what should I write in code to get cipher text zebbw and decrypted text hello

[wary socket]

<@&987246746478460948> please have a look, thanks.

[wary socket]

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[willow bramble]

mind sharing ur code as non-picture please

[upbeat surge]

Can you post the code

Ah beat me to it

[willow bramble]

also, what decryption is that? caesar/vigenere?

[prime lark]

package mini_project;
import java.util.Scanner;
public class AffineChiper {
public static void main(String[]Args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter the plaintext to be encrypted: ");
String plaintext = sc.nextLine();
String ciphertext = encrypt(plaintext);
System.out.println("Ciphertext: "+ciphertext);
System.out.println("Decrypted Plaintext: "+decrypt(ciphertext));
}
public static String encrypt(String plaintext) {
String ciphertext = "";
int k1 = 7;
int k2 = 2;
for(int i=0;i<plaintext.length();i++) {
int p = (int)plaintext.charAt(i);

            int c = ((p*k1)+k2)%26;
            ciphertext += (char)((c-k2)%26);
        }
        return ciphertext;
    }
    public static String decrypt(String ciphertext) {
        String plaintext = "";
        int k1 = 7;
        int k2 = 2;
        //i
       // 
        for(int i=0;i<ciphertext.length();i++) {
            int c = (int)ciphertext.charAt(i);
            int p = ((c-k2)*1/k1)%26;
            plaintext += (char)(p*1/k1)%26;
           
        }
        return plaintext;

}
}

[upbeat surge]

Looks like a Caesar, can you confirm what it is? @prime lark

[wary socket]

Detected code, here are some useful tools:

Formatted code

package mini_project;

import java.util.Scanner;

public class AffineChiper {
  public static void main(String[] Args) {
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter the plaintext to be encrypted: ");
    String plaintext = sc.nextLine();
    String ciphertext = encrypt(plaintext);
    System.out.println("Ciphertext: " + ciphertext);
    System.out.println("Decrypted Plaintext: " + decrypt(ciphertext));
  }
  public static String encrypt(String plaintext) {
    String ciphertext = "";
    int k1 = 7;
    int k2 = 2;
    for (int i = 0; i < plaintext.length(); i++) {
      int p = (int ) plaintext.charAt(i);
      int c = ((p * k1) + k2) % 26;
      ciphertext += (char ) ((c - k2) % 26);
    }
    return ciphertext;
  }
  public static String decrypt(String ciphertext) {
    String plaintext = "";
    int k1 = 7;
    int k2 = 2;
    //i
    // 
    for (int i = 0; i < ciphertext.length(); i++) {
      int c = (int ) ciphertext.charAt(i);
      int p = ((c - k2) * 1 / k1) % 26;
      plaintext += (char ) (p * 1 / k1) % 26;
    }
    return plaintext;
  }
}

[upbeat surge]

AffineChiper

nvm ;p

[prime lark]

In a multiplicative cipher, the encryption algorithm multiplies the plaintext characters by the
key and the decryption algorithm divides the ciphertext characters by the key as shown in
the following Figure.
In this project our goal is to design a Affine chiper. Here, we can combine the additive and
multiplicative ciphers with a pair of keys to get what is called the affine cipher. The first
key is used with the multiplicative cipher; the second key is used with the additive cipher.
The following figure shows that the affine cipher is actually two ciphers, applied one after
another.
In the affine cipher, the relationship between the plaintext P and the ciphertext C is
C = (P k1 + k2) mod 26 × k1 + k2) mod 26
P = ((C k2) k1 − k2) × k1 × k1 + k2) mod 26 − k2) × k11
) mod 26
Representation of plaintext and ciphertext characters in Z26 is shown below.
Example: Using affine cipher encrypt the message “hello” then decript with key pair (7, 2).
Encryption:
We use 7 for the multiplicative key and 2 for the additive key. We get “ZEBBW”.
Decryption:
Add the additive inverse of 2 24 (mod 26) to the received ciphertext. Then multiply the − k2) × k1 ≡ 24 (mod 26) to the received ciphertext. Then multiply the
result by the multiplicative inverse of 7− k2) × k11 15 (mod 26) to find the plaintext “hello”. ≡ 24 (mod 26) to the received ciphertext. Then multiply the
Definitions:

