High-Level Synthesis: Creating Custom Circuits from High-Level Code

High-Level Synthesis: Creating Custom Circuits from High-Level Code
Greg Stitt ECE Department University of Florida

Existing FPGA Tool Flow
Register-transfer (RT) synthesis Specify RT structure (muxes, registers, etc) + Allows precise specification - But, time consuming, difficult, error prone HDL RT Synthesis Netlist Technology Mapping Physical Design Placement Bitfile Routing Processor FPGA

Future FPGA Tool Flow? High-level Synthesis RT Synthesis
C/C++, Java, etc. High-level Synthesis HDL RT Synthesis Netlist Technology Mapping Physical Design Placement Bitfile Routing Processor FPGA

High-level Synthesis Wouldn’t it be nice to write high-level code?
Ratio of C to VHDL developers (10000:1 ?) + Easier to specify + Separates function from architecture + More portable - Hardware potentially slower Similar to assembly code era Programmers could always beat compiler But, no longer the case Hopefully, high-level synthesis will catch up to manual effort

High-level Synthesis More challenging than compilation
Compilation maps behavior into assembly instructions Architecture is known to compiler High-level synthesis creates a custom architecture to execute behavior Huge hardware exploration space Best solution may include microprocessors Should handle any high-level code Not all code appropriate for hardware

High-level Synthesis First, consider how to manually convert high-level code into circuit Steps 1) Build FSM for controller 2) Build datapath based on FSM acc = 0; for (i=0; i < 128; i++) acc += a[i];

Manual Example Build a FSM (controller) Decompose code into states
acc = 0; for (i=0; i < 128; i++) acc += a[i]; acc=0, i = 0 if (i < 128) load a[i] Done acc += a[i] i++

Manual Example Build a datapath Allocate resources for each state + +
acc=0, i = 0 if (i < 128) load a[i] Done i acc a[i] addr 1 128 1 acc += a[i] + + < + i++ acc = 0; for (i=0; i < 128; i++) acc += a[i];

Manual Example Build a datapath Determine register inputs + + < +
In from memory acc=0, i = 0 &a if (i < 128) 2x1 2x1 2x1 load a[i] Done i acc a[i] addr 1 128 1 acc += a[i] + < + + i++ acc = 0; for (i=0; i < 128; i++) acc += a[i];

Manual Example Build a datapath Add outputs + + + < In from memory
acc=0, i = 0 &a if (i < 128) 2x1 2x1 2x1 load a[i] Done i acc a[i] addr 1 128 1 acc += a[i] + + < + i++ acc = 0; for (i=0; i < 128; i++) acc += a[i]; acc Memory address

Manual Example Build a datapath Add control signals + + + <
In from memory acc=0, i = 0 &a if (i < 128) 2x1 2x1 2x1 load a[i] Done i acc a[i] addr 1 128 1 acc += a[i] + + < + i++ acc = 0; for (i=0; i < 128; i++) acc += a[i]; acc Memory address

Manual Example Combine controller+datapath Controller + + + <
In from memory Controller &a 2x1 2x1 2x1 i acc a[i] addr 1 128 1 + + < + acc = 0; for (i=0; i < 128; i++) acc += a[i]; Done Memory Read acc Memory address

Manual Example Alternatives Use one adder (plus muxes) < +
In from memory &a 2x1 2x1 2x1 i acc a[i] addr 1 128 < MUX MUX + acc Memory address

Manual Example Comparison with high-level synthesis
Determining when to perform each operation => Scheduling Allocating resource for each operation => Resource allocation Mapping operations onto resources => Binding

Another Example Your turn Steps
for (i=0; i < 100; i++) { if (a[i] > 0) x ++; else x --; a[i] = x; } //output x Steps 1) Build FSM (do not perform if conversion) 2) Build datapath based on FSM

High-Level Synthesis Could be C, C++, Java, Perl, Python, SystemC, ImpulseC, etc. High-level Code High-Level Synthesis Custom Circuit Usually a RT VHDL description, but could as low level as a bit file

