From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: from mail-ej1-x631.google.com (mail-ej1-x631.google.com [IPv6:2a00:1450:4864:20::631]) by sourceware.org (Postfix) with ESMTPS id 99B4B385AC22 for ; Fri, 23 Sep 2022 11:59:40 +0000 (GMT) DMARC-Filter: OpenDMARC Filter v1.4.1 sourceware.org 99B4B385AC22 Authentication-Results: sourceware.org; dmarc=pass (p=none dis=none) header.from=linaro.org Authentication-Results: sourceware.org; spf=pass smtp.mailfrom=linaro.org Received: by mail-ej1-x631.google.com with SMTP id z13so162700ejp.6 for ; Fri, 23 Sep 2022 04:59:40 -0700 (PDT) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=linaro.org; s=google; h=to:subject:message-id:date:from:in-reply-to:references:mime-version :from:to:cc:subject:date; bh=ECJ19kv4X0nbE16DRknmTaLXJJInvOZAQaD9xgQI17s=; b=Ad+glpUIyZD68W/AFOhVQEKS/yhQ11llZLlng+7wfcuXns5VDHMqiHiFya35qhbB70 sGGWVbxQ5Z/jDemoL3ZaoXSzlzbHVQZBaHjAaakuW74RvKdB20iAQCOXDiZdjG5Yd2o1 ssI4+YGWmU9oveeBCSebCskLJbqgGK0I7eMWFZKdT4EdRashL7kgIGxhNvvoThuJ67EB O5xOeOzOEBjyGFeq2ViJAd4guyptgNyKT3BtdEMNcZH22k4fTiYhggqesIpfiSMBMyyP 3ySNIuBSnEoF4lBeD3kqCEF2CIDTSAnycBiJ/CJZnQNJ0CwyTs5YR/9Om+fT296vHqKi vGDw== X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20210112; h=to:subject:message-id:date:from:in-reply-to:references:mime-version :x-gm-message-state:from:to:cc:subject:date; bh=ECJ19kv4X0nbE16DRknmTaLXJJInvOZAQaD9xgQI17s=; b=s5ZPy3F21+V9cJ5oDWP4GJwTls6W0huPovQVHbHIwzAvCu+2V+CrychUEw197PIo3S xrwEQjPOYR2wc/gnRcQOLBn7ZERN+j8XI1YsZ2lpVMC3A5YyuykHo79PNaST7useIxyl f6kQFT9nN1v23/IckJz106cyu1Gf40piCpG5S8lOOUB/Q4P/C2ymBuUJZLZY2IVa/6wF toabRKBhvZuQ1EbEP+8MwGFx2NdAZQ8R81uNMMZ6WEhwz7fE7pDByjoAhPciPFLH6Kzc /ZkdRn7p7yuFyxGPfzsCDomCrFMD9Ai3rAOlqD5GFa1w4976VHWsY6grujsWS8ayPXCN H6oQ== X-Gm-Message-State: ACrzQf30Na8CeZaNDZaICF8yfefMOUMb/bp4/g5mK9uiqPD4snN0zQho DjQq9o/B8iT49C+k7RdtzkrTngAcrRKnQbEXDB/IMO/aUC4= X-Google-Smtp-Source: AMsMyM40Mhd/E29cuNkFsLhQymI9meHCI5+nllIbo0ElqfQICrh4N/k9ogTaCuTSi/E7dMhmgTArnt2vDFsdep8jDoc= X-Received: by 2002:a17:907:d27:b0:782:cc3:8e87 with SMTP id gn39-20020a1709070d2700b007820cc38e87mr6418450ejc.94.1663934379201; Fri, 23 Sep 2022 04:59:39 -0700 (PDT) MIME-Version: 1.0 References: In-Reply-To: From: Prathamesh Kulkarni Date: Fri, 23 Sep 2022 17:29:03 +0530 Message-ID: Subject: Re: Extend fold_vec_perm to fold VEC_PERM_EXPR in VLA manner To: Prathamesh Kulkarni , gcc Patches , Richard Biener , richard.sandiford@arm.com Content-Type: text/plain; charset="UTF-8" X-Spam-Status: No, score=-3.3 required=5.0 tests=BAYES_00,DKIM_SIGNED,DKIM_VALID,DKIM_VALID_AU,DKIM_VALID_EF,RCVD_IN_DNSWL_NONE,SPF_HELO_NONE,SPF_PASS,TXREP autolearn=ham autolearn_force=no version=3.4.6 X-Spam-Checker-Version: SpamAssassin 3.4.6 (2021-04-09) on server2.sourceware.org List-Id: On Tue, 20 Sept 2022 at 18:09, Richard Sandiford wrote: > > Prathamesh Kulkarni writes: > > On Mon, 12 Sept 2022 at 19:57, Richard Sandiford > > wrote: > >> > >> Prathamesh Kulkarni writes: > >> >> The VLA encoding encodes the first N patterns explicitly. The > >> >> npatterns/nelts_per_pattern values then describe how to extend that > >> >> initial sequence to an arbitrary number of elements. So when performing > >> >> an operation on (potentially) variable-length vectors, the questions is: > >> >> > >> >> * Can we work out an initial sequence and npatterns/nelts_per_pattern > >> >> pair that will be correct for all elements of the result? > >> >> > >> >> This depends on the operation that we're performing. E.g. it's > >> >> different for unary operations (vector_builder::new_unary_operation) > >> >> and binary operations (vector_builder::new_binary_operations). It also > >> >> varies between unary operations and between binary operations, hence > >> >> the allow_stepped_p parameters. > >> >> > >> >> For VEC_PERM_EXPR, I think the key requirement is that: > >> >> > >> >> (R) Each individual selector pattern must always select from the same vector. > >> >> > >> >> Whether this condition is met depends both on the pattern itself