FASM – Serpent-1 Encryption

The Serpent-1 Encryption Algorithm was a candidate for the AES and came second to the eventual winner Rijndael.

The homepage for Serpent is here:

http://www.cl.cam.ac.uk/~rja14/serpent.html

Serpent uses a 256 bit key to encrypt 128 bit blocks. The 128 bit input block is encrypted in 32 rounds.

Related Post

The Javascript version can be found here: Serpent-1 Encryption Algorithm

The Key Schedule

The 256 bit input key is used to generate 33 x 128 bit sub-keys. The first 32 sub-keys are used in the 32 rounds of the encryption process. The 33rd sub-key is used in the final round.

The following pseudo-code shows the sub-key generation process. An array of 140 32-bit words is created. The first 8 elements are filled with the original 256 bit input key. Then the remaining 132 words are generated as follows:

for(i = 8 ; i < 140 ; i++) { Key[i] = Key[i-8] xor Key[i-5] xor Key[i-3] xor Key[i-1] xor 0x9e3779b9 xor i; Key[i] = ROTL(Key[i],11); }

The 33 sub-keys (of 128 bits each) are then put through the s-boxes as follows:

s_index = 3; for(i = 8; i < 136 ; i += 4) { Key[i],Key[i+1],Key[i+2],Key[i+3] = S-Box[s_index](Key[i],Key[i+1],Key[i+2],Key[i+3]); s_index--; if(s_index < 0) s_index = 7; } Key[i],Key[i+1],Key[i+2],Key[i+3] = S-Box[s_index](Key[i],Key[i+1],Key[i+2],Key[i+3]);

Encryption

The 128 bit input block is encrypted in 32 rounds as follows:

C = input block; for(round = 0 ; round < 32 ; round++) { C = C xor SubKey[round]; C = S-Box[round mod 8](C); if(round==31) break; C = LT(C); } C = C xor SubKey[32];

C is the 128 bit cipher text. SubKey[round] is one of the 33 x 128 bit sub-keys, indexed from 0 to 32. LT is the linear transform as described below.

Decryption

Decryption is the exact reverse of the encryption process:

C = C xor SubKey[32]; for(round = 31 ; round >= 0 ; round--) { if(round == 31) goto .sub C = InvLT(C); .sub: C = InvS-Box[round mod 8](C); C = C xor SubKey[round]; }

The Linear Transform

The linear transform is applied to a 128 bit block. The block is divided into 4 x 32 bit words as follows: |w0|w1|w2|w3|

w0 = RotLeft(w0,13); w2 = RotLeft(w2,3); w1 = w1 xor w0 xor w2; w3 = w3 xor w2 xor ShiftLeft(w0,3); w1 = RotLeft(w1,1); w3 = RotLeft(w3,7); w0 = w0 xor w1 xor w3; w2 = w2 xor w3 xor ShiftLeft(w1,7); w0 = RotLeft(w0,5); w2 = RotLeft(w2,22);

The Inverse Linear Transform

The inverse linear transform is just the linear transform operations in reverse order, but with ROTR instead of ROTL:

w2 = RotRight(w2,22); w0 = RotRight(w0,5); w2 = w2 xor w3 xor ShiftLeft(w1,7); w0 = w0 xor w1 xor w3; w3 = RotRight(w3,7); w1 = RotRight(w1,1); w3 = w3 xor w2 xor ShiftLeft(w0,3); w1 = w1 xor w0 xor w2; w2 = RotRight(w2,3); w0 = RotRight(w0,13);

The S-Boxes

Serpent uses 8 s-boxes. Each s-box contains 16 elements. The input and output to a s-box is a 4 bit value, as follows:

n’ = s-box[n], (n = 0-15)

My program implements the bitslice mode of the serpent cipher algorithm. The diagram below shows how the s-boxes are implemented in bitslice mode. The 128 bit input block is divided up into 4 x 32 bit words: |w0|w1|w2|w3|. These are then arranged as shown in the diagram. The substitution is then applied by taking 32 slices of 4 bits each and putting then through 32 copies of the s-box in parallel. Bit 0 is the lsb in the 4 bit value.

