`} dolgozva és megfelelő hatványra emelve, visszakapjuk ugyanazt az ${tex`a`} értéket, amiből kiindultunk.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":534,"value":"md`\nInduljunk el most a titkosítás irányából! Tegyük fel, hogy az ${tex`m`} nyílt szöveget szeretnénk titkosítani. Keresünk ekkor olyan ${tex`e`} és ${tex`d`} értékeket, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":540,"value":"tex.block`\nm^{ed} \\equiv m \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":544,"value":"md`\nFigyeljük meg, hogy az Euler-Fermat tételből kiindulva egy nagyon hasonló alakú kifejezéshez jutottunk. Kérdés, hogy összeér-e a két oldal és valóban egy aszimmetrikus titkosítási módszert fogunk-e kapni?\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":556,"value":"md`\nIsmert, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":558,"value":"tex.block`\nm^{\\varphi(n)} \\equiv 1 \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":561,"value":"md`\nVálasszuk meg ${tex`e`}-t úgy, hogy ${tex`e`} legyen kisebb, mint ${tex`\\varphi(n)`}, valamint ${tex`e`} és ${tex`\\varphi(n)`} legyenek relatív prímek! Ekkor tetszőleges ${tex`0 \\leq m < n`}-re a titkos szöveg\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":567,"value":"tex.block`\nm^e \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":573,"value":"md`\nMivel ${tex`e`} és ${tex`\\varphi(n)`} relatív prímek (azaz a legnagyobb közös osztójuk ${tex`1`}), ezért a Bézout-tételt felhasználva kapjuk, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":576,"value":"tex.block`\n1 = x \\cdot e + y \\cdot \\varphi(n)\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":587,"value":"md`\nEz pontosan azt jelenti, hogy létezik olyan ${tex`x`} érték, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":589,"value":"tex.block`\nx \\cdot e \\equiv 1 \\;\\mathrm{mod}\\; \\varphi(n)\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":593,"value":"md`\nEzt az ${tex`x`} értéket nevezzük ${tex`e`} multiplikatív inverzének modulo ${tex`\\varphi(n)`}, és a továbbiakban jelöljük ${tex`d`}-vel, azaz\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":598,"value":"tex.block`\nd \\cdot e \\equiv 1 \\;\\mathrm{mod}\\; \\varphi(n)\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":613,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":601,"value":"md`\nIdézzük most fel a maradékos osztás definícióját:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":604,"value":"md`\n**Definíció (Maradékos osztás)**:\nHa ${tex`a, b \\in \\mathbb{Z}`} és ${tex`b \\neq 0`}, akkor egyértelműen léteznek olyan ${tex`q, r \\in \\mathbb{Z}`} számok, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":607,"value":"tex.block`\n a = b \\cdot q + r,\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":611,"value":"md`\nahol ${tex`0 \\leq r < |\\,b\\,|`}.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":618,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":621,"value":"md`\nKombinálva a maradékos osztást és a fenti kongruenciát, azt kapjuk, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":623,"value":"tex.block`\ned = k \\cdot \\varphi(n) + 1\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":632,"value":"md`\nami pedig azt jelenti, hogy ezzel az ${tex`e, d`} és ${tex`n`} választással, tetszőleges ${tex`0 \\leq m < n`} értékre\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":634,"value":"tex.block`\nm^{ed} = m^{k\\cdot\\varphi(n) + 1} \\equiv m \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":638,"value":"md`\nA továbbiakban az ${tex`(n, e)`} pár lesz a publikus kulcs, ${tex`d`} pedig a privát kulcs.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":650,"value":"md`\nMi biztosítja azonban, hogy a publikus kulcsból praktikusan nem lehetséges a privát kulcs kiszámítása?\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":653,"value":"md`${tex`d`} értékét ${tex`e`} és ${tex`\\varphi(n)`} felhasználásával határoztuk meg. Mivel ${tex`e`} és ${tex`n`} nyilvánosak, ezért arról kell megbizonyosodnunk, hogy ${tex`\\varphi(n)`} kiszámítása még ezek ismeretében is rendkívül nehéz.