and on > >> >> the number of patterns that it's combined with. > >> >> > >> >> E.g. suppose we had the selector pattern: > >> >> > >> >> { 0, 1, 4, ... } i.e. 3x - 2 for x > 0 > >> >> > >> >> If the arguments and selector are n elements then this pattern on its > >> >> own would select from more than one argument if 3(n-1) - 2 >= n. > >> >> This is clearly true for large enough n. So if n is variable then > >> >> we cannot represent this. > >> >> > >> >> If the pattern above is one of two patterns, so interleaved as: > >> >> > >> >> { 0, _, 1, _, 4, _, ... } o=0 > >> >> or { _, 0, _, 1, _, 4, ... } o=1 > >> >> > >> >> then the pattern would select from more than one argument if > >> >> 3(n/2-1) - 2 + o >= n. This too would be a problem for variable n. > >> >> > >> >> But if the pattern above is one of four patterns then it selects > >> >> from more than one argument if 3(n/4-1) - 2 + o >= n. This is not > >> >> true for any valid n or o, so the pattern is OK. > >> >> > >> >> So let's define some ad hoc terminology: > >> >> > >> >> * Px is the number of patterns in x > >> >> * Ex is the number of elements per pattern in x > >> >> > >> >> where x can be: > >> >> > >> >> * 1: first argument > >> >> * 2: second argument > >> >> * s: selector > >> >> * r: result > >> >> > >> >> Then: > >> >> > >> >> (1) The number of elements encoded explicitly for x is Ex*Px > >> >> > >> >> (2) The explicit encoding can be used to produce a sequence of N*Ex*Px > >> >> elements for any integer N. This extended sequence can be reencoded > >> >> as having N*Px patterns, with Ex staying the same. > >> >> > >> >> (3) If Ex < 3, Ex can be increased by 1 by repeating the final Px elements > >> >> of the explicit encoding. > >> >> > >> >> So let's assume (optimistically) that we can produce the result > >> >> by calculating the first Pr*Er elements and using the Pr,Er encoding > >> >> to imply the rest. Then: > >> >> > >> >> * (2) means that, when combining multiple input operands with potentially > >> >> different encodings, we can set the number of patterns in the result > >> >> to the least common multiple of the number of patterns in the inputs. > >> >> In this case: > >> >> > >> >> Pr = least_common_multiple(P1, P2, Ps) > >> >> > >> >> is a valid number of patterns. > >> >> > >> >> * (3) means that the number of elements per pattern of the result can > >> >> be the maximum of the number of elements per pattern in the inputs. > >> >> (Alternatively, we could always use 3.) In this case: > >> >> > >> >> Er = max(E1, E2, Es) > >> >> > >> >> is a valid number of elements per pattern. > >> >> > >> >> So if (R) holds we can compute the result -- for both VLA and VLS -- by > >> >> calculating the first Pr*Er elements of the result and using the > >> >> encoding to derive the rest. If (R) doesn't hold then we need the > >> >> selector to be constant-length. We should then fill in the result > >> >> based on: > >> >> > >> >> - Pr == number of elements in the result > >> >> - Er == 1 > >> >> > >> >> But this should be the fallback option, even for VLS. > >> >> > >> >> As far as the arguments go: we should reject CONSTRUCTORs for > >> >> variable-length types. After doing that, we can treat a CONSTRUCTOR > >> >> for an N-element vector type by setting the number of patterns to N > >> >> and the number of elements per pattern to 1. > >> > Hi Richard, > >> > Thanks for the suggestions, and sorry for late response. > >> > I have a couple of very elementary questions: > >> > > >> > 1: Consider following inputs to VEC_PERM_EXPR: > >> > op1: P_op1 == 4, E_op1 == 1 > >> > {1, 2, 3, 4, ...} > >> > > >> > op2: P_op2 == 2, E_op2 == 2 > >> > {11, 21, 12, 22, ...} > >> > > >> > sel: P_sel == 3, E_sel == 1 > >> > {0, 4, 5, ...