High-Level Synthesis Controller < + High-Level Synthesis acc i addr
for (i=0; i < 128; i++) acc += a[i]; High-Level Synthesis acc i < addr a[i] + 1 128 2x1 &a In from memory Memory address Done Memory Read Controller

Main Steps Controller + Datapath
High-level Code Converts code to intermediate representation - allows all following steps to use common format. Front-end Syntactic Analysis Intermediate Representation Optimization Determines when each operation will execute, and resources used Scheduling/Resource Allocation Back-end Binding/Resource Sharing Maps operations onto physical resources Controller + Datapath

Intermediate Representation
Syntactic Analysis Definition: Analysis of code to verify syntactic correctness Converts code into intermediate representation 2 steps 1) Lexical analysis (Lexing) 2) Parsing High-level Code Lexical Analysis Syntactic Analysis Parsing Intermediate Representation

Lexical Analysis Lexical analysis (lexing) breaks code into a series of defined tokens Token: defined language constructs x = 0; if (y < z) x = 1; Lexical Analysis ID(x), ASSIGN, INT(0), SEMICOLON, IF, LPAREN, ID(y), LT, ID(z), RPAREN, ID(x), ASSIGN, INT(1), SEMICOLON

Lexing Tools Define tokens using regular expressions - outputs C code that lexes input Common tool is “lex” /* braces and parentheses */ "[" { YYPRINT; return LBRACE; } "]" { YYPRINT; return RBRACE; } "," { YYPRINT; return COMMA; } ";" { YYPRINT; return SEMICOLON; } "!" { YYPRINT; return EXCLAMATION; } "{" { YYPRINT; return LBRACKET; } "}" { YYPRINT; return RBRACKET; } "-" { YYPRINT; return MINUS; } /* integers [0-9]+ { yylval.intVal = atoi( yytext ); return INT; }

Parsing Analysis of token sequence to determine correct grammatical structure Languages defined by context-free grammar Correct Programs Grammar x = 0; x = 0; y = 1; Program = Exp Exp = Stmt SEMICOLON | IF LPAREN Cond RPAREN Exp | Exp Exp if (a < b) x = 10; if (var1 != var2) x = 10; Cond = ID Comp ID x = 0; if (y < z) x = 1; x = 0; if (y < z) x = 1; y = 5; t = 1; Stmt = ID ASSIGN INT Comp = LT | NE

Parsing Incorrect Programs Grammar Program = Exp Exp = S SEMICOLON |
IF LPAREN Cond RPAREN Exp | Exp Exp x = 5;; x = 5 if (x+5 > y) x = 2; Cond = ID Comp ID x = y; S = ID ASSIGN INT Comp = LT | NE

Parsing Tools Define grammar in special language
Automatically creates parser based on grammar Popular tool is “yacc” - yet-another-compiler-compiler program: functions { $$ = $1; } ; functions: function { $$ = $1; } | functions function { $$ = $1; } function: HEXNUMBER LABEL COLON code { $$ = $2; }

Parser converts tokens to intermediate representation Usually, an abstract syntax tree Assign x = 0; if (y < z) x = 1; d = 6; x if assign cond assign y < z x 1 d 6

Why use intermediate representation? Easier to analyze/optimize than source code Theoretically can be used for all languages Makes synthesis back end language independent C Code Java Perl Syntactic Analysis Syntactic Analysis Syntactic Analysis Intermediate Representation Scheduling, resource allocation, binding, independent of source language - sometimes optimizations too Back End

Different Types Abstract Syntax Tree Control/Data Flow Graph (CDFG) Sequencing Graph Etc. We will focus on CDFG Combines control flow graph (CFG) and data flow graph (DFG)

Control flow graphs CFG
Represents control flow dependencies of basic blocks Basic block is section of code that always executes from beginning to end I.e. no jumps into or out of block acc=0, i = 0 acc = 0; for (i=0; i < 128; i++) acc += a[i]; if (i < 128) acc += a[i] i ++ Done

