Decrypt Hidden Message with bitwise operator - c

Question

I encountered this exercise where it asks you to Decrypt a hidden message in the encrypted code in the picture below.
This exercise is related to bitwise operators , so that I assume we need to use " XOR " operator to decrypt the code with 2-digit keys ranging from 00 to 99.
Here's my code.
char input[9] = "dEsxDI^I" , decrypt[9]; // Variable Declarations and Initializations.
int key;
printf("\n\n\nDecrpytion for the Encrypted code \"dEsxDI^I\" :\n\n\n");
for(key = 0 ; key <= 99 ; key++) // 00 to 99 Key Testing.
{
for (int i = 0; i < 8; i++) // Decryption Process.
{
decrypt[i] = input[i] ^ key;
printf("%c", decrypt[i]);
}
printf(" = Key No (%d)\n\n\n", key); // Displays Key No for each decryption.
}
printf("\n\n");
return 0;
The message I decrypted is " Hi_There " with the encryption key of " 44 ".
Just wondering is it the correct message or not? Since that's the most logical message of all the messages I decrypted.
There's no solution provided for this exercise , so I'm curious.
So , what are your output guys?

Answer 1

The message I decrypted is " Hi_There " with the encryption key of "
44 ". Just wondering is it the correct message or not?
There are no firm directions about whether your solution to the 8-character cipher must be an 8-character solution, but that is the general case. Absent clear instruction ("Your solution may, or may not, contain all characters in the cipher."), I would look for a solution that included 8-chars.
Though you have not provided specifics of your code, a simple brute-force loop over all possible keys is about as good as anything else. Since you are not provided with the solution, you cannot do any comparisons along the way. The best you can do is loop over all keys, applying it to each character in your cipher and outputting the result -- and then using you "good ole eyeballs" to pick out a solution. (If you knew the solution would be all upper/lower case characters -- you could impose that check before outputting each result for manual inspection).
It's not clear if you were told to XOR the key with each char, or whether you were just supposed to try all reasonable bitwise operations, but a general OR or XOR is a good place to start.
Putting that together for XOR, you could do something similar to:
#include <stdio.h>
int main (void) {
char *buf = "dEsxDI^I";
for (int key = 0; key < 100; key++) { /* loop over all keys */
char *p = buf; /* output key (and char) */
printf ("key: %2d ('%c') ", key, key >= ' ' ? key : 0);
while (*p) /* loop over all chars */
putchar (*p++ ^ key); /* output decrypted char */
putchar ('\n'); /* tidy up with '\n' */
}
return 0;
}
Example Use/Output
$ ./bin/keyfind2
key: 0 ('') dEsxDI^I
key: 1 ('') eDryEH_H
key: 2 ('') fGqzFK\K
key: 3 ('') gFp{GJ]J
key: 4 ('') `Aw|#MZM
key: 5 ('') a#v}AL[L
<snip>
key: 12 ('') hItHERE
<snip>
key: 43 ('+') OnXSobub
key: 44 (',') Hi_There
key: 45 ('-') Ih^Uidsd
<snip>
Using XOR to appy to the key and using the manual eyeballs to scan for solutions, key: 12 ('') hItHERE looks promising, but then it is only a 7-char solution, continuing key: 44 (',') Hi_There provides an 8-char solution and would be the better of the two answers.
So, "Yes", I think you found the correct solution.
Without Pointers using String Indexes
Since the use of the pointer p to iterate over your cipher string in buf seems to have left you scratching your head (don't worry, it will all soak in over time), you can easily do exactly the same thing with string indexes. A for loop is just as easily used with indexes. Eliminating pointers and using string indexes, you could do the following:
#include <stdio.h>
int main (void) {
char *buf = "dEsxDI^I";
for (int key = 0; key < 100; key++) { /* loop over all keys */
printf ("key: %2d ('%c') ", key, key >= ' ' ? key : 0);