Input Byte Ordering

The input/output is represented as a hexadecimal number, with each pair of hex digits corresponding directly to a byte in an internal byte array. Let the internal array be called input_array[]. Then the first 2 digits in the input correspond to the byte at input_array[0], the next 2 digits correspond to the byte at input_array[1], and so on. The input bytes are organised internally into 32 bit words, with input_array[0] – input_array[3] being the first word, input_array[4] – input_array[7] being the second word, and so on.

For example:

Byte array (NESSIE) input/output: 16 byte input = 11223344 55667788 9900aabb ccddeeff Stored as: input_array[16] = [11,22,33,44,55,66,77,88,99,00,aa,bb,cc,dd,ee,ff];

Test vectors for Serpent-1 with 256 bit keys are specified in byte array (NESSIE) format here:

http://www.cs.technion.ac.il/~biham/Reports/Serpent/Serpent-256-128.verified.test-vectors

The FASM x86 Source Code

To use the program enter a 256 bit key (64 hex digits) and press [Set Key]. This will display the key schedule. Note that the program expects a full 256 bit key, it will not pad keys shorter than this in the way specified by the algorithm.

Then enter the 128 bit input text (32 hex digits) and press [Encrypt] or [Decrypt]. The program will display the result from each round of the 32 rounds of encryption or decryption, with the final round being the result.

