An exercise on polyglossy: the same problem solved on multiple languages
Merge pull request #2 from Androthi/master
Integrated a 32 bit assembly HLA version into the repo.
2 files changed, 126 insertions, 10 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Added | hla/conjecture.hla | 116 | 116 | 0 |
| Modified | x86/PROGRAM.ASM | 20 | 10 | 10 |
diff --git a/hla/conjecture.hla b/hla/conjecture.hla @@ -0,0 +1,116 @@ +program conjecture; + + #includeonce( "stdlib.hhf" ) + + ?@nodisplay := true; + ?@noalignstack := true; + +// This program is a simple test for the following conjecture: + +// Let S: N -> N be the sum of the digits of a positive integer. +// For all A and B in N, S(A + B) = S(A) + S(B) - 9k, where k is an interger. + +// data segments +storage + max :uns32; + sumAB :uns32; + sumA :uns32; + sumB :uns32; + +static + err :boolean := false; + + +// input eax = number to sum +// returns sum in ebx +procedure sum_digits; + + // don't build a stack frame for this procedure + @nodisplay; @noframe; + +begin sum_digits; + + mov( 10, ecx ); + xor( ebx, ebx ); + while( eax ) do + xor( edx, edx ); + div( ecx ); + add( edx, ebx ); + endwhile; + ret; + +end sum_digits; + + + +begin conjecture; + + // read an unsigned integer from the user + forever + try + stdout.put ("enter an unsigned integer :>" ); + stdin.getu32(); + mov( eax, max ); + break; + anyexception; + stdout.newln(); + continue; + endtry; + endfor; + + + stdout.put ("Loading ..." nl ); + + for( xor(esi, esi); esi <= max; inc( esi ) ) do + + // outer loop esi = 0 to max iterations + + for( mov(esi, edi); edi <= max; inc( edi ) ) do + + // inner loop. edi = esi to max iterations + + // get S(a+b) + mov( esi, eax ); + add( edi, eax ); + call sum_digits; + mov( ebx, sumAB ); + + // get S(a) + mov( esi, eax ); + call sum_digits; + mov( ebx, sumA ); + + // get S(b) + mov( edi, eax ); + call sum_digits; + mov( ebx, sumB ); + + // get S(a+b) - S(a) - S(b) + mov( sumAB, eax ); + sub( sumA, eax ); + sub( sumB, eax ); + + // sign extend into edx + cdq(); + mov( 9, ecx ); + idiv( ecx ); // signed division of edx:eax + test( edx, edx ); // is remainder zero? + if( @nz ) then + // if remainder is not zero, the conjecture is false + stdout.put ("Conjecture is disproved, here is the counter example :", (type uns32 eax), nl ); + mov( true, err ); + break; + endif; + endfor; + + breakif( err ); + + endfor; + + // if we get here, we looped through all the values and the + // conjecture is proved. + if( !err ) then + stdout.put ("The conjecture is proved for all natural numbers smaller or equals to ", max, "!", nl ); + endif; + +end conjecture;
diff --git a/x86/PROGRAM.ASM b/x86/PROGRAM.ASM @@ -3,7 +3,7 @@ .stack 1000h .data - buffer db 22 DUP(0) + input db 22 DUP(0) invalid db "' is not a natural number!", 0dh, 0ah, 0dh, 0ah, "$" max dw 0 @@ -44,11 +44,11 @@ start: call iter read_uint: - ; Load the user input into the buffer + ; Load the user input into the input mov ax, cs mov ds, ax - mov buffer[0], 20 ; Configure the buffer length - mov dx, offset buffer + mov input[0], 20 ; Configure the input length + mov dx, offset input mov ah, 0ah int 21h @@ -59,8 +59,8 @@ read_uint: mov dx, 0ah int 21h - ; Move si to the to adress of the first character in the buffer - mov si, offset buffer + ; Move si to the to adress of the first character in the input + mov si, offset input add si, 2 mov ax, 0 ; Initialize the register where the result will be stored @@ -88,8 +88,8 @@ read_loop: pop cx add ax, bx - ; Jump if the counter is still smaller than the buffer length - cmp cl, buffer[1] + ; Jump if the counter is still smaller than the input length + cmp cl, input[1] jbe short read_loop mov max, ax @@ -104,7 +104,7 @@ invalid_input: mov dx, 27h int 21 - mov si, offset buffer + mov si, offset input add si, 2 mov cx, 1 @@ -115,7 +115,7 @@ invalid_input_loop: inc cx inc si - cmp cl, buffer[1] + cmp cl, input[1] jbe short invalid_input_loop invalid_input_end: