I am new to assembly programming, currently taking online course.
Original problem was to count number of paths from top left corner to bottom right corner. But I found a good solution to that here:
https://www.geeksforgeeks.org/count-possible-paths-top-left-bottom-right-nxm-matrix/
Based on the combinatorics solution I should be able to find all paths in a binary manner.
First question, do you know a faster way to count paths?
Searched for the solution to print all paths in:
https://www.geeksforgeeks.org/print-all-possible-paths-from-top-left-to-bottom-right-of-a-mxn-matrix/
But did not notice any using the binary approach with seemed adequate for assembly.
Searching a bit more online I found:
https://www.baeldung.com/cs/generate-k-combinations
Revolving door algorithm was well detailed, and I calculate it to be O (number of combinations) * O (width or height of matrix (for printing) -1) * O (branching loops) on time complexity and O (width or height + 1) on space. Second question is this a correct assumption? If not, what is the correct complexity? Is it faster than the other solutions posted for finding all paths to this problem? Those are stated to be O(2^(width*height))
Third question: Who wrote this algorithm? Where can I find more like it?
And lastly, I will post my newbie 32-bit assembly pasta code for fasm, should work on matrixes larger than 3 x 3 smaller than 32 x 32(not recommended to go above 16 x 16 that is already a lot of combinations and only omitting the print instructions), any improvements are more than welcome. Thank you.
format PE console
entry start
include 'win32a.inc'
; ===============================================
struct MAT
h db ? ; X coordinate.
w db ? ; Y coordinate.
ends
; ===============================================
section '.bss' readable writeable
; Declare the uninitialized table in memory:
path_matrix MAT ?
count dd ?
indices db path_matrix.w - 1 dup ?
end_indices:
; ===============================================
section '.text' code readable executable
start:
call read_hex
mov [path_matrix.h], al ; down will be 0
call read_hex
mov [path_matrix.w], al ; right will be 1
dec eax
mov ecx, eax
initialize:
mov ebx, ecx
dec ebx
mov byte[indices+ecx], bl
loop initialize
movzx ebx, [path_matrix.h]
dec ebx
add ebx, eax
mov byte[indices+eax+1], bl
mov eax, ebx
print_combination:
inc [count]
movzx ebx, [end_indices - indices]
dec ebx
xor eax, eax
print_loop:
xor esi, esi
inc esi
mov cl, byte[indices + ebx ]
shl esi, cl
xor eax, esi
dec ebx
cmp ebx, 0
jnz print_loop
call print_eax_binary
branch:
lea edi, [indices +1]
movzx eax, [path_matrix.w] ; check if withd is eaven, if true matrix is odd (w -1)
shr eax, 1
jnc odd
eaven:
movzx eax, byte[edi]
cmp eax, 0
jle eaven_indice
dec eax
mov byte[edi], al
jmp print_combination
eaven_indice:
inc edi
try_to_increase:
movzx ebx, byte[edi]
inc ebx
cmp bl, [edi+1]
jl increase
lea ecx, [edi-indices+1]
cmp cl, [path_matrix.w]
jl increase_indice
jmp fin
increase:
mov byte[edi], bl
dec ebx
mov byte[edi-1], bl
jmp print_combination
odd:
movzx eax, byte[edi]
inc eax
cmp al, [edi+1]
jge increase_indice
mov byte[edi], al
jmp print_combination
increase_indice:
inc edi
try_decrease:
lea eax, [edi - indices]
cmp byte[edi], al
jl eaven_indice
decrease:
movzx ebx, byte[edi-1]
mov byte[edi], bl
sub eax, 2
mov byte[edi-1], al
jmp print_combination
fin:
mov eax, [count]
call print_eax
; Exit the process:
push 0
call [ExitProcess]
include 'training.inc'
The solution is not binary because paths from the top or left can overlap - creating duplicates. Working backward the solution is additive - sum of paths from left and paths from top.