; ------------------------------------------------------------------------------------- format PE GUI 4.0 entry start include 'win32a.inc' ; ------------------------------------------------------------------------------------- IDD_THE_DIALOG = 102 IDC_INPUT = 1000 IDC_KEY = 1001 IDC_OUTPUT = 1002 IDC_BTN_ENCRYPT = 1003 IDC_BTN_DECRYPT = 1004 IDC_BTN_KEY_256 = 1005 IDC_BTN_RESET = 1006 ; ------------------------------------------------------------------------------------- section '.code' code readable executable start: invoke GetModuleHandle,0 invoke DialogBoxParam,eax,IDD_THE_DIALOG,0,DialogProc,0 exit: invoke ExitProcess,0 ; ------------------------------------------------------------------------------------- proc DialogProc uses esi edi ebx,hwnddlg,msg,wparam,lparam cmp [msg],WM_INITDIALOG je .wminitdialog cmp [msg],WM_COMMAND je .wmcommand cmp [msg],WM_CLOSE je .wmclose xor eax,eax jmp .quit .wminitdialog: invoke SetDlgItemText,[hwnddlg],IDC_INPUT,szInputMessage invoke SetDlgItemText,[hwnddlg],IDC_KEY,szKeyMessage invoke SetDlgItemText,[hwnddlg],IDC_OUTPUT,"" jmp .done .wmcommand: cmp [wparam], BN_CLICKED shl 16 + IDC_BTN_ENCRYPT je .ENCRYPT cmp [wparam], BN_CLICKED shl 16 + IDC_BTN_DECRYPT je .DECRYPT cmp [wparam], BN_CLICKED shl 16 + IDC_BTN_KEY_256 je .SET_KEY_256 cmp [wparam], BN_CLICKED shl 16 + IDC_BTN_RESET je .RESET jmp .done .ENCRYPT: invoke GetDlgItemText,[hwnddlg],IDC_INPUT,bfDisplay,800 stdcall onNullInput stdcall StrToInputBlock stdcall PrintInputBlock invoke SetDlgItemText,[hwnddlg],IDC_INPUT,bfDisplay stdcall EncryptBlock stdcall Print_Rounds invoke SetDlgItemText,[hwnddlg],IDC_OUTPUT,bfDisplay jmp .done .DECRYPT: invoke GetDlgItemText,[hwnddlg],IDC_INPUT,bfDisplay,800 stdcall onNullInput stdcall StrToInputBlock stdcall PrintInputBlock invoke SetDlgItemText,[hwnddlg],IDC_INPUT,bfDisplay stdcall DecryptBlock stdcall Print_Rounds invoke SetDlgItemText,[hwnddlg],IDC_OUTPUT,bfDisplay jmp .done .SET_KEY_256: invoke GetDlgItemText,[hwnddlg],IDC_KEY,bfDisplay,800 stdcall onNullInput stdcall StrToKeySchedule .EXPAND_256: stdcall Expand_Key_256 ; print the key schedule stdcall PrintKeySchedule invoke SetDlgItemText,[hwnddlg],IDC_OUTPUT,bfDisplay ; print the key - null terminate bfDisplay at 71 bytes lea esi,[bfDisplay] mov [esi+71],byte 0 invoke SetDlgItemText,[hwnddlg],IDC_KEY,bfDisplay jmp .done .RESET: invoke SetDlgItemText,[hwnddlg],IDC_INPUT,szInputMessage invoke SetDlgItemText,[hwnddlg],IDC_KEY,szKeyMessage invoke SetDlgItemText,[hwnddlg],IDC_OUTPUT,"" jmp .done .wmclose: invoke EndDialog,[hwnddlg],0 .done: mov eax,1 .quit: ret endp ; ------------------------------------------------------------------------------------- proc Expand_Key_256 ; Key = 256 bits = 32 bytes ; Key Schedule = 33 x 128 bit sub-keys ; Total storage (key + key schedule) = 560 bytes ; generate pre-keys w(i) lea esi,[KeySchedule] lea edi,[KeySchedule] ; the first 32 bytes of the key schedule store the input key add edi,32 xor ecx,ecx .EXPAND: ; w[i] = w[i-8] xor w[i-5] xor w[i-3] xor w[i-1] xor 0x9e3779b9 xor i ; w[i] = ROTL(w[i],11) ; w[i] is a 32 bit word mov eax,[esi] ; xor w[i-5] xor eax,[esi+12] ; xor w[i-3] xor eax,[esi+20] ; xor w[i-1] xor eax,[esi+28] ; xor 0x9e3779b9 xor eax,0x9e3779b9 ; xor i xor eax,ecx ; rotl(w[i],11) rol eax,11 mov [edi],eax add edi,4 add esi,4 inc cx cmp cx,132 jl .EXPAND ; transform w(i) into words of round key k(i) by applying the s-boxes lea edi,[KeySchedule] ; the first 32 bytes are the input key - skip these