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Wednesday, May 30, 2018

Cara Flashing Ulang Samsung J320G Dengan Odin

By Admin May 30, 2018   No comments

Firmware SAMSUNG GALAXY J3 SM-J320G


Model SM-J320G
Country Indonesia
Version Android 5.1.1 (Lollipop)
Security Patch Level 2016-08-01
Product code XID (Indonesia)
PDA J320GXXU0APH1
CSC J320GOLE0APG2
Model: SM-J320G
Version: J320GXXU0APH1/J320GOLE0APG2/J320GXXU0APG2/J320GXXU0APH1
OS: Lollipop(Android 5.1.1)
Size: 968868128 bytes
Firmware name : Samsung Galaxy J3 (SM-J320G) Lollipop.zip
Link Download https://drive.google.com/uc?id=0B2GF4Ufc7jHkd3UwVkhSTlpQVFU&export=download

Download Odin

Panduan pemasangan:
1. Unduh dan ekstraksi file zip yang berisi seluruh firmware yang Anda inginkan.
2. Buka Program Odin.
3. Nyalakan peranti dalam "Moda Unduh": Tekan tombol Volume Kecil, Daya dan Utama pada saat yang bersamaan selama 5-8 detik sampai moda unduh aktif.
4. Sambungkan peranti Anda ke PC dengan kabel USB saat sedang dalam moda unduh.
5. Selanjutnya, beri tanda centang pada opsi "Auto Reboot" dan "F. Reset Time" dalam Program Odin.
6. Tekan tombol AP/PDA lalu cari dan pilih sebuah file tar.md5 dari folder tempat menyimpan file-file yang diekstraksi.
7. Terakhir, tekan tombol start untuk mulai memasang pembaruan firmware bagi peranti Anda.


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Tuesday, March 27, 2018

Simpan perubahanCara Mendeteksi Status Koneksi Jaringan Dengan Delphi 7

By  March 27, 2018  , ,  No comments
Mendeteksi koneksi jaringan secara realtime dengan timer delphi 7. aplikasi akan secara otomatis memberitahukan jika koneksi jaringan terputus atau terhubung


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Monday, March 26, 2018

How to manually solve the DES algorithm

By  March 26, 2018  ,  No comments

After converting to 64-bit representation, key is permuted according to the PC-1 table. Idea of discussed permutation is following: the first value in the table equals "57". It means that the 57th bit of the given key K is the first bit of the permuted key K+. The 41st bit of the original key becomes the 3rd bit of the permuted key and so on. This rule is applied to any further permutation. It's important to notice that only 56 bits of the original key remains in the newly created permuted key. As written before, every 8th bit of each byte is ignored.


Given Key (in hex):

13 34 57 79 9B BC DF F1

Givenkey K:

00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001

PC-1:
57
49
41
33
25
17
9
1
58
50
42
34
26
18
10
2
59
51
43
35
27
19
11
3
60
52
44
36
63
55
47
39
31
23
15
7
62
54
46
38
30
22
14
6
61
53
45
37
29
21
13
5
28
20
12
4


K+:
1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111


Step 2: K+ splitting
The next step is splitting K+ into halves: C0 and D0. Each half has 28 bits.


K+:
1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111

C0:
1111000 0110011 0010101 0101111


D0:
0101010 1011001 1001111 0001111

Step 3: Creating 16 subkeys using shiftingNext we create sixteen pairs Cn and Dn for 1<=n<=16. Every pair Cn, Dn is created from the previous pairCn-1, Dn-1 for 1<=n<=16 through certain number of "left shifts" of the previous block of bits. To perform a left shit we need to move each bit of a block of data one place to the left. The first bit goes to the end of the block. The left shifts schedule and all pairs Cn, Dn are depicted in the following table:
Iteration
number
Number of
shifts
Created pairs


C0: 1111000011001100101010101111
D0: 0101010101100110011110001111
1
1
C1: 1110000110011001010101011111
D1: 1010101011001100111100011110
2
1
C2: 1100001100110010101010111111
D2: 0101010110011001111000111101
3
2
C3: 0000110011001010101011111111
D3: 0101011001100111100011110101
4
2
C4: 0011001100101010101111111100
D4: 0101100110011110001111010101
5
2
C5: 1100110010101010111111110000
D5: 0110011001111000111101010101
6
2
C6: 0011001010101011111111000011
D6: 1001100111100011110101010101
7
2
C7: 1100101010101111111100001100
D7: 0110011110001111010101010110
8
2
C8: 0010101010111111110000110011
D8: 1001111000111101010101011001
9
1
C9: 0101010101111111100001100110
D9: 0011110001111010101010110011
10
2
C10: 0101010111111110000110011001
D10: 1111000111101010101011001100
11
2
C11: 0101011111111000011001100101
D11: 1100011110101010101100110011
12
2
C12: 0101111111100001100110010101
D12: 0001111010101010110011001111
13
2
C13: 0111111110000110011001010101
D13: 0111101010101011001100111100
14
2
C14: 1111111000011001100101010101
D14: 1110101010101100110011110001
15
2
C15: 1111100001100110010101010111
D15: 1010101010110011001111000111
16
1
C16: 1111000011001100101010101111
D16: 0101010101100110011110001111


Step 4: PC-2 permutation
Before the final permutation we need to merge each pair of data. After that each block of bits CnDn for 1<=n<=16 is permuted according to the PC-2 table forming the keys Kn. Only 48 bits of each concatenated pair remain in the permuted key.