Which leads to a simple solution:
_m = 16
_n = 16
ddp rq _m
; initialize first position
mov [ddp + (_n-1)*8], 1
mov ecx, _m
outer:
push rcx
mov ecx, _n
xor eax, eax
inner:
add rax, [ddp + (rcx-1)*8]
mov [ddp + (rcx-1)*8], rax
loop inner
pop rcx
loop outer
There is a closed form solution, but it's a little tricky to reduce the expression in such a way as to cover a large number of inputs:
; number of paths from one corner to opposite diagonal corner of M x N matrix
; RCX : N, RDX : M
numberOfPaths:
mov eax,1
mov r9,1
lea r8,[rcx+rdx-1]
jmp .try
.more:
mul rcx
inc ecx
div r9
inc r9
.try:
cmp rcx,r8
jc .more
retn
It can be reduced further with more code.
Related
I am trying to implement bubble sort in assembly. Here is my code. I keep getting segmentation fault. I have a function down below. I have been trying to figure this out but I couldn't find a compiler for x86 and I have been checking with my code to check what is wrong but to no avail.
here is my function code:
bubble:
push ebp
mov ebp, esp
push ecx
push edx
push edi
push esi
push ebx
push eax
mov eax, 0
mov ecx, [ebp+12] ; number of elements in the array
mov edx, [ebp+8]; address of array
mov edi, 0
mov esi, 0
dec ecx; n - 1
mov esi, 1
sortOuter:
cmp esi, 0
jg sort
sort:
mov esi, 0 ;count
check:
cmp edi, ecx ; i < n - 1
jl sortInner
sortInner:
mov ebx, [edx+edi*4] ; mov array[i+1] to ebx
cmp [edx], ebx ; cmp array[i] to array[i+1]
jle noswap
swap:
mov eax, ebx ; mov array[i+1] to eax
mov ebx, [edx] ; mov array[i] to array[i+1]
mov [edx], eax ; mov array[i+1] to array[i]
inc esi ; count++
noswap:
inc edi ; i++
jmp check
jmp sortOuter
done:
call print_nl
pop ebx
pop esi
pop edi
pop edx
pop ecx
mov esp, ebp
pop ebp
ret
The segmentation error comes from the infinite loop that you have created with
check:
cmp edi, ecx ; i < n - 1
jl sortInner
sortInner:
If EDI is less than ECX you jump to sortInner, but if it isn't you fall-through into sortInner. No matter, you always end up running the code at sortInner and because the memory addresses that the code uses keep growing, at some point you will be trying to read memory that you don't have access to, hence the segmentation error.
Now if you were to correct this, then there's a second infinite loop waiting.
sortOuter:
cmp esi, 0
jg sort
sort:
Other errors include:
Resetting ESI instead of EDI at the start of the inner loop
Not swapping elements at all but always writing the smallest value in the first array element
Forgetting to restore the register EAX
This is a working BubbleSort. Don't just copy it, but find out how it functions!
In an array with N elements we have to do N-1 comparisons at most (to have the greatest value bubble towards the rear).
Because the InnerLoop uses a counter that gets initialized from the ever decreasing OuterLoop counter, with each iteration of the OuterLoop the portion of the array that is processed in the InnerLoop gets smaller. The portion of the array that we then no longer process contains the greatest elements that have bubbled towards the end of the array, hence the name BubbleSort.
Provisions have been made for an empty array or an array that has but 1 element. Always include some code for the special cases!
bubble:
push ebp
mov ebp, esp
push ...
mov ecx, [ebp + 12] ; Number of elements in the array
sub ecx, 1 ; First time we do n = N-1
jbe Done ; Array is empty or has but 1 element
OuterLoop:
mov edx, ecx ; Current value of the OuterLoop counter (n)
mov esi, [ebp + 8] ; Address of array
InnerLoop:
mov eax, [esi]
mov ebx, [esi + 4]
cmp eax, ebx
jng NoSwap
mov [esi], ebx
mov [esi + 4], eax
NoSwap:
add esi, 4
dec edx
jnz InnerLoop
dec ecx ; Next times we do n = n-1
jnz OuterLoop
Done:
pop ...