add edi,32 xor ecx,ecx ; the 8 s-boxes are stored in the one array, each taking up 16 bytes ; ESI points to the relevant s-box .SUB: .S3: lea esi,[S_Box] add esi,48 stdcall Substitute_Block add edi,16 .S2: lea esi,[S_Box] add esi,32 stdcall Substitute_Block add edi,16 .S1: lea esi,[S_Box] add esi,16 stdcall Substitute_Block add edi,16 .S0: lea esi,[S_Box] stdcall Substitute_Block add edi,16 .S7: lea esi,[S_Box] add esi,112 stdcall Substitute_Block add edi,16 .S6: lea esi,[S_Box] add esi,96 stdcall Substitute_Block add edi,16 .S5: lea esi,[S_Box] add esi,80 stdcall Substitute_Block add edi,16 .S4: lea esi,[S_Box] add esi,64 stdcall Substitute_Block add edi,16 inc cx cmp cx,4 jl .SUB ; s-box 3 lea esi,[S_Box] add esi,48 stdcall Substitute_Block ret endp ; ------------------------------------------------------------------------------------- proc EncryptBlock uses esi edi ; Key = 256 bits = 32 bytes ; Key Schedule = 33 x 128 bit sub-keys ; Total storage (key + key schedule) = 560 bytes ; for round = 0 to 31 - XOR the sub-key[round] into the 128 bit data block ; then apply the s-boxes ; then apply the linear transform - except for the last round ; for the last round - XOR the sub-key[32] into the data block lea edi,[InputBlock] ; ECX gives the offset in bytes from the start of the key schedule array ; the first 32 bytes store the 256 bit input key - skip these bytes mov ecx,32 .ROUNDS: ; 32 rounds lea esi,[KeySchedule] add esi,ecx ; xor the sub-key into the cipher text mov eax,[esi] xor [edi],eax mov eax,[esi+4] xor [edi+4],eax mov eax,[esi+8] xor [edi+8],eax mov eax,[esi+12] xor [edi+12],eax ; apply the S-box lea esi,[S_Box] ; calculate the s-box number = (ECX-32)/16 mod 8 mov eax,ecx sub eax,32 shr eax,4 and eax,0x7 ; multiply by 16 to get the byte offset from the start of the s-box array shl eax,4 add esi,eax stdcall Substitute_Block add ecx,16 ; the sub-keys are indexed from 0 to 32 ; the last sub-key (32) is at the byte offset of 544 in the key schedule ; at ECX = 544, skip the linear transform and do the final sub-key XOR cmp ecx,544 je .FINAL stdcall Transform ; store the round output ; EDX gives the byte offset from the start of Roundsx32 mov edx,ecx sub edx,48 stdcall StoreRoundOutput jmp .ROUNDS .FINAL: ; XOR the last sub-key into the data block ; the index of the last sub-key is 32 lea esi,[KeySchedule] add esi,ecx mov eax,[esi] xor [edi],eax mov eax,[esi+4] xor [edi+4],eax mov eax,[esi+8] xor [edi+8],eax mov eax,[esi+12] xor [edi+12],eax ; store the round output ; EDX gives the byte offset from the start of Roundsx32 mov edx,31 shl edx,4 stdcall StoreRoundOutput ret endp ; ------------------------------------------------------------------------------------- proc Transform ; apply the linear transform to the cipher text block ; address of the block is passed in EDI ; x0 = rotl(x0,13) mov eax,[edi] rol eax,13 mov [edi],eax ; x2 = rotl(x2,3) mov eax,[edi+8] rol eax,3 mov [edi+8],eax ; x1 = x1 xor x0 xor x2 ; x2 already in EAX xor eax,[edi] xor eax,[edi+4] mov [edi+4],eax ; x3 = x3 xor x2 xor shl(x0,3) mov eax,[edi] shl eax,3 xor eax,[edi+8] xor eax,[edi+12] mov [edi+12],eax ; x1 = rotl(x1,1) mov eax,[edi+4] rol eax,1 mov [edi+4],eax ; x3 = rotl(x3,7) mov eax,[edi+12] rol eax,7 mov [edi+12],eax ; x0 = x0 xor x1 xor x3 ; x3 already in EAX xor eax,[edi] xor eax,[edi+4] mov [edi],eax ; x2 = x2 xor x3 xor shl(x1,7) mov eax,[edi+4] shl eax,7 xor eax,[edi+8] xor eax,[edi+12] mov [edi+8],eax ; x0 = rotl(x0,5) mov eax,[edi] rol