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":664,"value":"md`\nLegyen ${tex`n = pq`}, ahol ${tex`p`} és ${tex`q`} megfelelően nagy, különböző prímek. Ekkor ${tex`\\varphi(n)=(p - 1)(q - 1)`}. Ehhez azonban ismerni kell ${tex`n`} prímtényezős felbontását. Ha viszont ${tex`p`} és ${tex`q`} jól megválasztott, elegendően nagy prímek, akkor ${tex`n`} faktorizációja egy nagyon nehéz feladat, hiszen nagy értékekre a prímfaktorizáció rendkívül nehéz.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":677,"value":"md`## Az RSA komponensei`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":679,"value":"md`### Kulcsgenerálás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":683,"value":"md`\nA kulcsgenerálás során előállítunk egy publikus (nyilvános) és egy privát (titkos) kulcsból álló kulcspárt. Ennek lépései a következők.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":690,"value":"md`\n 1. Véletlenszerűen válasszunk két nagy, egymástól különböző ${tex`p`} és ${tex`q`} prímet. Ehhez használhatjuk a [Miller-Rabin prímtesztet](https://observablehq.com/@battila7/miller-rabin-primteszt-miller-rabin-primality-test).\n 1. Ekkor ${tex.block`n = pq`} és ${tex.block`\\varphi(n) = (p - 1)(q - 1)`}\n 1. Válasszunk egy ${tex`1 < e < \\varphi(n)`} értéket úgy, hogy ${tex`e`} és ${tex`\\varphi(n)`} relatív prímek legyenek.\n 1. Számítsuk ki ${tex`d`} értékét, ismerve, hogy ${tex.block`d \\cdot e \\equiv 1 \\;\\mathrm{mod}\\; \\varphi(n)`} Ehhez használhatjuk a [kibővített euklideszi algoritmust](https://observablehq.com/@battila7/kibovitett-euklideszi-algoritmus).\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":707,"value":"md`\nA kulcsgenerálás ezzel véget ért:\n * **Publikus kulcs**: ${tex`(n, e)`}\n * **Privát kulcs**: ${tex`d`}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":830,"value":"md`\n${tex`p, q`} és ${tex`\\varphi(n)`} értéke pedig titokban tartandó.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":712,"value":"md`### Titkosítás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":716,"value":"md`\nLegyen ${tex`0 \\leq m < n`} a titkosítandó érték. Ekkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":723,"value":"tex.block`\nc \\equiv m^{e} \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":732,"value":"md`\nahol ${tex`c`} jelöli a titkosított üzenetet. ${tex`c`} értékének hatékony kiszámításához használhatjuk a [gyorshatványozást](https://observablehq.com/@battila7/gyorshatvanyozas-fast-modular-exponentiation).\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":737,"value":"md`### Visszafejtés`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":742,"value":"md`\nA visszafejtés, azaz a titkosított üzenetből a nyílt szöveg előállítása kétféle módon is történhet.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":745,"value":"md`\n#### Visszafejtés egyszerű gyorshatványozással\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":749,"value":"md`\nLegyen ${tex`c`} a visszafejtendő titkosított üzenet. Ekkor\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":752,"value":"tex.block`\nm \\equiv c^{d} \\;\\mathrm{mod}\\; n\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":758,"value":"md`\nahol ${tex`m`} jelöli a nyílt szöveget. ${tex`m`} értékének hatékony kiszámításához használhatjuk a [gyorshatványozást](https://observablehq.com/@battila7/gyorshatvanyozas-fast-modular-exponentiation).\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":761,"value":"md`#### Visszafejtés kínai maradéktétellel`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":764,"value":"md`\nBár a gyorshatványozás önmagában is elég hatékony, azonban a visszafejtés folyamata tovább javítható a kínai maradéktétel alkalmazásával.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":771,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":770,"value":"md`\n**Tétel (Kínai maradéktétel)**: Adott ${tex`k`} esetén, legyenek az ${tex`n_{i}`} (${tex`1 \\leq i\\leq k`}) értékek páronként relatív prímek, az ${tex`a_{i}`} (${tex`1 \\leq i\\leq k`}) értékek pedig tetszőlegesek. Ekkor létezik olyan ${tex`x`} egész, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":806,"value":"tex.block`\n\\begin{aligned}\nx &\\equiv a_{1} \\;\\mathrm{mod}\\; n_{1} \\\\\nx &\\equiv a_{2} \\;\\mathrm{mod}\\; n_{2} \\\\\n&\\;\\:\\vdots \\\\\nx &\\equiv a_{k} \\;\\mathrm{mod}\\; n_{k}\n\\end{aligned}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":820,"value":"md`\nés bármely ilyen ${tex`x`} kongruens modulo ${tex`N`} (ahol ${tex`N = n_{1}\\cdot n_{2} \\cdot \\ldots \\cdot n_{k}`}).