pop ebp
ret
My goal in this code is to find the smallest number in the list. I used bubble sort method in this case; unfortunately, the code is not giving me the smallest/minimum number. Please take a look, Thanks:
include irvine32.inc
.data
input byte 100 dup(0)
stringinput byte "Enter any string: ",0
totallength byte "The total length is: ",0
minimum byte "The minimum value is: ",0
.code
stringLength proc
push ebp
mov ebp, esp
push ebx
push ecx
mov eax, 0
mov ebx, [ebp+8]
L1:
mov ecx, [ebx] ;you can use ecx, cx, ch, cl
cmp ecx, 0 ;you can use ecx, cx, ch, cl
JE L2
add ebx, 1
add eax, 1
jmp L1
L2:
pop ecx
pop ebx
mov ebp, esp
pop ebp
ret 4
stringLength endp
BubbleSort PROC uses ECX
push edx
xor ecx,ecx
mov ecx, 50
OUTER_LOOP:
push ecx
xor ecx,ecx
mov ecx,14
mov esi, OFFSET input
COMPARE:
xor ebx,ebx
xor edx,edx
mov bl, byte ptr ds:[esi]
mov dl, byte ptr ds:[esi+1]
cmp bl,dl
jg SWAP1
CONTINUE:
add esi,2
loop COMPARE
mov esi, OFFSET input
pop ecx
loop OUTER_LOOP
jmp FINISHED
SWAP1:
xchg bl,dl
mov byte ptr ds:[esi+1],dl
mov byte ptr ds:[esi],bl
jmp CONTINUE
FINISHED:
pop edx
ret 4
BubbleSort ENDP
main proc
call clrscr
mov edx, offset stringinput
call writeString
mov edx, offset input
call writeString
call stringLength
mov edx, offset input
mov ecx, sizeof input
call readstring
call crlf
mov edx,offset totallength
call writestring
call writedec
call crlf
mov edx, offset minimum
call crlf
call writeString
push offset input
call BubbleSort
mov edx, offset input
call writeString
call crlf
exit
main endp
end main
I haven't looked over your code, because sorting is an over complicated method for what you want to do. Not only that, but most of us don't pay too much attention to uncommented code. Just takes to long to figure out what you're trying to do.
Simply iterate through the entire list and start with 255 (FFH) in AL let's say. Each time you come across a number that is smaller than the one in AL, then replace it with that value and then when loop is finished, AL will have the lowest value.
If you need to know where it is in the list, you could maybe use AH which would be the difference between start address and current address. Knowledge of the instruction set is essential as finding the length of the string can be simplified by;
mov di, input ; Point to beginning of buffer
mov cx, -1 ; for a maximum of 65535 characters
xor al, al ; Looking for NULL
rep scasb
neg cx
dec cx ; CX = length of string.
Remember, ES needs to point to #DATA
I am looking for a way to print an integer in assembler (the compiler I am using is NASM on Linux), however, after doing some research, I have not been able to find a truly viable solution. I was able to find a description for a basic algorithm to serve this purpose, and based on that I developed this code:
global _start
section .bss
digit: resb 16
count: resb 16
i: resb 16
section .data
section .text
_start:
mov dword[i], 108eh ; i = 4238
mov dword[count], 1
L01:
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[digit], edx
add dword[digit], 30h ; add 48 to digit to make it an ASCII char
call write_digit
inc dword[count]
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[i], eax
cmp dword[i], 0Ah
jg L01
add dword[i], 48 ; add 48 to i to make it an ASCII char
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, i ; store *address* of i into ecx
mov edx, 16 ; byte size of 16
int 80h
jmp exit
exit:
mov eax, 01h ; exit()
xor ebx, ebx ; errno
int 80h
write_digit:
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, digit ; store *address* of digit into ecx
mov edx, 16 ; byte size of 16
int 80h
ret
C# version of what I want to achieve (for clarity):
static string int2string(int i)
{
Stack<char> stack = new Stack<char>();
string s = "";
do
{
stack.Push((char)((i % 10) + 48));
i = i / 10;
} while (i > 10);
stack.Push((char)(i + 48));
foreach (char c in stack)
{
s += c;
}
return s;
}
The issue is that it outputs the characters in reverse, so for 4238, the output is 8324. At first, I thought that I could use the x86 stack to solve this problem, push the digits in, and pop them out and print them at the end, however when I tried implementing that feature, it flopped and I could no longer get an output.