eax,5 mov [edi],eax ; x2 = rotl(x2,22) mov eax,[edi+8] rol eax,22 mov [edi+8],eax ret endp ; ------------------------------------------------------------------------------------- proc DecryptBlock uses esi edi ; Key = 256 bits = 32 bytes ; Key Schedule = 33 x 128 bit sub-keys ; Total storage (key + key schedule) = 560 bytes ; the sub-keys are indexed from 0 to 32 ; for the first round - XOR the 128 bit sub-key[32] into the data block ; then apply the inverse s-box and XOR the sub-key[31] into the data block ; for the next 31 rounds: ; (i) apply the inverse linear transform ; (ii) apply the inverse s-box ; (iii) XOR the sub-key into the data block locals round db 0 endl lea edi,[InputBlock] ; ECX gives the offset in bytes from the start of the key schedule array ; the first 32 bytes store the 256 bit input key - skip these bytes ; so start from the last sub-key at (32 + (32 x 16)) = 544 bytes mov ecx,544 lea esi,[KeySchedule] add esi,ecx ; XOR this sub-key into the data block mov eax,[esi] xor [edi],eax mov eax,[esi+4] xor [edi+4],eax mov eax,[esi+8] xor [edi+8],eax mov eax,[esi+12] xor [edi+12],eax sub ecx,16 jmp .ISUB .ROUNDS: ; 31 rounds ; apply the inverse linear transform stdcall InverseTransform .ISUB: ; apply the inverse S-box lea esi,[IS_Box] ; calculate the inverse s-box number = (ECX-32)/16 mod 8 mov eax,ecx sub eax,32 shr eax,4 and eax,0x7 ; multiply by 16 to get the byte offset from the start of the inverse s-box array shl eax,4 add esi,eax stdcall Substitute_Block ; xor the sub-key into the cipher text lea esi,[KeySchedule] add esi,ecx mov eax,[esi] xor [edi],eax mov eax,[esi+4] xor [edi+4],eax mov eax,[esi+8] xor [edi+8],eax mov eax,[esi+12] xor [edi+12],eax ; store the round output ; EDX gives the byte offset from the start of Roundsx32 xor edx,edx mov dl,[round] shl edx,4 stdcall StoreRoundOutput inc byte [round] ; the first sub-key is at a byte offset of 32 from the start of the key schedule ; exit the loop when ECX becomes less then 32 sub ecx,16 cmp ecx,32 jge .ROUNDS ret endp ; ------------------------------------------------------------------------------------- proc InverseTransform ; apply the inverse linear transform to the cipher text block ; address of the block is passed in EDI ; x2 = rotr(x2,22) mov eax,[edi+8] ror eax,22 mov [edi+8],eax ; x0 = rotr(x0,5) mov eax,[edi] ror eax,5 mov [edi],eax ; x2 = x2 xor x3 xor shl(x1,7) mov eax,[edi+4] shl eax,7 xor eax,[edi+8] xor eax,[edi+12] mov [edi+8],eax ; x0 = x0 xor x1 xor x3 mov eax,[edi+12] xor eax,[edi] xor eax,[edi+4] mov [edi],eax ; x3 = rotr(x3,7) mov eax,[edi+12] ror eax,7 mov [edi+12],eax ; x1 = rotr(x1,1) mov eax,[edi+4] ror eax,1 mov [edi+4],eax ; x3 = x3 xor x2 xor shl(x0,3) mov eax,[edi] shl eax,3 xor eax,[edi+8] xor eax,[edi+12] mov [edi+12],eax ; x1 = x1 xor x0 xor x2 mov eax,[edi+8] xor eax,[edi] xor eax,[edi+4] mov [edi+4],eax ; x2 = rotr(x2,3) mov eax,[edi+8] ror eax,3 mov [edi+8],eax ; x0 = rotr(x0,13) mov eax,[edi] ror eax,13 mov [edi],eax ret endp ; ------------------------------------------------------------------------------------- proc Substitute_Block uses ecx edx ; use the S-Box to substitute each bit in a 128 bit block ; the address of the block is passed in EDI ; the address of the S-Box is passed in ESI ; this proc is used for both the s-boxes and inverse s-boxes locals bitslice db 0 endl xor ecx,ecx ; the bitslice is constructed by taking the lsb from each of the 32 bit words ; each of the 4 input words are then right shifted by 