} > >> > > >> > What shall be the result in this case ? > >> > P_res = lcm(4, 2, 3) == 12 > >> > E_res = max(1, 2, 1) == 2. > >> > >> Yeah, that looks right. Of course, since sel is just repeating > >> every three elements, it could just be P_res==3, E_sel==1, > >> but the vector_builder would do that optimisation for us. > >> > >> (I'm not sure whether we'd see a P==3 encoding in practice, > >> but perhaps it's possible.) > >> > >> If sel was P_sel==1, E_sel==3 (so a stepped encoding rather than > >> repeating every three elements) then: > >> > >> P_res = lcm(4, 2) == 4 > >> E_res = max(1, 2, 3) == 3 > >> > >> which also looks like it would give the right encoding. > >> > >> > 2. How should we specify index of element in sel when it is not > >> > explicitly encoded in the operand ? > >> > For eg: > >> > op1: npatterns == 2, nelts_per_pattern == 3 > >> > { 1, 0, 2, 0, 3, 0, ... } > >> > op2: npatterns == 6, nelts_per_pattern == 1 > >> > { 11, 12, 13, 14, 15, 16, ...} > >> > > >> > In sel, how do we refer to element with value 4, that would be 4th element > >> > of first pattern in op1, but not explicitly encoded ? > >> > In op1, 4 will come at index == 6. > >> > However in sel, index 6 would refer to 11, ie op2[0] ? > >> > >> What index 6 refers to depends on the length of op1. > >> If the length of op1 is 4 at runtime the index 6 refers to op2[2]. > >> If the length of op1 is 6 then index 6 refers to op2[0]. > >> If the length of op1 is 8 then index 6 refers to op1[6], etc. > >> > >> This comes back to (R) above. We need to be able to prove at compile > >> time that each pattern selects from the same input vectors (for all > >> elements, not just the encoded elements). If we can't prove that > >> then we can't fold for variable-length vectors. > > Hi Richard, > > Thanks for the clarification! > > I have come up with an approach to verify R: > > > > Consider following pattern: > > a0, a1, a1 + S, ..., > > nelts_per_pattern would be n / Psel, where n == actual length of the vector. > > And last element of pattern will be given by: > > a1 + (n/Psel - 2) * S > > (I think this is just a terminology thing, but in the source, > nelts_per_pattern is a compile-time constant that describes the > encoding. It always has the value 1, 2 or 3, regardless of the > runtime length.) > > > Rearranging the above term, we can think of pattern > > as a line with following equation: > > y = (S/Psel) * n + (a1 - 2S) > > where (S/Psel) is the slope, and (a1 - 2S) is the y-intercept. > > > > At, > > n = 2*Psel, y = a1 > > n = 3*Psel, y = a1 + S, > > n = 4*Psel, y = a1 + 2S ... > > > > To compare with n, we compare the following lines: > > y1 = (S/Psel) * n + (a1 - 2S) > > y2 = n > > > > So to check if elements always come from first vector, > > we want to check y1 < y2 for n > 0. > > Likewise, if elements always come from second vector, > > we want to check if y1 >= y2, for n > 0. > > One difficulty here is that the indices wrap around, so an index value of > 2n selects from the first vector rather than the second. (This is pretty > awkward for VLA and doesn't match the native SVE TBL behaviour.) So... > > > If both lines are parallel, ie S/PSel == 1, > > then we choose first or second vector depending on the y-intercept a1 - 2S. > > If a1 - 2S >= 0, then y1 >= y2 for all values of n, so select second vector. > > If a1 - 2S < 0, then y1 < y2 for all values of n, so select first vector. > > > > For eg, if we have following pattern: > > {0, 1, 3, ...} > > where a1 = 1, S = 2, and consider PSel = 2. > > > > y1 = n - 3 > > y2 = n > > > > In this case, y1 < y2 for all values of n, so we select first vector. > > > > Since y2 = n, passes thru origin with slope = 1, > > a line can intersect it either in 1st or 3rd quadrant. > > Calculate point of intersection: > > n_int = Psel * (a1 - 2S) / (Psel - S); > > > > (a) n_int > 0 > > n_int > 0 => intersecting in 1st quadrant. > > In this case there will be a cross-over at n_int. > > > > For eg, consider pattern { 0, 1, 4, ...