for (int i = 0; buf[i]; i++) /* loop over all chars */
putchar (buf[i] ^ key); /* output decrypted char */
putchar ('\n'); /* tidy up with '\n' */
}
return 0;
}
(output is exactly the same)
Your question regarding the use of the ternary operator in the printf statement is a good question for someone not familiar with it. The ternary operator is a shorthand for an if/else statement where the result of the expression is based on a given test condition. Essentially the ternary operator is:
condition ? if_true_value : if_false_value
In my printf statement I simply do:
printf ("key: %2d ('%c') ", key, key >= ' ' ? key : 0);
Obviously the integer value of key is used with the %2d conversion specifier to insure all key values are printed two characters wide (that's the purpose of the 2 field-width specifier. Now let's just look at what character I want to print through %c.
key >= ' ' ? key : 0 // huh?
It's easy. The first printable ASCII value is 0x20 (hex) or 32 decimal which corresponds to the ' ' (space) character. See ASCII table below. So what is my test condition? If (key >= ' ') greater or equal to space. If it is, just print the character represented by key. If it isn't, then print the nul-character (which is decimal 0 or the equivalent character '\0' -- which just takes more typing). You could do the same thing with:
if (key >= ' ')
printf ("key: %2d ('%c') ", key, key);
else
printf ("key: %2d ('') ", key);
(it's just more typing :)
ASCII Table
Char Dec Oct Hex | Char Dec Oct Hex | Char Dec Oct Hex | Char Dec Oct Hex
-------------------------------------------------------------------------------------
(nul) 0 0000 0x00 | (sp) 32 0040 0x20 | # 64 0100 0x40 | ` 96 0140 0x60
(soh) 1 0001 0x01 | ! 33 0041 0x21 | A 65 0101 0x41 | a 97 0141 0x61
(stx) 2 0002 0x02 | " 34 0042 0x22 | B 66 0102 0x42 | b 98 0142 0x62
(etx) 3 0003 0x03 | # 35 0043 0x23 | C 67 0103 0x43 | c 99 0143 0x63
(eot) 4 0004 0x04 | $ 36 0044 0x24 | D 68 0104 0x44 | d 100 0144 0x64
(enq) 5 0005 0x05 | % 37 0045 0x25 | E 69 0105 0x45 | e 101 0145 0x65
(ack) 6 0006 0x06 | & 38 0046 0x26 | F 70 0106 0x46 | f 102 0146 0x66
(bel) 7 0007 0x07 | ' 39 0047 0x27 | G 71 0107 0x47 | g 103 0147 0x67
(bs) 8 0010 0x08 | ( 40 0050 0x28 | H 72 0110 0x48 | h 104 0150 0x68
(ht) 9 0011 0x09 | ) 41 0051 0x29 | I 73 0111 0x49 | i 105 0151 0x69
(nl) 10 0012 0x0a | * 42 0052 0x2a | J 74 0112 0x4a | j 106 0152 0x6a
(vt) 11 0013 0x0b | + 43 0053 0x2b | K 75 0113 0x4b | k 107 0153 0x6b
(np) 12 0014 0x0c | , 44 0054 0x2c | L 76 0114 0x4c | l 108 0154 0x6c
(cr) 13 0015 0x0d | - 45 0055 0x2d | M 77 0115 0x4d | m 109 0155 0x6d
(so) 14 0016 0x0e | . 46 0056 0x2e | N 78 0116 0x4e | n 110 0156 0x6e
(si) 15 0017 0x0f | / 47 0057 0x2f | O 79 0117 0x4f | o 111 0157 0x6f
(dle) 16 0020 0x10 | 0 48 0060 0x30 | P 80 0120 0x50 | p 112 0160 0x70
(dc1) 17 0021 0x11 | 1 49 0061 0x31 | Q 81 0121 0x51 | q 113 0161 0x71
(dc2) 18 0022 0x12 | 2 50 0062 0x32 | R 82 0122 0x52 | r 114 0162 0x72
(dc3) 19 0023 0x13 | 3 51 0063 0x33 | S 83 0123 0x53 | s 115 0163 0x73
(dc4) 20 0024 0x14 | 4 52 0064 0x34 | T 84 0124 0x54 | t 116 0164 0x74
(nak) 21 0025 0x15 | 5 53 0065 0x35 | U 85 0125 0x55 | u 117 0165 0x75
(syn) 22 0026 0x16 | 6 54 0066 0x36 | V 86 0126 0x56 | v 118 0166 0x76
(etb) 23 0027 0x17 | 7 55 0067 0x37 | W 87 0127 0x57 | w 119 0167 0x77
(can) 24 0030 0x18 | 8 56 0070 0x38 | X 88 0130 0x58 | x 120 0170 0x78
(em) 25 0031 0x19 | 9 57 0071 0x39 | Y 89 0131 0x59 | y 121 0171 0x79
(sub) 26 0032 0x1a | : 58 0072 0x3a | Z 90 0132 0x5a | z 122 0172 0x7a
(esc) 27 0033 0x1b | ; 59 0073 0x3b | [ 91 0133 0x5b | { 123 0173 0x7b
(fs) 28 0034 0x1c | < 60 0074 0x3c | \ 92 0134 0x5c | | 124 0174 0x7c
(gs) 29 0035 0x1d | = 61 0075 0x3d | ] 93 0135 0x5d | } 125 0175 0x7d
(rs) 30 0036 0x1e | > 62 0076 0x3e | ^ 94 0136 0x5e | ~ 126 0176 0x7e
(us) 31 0037 0x1f | ? 63 0077 0x3f | _ 95 0137 0x5f | (del) 127 0177 0x7f