\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":834,"value":"md`**Megoldás ${tex`k=2`}-re**`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":838,"value":"md`Ekkor`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":843,"value":"tex.block`\n\\begin{aligned}\nx &\\equiv a_{1} \\;\\mathrm{mod}\\; n_{1} \\\\\nx &\\equiv a_{2} \\;\\mathrm{mod}\\; n_{2}\n\\end{aligned}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":847,"value":"md`Ismert, hogy ${tex`n_{1}`} és ${tex`n_{2}`} relatív prímek. A Bézout-tétel szerint ebben az esetben`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":851,"value":"tex.block`\n1 = y_{1}n_{1} + y_{2}n_{2}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":855,"value":"md`ahol ${tex`y_{1}`} és ${tex`y_{2}`} meghatározható a [kibővített euklideszi algoritmussal](https://observablehq.com/@battila7/kibovitett-euklideszi-algoritmus).`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":860,"value":"md`\nEbből már következik, hogy\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":862,"value":"tex.block`\nx = a_{1}y_{2}n_{2} + a_{2}y_{1}n_{1} \\;\\mathrm{mod}\\; N\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":866,"value":"md`hiszen`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":868,"value":"tex.block`\n\\begin{aligned}\nx &= a_{1}y_{2}n_{2} + a_{2}y_{1}n_{1} \\\\\n&= a_{1}(1 - y_{1}n_{1}) + a_{2}y_{1}n_{1} \\\\\n&= a_{1} -a_{1}y_{1}n_{1} + a_{2}y_{1}n_{1} \\\\\n&= a_{1} + (a_{2} - a_{1})y_{1}n_{1}\n\\end{aligned}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":877,"value":"md`amiből modulo ${tex`n_{1}`} adódik, hogy`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":879,"value":"tex.block`\nx \\equiv a_{1} \\;\\mathrm{mod}\\; n_{1}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":886,"value":"md`Analóg módon bizonyítható, hogy fennáll a második konguencia is.`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":774,"value":"md`----`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":892,"value":"md`\nA visszafejtés folyamata a következő:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":894,"value":"md`\n 1. ${tex`d_{p}`} és ${tex`d_{q}`} kiszámítása:\n ${tex.block`\n \\begin{aligned}\n d_{p} &\\equiv d \\;\\mathrm{mod}\\; p - 1 \\\\\n d_{q} &\\equiv d \\;\\mathrm{mod}\\; q - 1\n \\end{aligned}\n `}\n 2. ${tex`m_{p}`} és ${tex`m_{q}`} kiszámítása:\n ${tex.block`\n \\begin{aligned}\n m_{p} &\\equiv c^{d_{p}} \\;\\mathrm{mod}\\; p \\\\\n m_{q} &\\equiv c^{d_{q}} \\;\\mathrm{mod}\\; q\n \\end{aligned}\n `}\n 3. Az ${tex`m`}-re vonatkozó kongruenciarendszer felírása:\n ${tex.block`\n \\begin{aligned}\n m &\\equiv m_{p} \\;\\mathrm{mod}\\; p \\\\\n m &\\equiv m_{q} \\;\\mathrm{mod}\\; q\n \\end{aligned}\n `}\n 4. Megoldás ${tex`m`}-re a kínai maradéktétel alkalmazásával:\n ${tex.block` \n 1 = y_{p}p + y_{q}q\n `}\n amiből adódik, hogy \n${tex.block` \n m \\equiv m_{p}y_{q}q + m_{q}y_{p}p \\;\\mathrm{mod}\\; n\n `}\n hiszen ${tex`n = pq`}.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":6,"value":"md`## Gyakorlás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":949,"value":"md`\nA következőkben lépésről lépésre végigszámoljuk az RSA komponenseit.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":53,"value":"md`### Kulcsgenerálás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":952,"value":"md`\nKezdjük a kulcsgenerálással! ${tex`p`} és ${tex`q`} legyenek véletlenszerűen választott, különböző prímek!\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":37,"value":"mutable inputP = number({\n title: md`${tex`p`} értéke`,\n value: 463\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":39,"value":"mutable inputQ = number({\n title: md`${tex`q`} értéke`,\n value: 547\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":44,"value":"viewof initiateKeyGeneration = button({\n value: \"Kulcsgenerálás!