1 bit ; the 4-bit bitslice is put through the s-box ; the 4 output bits are used to set the msb of each of the 4 input words ; after 32 iterations all of the input bits will have been substituted .SUB: ; construct the bit-slice for input into the s-box ; bit 3: mov al,[edi+12] and al,1 shl al,3 mov [bitslice],al ; bit 2: mov al,[edi+8] and al,1 shl al,2 or [bitslice],al ; bit 1: mov al,[edi+4] and al,1 shl al,1 or [bitslice],al ; bit 0: mov al,[edi] and al,1 or [bitslice],al ; right shift the bits in each of the 4 x 32 bit words of the input block mov eax,[edi] shr eax,1 mov [edi],eax mov eax,[edi+4] shr eax,1 mov [edi+4],eax mov eax,[edi+8] shr eax,1 mov [edi+8],eax mov eax,[edi+12] shr eax,1 mov [edi+12],eax ; substitute the bits xor eax,eax mov al,[bitslice] mov dl,[esi+eax] ; insert the output bits back into the input block ; bit 0: mov al,dl and al,1 shl eax,31 or [edi],eax ; bit 1: mov al,dl and al,2 shl eax,30 or [edi+4],eax ; bit 2: mov al,dl and al,4 shl eax,29 or [edi+8],eax ; bit 3: mov al,dl and al,8 shl eax,28 or [edi+12],eax inc cx cmp cx,32 jl .SUB ret endp ; ------------------------------------------------------------------------------------- proc StoreRoundOutput uses esi edi ; store the 16 byte output from each of the 32 rounds ; EDX gives the offset in bytes from the start of [Roundsx32] lea esi,[InputBlock] lea edi,[Roundsx32] add edi,edx mov eax,[esi] mov [edi],eax mov eax,[esi+4] mov [edi+4],eax mov eax,[esi+8] mov [edi+8],eax mov eax,[esi+12] mov [edi+12],eax ret endp ; ------------------------------------------------------------------------------------- proc Print_Rounds uses esi edi ; print the data in the Roundsx32 buffer to the bfDisplay string locals count db 0 endl lea esi,[Roundsx32] lea edi,[bfDisplay] mov ecx,512 .PRINT: ; byte to digits mov al,[esi] stdcall ByteToDigits ; add the 2 digits to the string mov [edi],ah inc edi mov [edi],al inc edi dec cx jcxz .DONE ; update ESI inc esi ; add spaces and CRLF to the output string ; a space after every 8 chars ; a CRLF after every 32 chars add [count],2 test [count],7 jne .PRINT test [count],31 je .ADD_CRLF mov [edi],byte 32 inc edi jmp .PRINT .ADD_CRLF: mov [edi],byte 13 inc edi mov [edi],byte 10 inc edi jmp .PRINT .DONE: ; zero terminate the string mov [edi],byte 0 ret endp ; ------------------------------------------------------------------------------------- proc StrToBuffer ; load the buffer in memory from the string bfDisplay ; address of buffer is passed in EAX ; max number of hex digits to read passed in EDX locals count dw 0 endl lea esi,[bfDisplay] mov edi,eax mov ecx,0 .GET_CHARS: mov al,byte [esi] cmp al,0 je .DONE stdcall HexDigitToValue ; just skip and get the next char if not a hex digit cmp eax,0xffff je .NEXT ; AL contains a nibble, determine where to place it in the dest byte ; if count is even, the nibble is mapped to the lower 4 bits in dest ; if count is odd, the nibble is mapped to the upper 4 bits in dest inc word [count] ; even if count AND 1 == 0 test [count],1 jz .EVEN .ODD: inc cx shl al,4 mov [edi],al jmp .NEXT .EVEN: inc cx or [edi],al inc edi .NEXT: inc esi cmp cx,dx jge .DONE jmp .GET_CHARS .DONE: ; CX contains the number of hex digits that were read, return in EDX mov edx,ecx ret endp ; ------------------------------------------------------------------------------------- proc ResetInputBlock uses edi ecx ; clear the 16 byte input block buffer lea edi,[InputBlock] mov cx,0 .CLEAR: mov [edi],byte 0 inc edi inc cx cmp cx,16 