Answer 2

I did XOR, ~XOR, AND, NAND, OR, NOR, left-shift, right-shift operations on your decrypted message.
naturally as any of us would have thought the answer key is 44.
I did not get any other readable message from that String.
although If you consider all possibilities :
XOR:
key 12 : hItHERE
key 44 : Hi_There
key 112 : 49.9____
replace underscores with spaces(not unprintable chars, just empty space)
OR:
key 14 : no~no~o
key 46 : no~no~o
yes they both are same, in fact OR produces a lot of Strings with repeating chars like '{{{{{{{{'
~XOR:
key 243: hItHERE
key 211: Hi_There
Now the definition of readable message depends on every individual.
hItHERE might be some marker for aerial strike.
no~no~o might be a call for help or denial of something.
49.9 might be the quantity of some secret ingredient in a bio-weapons lab.
0~99 is quite a limited range.

Answer 3

A goal should be to to automate the detection mechanism. Rather than find the decrypted phrase from an exhaustive list, form code than does it.
With some minor modifications to OP's code, the below calls various valid() to assess the decryption validity.
Recommend that OP further adjust valid() until it provides 1 "best" decryption. Then the "formula" for decryption is found.
For fun: Try finding the phone number of this spy and Google to determine the name. Her encrypted data is "0>?%=;81". Adjust valid() accordingly.
#include <stdio.h>
#if 1
int valid(const char *decrypt) {
size_t end = strspn(decrypt, " ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz");
return end == 8;
}
#else
int valid(const char *decrypt) {
size_t end = strspn(decrypt, " ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz");
if (end != 8) return 0;
for (int i=1; i<8; i++) {
if (isupper(decrypt[i-1]) && isupper(decrypt[i])) { //reject paired capital letters
return 0;
}
}
return 1;
}
#endif
int main(void) {
const char input[9] = "dEsxDI^I";
printf("Decryption for the Encrypted code \"dEsxDI^I\" :\n\n");
for (int key = 0; key <= 99; key++) {
char decrypt[9] = ""; // Variable Declarations and Initializations.
for (int i = 0; i < 8; i++) {
decrypt[i] = (char) (input[i] ^ key);
}
if (valid(decrypt)) {
printf("Key No %2d Decrypt:<%s>\n", key, decrypt);
}
}
return 0;
}
Output - 1st valid()
Decryption for the Encrypted code "dEsxDI^I" :
Key No 1 Decrypt:<eDryEH_H>
Key No 10 Decrypt:<nOyrNCTC>
Key No 11 Decrypt:<oNxsOBUB>
Key No 16 Decrypt:<tUchTYNY>
Key No 17 Decrypt:<uTbiUXOX>
Key No 22 Decrypt:<rSenR_H_>
Key No 28 Decrypt:<xYodXUBU>
Key No 29 Decrypt:<yXneYTCT>
Key No 39 Decrypt:<CbT_cnyn>
Key No 42 Decrypt:<NoYRnctc>
Key No 43 Decrypt:<OnXSobub>
Key No 44 Decrypt:<Hi_There>
Key No 48 Decrypt:<TuCHtyny>
Key No 49 Decrypt:<UtBIuxox>
Key No 60 Decrypt:<XyODxubu>
Key No 61 Decrypt:<YxNEytct>
Output - 2nd valid()
Decryption for the Encrypted code "dEsxDI^I" :
Key No 22 Decrypt:<rSenR_H_>
Key No 39 Decrypt:<CbT_cnyn>
Key No 44 Decrypt:<Hi_There>