PC-2:
14
17
11
24
1
5
3
28
15
6
21
10
23
19
12
4
26
8
16
7
27
20
13
2
41
52
31
37
47
55
30
40
51
45
33
48
44
49
39
56
34
53
46
42
50
36
29
32


C1D1:
1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110

K1 setelah permutation:
000110 110000 001011 101111 111111 000111 000001 110010


Semuasubkeys:
K1: 000110 110000 001011 101111 111111 000111 000001 110010
K2: 011110 011010 111011 011001 110110 111100 100111 100101
K3: 010101 011111 110010 001010 010000 101100 111110 011001
K4: 011100 101010 110111 010110 110110 110011 010100 011101
K5: 011111 001110 110000 000111 111010 110101 001110 101000
K6: 011000 111010 010100 111110 010100 000111 101100 101111
K7: 111011 001000 010010 110111 111101 100001 100010 111100
K8: 111101 111000 101000 111010 110000 010011 101111 111011
K9: 111000 001101 101111 101011 111011 011110 011110 000001
K10: 101100 011111 001101 000111 101110 100100 011001 001111
K11: 001000 010101 111111 010011 110111 101101 001110 000110
K12: 011101 010111 000111 110101 100101 000110 011111 101001
K13: 100101 111100 010111 010001 111110 101011 101001 000001
K14: 010111 110100 001110 110111 111100 101110 011100 111010
K15: 101111 111001 000110 001101 001111 010011 111100 001010
K16: 110010 110011 110110 001011 000011 100001 011111 110101 

Message Encoding

Step 1: IP permutation

Initial permutation IP of the plaintext M is the first step of message encrypting stage. It permutes bits ofM according to the IP table creating rearranged block of bits IP.


Given message (in hex):
0123456789ABCDEF

Given message M:
00000001 00100011 01000101 01100111 10001001 10101011 11001101 11101111

IP:
58
50
42
34
26
18
10
2
60
52
44
36
28
20
12
4
62
54
46
38
30
22
14
6
64
56
48
40
32
24
16
8
57
49
41
33
25
17
9
1
59
51
43
35
27
19
11
3
61
53
45
37
29
21
13
5
63
55
47
39
31
23
15
7


IP:
11001100 00000000 11001100 11111111 11110000 10101010 11110000 10101010



Step 2: IP splitting
The next step is splitting IP into halves: L0 and R0. Both halves have 32 bits.


IP:
11001100 00000000 11001100 11111111 11110000 10101010 11110000 10101010

L0:
11001100 00000000 11001100 11111111

R0:
11110000 10101010 11110000 10101010

Part 3: Iterations
Next step is 16 iterations. For 1<=n<=16 we calculate:
Ln = Rn-1
Rn = Ln-1 XOR f(Rn-1,Kn)
Function f operates on two blocks of data: Rn-1 and Kn. It produces 32-bit long block of data. Process of calculating f function consists of 4 steps:

1. E permutation
2. XOR with a subkey
3. S box transformation
4. P permutation

f function executing - E permutation
Let's start the first iteration! In the beginning we expand block Rn-1 from 32 to 48 bits. It is made by permuting the input block according to the E table. In this operation some of bits are repeated. Output block is marked E(R0).


R0:
1111 0000 1010 1010 1111 0000 1010 1010


E:
32
1
2
3
4
5
4
5
6
7
8
9
8
9
10
11
12
13
12
13
14
15
16
17
16
17
18
19
20
21
20
21
22
23
24
25
24
25
26
27
28
29
28
29
30
31
32
1


E(R0):
011110 100001 010101 010101 011110 100001 010101 010101

f function executing - XOR with the key
In this step we XOR E(Rn-1) with the key Kn. In the first iteration it is K1 XOR E(R0).
K1:
000110 110000 001011 101111 111111 000111 000001 110010

E(R0):
011110 100001 010101 010101 011110 100001 010101 010101

K1 XOR E(R0):
011000 010001 011110 111010 100001 100110 010100 100111

f function executing - S box transformation
On this stage we have 48-bit long block of data to work with. We can present this also as eight groups of six bits. The groups of six are used now as addresses in tables called "S boxes". At the specific location is placed a 4-bit number

S1:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

14
4
13
1
2
15
11
8
3
10
6
12
5
9
0
7
1

0
15
7
4
14
2
13
1
10
6
12
11
9
5
3
8
2

4
1
14
8
13
6
2
11
15
12
9
7
3
10
5
0
3

15
12
8
2
4
9
1
7
5
11
3
14
10
0
6
13
S2:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

