Question

"Four".Q2: Using Caesar Cipher please do this exercise. You need to write a step-by-step description of the process. (10 Marks) Ciphertext is "Ju yuskznotm zujge zngz euax lazaxk ykrl corr zngtq eua lux”.

Answer 1

All the explanation is in the code comments. Hope this helps!

Code:

#include <iostream>
using namespace std;

// function to decrypt text given a caesarcipher encryption key
string decrypt(string s, int key)
{
// decrypted text
string res = "";
// loop for all characters in s
for(int i=0; i<s.size(); i++)
{
// if s[i] is a letter
// to get decryption key, subtract key by 26
// do modulus for connecting letters in a loop
if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z'))
{
if(s[i] >= 'a' && s[i] <= 'z')
res += ((s[i] - 'a') + 26 - key) % 26 + 'a';
else
res += ((s[i] - 'A') + 26 - key) % 26 + 'A';
}
// for other characters
else
{
res += s[i];
}
}
// return result
return res;
}

int main()
{
// given string
string s = "Ju yuskznotm zujge zngz euax lazaxk ykrl corr zngtq eua lux";
  
// to find the decrypted text, try with every key from 1 to 25
for(int i=1; i<=25; i++)
{
cout << "For key " << i << ": " << decrypt(s, i) << endl;
}
  
// the encryption key is 6 and the decrypted message is -
// Do something today that your future self will thank you for
return 0;
}

Sample run:

For key 1: It xtrjymnsl ytifd ymfy dtzw kzyzwj xjqk bngg ymfsp dtz ktw For key 2: Hs wsqixlmrk xshec xlex csyv jyxyvi wipj ampp xlero csy jsv For key 3: Gr vrphwkląj wrgdb wkdw_brxu ixwxuh vhoi zloo wkdan brx iru For key 4: Fq ugogvjkpi vqfca vjcv_aqwt hwvwtg ugnh yknn vjcpm aqw hqt For key 5: Ep tpnfuijoh upebz uibu_zpvs gvuvsf tfmg xjmm uibol zpv gps 'or key 6: Do something today that your future self will thank you for For key 7: Cn rnldsghmf snczx sgzs xntq etstad rdke vhkk sgzmj xnt enq For key 8: Bm qmkcrfgle rmbyw rfyr_wmsp dsrspc acid ugjj rfyli wms dmp For key 9: Al pljbqefkd qlaxv gexq vlro crqrob pbic tfii qexkh vlr clo For key 10: Zk okiapdejc pkzwu pdwpukan bapana oahb sehh pdwjg ukq bkn For key 11: Yj njhzocdib ojyvt ocvo tjpm apopmz nzga rdgg ocvif tjp ajm For key 12: Xi migynbcha nixus nbun siol zonoly myfz qcff nbuhe sio zil For key 13: Wh lhfxmabgz mhwtr matm rhnk ynmnkx lxey pbee matgd rhn yhk For key 14: Vg kgewlzafy lgvsą lzsl agmj xmlmjw kwdx oadd lzsfc agm xgj For key 15: Uf jfdvkyzex kfurp kyrk pfli wlkliv jvcw nzcc kyreb pfl wfi For key 16: Te iecujxydw jetqo jxqj oekh vkjkhu iuby mybb jxqda oek veh For key 17: Sd hdbtiwxcv idspn iwpi ndjg ujijgt htau lxaa iwpcz ndj udg For key 18: Rc gcashvwbu hcrom hvoh_mcif tihifs gszt kwzz hvoby mci tcf For key 19: Qb fbzrguvat gbqnl gung lbhe shgher frys jvyy gunax lbh sbe For key 20: Pa eayqftuzs fapmk ftmf kagd rgfgdg eqxr iuxx ftmzw kag rad For key 21: Oz dzxpestyr ezolj esle jzfc qfefcp dpwg htww eslyv jzf qzc For key 22: Ny cywodrsxq dynki drkd iyeb pedebo covp gsvv drkxu iye pyb For key 23: Mx bxvncqrwp cxmjh cajc_hxda odcdan bnuo fruu cqjwt hxd oxa For key 24: Lw awumbpqvo bwlig bpib gwcz ncbczm amtn eqtt bpivs gwc nwz For key 25: Kv zvtlaopun avkhf aoha fvby mbabyl zlsm dpss aohur fvb mvy

Code screenshot:

#include <iostream> using namespace std; // function to decrypt text given a caesarcipher encryption key string decrypt(string s, int key) // decrypted text string res = ""; // loop for all characters in s for(int i=0; i<s.size(); i++) // if s[i] is a letter // to get decryption key, subtract key by 26, // do modulus for connecting letters in a loop if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z')). if(s[i] >= 'a' && s[i] <= 'z') res += ((s[i] - 'a') + 26 - key) % 26 + 'a'; else res += ((s[i] - 'A') + 26 - key) % 26 + 'A'; i // for other characters else res += s[i]; // return result return res; int main() // given string string s = "Ju yuskznotm zujge zngz euax lazaxk ykrl corr zngtq eua lux"; // to find the decrypted text, try with every key from 1 to 25 for(int i=1; i<=25; i++) cout << "For key " « i «": " << decrypt(s, i) << endl; // the encryption key is 6 and the decrypted message is - // Do something today that your future self will thank you for return 0;