Organization of a Modern Compiler. Front-end. Middle1. Back-end DO I = 1, N DO J = 1,M S 1 1 N. Source Program

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1 Organization of a Modern Compiler ource Program Front-end snta analsis + tpe-checking + smbol table High-level ntermediate Representation (loops,arra references are preserved) Middle loop-level transformations Low-level ntermediate Representation (arra references converted into low level operations, loops converted to control flow) Low-level intermediate Representation Middle2 conventional optimizations Back-end register allocation instruction selection Assembl Code DO, DO,M M

2 DO, DO,M M 0 0 L K DO K, M DO L, L M K DO 2, DO, M A[,] A[-,+] + 2 Assume that arra has s stored everwhere before loop begins. After loop permutation: DO, M DO 2, A[,] A[-,+] + Transformed loop will produce different values (A[3,] for eample) > permutation is illegal for this loop. Question: How do we determine when loop permutation is legal? DO, DO, 0 0 L K DO K, DO L, L K Question: How do we generate loop bounds for transformed loop nest?

3 Two problems: Given a sstem of linear inequalities where A is a m X n matri of integers, b is an m vector of integers, is an n vector of unknowns, A < b (i) Are there integer solutions? (ii) Enumerate all integer solutions. Most problems regarding correctness of transformations and code generation can be reduced to these problems. ntuition about sstems of linear inequalities: Equalit: line (2D), plane (3D), hperplane (> 3D) nequalit: half-plane (2D), half-space(>2d) ntuition about sstems of linear inequalities: Conjunction of inequalties intersection of half-spaces > some conve region < > < < < 2 Region described b inequalit is conve (if two points are in region, all points in between them are in region) Region described b inequalities is a conve polhedron (if two points are in region, all points in between them are in region)

4 Dependences: data control flow anti output Flow dependence: -> 2 (i) eecutes before 2 in program order (ii) writes into a location that is read b 2 Anti-dependence: -> 2 (i) eecutes before 2 (ii) reads from a location that is overwritten later b 2 Output dependence: -> 2 (i) eecutes before 2 (ii) and 2 write to the same location nput dependence: -> 2 (i) eecutes before 2 (ii) and 2 both read from the same location output : 2 flow : + anti : 3 output : 7 Conservative Approimation: Loop level Analsis: granularit is a loop iteration - Real programs: imprecise information > need for safe approimation Eample: When ou are not sure whether a dependence eists, ou must assume it does. procedure f (X,i,j) begin X(i) 0; X(j) ; end Question: s there an output dependence from the first assignment to the second? Answer: f (i j), there is a dependence; otherwise, not. > Unless we know from interprocedural analsis that the parameters i and j are alwas distinct, we must pla it safe and insert the dependence. Ke notion: Aliasing : two program names ma refer to the same location (like X(i) and X(j)) Ma-dependence vs must-dependence: More precise analsis ma eliminate ma-dependences DO, 00 DO, 00 each (,) value of loop indices corresponds to one point in picture Dnamic instance of a statement: Eecution of a statement for given loop inde values Dependence between iterations: teration (,) is said to be dependent on iteration (2,2) if a dnamic instance (,) of a statement in loop bod is dependent on a dnamic instance (2,2) of a statement in the loop bod. How do we compute dependences between iterations of a loop nest?

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6 Min s and ma s in loop bounds maseem weird, but actuall the describe general polhedral iteration spaces! U L L2 U2 For a given, the co-ordinate of a point in the iteration space of the loop nest satisfies ma(l(),l2()) < < min(u(),u2())