Control flow graphs Your turn Create a CFG for this code i = 0;
while (j < 10) { if (x < 5) y = 2; else if (z < 10) y = 6; }

Data Flow Graphs DFG Represents data dependencies between operations a
x = a+b; y = c*d; z = x - y; + * - x y z

Control/Data Flow Graph
Combines CFG and DFG Maintains DFG for each node of CFG acc = 0; for (i=0; i < 128; i++) acc += a[i]; acc i acc=0; i=0; if (i < 128) acc a[i] i 1 acc += a[i] i ++ Done + + acc i

High-Level Synthesis: Optimization

Synthesis Optimizations
After creating CDFG, high-level synthesis optimizes graph Goals Reduce area Improve latency Increase parallelism Reduce power/energy 2 types Data flow optimizations Control flow optimizations

Data Flow Optimizations
Tree-height reduction Generally made possible from commutativity, associativity, and distributivity a b c d a b c d + + + + + + a b c d a b c d * + * + + +

Operator Strength Reduction Replacing an expensive (“strong”) operation with a faster one Common example: replacing multiply/divide with shift 1 multiplication 0 multiplications b[i] = a[i] * 8; b[i] = a[i] << 3; a = b * 5; c = b << 2; a = b + c; a = b * 13; c = b << 2; d = b << 3; a = c + d + b;

Constant propagation Statically evaluate expressions with constants x = 0; y = x * 15; z = y + 10; x = 0; y = 0; z = 10;

Function Specialization Create specialized code for common inputs Treat common inputs as constants If inputs not known statically, must include if statement for each call to specialized function int f (int x) { y = x * 15; return y + 10; } int f (int x) { y = x * 15; return y + 10; } int f_opt () { return 10; } Treat frequent input as a constant for (I=0; I < 1000; I++) f(0); … } for (I=0; I < 1000; I++) f_opt(0); … }

Common sub-expression elimination If expression appears more than once, repetitions can be replaced a = x + y; b = c * 25 + x + y; a = x + y; b = c * 25 + a; x + y already determined

Dead code elimination Remove code that is never executed May seem like stupid code, but often comes from constant propagation or function specialization int f (int x) { if (x > 0 ) a = b * 15; else a = b / 4; return a; } int f_opt () { a = b * 15; return a; } Specialized version for x > 0 does not need else branch - “dead code”

Code motion (hoisting/sinking) Avoid repeated computation for (I=0; I < 100; I++) { z = x + y; b[i] = a[i] + z ; } z = x + y; for (I=0; I < 100; I++) { b[i] = a[i] + z ; }

Control Flow Optimizations
Loop Unrolling Replicate body of loop May increase parallelism for (i=0; i < 128; i++) a[i] = b[i] + c[i]; for (i=0; i < 128; i+=2) { a[i] = b[i] + c[i]; a[i+1] = b[i+1] + c[i+1] }

Function Inlining Replace function call with body of function Common for both SW and HW SW - Eliminates function call instructions HW - Eliminates unnecessary control states for (i=0; i < 128; i++) a[i] = f( b[i], c[i] ); int f (int a, int b) { return a + b * 15; } for (i=0; i < 128; i++) a[i] = b[i] + c[i] * 15;

Conditional Expansion Replace if with logic expression Execute if/else bodies in parallel y = ab if (a) x = b+d else x =bd y = ab x = a(b+d) + a’bd [DeMicheli] Can be further optimized to: y = ab x = y + d(a+b)

Example Optimize this x = 0; y = a + b; if (x < 15) z = a + b - c;
else z = x + 12; output = z * 12;

High-Level Synthesis: Scheduling

Scheduling Scheduling assigns a start time to each operation in DFG
Start times must not violate dependencies in DFG Start times must meet performance constraints Alternatively, resource constraints Performed on the DFG of each CFG node => Can’t execute multiple CFG nodes in parallel