jl .CLEAR ret endp ; ------------------------------------------------------------------------------------- proc ResetKeySchedule uses edi ecx ; clear the 560 byte key schedule buffer lea edi,[KeySchedule] mov cx,0 .CLEAR: mov [edi],byte 0 inc edi inc cx cmp cx,560 jl .CLEAR ret endp ; ------------------------------------------------------------------------------------- proc StrToInputBlock ; load the input block buffer from the input string ; input string is a hex value ; zero the input text value stdcall ResetInputBlock stdcall StrToHexStr lea eax,[InputBlock] ; read a max of 32 hex digits mov edx,32 stdcall StrToBuffer ret endp ; ------------------------------------------------------------------------------------- proc StrToKeySchedule ; load the key schedule buffer from the input string ; input string is a hex value ; zero the cipher text value stdcall ResetKeySchedule stdcall StrToHexStr lea eax,[KeySchedule] ; read a max of 64 hex digits: 64 x 4 = 256 bits mov edx,64 stdcall StrToBuffer ret endp ; ------------------------------------------------------------------------------------- proc PrintInputBlock ; print the 16 byte input block to the display string lea eax,[InputBlock] xor edx,edx mov dx,16 stdcall PrintBuffer ret endp ; ------------------------------------------------------------------------------------- proc PrintKeySchedule ; print the key schedule to the display string lea eax,[KeySchedule] xor edx,edx mov dx,560 stdcall PrintBuffer ret endp ; ------------------------------------------------------------------------------------- proc PrintBuffer uses esi edi ; print the hex value in the buffer to the bfDisplay string ; address of buffer passed in EAX ; number of bytes in buffer passed in EDX locals count db 0 endl mov esi,eax lea edi,[bfDisplay] mov ecx,edx .PRINT: ; byte to digits mov al,[esi] stdcall ByteToDigits ; add the 2 digits to the string mov [edi],ah inc edi mov [edi],al inc edi dec cx jcxz .DONE ; update ESI inc esi ; add spaces and CRLF to the output string ; a space after every 8 chars ; a CRLF after every 64 chars add [count],2 test [count],7 jne .PRINT test [count],63 je .ADD_CRLF mov [edi],byte 32 inc edi jmp .PRINT .ADD_CRLF: mov [edi],byte 13 inc edi mov [edi],byte 10 inc edi jmp .PRINT .DONE: ; zero terminate the string mov [edi],byte 0 ret endp ; ------------------------------------------------------------------------------------- proc StrToHexStr uses esi edi ; filter out any non hex digit characters from the bfDisplay string lea esi,[bfDisplay] lea edi,[bfDisplay] .FILTER: mov al,[esi] cmp al,0 je .DONE cmp al,48 jl .NOT_HEX cmp al,57 jle .COPY cmp al,65 jl .NOT_HEX cmp al,70 jle .COPY cmp al,97 jl .NOT_HEX cmp al,102 jg .NOT_HEX .COPY: mov [edi],al inc edi .NOT_HEX: inc esi jmp .FILTER .DONE: ; null terminate the new string mov [edi],byte 0 ret endp ; ------------------------------------------------------------------------------------- proc HexDigitToValue ; convert a hex digit to a 4 bit value ; input in AL - output in AL ; is AL in the range: 48 - 57 (0 - 9) ? cmp al,48 jl .NOT_HEX cmp al,57 jg .A_Z sub al,48 jmp .DONE .A_Z: ; is AL in the range: 65 - 70 (A - B) ? cmp al,65 jl .NOT_HEX cmp al,70 jg .a_z sub al,55 jmp .DONE .a_z: ; is AL in the range: 97 - 102 (a - b) ? cmp al,97 jl .NOT_HEX cmp al,102 jg .NOT_HEX sub al,87 jmp .DONE .NOT_HEX: ; set EAX to 0xffff if input is not a valid hex digit mov eax,0xffff .DONE: ret endp ; ------------------------------------------------------------------------------------- proc