7 Presentation sequence: - one equation, several variables several equations, several variables z equations & inequalities < < -9 Diophatine equations: use integer Gaussian elimination olve equalities first then use Fourier-Motzkin elimination One equation, man variables: Thm: The linear Diophatine equation a + a an n c has integer solutions iff gcd(a,a2,...,an) divides c. Eamples: () 2 3 o solutions (2) 2 6 One solution: 3 (3) GCD(2,) which divides 3. olutions: t, (3-2t) (4) GCD(2,3) which divides 3. Let z + floor(3/2) + Rewrite equation as 2z + 3 olutions: z t > (3t - 3) (3-2t) (3-2t) ntuition: Think of underdetermined sstems of eqns over reals. Caution: nteger constraint > Diophantine sstem ma have no solns Thm: The linear Diophatine equation a + a an n c has integer solutions iff gcd(a,a2,...,an) divides c. Proof: WLOG, assume that all coefficients a,a2,...an are positive. We prove onl the F case b induction, the proof in the other direction is trivial. nduction is on min(smallest coefficient, number of variables). Base case: f (# of variables ), then equation is a c which has integer solutions if a divides c. f (smallest coefficient ), then gcd(a,a2,...,an) which divides c. Wlog, assume that a, and observe that the equation has solutions of the form (c - a2 t2 - a3 t an tn, t2, t3,...tn). nductive case: ummar: Eqn: a + a an n c uppose smallest coefficient is a, and let t + floor(a2/a) floor(an/a) n n terms of this variable, the equation can be rewritten as (a) t + (a2 mod a) (an mod a) n c () where we assume that all terms with zero coefficient have been deleted. Observe that () has integer solutions iff original equation does too. ow gcd(a,b) gcd(a mod b, b) > gcd(a,a2,...,an) gcd(a, (a2 mod a),..,(an mod a)) > gcd(a, (a2 mod a),..,(an mod a)) divides c. f a is the smallest co-efficient in (), we are left with variable base case. Otherwise, the size of the smallest co-efficient has decreased, so we have made progress in the induction. - Does this have integer solutions? Does gcd(a,a2,...,an) divide c? t is useful to consider solution process in matri-theoretic terms. We can write single equation as T (3 8)( z) 6 t is hard to read off solution from this, but for special matrices, it is eas. (2 0)(a b) T 8 olution is a 4, b t looks lower triangular, right? Ke concept: column echelon form - "lower triangular form for underdetermined sstems" For a matri with a single row, column echelon form is ( ) z 6 ubstitution: t + + 2z ew equation: 3t z 6 ubstitution: u +z+t ew equation: 2u + t 6 olution: u p t (6-2p) Backsubstitution: p2 t (6-2p) z (3p-p2-6) Backsubstitution: (8-8p+p2) p2 z (3p-p2-6) (3 8) (3 2 2) ( 2 0) (3 8) (3 2 2) ( 0 0) T olution: (6 a b) U3 ( 2 0) Product of matrices olution to original sstem: U*U2*U3*(6 a b) T U U2 U*U2*U a-b -6+3a-b b

8 stems of Diophatine Equations: Ke idea: use integer Gaussian elimination Eample: z - + 2z > t is not eas to determine if this Diophatine sstem has solutions. Eas special case: lower triangular matri z > Question: Can we convert general integer matri into equivalent lower triangular sstem? z TEGER GAUA ELMATO 3 z arbitrar integer nteger gaussian Elimination - Use row/column operations to get matri into triangular form - For us, column operations are more important because we usuall have more unknowns than equations Overall strateg: Given A b Find matrices U, U2,...Uk such that Proof: A*U*U2*...*Uk is lower triangular (sa L) olve L b (eas) Compute (U*U2*...*Uk)* (A*U*U2...*Uk) b > A(U*U2*...*Uk) b > (U*U2...*Uk) Caution: ot all column operations preserve integer solutions olution: -8, 7 which has no integer solutions! ntuition: With some column operations, recovering solution of original sstem requires solving lower triangular sstem using rationals. Question: Can we sta purel in the integer domain? One solution: Use onl unimodular column operations Unimodular Column Operations: (a) nterchange two columns (b) egate a column Check Let, satisf first eqn. Let, satisf second eqn., Check, - (c) Add an integer multiple of one column to another n - Check + n Eample: z > > > > Facts:. The three unimodular column operations - interchanging two columns - negating a column - adding an integer multiple of one column to another on the matri A of the sstem A b preserve integer solutions, as do sequences of these operations > > 3 z z t z t 4-2t - t 2. Unimodular column operations can be used to reduce a matri A into lower triangular form. 3. A unimodular matri of + or -. has integer entries and a determinant 4. The product of two unimodular matrices is also unimodular.

9 Algorithm: Given a sstem of Diophantine equations A b. Use unimodular column operations to reduce matri A to lower triangular form L. 2. f L b has integer solutions, so does the original sstem. 3. f eplicit form of solutions is desired, let U be the product of unimodular matrices corresponding to the column operations. U where is the solution of the sstem L b Detail: nstead of lower triangular matri, ou should to compute column echelon form of matri. Column echelon form: Let rj be the row containing the first non-zero in column j. (i) r(j+) > rj if column j is not entirel zero. (ii) column (j+) is zero if column j is is lower triangular but not column echelon. Point: writing down the solution for this sstem requires additional work with the last equation ( equation, 2 variables). This work is precisel what is required to produce the column echelon form. ote: Even in regular Gaussian elimination, we want column echelon form rather than lower triangular form when we have under-determined sstems.