Examples a b c d a b c d + Cycle1 Cycle1 + + Cycle2 Cycle2 + Cycle3 +

Scheduling Problems Several types of scheduling problems Problems:
Usually some combination of performance and resource constraints Problems: Unconstrained Not very useful, every schedule is valid Minimum latency Latency constrained Mininum-latency, resource constrained i.e. find the schedule with the shortest latency, that uses less than a specified # of resources NP-Complete Mininum-resource, latency constrained i.e. find the schedule that meets the latency constraint (which may be anything), and uses the minimum # of resources

Minimum Latency Scheduling
ASAP (as soon as possible) algorithm Find a node whose predecessors are scheduled Schedule node one cycle later than max cycle of predecessor Repeat until all nodes scheduled a b c d e f g h Cycle1 + + - < Cycle2 * Cycle3 * Cycle4 + Minimum possible latency - 4 cycles

ALAP (as late as possible) algorithm Run ASAP, get minimum latency L Find a node whose successors are scheduled Schedule node one cycle before min cycle of predecessor Nodes with no successors scheduled to cycle L Repeat until all nodes scheduled a b c d e f g h Cycle1 + + - < Cycle4 Cycle3 Cycle2 * Cycle3 * Cycle4 + L = 4 cycles

ALAP (as late as possible) algorithm Run ASAP, get minimum latency L Find a node whose successors are scheduled Schedule node one cycle before min cycle of predecessor Nodes with no successors scheduled to cycle L Repeat until all nodes scheduled a b c d e f g h Cycle1 + + Cycle2 * Cycle3 - * Cycle4 + < L = 4 cycles

ALAP Has to run ASAP first, seems pointless But, many heuristics need the mobility/slack of each operation ASAP gives the earliest possible time for an operation ALAP gives the latest possible time for an operation Slack = difference between earliest and latest possible schedule Slack = 0 implies operation has to be done in the current scheduled cycle The larger the slack, the more options a heuristic has to schedule the operation

Latency-Constrained Scheduling
Instead of finding the minimum latency, find latency less than L Solutions: Use ASAP, verify that minimum latency less than L Use ALAP starting with cycle L instead of minimum latency (don’t need ASAP)

Scheduling with Resource Constraints
Schedule must use less than specified number of resources Constraints: 1 ALU (+/-), 1 Multiplier a b c d e f g Cycle1 + + - Cycle2 Cycle3 * * + Cycle4 Cycle5 +

Scheduling with Resource Constraints
Schedule must use less than specified number of resources Constraints: 2 ALU (+/-), 1 Multiplier a b c d e f g Cycle1 + + - Cycle2 * * + Cycle3 Cycle4 +

Mininum-Latency, Resource-Constrained Scheduling
Definition: Given resource constraints, find schedule that has the minimum latency Example: Constraints: 1 ALU (+/-), 1 Multiplier a b c d e f g Cycle1 + + - Cycle2 Cycle4 Cycle3 + * Cycle5 + Cycle6

Definition: Given resource constraints, find schedule that has the minimum latency Example: Constraints: 1 ALU (+/-), 1 Multiplier a b c d e f g Cycle1 + + - Cycle2 Cycle3 Cycle4 + * + Cycle5 Different schedules may use same resources, but have different latencies

Hu’s Algorithm Assumes one type of resource Basic Idea Input: graph, # of resources r 1) Label each node by max distance from output i.e. Use path length as priority 2) Determine C => set of unscheduled nodes that are candidates to be scheduled Candidate if either no predecessor, or predecessors scheduled 3) From C, schedule up to r nodes to current cycle, using label as priority 4) Increment current cycle, repeat from 2) until all nodes scheduled

Hu’s Algorithm Example a b c d e f g j k + + - - * * + + r = 3

Hu’s Algorithm Step 1 - Label each node by max distance from output i.e. use path length as priority a b c d e f g j k 3 4 1 4 3 2 2 1 r = 3

Hu’s Algorithm Step 2 - Determine C => set of unscheduled nodes that are candidates to be scheduled a b c d e f g j k C = 3 4 1 4 3 2 2 Cycle 1 1 r = 3

Hu’s Algorithm Step 3 - From C, schedule up to r nodes to current cycle, using label as priority a b c d e f g j k Cycle1 3 4 1 4 3 2 Not scheduled due to lower priority 2 Cycle 1 1 r = 3

Hu’s Algorithm Step 2 a b c d e f g j k Cycle1 3 4 1 4 3 2 C = 2 Cycle 2 1 r = 3

Hu’s Algorithm Step 3 a b c d e f g j k Cycle1 3 4 1 4 Cycle2 3 2 2 Cycle 2 1 r = 3

Hu’s Algorithm Skipping to finish a b c d e f g j k Cycle1 3 4 1 4 Cycle2 3 2 Cycle3 2 Cycle4 1 r = 3

Hu’s is simplified problem Common Extensions: Multiple resource types Multi-cycle operation a b c d Cycle1 + - * Cycle2 /

List Scheduling - (minimum latency, resource-constrained version) Basic Idea - Hu’s algorithm for each resource type Input: graph, set of constraints R for each resource type 1) Label nodes based on max distance to output 2) For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to Rt operations from C based on priority, to current cycle Rt is the constraint on resource type t Increment cycle, repeat from 2) until all nodes scheduled

List scheduling - minimum latency Step 1 - Label nodes based on max distance to output (not shown, so you can see operations) *nodes given IDs for illustration purposes 2 ALUs (+/-), 2 Multipliers a b c d e f g j k + 4 1 * 2 3 - + * 6 * 5 + 7 - 8

List scheduling - minimum latency For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to Rt operations from C based on priority, to current cycle Rt is the constraint on resource type t 2 ALUs (+/-), 2 Multipliers Candidates a b c d e f g j k Mult ALU Cycle ,3, + 2 3 4 - 1 * + * 6 * 5 + 7 - 8

List scheduling - minimum latency For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to Rt operations from C based on priority, to current cycle Rt is the constraint on resource type t 2 ALUs (+/-), 2 Multipliers Candidates a b c d e f g j k Mult ALU Cycle ,3, * 2 3 + 4 - 1 + Cycle1 * 6 * Candidate, but not scheduled due to low priority 5 + 7 - 8

List scheduling - minimum latency For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to Rt operations from C based on priority, to current cycle Rt is the constraint on resource type t 2 ALUs (+/-), 2 Multipliers Candidates a b c d e f g j k Mult ALU Cycle 2,3, 5, + 2 3 4 * - 1 + Cycle1 * 6 * 5 + 7 - 8

List scheduling - minimum latency For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to Rt operations from C based on priority, to current cycle Rt is the constraint on resource type t 2 ALUs (+/-), 2 Multipliers Candidates a b c d e f g j k Mult ALU Cycle 2,3, 5, + 1 * 2 3 4 - + Cycle1 * 6 * 5 Cycle2 + 7 - 8

List scheduling - (minimum latency) Final schedule Note - ASAP would require more resources ALAP wouldn’t but in general, it would a b c d e f g j k Cycle1 3 + 1 * 2 + Cycle2 * 6 * 4 - 5 Cycle3 + 7 Cycle4 - 8

Extensions: Multicycle operations Same idea (differences shown in red) Input: graph, set of constraints R for each resource type 1) Label nodes based on max distance to output 2) For each resource type t 3) Determine candidate nodes, C (those w/ no predecessors or w/ scheduled predecessors) 4) Schedule up to (Rt - nt) operations from C based on priority, one cycle after predecessor Rt is the constraint on resource type t nt is the number of resource t in use from previous cycles Repeat from 2) until all nodes scheduled

Example: 2 ALUs (+/-), 2 Multipliers a b c d e f g j k Cycle1 * 2 3 + 1 + Cycle2 * 4 Cycle3 6 * Cycle4 5 * Cycle5 Cycle6 + 7 Cycle7 - 8