As a result, I am a little bit perplexed about how I can implement a stack in to this algorithm in order to accomplish my goal, aka printing an integer. I would also be interested in a simpler/better solution if one is available (as it's one of my first assembler programs).
One approach is to use recursion. In this case you divide the number by 10 (getting a quotient and a remainder) and then call yourself with the quotient as the number to display; and then display the digit corresponding to the remainder.
An example of this would be:
;Input
; eax = number to display
section .data
const10: dd 10
section .text
printNumber:
push eax
push edx
xor edx,edx ;edx:eax = number
div dword [const10] ;eax = quotient, edx = remainder
test eax,eax ;Is quotient zero?
je .l1 ; yes, don't display it
call printNumber ;Display the quotient
.l1:
lea eax,[edx+'0']
call printCharacter ;Display the remainder
pop edx
pop eax
ret
Another approach is to avoid recursion by changing the divisor. An example of this would be:
;Input
; eax = number to display
section .data
divisorTable:
dd 1000000000
dd 100000000
dd 10000000
dd 1000000
dd 100000
dd 10000
dd 1000
dd 100
dd 10
dd 1
dd 0
section .text
printNumber:
push eax
push ebx
push edx
mov ebx,divisorTable
.nextDigit:
xor edx,edx ;edx:eax = number
div dword [ebx] ;eax = quotient, edx = remainder
add eax,'0'
call printCharacter ;Display the quotient
mov eax,edx ;eax = remainder
add ebx,4 ;ebx = address of next divisor
cmp dword [ebx],0 ;Have all divisors been done?
jne .nextDigit
pop edx
pop ebx
pop eax
ret
This example doesn't suppress leading zeros, but that would be easy to add.
I think that maybe implementing a stack is not the best way to do this (and I really think you could figure out how to do that, saying as how pop is just a mov and a decrement of sp, so you can really set up a stack anywhere you like by just allocating memory for it and setting one of your registers as your new 'stack pointer').
I think this code could be made clearer and more modular if you actually allocated memory for a c-style null delimited string, then create a function to convert the int to string, by the same algorithm you use, then pass the result to another function capable of printing those strings. It will avoid some of the spaghetti code syndrome you are suffering from, and fix your problem to boot. If you want me to demonstrate, just ask, but if you wrote the thing above, I think you can figure out how with the more split up process.
; Input
; EAX = pointer to the int to convert
; EDI = address of the result
; Output:
; None
int_to_string:
xor ebx, ebx ; clear the ebx, I will use as counter for stack pushes
.push_chars:
xor edx, edx ; clear edx
mov ecx, 10 ; ecx is divisor, devide by 10
div ecx ; devide edx by ecx, result in eax remainder in edx
add edx, 0x30 ; add 0x30 to edx convert int => ascii
push edx ; push result to stack
inc ebx ; increment my stack push counter
test eax, eax ; is eax 0?
jnz .push_chars ; if eax not 0 repeat
.pop_chars:
pop eax ; pop result from stack into eax
stosb ; store contents of eax in at the address of num which is in EDI
dec ebx ; decrement my stack push counter
cmp ebx, 0 ; check if stack push counter is 0
jg .pop_chars ; not 0 repeat
mov eax, 0x0a
stosb ; add line feed
ret ; return to main
; eax = number to stringify/output
; edi = location of buffer
intToString:
push edx
push ecx
push edi
push ebp
mov ebp, esp
mov ecx, 10
.pushDigits:
xor edx, edx ; zero-extend eax
div ecx ; divide by 10; now edx = next digit
add edx, 30h ; decimal value + 30h => ascii digit
push edx ; push the whole dword, cause that's how x86 rolls
test eax, eax ; leading zeros suck
jnz .pushDigits
.popDigits:
pop eax
stosb ; don't write the whole dword, just the low byte
cmp esp, ebp ; if esp==ebp, we've popped all the digits
jne .popDigits
xor eax, eax ; add trailing nul
stosb
mov eax, edi
pop ebp
pop edi
pop ecx
pop edx
sub eax, edi ; return number of bytes written
ret
I need to translate the next pseudo-code to assembly. only need to read 5 numbers in array and compare with the last value. Params are taken well.
