// ************************************************************************** // // // // eses eses // // eses eses // // eses eseses esesese eses Embedded Systems Group // // ese ese ese ese ese // // ese eseseses eseseses ese Department of Computer Science // // eses eses ese eses // // eses eseses eseseses eses University of Kaiserslautern // // eses eses // // // // ************************************************************************** // // ** Note: We are only interested in even values of B ** // // -------------------------------------------------------------------------- // // This version of the B-complement division algorithm is an adaptation of // // of NatDivModSeq1 for B-complement numbers. Analogous to NatDivModSeq1, it // // is an optimization of NIntDivModSeq0 in that the partial remainders // //z[N+i..i] used there now are stored in r[N-1..0]@x[i..0] to compute digit // // [i] of the quotient. // // ************************************************************************** // macro B = 4; // base of the radix numbers macro M = 4; // number of digits used for x macro N = 3; // number of digits used for y macro alpha(x) = (x<B/2 ? +x : +x-B); macro gamma(y) = (y<0 ? y+B : y); macro dval(x,i,k) = (i==k-1 ? alpha(x[i]) : +x[i]); macro natval(x,m) = sum(i=0..m-1) (x[i] * exp(B,i)); macro intval(x,k) = sum(i=0..k-1) (dval(x,i,k) * exp(B,i)); macro z(i,j) = (j==0 ? x[i] : r[j-1]); // should be z[j+i] of IntDivModSeq0 module IntDivSeq1([M]nat{B} ?x,[N]nat{B} ?y,[M+1]nat{B} q,[N]nat{B} r,event !rdy) { [N]nat{B} s; // auxiliary variable for sums [N+1]int{B} c; // auxiliary variable for carry digits int{2} sgn_y; // sign of y event isNegative; // is set when subtraction yields negative result //------------------------------------------------------------------------- // apply digit extension to x and put the extended digits in r //------------------------------------------------------------------------- for(j=0..N-1) r[j] = (x[M-1]<B/2 ? 0 : B-1); //------------------------------------------------------------------------- // determine leftmost digit of the quotient and the first partial remainder //------------------------------------------------------------------------- sgn_y = (y[N-1]<B/2 ? +1 : -1); q[M] = (x[M-1]<B/2 ? 0 : B-1); // compute r[N-1..0] = r[N-1..0] - alpha(q[M]) * sgn_y * y c[0] = 0; for(j=0..N-1) // note that -(B-1)<=sm<=2*B-1, thus -1<=sm/B<=1 let(yj = (j==N-1 ? alpha(y[j]) : y[j])) let(sm = r[j] + c[j] - alpha(q[M]) * sgn_y * yj) { next(r[j]) = sm % B; c[j+1] = sm / B; } w0: pause; //------------------------------------------------------------------------- // computing the remaining quotient digits //------------------------------------------------------------------------- for(iup=0..M-1) let(i=M-1-iup) {// thus i=M-1..0 next(q[i]) = B-1; do { w1: pause; // compute s[N..0] = z[N+i..i] - q[i] * sgn_y * y[N-1..0] for digits // q[i] = B-1,... until s>=0; z[N+i..i] is stored in r[N-1..0]@x[i] c[0] = 0; for(j=0..N-1) // note that 1-B*(B+1)<=sm<=(B-1)*(B+1), thus -(B-1)<=sm/B<=B-1 // hence c[j+1] fits into int{B} let(yj = (j==N-1 ? alpha(y[j]) : y[j])) let(sm = z(i,j) + c[j] - q[i] * sgn_y * yj) { s[j] = sm % B; c[j+1] = sm / B; } isNegative = (z(i,N) + c[N] < 0); if(isNegative) next(q[i]) = q[i] - 1; } while(isNegative); // commit the subtraction in that s is copied to r w2: pause; for(j=0..N-1) r[j] = s[j]; } //------------------------------------------------------------------------- // here we have x = q * |y| + r; so if y<0 holds, we have to negate q //------------------------------------------------------------------------- if(y[N-1]>=B/2) { [M+1]nat{2} cq; cq[0]=0; for(i=0..M-1) let(sm = -(cq[i] + q[i])) { cq[i+1] = -(sm / B); next(q[i]) = sm % B; } let(sm = -(cq[M] + alpha(q[M]))) next(q[M]) = gamma(sm % B); w3:pause; } emit(rdy); //------------------------------------------------------------------------- // final assertion //------------------------------------------------------------------------- if(intval(y,N)!=0) { assert(intval(x,M) == intval(q,M+1) * intval(y,N) + intval(r,N)); assert(intval(r,N) < abs(intval(y,N))); } }