List Scheduling (Min Latency)
Your turn (2 ALUs, 1 Mult) Steps (will be on test) 1) Label nodes with priority 2) Update candidate list for each cycle 3) Redraw graph to show schedule 6 2 3 * - 5 + + 1 + * 4 * 8 + 7 + - 10 9 - 11

List Scheduling (Min Latency)
Your turn (2 ALUs, 1 Mult, Mults take 2 cycles) a b c d e f g 2 3 * * 1 + + 4 * 5 + 7

Minimum-Resource, Latency-Constrained
Note that if no resource constraints given, scheduling determines number of required resources Max # of each resource type used in a single cycle a b c d e f g 3 ALUs Cycle1 + + - 2 Mults Cycle2 * * + Cycle3 Cycle4 +

Minimum-Resource Latency-Constrained Scheduling: For all schedules that have latency less than the constraint, find the one that uses the fewest resources Latency Constraint <= 4 Latency Constraint <= 4 a b c d e f g a b c d e f g Cycle1 + + - Cycle1 + + - Cycle2 * * Cycle2 * * + Cycle3 + Cycle3 Cycle4 + Cycle4 + 2 ALUs, 1 Mult 3 ALUs, 2 Mult

List scheduling (Minimum resource) Basic Idea 1) Compute latest start times for each op using ALAP with specified latency constraint 2) For each resource type 3) Determine candidate nodes 4) Compute slack for each candidate Slack = current cycle - latest possible cycle 5) Schedule ops with 0 slack Update required number of resources 6) Schedule ops that require no extra resources 7) Repeat from 2) until all nodes scheduled

1) Find ALAP schedule a b c d e f g j k + 1 * 2 + 3 4 - Last Possible Cycle 6 * 5 * Node LPC - 7 1 3 2 Latency Constraint = 3 cycles a b c d e f g j k Cycle1 1 * 2 + 3 + Cycle2 * 6 5 * Cycle3 4 7 - - Defines latest possible cycle for each operation

2) For each resource type 3) Determine candidate nodes C 4) Compute slack for each candidate Slack = current cycle - latest possible cycle Candidates = {1,2,3,4} Node LPC Slack Cycle 1 3 2 2 a b c d e f g j k * 2 + 3 + 4 1 - * 6 * 5 - 7 Cycle 1

5)Schedule ops with 0 slack Update required number of resources 6) Schedule ops that require no extra resources Node LPC Slack Cycle Candidates = {1,2,3,4} 1 3 2 X Resources = 1 Mult, 2 ALU a b c d e f g j k * 2 + 3 + 4 1 - * 6 * 5 4 requires 1 more ALU - 7 Cycle 1

2)For each resource type 3) Determine candidate nodes C 4) Compute slack for each candidate Slack = current cycle - latest possible cycle Node LPC Slack Cycle Candidates = {4,5,6} 1 3 2 1 Resources = 1 Mult, 2 ALU a b c d e f g j k * 2 + 3 + 4 1 - * 6 * 5 - 7 Cycle 2

5)Schedule ops with 0 slack Update required number of resources 6) Schedule ops that require no extra resources Node LPC Slack Cycle Candidates = {4,5,6} 1 3 2 1 Resources = 2 Mult, 2 ALU a b c d e f g j k + 1 * 2 + 3 4 - * 6 * 5 - 7 Already 1 ALU - 4 can be scheduled Cycle 2

2)For each resource type 3) Determine candidate nodes C 4) Compute slack for each candidate Slack = current cycle - latest possible cycle Node LPC Slack Cycle Candidates = {7} 1 3 2 1 2 Resources = 2 Mult, 2 ALU a b c d e f g j k * 2 + 3 + 4 1 - * 6 * 5 - 7 Cycle 3

Final Schedule Required Resources = 2 Mult, 2 ALU Node LPC Slack Cycle a b c d e f g j k 1 3 2 1 2 3 Cycle1 * 2 + 3 1 + Cycle2 * 6 * 4 - 5 Cycle3 - 7

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