\"\n});","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":78,"value":"isKeyGenerationInputsPrimes \n ? md``\n : md`${tex`p`} és ${tex`q`} is prím kell, hogy legyen!`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":57,"value":"md`#### 1. ${tex`n`} meghatározása`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":956,"value":"md`Ekkor`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":60,"value":"tex.block`\nn = pq = ${keyGenerationInputData.p} \\cdot ${keyGenerationInputData.q} = ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":959,"value":"md`és`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":75,"value":"tex.block`\n\\varphi(n) = (p - 1)(q - 1) = ${keyGenerationInputData.p - 1} \\cdot ${keyGenerationInputData.q - 1} = ${phin}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":67,"value":"md`#### 2. ${tex`e`} meghatározása`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":962,"value":"md`\nIsmerve ${tex`\\varphi(n)`} értékét, válasszunk most egy megfelelő ${tex`e`}-t.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":87,"value":"mutable inputE = number({\n title: md`${tex`e`} értéke`,\n value: 47\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":93,"value":"viewof setE = button({\n value: \"Beállítás\"\n});","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":115,"value":"isECorrect\n ? md``\n : md`${tex`e`} nem teljesíti a feltételeket!`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":129,"value":"md`#### 3. ${tex`d`} meghatározása`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":966,"value":"md`\nA [kibővített euklideszi algoritmus](https://observablehq.com/@battila7/kibovitett-euklideszi-algoritmus) használatával számítsuk ki ${tex`d`}-t! Ha az algoritmust végigszámolva negatív értéket kaptunk akkor sem kell megijednünk, hiszen modulo ${tex`${phin}`} dolgozunk.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":155,"value":"tex.block`d \\equiv ${calculatedD} \\;\\mathrm{mod}\\; ${phin}`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":160,"value":"tex.block`\nd = ${d}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":164,"value":"md`#### 4. Kulcspár`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":974,"value":"md`\nMost már felírhatjuk a kulcspár mindkét részét.\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":167,"value":"md`\nPublikus kulcs:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":171,"value":"tex.block`\n(n, e) = (${n}, ${e})\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":174,"value":"md`Privát kulcs:`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":180,"value":"tex.block`\nd = ${d}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":185,"value":"md`### Titkosítás`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":977,"value":"md`\nA kulcsgenerálással létrehozott nyilvános kulccsal titkosítsunk valamilyen ${tex`m`} értéket. \n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":188,"value":"mutable inputM = number({\n title: md`${tex`m`} értéke`,\n value: 47\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":191,"value":"viewof initiateEncryption = button({\n value: \"Titkosítás\"\n});","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":208,"value":"isMCorrect\n ? md``\n : md`${tex`m`} nem felel meg a feltételeknek!`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":211,"value":"md`#### 1. ${tex`m`} titkosítása`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":984,"value":"md`\nA titkosítás egyetlen [gyorshatványozást](https://observablehq.com/@battila7/gyorshatvanyozas-fast-modular-exponentiation) jelent:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":214,"value":"tex.block`\nc \\equiv ${m}^{${e}} \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":221,"value":"tex.block`\nc \\equiv ${ciphertext} \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":234,"value":"md`### Visszafejtés`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":988,"value":"md`\nA kulcsgenerálással létrehozott privát kulccsal fejtsünk vissza valamilyen ${tex`c`} értéket. \n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":239,"value":"mutable inputC = number({\n title: md`${tex`c`} értéke`,\n value: 47\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":242,"value":"viewof initiateDecryption = button({\n value: \"Visszafejtés\"\n});","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":259,"value":"isCCorrect\n ? md``\n : md`${tex`c`} nem teljesíti a feltételeket!