} > > a1 = 1, S = 3, and let's take PSel = 2 > > > > y1 = (3/2)n - 5 > > y2 = n > > > > Both intersect at (10, 10). > > So for n < 10, y1 < y2 > > and for n > 10, y1 > y2. > > so in this case we can't fold since we will select elements from both vectors. > > > > (b) n_int <= 0 > > In this case, the lines will intersect in 3rd quadrant, > > so depending upon the slope we can choose either vector. > > If (S/Psel) < 1, ie y1 has a gentler slope than y2, > > then y1 < y2 for n > 0 > > If (S/Psel) > 1, ie, y1 has a steeper slope than y2, > > then y1 > y2 for n > 0. > > > > For eg, in the above pattern {0, 1, 4, ...} > > a1 = 1, S = 3, and let's take PSel = 4 > > > > y1 = (3/4)n - 5 > > y2 = n > > Both intersect at (-20, -20). > > y1's slope = (S/Psel) = (3/4) < 1 > > So y1 < y2 for n > 0. > > Graph: https://www.desmos.com/calculator/ct7edqbr9d > > So we pick first vector. > > > > The following pseudo code attempts to capture this: > > > > tree select_vector_for_pattern (op1, op2, a1, S, Psel) > > { > > if (S == Psel) > > { > > /* If y1 intercept >= 0, then y1 >= y2 > > for all values of n. */ > > if (a1 - 2*S >= 0) > > return op2; > > return op1; > > } > > > > n_int = Psel * (a1 - 2*S) / (Psel - S) > > /* If intersecting in 1st quadrant, there will be cross over, > > bail out. */ > > if (n_int > 0) > > return NULL_TREE; > > /* If S/Psel < 1, ie y1 has gentler slope than y2, > > then y1 < y2 for n > 0. */ > > if (S < Psel) > > return op1; > > /* If S/Psel > 1, ie y1 has steeper slope than y2, > > then y1 > y2 for n > 0. */ > > return op2; > > } > > > > Does this look reasonable ? > > ...I think we need to be more conservative. I think we also need to > distinguish n1 (the number of elements in the input vectors) and > nsel (the number of elements in the selector). > > If nsel is a multiple of Psel and nsel >= 2 * Psel then like you say > there will be (nsel /exact Psel) - 1 index elements from the stepped > encoding and the final index value will be: > > ae = a1 + (nsel /exact Psel - 2) * S > > Because of wrap-around, we need to ensure that that doesn't run > into an adjoining vector. I think the easiest way of doing that > is to calculate a1 /trunc n1 and ae /trunc n1 (using can_div_trunc_p) > and check that the quotients are equal. IIUC, If a1/n1 == ae/n1, then the sequence will choose from the same vector since ae is last elem, and the quotient can choose the vector because it will be either 0 or 1 (since indices wrap around after 2n). Um, could you please elaborate a bit on how will can_div_trunc_p calculate quotients, when n1 and nsel are unknown at compile time ? To calculate the quotients for a hard coded pattern, with a1 = 1, nsel = n1 = len(VNx4SI), S = 3, Psel = 4, I tried the following: poly_uint64 n1 = GET_MODE_NUNITS (VNx4SImode); poly_uint64 nsel = n1; poly_uint64 a1 = 1 poly_uint64 Esel = exact_div (nsel / Psel); poly_uint64 ae = a1 + (Esel - 2) * S; int q1, qe; poly_uint64 r1, re; bool div1_p = can_div_trunc_p (a1, n1, &q1, &r1); bool dive_p = can_div_trunc_p (ae, n1, &qe, &re); Which gave strange values for qe and 0 for q1, with first call succeeding, and second call returning false. Am I calling it incorrectly ? Thanks, Prathamesh > > However, I now realise that there's a wrinkle. If S < 0 then we > also need to check that either: > > (a) the chosen input vector (given by the quotient above) has either: > > (i) nelts_per_pattern == 1 > (ii) nelts_per_pattern == 3 and the difference between the > first and second elements in each pattern is the same as > the difference between the second and third elements > (i.e. every pattern is a natural stepped one). > > (b) ae % n1 >= the number of patterns in the input vector. > (ae % n1 is calculated as a side-effect of can_div_trunc_p). > > Otherwise the index vector has the effect of moving the "foreground" > from the front of the input vector to the end of the result vector. > > If nsel == Psel then the stepped part of the sequence doesn't matter. > Thus, the same condition works whenever nsel is a multiple of Psel. > > If nsel is not a multiple of Psel then I think we should punt for now. > There are some cases that we could handle when n1 == nsel, but "nsel > is a multiple of Psel" will be the normal case. > > Thanks, > Richard