PSEUDO-CODE:
tmp= not L0
tmp = not ( not tmp and not L1)
tmp= tmp or L2
tmp = not (not tmp or not L3)
tmp= tmp and L4
if(tmp == L5 )
valido
sino
invalido
and its my "solution" :
EsLicenciaValida PROC licencia: dword
mov ebx, licencia
mov edx,[ebx]
mov ecx, edx
neg ecx
add ebx, 4
mov edx,[ebx]
not edx
not ecx
and ecx, edx
not ecx
add ebx, 4
mov edx,[ebx]
or ecx, edx
add ebx, 4
mov edx,[ebx]
not edx
not ecx
or ecx,edx
not ecx
add ebx, 4
mov edx,[ebx]
and ecx, edx
add ebx, 4
mov edx,[ebx]
cmp edx, ecx
jne invalido
valido:
mov eax, ecx
jmp salir
invalido:
mov eax, 0000h
salir:
ret
EsLicenciaValida ENDP
ECX register have the tmp value of the pseudo-code. EBX the value in actual position of array.
Some one knows whats wrong? Thanks for all =)
an array with values in hexadecimal: A-A-A-A-A-32
Why do you expect a sum is calculated?
Why do you expect the result to be 50?
The number in L5 is 10! (derived from the presented pseudocode)
Suppose that I have a 2D array, and I want to check whether one slot is adjacent and touching with another.
Suppose that the coordinates are in the 4-byte variables: OneX, OneY, TwoX, TwoY.
The solution I had for a while was that if you have the differences OneX - OneY and TwoX - TwoY and add them, if the result is either 1 or -1, then yes, the slots are adjacent and touching.
mov EBX,[oneX]
sub EBX,[oneY]
mov ECX,[twoX]
sub ECX,[twoY]
add EBX,ECX
; Compare EBX with 1 or -1.......
This almost works. But no - given a format (x,y), take: (3,3) and (0,1). They're clearly not adjacent nor touching, but the function will say they are.
The question at Get adjacent elements in a two-dimensional array? is somewhat useful, but it focuses on finding all adjacent matches, whereas I want to check for two specific slots instead.
The structure of my array is like this:
map: dd 'a','b','c','d' ; Double words just to make my life easier
Which is interpreted like
a b
c d
It's a square map.
There is no reason to Add them! And you are also substracting incorrect variables :).
You have to have two conditions OneX - TwoX and OneY - TwoY : both has to be 1, 0 or -1.
For example One is [4,5] and Two is [5,5] >= OneX - TwoX = -1 and OneY - TwoY = 0 => it is adjanced tile.
EDIT : For non-diagonal, there are two approaches :
a)One of condition must be 0 and the other one must be 1 or -1
b)Adding absolute value of OneX - TwoX and absolute value of OneY - TwoY must be 1
Here is some practical code, based on #libik answer:
Optimized for speed:
; eax = OneX-TwoX; ecx = OneY-TwoY
mov eax, [OneX]
mov ecx, [OneY]
sub eax, [TwoX]
sub ecx, [TwoY]
; eax = abs(eax); ecx=abs(ecx)
mov ebx, eax
mov edx, ecx
sar ebx, 31
sar edx, 31
xor eax, ebx
xor ecx, edx
sub eax, ebx
sub ecx, edx
; eax=abs(delta1)+abs(delta2); if eax = 1, jump to where needed
add eax, ecx
dec eax
lz .adjacent
Optimized for size:
; eax = abs(OneX-TwoX); ecx = abs(OneY-TwoY)
mov eax, [OneX]
mov ecx, [OneY]
sub eax, [TwoX]
jns #f
neg eax
##:
sub ecx, [TwoY]
jns #f
neg ecx
##:
; eax=abs(delta1)+abs(delta2); if eax = 1, jump to where needed
add eax, ecx
dec eax
lz .adjacent
Including diagonal cases
Replace add eax, ecx with or eax, ecx