`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":262,"value":"md`#### A. Visszafejtés egyszerű gyorshatványozással`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":991,"value":"md`\nAz egyik lehetséges módszer a [gyorshatványozás](https://observablehq.com/@battila7/gyorshatvanyozas-fast-modular-exponentiation) alkalmazása:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":268,"value":"tex.block`\n m \\equiv ${c}^{${d}} \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":276,"value":"tex.block`\nm \\equiv ${plaintextByExponentiation} \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":265,"value":"md`#### B. Visszafejtés kínai maradéktétellel`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":995,"value":"md`\nHatékonyabban számolható a megoldás a kínai maradéktétel segítségével:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":319,"value":"tex.block`\nd_{p} \\equiv ${dp} \\;\\mathrm{mod}\\; ${keyGenerationInputData.p - 1}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":324,"value":"tex.block`\nd_{q} \\equiv ${dq} \\;\\mathrm{mod}\\; ${keyGenerationInputData.q - 1}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":315,"value":"tex.block`\nm_{p} \\equiv ${c}^{${dp}} \\;\\mathrm{mod}\\; ${keyGenerationInputData.p}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":331,"value":"tex.block`\nm_{q} \\equiv ${c}^{${dq}} \\;\\mathrm{mod}\\; ${keyGenerationInputData.q}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":998,"value":"md`\nAmiből a megoldandó kongruenciarendszer:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":334,"value":"tex.block`\n\\begin{aligned}\nm &\\equiv ${mp} \\;\\mathrm{mod}\\; ${keyGenerationInputData.p} \\\\\nm &\\equiv ${mq} \\;\\mathrm{mod}\\; ${keyGenerationInputData.q}\n\\end{aligned}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1001,"value":"md`\nA [kibővített euklideszi algoritmust](https://observablehq.com/@battila7/kibovitett-euklideszi-algoritmus) felhasználva:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":340,"value":"{\n const wrap = n => n < 0 ? `(${n})` : `${n}`\n \n return tex.block`\n 1 = y_{p} \\cdot ${wrap(keyGenerationInputData.p)} + y_{q}\\cdot ${wrap(keyGenerationInputData.q)}\n `\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":345,"value":"tex.block`\n\\begin{aligned}\ny_{p} &= ${yp} \\\\\ny_{q} &= ${yq}\n\\end{aligned}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1005,"value":"md`\nEzután a behelyettesítés:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":349,"value":"tex.block`\nm \\equiv (${mp} \\cdot ${yq} \\cdot ${keyGenerationInputData.q} + ${mq} \\cdot ${yp} \\cdot ${keyGenerationInputData.p}) \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":355,"value":"tex.block`\nm \\equiv ${calculatedPlaintextByChineseRemainderTheorem} \\;\\mathrm{mod}\\; ${n}\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1018,"value":"md`\nTehát a nyílt szöveg:\n`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1021,"value":"tex.block`m = ${plaintextByChineseRemainderTheorem}`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":4,"value":"md`## Függelék`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":16,"value":"bigInt = require('big-integer')","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":10,"value":"import {\n $drawing,\n $curves,\n $figure,\n $maths,\n $styling,\n $plot,\n $utility\n} from '@battila7/kompgraf-stdlib'","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":133,"value":"import { extendedEuclideanAlgorithm } from '@battila7/kibovitett-euklideszi-algoritmus'","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":12,"value":"$$styling.css.twoColumnContainer","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":14,"value":"import {button, number} from \"@jashkenas/inputs\"\n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":24,"value":"{\n initiateKeyGeneration;\n mutable keyGenerationInputData = {\n p: Number.parseInt((mutable inputP)[0].value),\n q: Number.parseInt((mutable inputQ)[0].value),\n }\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":97,"value":"{\n setE;\n mutable e = Number.parseInt((mutable inputE)[0].value)\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":194,"value":"{\n initiateEncryption;\n mutable m = Number.parseInt((mutable inputM)[0].value)\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":245,"value":"{\n initiateDecryption;\n mutable c = Number.parseInt((mutable inputC)[0].value)\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":247,"value":"mutable c = 252090","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":197,"value":"mutable m = 10","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":100,"value":"mutable e = 47","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":27,"value":"mutable keyGenerationInputData = ({\n p: 463,\n q: 547\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":31,"value":"isKeyGenerationInputsPrimes = bigInt(keyGenerationInputData.p).isProbablePrime() && bigInt(keyGenerationInputData.q).isProbablePrime()","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":63,"value":"n = keyGenerationInputData.p * keyGenerationInputData.q","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":71,"value":"phin = (keyGenerationInputData.p - 1) * (keyGenerationInputData.q - 1)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":105,"value":"isECoprimeWithPhin = bigInt.gcd(bigInt(e), bigInt(phin)).equals(1)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":109,"value":"isEBetweenOneAndPhin = 1 < e && e < phin","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":111,"value":"isECorrect = isEBetweenOneAndPhin && isECoprimeWithPhin","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":139,"value":"calculatedD = extendedEuclideanAlgorithm(e, phin).x","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":148,"value":"d = calculatedD < 0\n ? phin + (calculatedD % phin)\n : calculatedD","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":200,"value":"isMBetweenZeroAndN = 0 <= m && m < n ","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":204,"value":"isMCorrect = isMBetweenZeroAndN","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":218,"value":"ciphertext = bigInt(m).modPow(e, n).toJSNumber()","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":253,"value":"isCBetweenZeroAndN = 0 <= c && c < n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":256,"value":"isCCorrect = isCBetweenZeroAndN","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":269,"value":"plaintextByExponentiation = bigInt(c).modPow(d, n).toJSNumber()","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":281,"value":"dp = d % (keyGenerationInputData.p - 1)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":285,"value":"dq = d % (keyGenerationInputData.q - 1)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":288,"value":"mp = bigInt(c).modPow(dp, keyGenerationInputData.p).toJSNumber()","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":291,"value":"mq = bigInt(c).modPow(dq, keyGenerationInputData.q).toJSNumber()","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":294,"value":"epq = extendedEuclideanAlgorithm(keyGenerationInputData.p, keyGenerationInputData.q)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":300,"value":"yp = epq.x","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":308,"value":"yq = epq.y","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":312,"value":"calculatedPlaintextByChineseRemainderTheorem = (mp * yq * keyGenerationInputData.q + mq * yp * keyGenerationInputData.p) % n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1012,"value":"plaintextByChineseRemainderTheorem = calculatedPlaintextByChineseRemainderTheorem < 0\n ? (n + calculatedPlaintextByChineseRemainderTheorem) % n\n : calculatedPlaintextByChineseRemainderTheorem\n","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W30","stdlib":"1","titling":"auto"}} 2f:"$2e:body"