15
1
8
14
6
11
3
4
9
7
2
13
12
0
5
10
1

3
13
4
7
15
2
8
14
12
0
1
10
6
9
11
5
2

0
14
7
11
10
4
13
1
5
8
12
6
9
3
2
15
3

13
8
10
1
3
15
4
2
11
6
7
12
0
5
14
9


S3:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

10
0
9
14
6
3
15
5
1
13
12
7
11
4
2
8
1

13
7
0
9
3
4
6
10
2
8
5
14
12
11
15
1
2

13
6
4
9
8
15
3
0
11
1
2
12
5
10
14
7
3

1
10
13
0
6
9
8
7
4
15
14
3
11
5
2
12



S4:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

7
13
14
3
0
6
9
10
1
2
8
5
11
12
4
15
1

13
8
11
5
6
15
0
3
4
7
2
12
1
10
14
9
2

10
6
9
0
12
11
7
13
15
1
3
14
5
2
8
4
3

3
15
0
6
10
1
13
8
9
4
5
11
12
7
2
14


S5:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

2
12
4
1
7
10
11
6
8
5
3
15
13
0
14
9
1

14
11
2
12
4
7
13
1
5
0
15
10
3
9
8
6
2

4
2
1
11
10
13
7
8
15
9
12
5
6
3
0
14
3

11
8
12
7
1
14
2
13
6
15
0
9
10
4
5
3


S6:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

12
1
10
15
9
2
6
8
0
13
3
4
14
7
5
11
1

10
15
4
2
7
12
9
5
6
1
13
14
0
11
3
8
2

9
14
15
5
2
8
12
3
7
0
4
10
1
13
11
6
3

4
3
2
12
9
5
15
10
11
14
1
7
6
0
8
13


S7:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

4
11
2
14
15
0
8
13
3
12
9
7
5
10
6
1
1

13
0
11
7
4
9
1
10
14
3
5
12
2
15
8
6
2

1
4
11
13
12
3
7
14
10
15
6
8
0
5
9
2
3

6
11
13
8
1
4
10
7
9
5
0
15
14
2
3
12


S8:

0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0

13
2
8
4
6
15
11
1
10
9
3
14
5
0
12
7
1

1
15
13
8
10
3
7
4
12
5
6
11
0
14
9
2
2

7
11
4
1
9
12
14
2
0
6
10
13
15
3
5
8
3

2
1
14
7
4
10
8
13
15
12
9
0
3
5
6
11


K1 XOR E(R0) = B1B2B3B4B5B6B7B8:
011000010001011110111010100001100110010100100111

S(B1)S(B2)S(B3)S(B4)S(B5)S(B6)S(B7)S(B8):
0101 1100 1000 0010 1011 0101 1001 0111
f function executing - P permutation
The final step in f function calculating is P permutation.
S(B1)S(B2)S(B3)S(B4)S(B5)S(B6)S(B7)S(B8):
0101 1100 1000 0010 1011 0101 1001 0111

P:
16
7
20
21
29
12
28
17
1
15
23
26
5
18
31
10
2
8
24
14
32
27
3
9
19
13
30
6
22
11
4
25
f = P[S(B1)S(B2)S(B3)S(B4)S(B5)S(B6)S(B7)S(B8)]:
0010 0011 0100 1010 1010 1001 1011 1011

The final result of the first iteration:

L0:
1100 1100 0000 0000 1100 1100 1111 1111

f(R0,K1):
0010 0011 0100 1010 1010 1001 1011 1011

R1 = L0 XOR f(R0,K1):
1110 1111 0100 1010 0110 0101 0100 0100
and

L1 = R0:
1111 0000 1010 1010 1111 0000 1010 1010

After 16 iterations we get L16 and R16:

L1:
1111 0000 1010 1010 1111 0000 1010 1010

R1:
1110 1111 0100 1010 0110 0101 0100 0100

L16:
0100 0011 0100 0010 0011 0010 0011 0100

R16:
0000 1010 0100 1100 1101 1001 1001 0101
Step 4: Reverse connecting
Next we reverse order of the two blocks and merge them.
L16:
0100 0011 0100 0010 0011 0010 0011 0100

R16:
0000 1010 0100 1100 1101 1001 1001 0101

R16L16:
00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100

Step 5: IP-1 permutation
After the last step in message encrypting, which is IP-1 permutation, we obtain the output.
R16L16:
00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100

IP-1:
40
8
48
16
56
24
64
32
39
7
47
15
55
23
63
31
38
6
46
14
54
22
62
30
37
5
45
13
53
21
61
29
36
4
44
12
52
20
60
28
35
3
43
11
51
19
59
27
34
2
42
10
50
18
58
26
33
1
41
9
49
17
57
25

IP-1:
10000101 11101000 00010011 01010100 00001111 00001010 10110100 00000101

Output:
85E813540F0AB405

Decrypting process

Decoding algorithm is almost the same as encoding. The only action we need to do is to reverse order during applying subkeys. The last subkey becomes the first one, the last but one becomes the second one and so on.
Given message:
0123456789ABCDEF
Given key:
133457799BBCDFF1
Output:
EE0F7C12E0B09338

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