ByteToDigits ; convert 1 byte to 2 hex digits ; input in AL - output in AX ; copy AL to AH mov ah,al ; AH: get the upper 4 bits of the byte shr ah,4 ; nibble to hex digit add ah,48 cmp ah,57 jle .NEXT add ah,7 .NEXT: ; AL: get the lower 4 bits of the byte and al,0xf ; nibble to hex digit add al,48 cmp al,57 jle .DONE add al,7 .DONE: ; output is in AX ret endp ; ------------------------------------------------------------------------------------- proc onNullInput ; if the bfDisplay string is empty, set it to "0" lea edi,[bfDisplay] cmp [edi],byte 0 jne .DONE mov [edi],byte 48 mov [edi+1],byte 0 .DONE: ret endp ; ------------------------------------------------------------------------------------- section '.sbox' readable writeable S_Box db\ 3,8,15,1,10,6,5,11,14,13,4,2,7,0,9,12,\ 15,12,2,7,9,0,5,10,1,11,14,8,6,13,3,4,\ 8,6,7,9,3,12,10,15,13,1,14,4,0,11,5,2,\ 0,15,11,8,12,9,6,3,13,1,2,4,10,7,5,14,\ 1,15,8,3,12,0,11,6,2,5,4,10,9,14,7,13,\ 15,5,2,11,4,10,9,12,0,3,14,8,13,6,7,1,\ 7,2,12,5,8,4,6,11,14,9,1,15,13,3,10,0,\ 1,13,15,0,14,8,2,11,7,4,12,10,9,3,5,6 IS_Box db\ 13,3,11,0,10,6,5,12,1,14,4,7,15,9,8,2,\ 5,8,2,14,15,6,12,3,11,4,7,9,1,13,10,0,\ 12,9,15,4,11,14,1,2,0,3,6,13,5,8,10,7,\ 0,9,10,7,11,14,6,13,3,5,12,2,4,8,15,1,\ 5,0,8,3,10,9,7,14,2,12,11,6,4,15,13,1,\ 8,15,2,9,4,1,13,14,11,6,5,3,7,12,10,0,\ 15,10,1,13,5,3,6,0,4,9,14,7,2,12,8,11,\ 3,0,6,13,9,14,15,8,5,12,11,7,10,1,4,2 ; ------------------------------------------------------------------------------------- section '.idata' import data readable writeable library kernel,'KERNEL32.DLL',\ user,'USER32.DLL' import kernel,\ GetModuleHandle,'GetModuleHandleA',\ ExitProcess,'ExitProcess' import user,\ DialogBoxParam,'DialogBoxParamA',\ SetDlgItemText,'SetDlgItemTextA',\ GetDlgItemText,'GetDlgItemTextA',\ EndDialog,'EndDialog' ; ------------------------------------------------------------------------------------- section '.data' readable writeable bfDisplay rb 1600 KeySchedule rb 560 InputBlock rb 16 Roundsx32 rb 512 szInputMessage db "Enter 16 bytes of data:",0 szKeyMessage db "Enter 32 bytes of key data:",0 ; ------------------------------------------------------------------------------------- section '.rc' resource data readable directory RT_DIALOG,dialogs resource dialogs,IDD_THE_DIALOG,LANG_ENGLISH+SUBLANG_DEFAULT,the_dialog dialog the_dialog,\ 'FASM - Serpent One Encryption Algorithm',50,50,360,400,\ DS_MODALFRAME+WS_MINIMIZEBOX+WS_POPUP+WS_VISIBLE+WS_CAPTION+WS_SYSMENU,\ 0,0,"Lucida Console",11 dialogitem 'BUTTON','Key',-1,7,5,346,40,BS_GROUPBOX+WS_VISIBLE,0 dialogitem 'BUTTON','Input',-1,7,50,346,30,BS_GROUPBOX+WS_VISIBLE,0 dialogitem 'BUTTON',"Output",-1,7,86,346,288,BS_GROUPBOX+WS_VISIBLE,0 dialogitem 'EDIT',0,IDC_KEY,13,16,335,22,ES_MULTILINE+ES_AUTOVSCROLL+ES_WANTRETURN+WS_VSCROLL+WS_BORDER+WS_VISIBLE,0 dialogitem 'EDIT',0,IDC_INPUT,13,61,335,12,ES_MULTILINE+ES_AUTOVSCROLL+ES_WANTRETURN+WS_VSCROLL+WS_BORDER+WS_VISIBLE,0 dialogitem 'EDIT',0,IDC_OUTPUT,13,97,335,268,ES_MULTILINE+ES_AUTOVSCROLL+ES_WANTRETURN+WS_VSCROLL+WS_BORDER+WS_VISIBLE,0 dialogitem 'BUTTON',"Encrypt",IDC_BTN_ENCRYPT,7,380,50,14,BS_PUSHBUTTON+WS_VISIBLE,0 dialogitem 'BUTTON',"Decrypt",IDC_BTN_DECRYPT,59,380,50,14,BS_PUSHBUTTON+WS_VISIBLE,0 dialogitem 'BUTTON',"Set Key",IDC_BTN_KEY_256,111,380,50,14,BS_PUSHBUTTON+WS_VISIBLE,0 dialogitem 'BUTTON',"Reset",IDC_BTN_RESET,163,380,50,14,BS_PUSHBUTTON+WS_VISIBLE,0 enddialog ; 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