1. Additive Inverse : In Zn
   , two numbers a and b are additive inverses of each other if
   a + b 0 (mod n) ≡ 24 (mod 26) to the received ciphertext. Then multiply the
   For example, the additive inverse of 4 in Z10 is 10 4 = 6. − k2) × k1
2. Multiplicative Inverse : In Zn
   , two numbers a and b are the multiplicative inverse of
   each other if
   a b 1 (mod n) × k1 + k2) mod 26 ≡ 24 (mod 26) to the received ciphertext. Then multiply the
   For example, if the modulus is 10, then the multiplicative inverse of 3 is 7.
   In other words, we have (3 7) mod 10 = 1.

[prime lark]

It's like encrypting text then decrypting it

[upbeat surge]

I see the issue, just trying to think of a way to explain it

[upbeat surge]

your encrypt is a little wrong

[prime lark]

What should I write?

[upbeat surge]

for (int i = 0; i < plaintext.length(); i++) {
  int p = (int) plaintext.charAt(i) - 'a';
  int c = ((p * k1) + k2) % 26;
  ciphertext += (char) (c + 'a');
}

The - 'a' puts it within the range of 0-25

the int c is already the encoded ciper text

(char) (c + 'a') here we get the correct value back

the rest was correct

so the only bit I changed was int p and the ciphertext += bit

ciphertext += (char ) ((c - k2) % 26); you don't need this bit because int c = ((p * k1) + k2) % 26; is the encryption

[prime lark]

Thank u so much

[upbeat surge]

so knowing this, do you know how to fix the decrypt method now? 🙂

[prime lark]

No 🥲

What shd i do

[upbeat surge]

so start with the p and c

we need it to fall within the 26 letter alphabet

[prime lark]

Yes

[upbeat surge]

then with the ciphertext we want to put it back to what it was before

so + 'a'

but that's not the important bit

i mean it is, but the actual decryption int p = ((c - k2) * 1 / k1) % 26; this formula is wrong

int c = (int) ciphertext.charAt(i) - 'a';
int p = ((c - k2) * 1 / k1) % 26;
plaintext += (char) (p + 'a');

so yuo need to fix p

the reason it's wrong is because you're not reversing what the encrypt did

the encrypt does a % 26

so the decrypt needs to do the reverse

sorry mod inverse % 26

[prime lark]

Mod inverse ? How to write that?

[upbeat surge]

The term is called Multiplicative Inverses

[prime lark]

How to write that?

[upbeat surge]

if you want to learn about it, you can readup on google

or you can do what I did

[willow bramble]

essentially, as we all learned, the "inverse" of 5 is 1/5

cause 5 * 1/5 is 1

[willow bramble]

but if u add modulo to the deal, it becomes tricky

lets say we have mod 5

so the numbers 0, 1, 2, 3, 4

[prime lark]

What?

[willow bramble]

now, look at a number like 3

what do u need to multiple 3 with to get 1 (mod 5)

and thats what he was talking about

the answer is 2 btw

3 * 2 = 6 -> mod 5 -> 1

[prime lark]

But what change shd i make in code

[willow bramble]

if u have to ask that, u probably didnt really understand it yet

[prime lark]

I haven't even understand anything about the project tbh the students in our college hv already submitted

I am left behind i hope they just accept my project 🥲

[upbeat surge]

int p = ((c - k2) * 17) % 26; (k1 inverse mod 26) would the be exact solution

but I hardcoded 17

[willow bramble]

well, then u have worse problems than this. u should step back and learn

it wont really help u if we give u some code and u copy pasta it

then ur lost again 5 minutes after

[upbeat surge]

you can del my msg if you want don't want the solution out zabu right away ;p

[prime lark]

No student really learned in our college they just copy pasted it

I thought to think and do it myself but

They all hv copy pasted and submitted it

[prime lark]

It's not coming

I get biPPc in the decrypted plaintext

[willow bramble]

as said. u wont be able to effectively use snippets from others that u dont fully understand

u probably made a minor mistake here or there, or applied the code wrong

this will continue to happen until u actually fully understand what ur doing and whats going on

[prime lark]

What should I do then

[prime lark]

Thanks thanks a lot

I got it