Hello All,

Following is the einstein's puzzle solved by cadence Incisive92  (solved in less than 3 seconds -> FAST!!!!!!)

Thanks,

Vinay Honnavara

Verification engineer at Keyu Tech

vinayh@keyutech.com

 // Author: Vinay Honnavara

// Einstein formulated this problem : he said that only 2% in the world can solve this problem
// There are 5 different parameters each with 5 different attributes
// The following is the problem

// -> In a street there are five houses, painted five different colors (RED, GREEN, BLUE, YELLOW, WHITE)

// -> In each house lives a person of different nationality (GERMAN, NORWEGIAN, SWEDEN, DANISH, BRITAIN)

// -> These five homeowners each drink a different kind of beverage (TEA, WATER, MILK, COFFEE, BEER),

// -> smoke different brand of cigar (DUNHILL, PRINCE, BLUE MASTER, BLENDS, PALL MALL)

// -> and keep a different pet (BIRD, CATS, DOGS, FISH, HORSES)


///////////////////////////////////////////////////////////////////////////////////////
// *************** Einstein's riddle is: Who owns the fish? ***************************
///////////////////////////////////////////////////////////////////////////////////////

/*
Necessary clues:

1. The British man lives in a red house.
2. The Swedish man keeps dogs as pets.
3. The Danish man drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the center house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The Blends smoker lives next to the one who drinks water.
*/




typedef enum bit [2:0]  {red, green, blue, yellow, white} house_color_type;
typedef enum bit [2:0]  {german, norwegian, brit, dane, swede} nationality_type;
typedef enum bit [2:0]  {coffee, milk, water, beer, tea} beverage_type;
typedef enum bit [2:0]  {dunhill, prince, blue_master, blends, pall_mall} cigar_type;
typedef enum bit [2:0]  {birds, cats, fish, dogs, horses} pet_type;




class Einstein_problem;

    rand house_color_type house_color[5];
    rand nationality_type nationality[5];
    rand beverage_type beverage[5];
    rand cigar_type cigar[5];
    rand pet_type pet[5];
        rand int arr[5];
    
    constraint einstein_riddle_solver {
    
        
    
        foreach (house_color[i])
            foreach (house_color[j])
               if (i != j)
                house_color[i] != house_color[j];
        foreach (nationality[i])
            foreach (nationality[j])
               if (i != j)
                nationality[i] != nationality[j];
        foreach (beverage[i])
            foreach (beverage[j])
               if (i != j)
                beverage[i] != beverage[j];
        foreach (cigar[i])
            foreach (cigar[j])
               if (i != j)
                cigar[i] != cigar[j];
        foreach (pet[i])
            foreach (pet[j])
               if (i != j)
                pet[i] != pet[j];
    
    
        //1) The British man lives in a red house.
        foreach(nationality[i])
                (nationality[i] == brit) -> (house_color[i] == red);
                
        
        //2) The Swedish man keeps dogs as pets.
        foreach(nationality[i])
                (nationality[i] == swede) -> (pet[i] == dogs);
                
                
        //3) The Danish man drinks tea.        
        foreach(nationality[i])
                (nationality[i] == dane) -> (beverage[i] == tea);
        
        
        //4) The Green house is next to, and on the left of the White house.
        foreach(house_color[i])        
                 if (i<4)
                    (house_color[i] == green) -> (house_color[i+1] == white);
        
        
        //5) The owner of the Green house drinks coffee.
        foreach(house_color[i])
                (house_color[i] == green) -> (beverage[i] == coffee);
                
        
        //6) The person who smokes Pall Mall rears birds.
        foreach(cigar[i])
                (cigar[i] == pall_mall) -> (pet[i] == birds);
        
        
        //7) The owner of the Yellow house smokes Dunhill.
        foreach(house_color[i])
                (house_color[i] == yellow) -> (cigar[i] == dunhill);
        
        
        //8) The man living in the center house drinks milk.
        foreach(house_color[i])
                if (i==2) // i==2 implies the center house (0,1,2,3,4) 2 is the center
                    beverage[i] == milk;
        
        
        
        //9) The Norwegian lives in the first house.
        foreach(nationality[i])        
                if (i==0) // i==0 is the first house
                    nationality[i] == norwegian;
        
        
        
        //10) The man who smokes Blends lives next to the one who keeps cats.
        foreach(cigar[i])        
                if (i==0) // if the man who smokes blends lives in the first house then the person with cats will be in the second
                    (cigar[i] == blends) -> (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i>0 && i<4) // if the man is not at the ends he can be on either side
                    (cigar[i] == blends) -> (pet[i-1] == cats) || (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i==4) // if the man is at the last
                    (cigar[i] == blends) -> (pet[i-1] == cats);
        
        foreach(cigar[i])        
                if (i==4)
                    (pet[i] == cats) -> (cigar[i-1] == blends);
        
        
        //11) The man who keeps horses lives next to the man who smokes Dunhill.
        foreach(pet[i])
                if (i==0) // similar to the last case
                    (pet[i] == horses) -> (cigar[i+1] == dunhill);
        
        foreach(pet[i])        
                if (i>0 & i<4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill) || (cigar[i+1] == dunhill);
                    
        foreach(pet[i])        
                if (i==4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill);
                    


        //12) The man who smokes Blue Master drinks beer.
        foreach(cigar[i])
                (cigar[i] == blue_master) -> (beverage[i] == beer);
        
        
        //13) The German smokes Prince.
        foreach(nationality[i])        
                (nationality[i] == german) -> (cigar[i] == prince);
        

        //14) The Norwegian lives next to the blue house.
        foreach(nationality[i])
                if (i==0)
                    (nationality[i] == norwegian) -> (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i>0 & i<4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue) || (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i==4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue);
        

        //15) The Blends smoker lives next to the one who drinks water.            
        foreach(cigar[i])            
                if (i==0)
                    (cigar[i] == blends) -> (beverage[i+1] == water);
        
        foreach(cigar[i])        
                if (i>0 & i<4)
                    (cigar[i] == blends) -> (beverage[i-1] == water) || (beverage[i+1] == water);
                    
        foreach(cigar[i])        
                if (i==4)
                    (cigar[i] == blends) -> (beverage[i-1] == water);
        
    } // end of the constraint block
    


    // display all the attributes
    task display ;
        foreach (house_color[i])
            begin
                $display("HOUSE : %s",house_color[i].name());
            end
        foreach (nationality[i])
            begin
                $display("NATIONALITY : %s",nationality[i].name());
            end
        foreach (beverage[i])
            begin
                $display("BEVERAGE : %s",beverage[i].name());
            end
        foreach (cigar[i])
            begin
                $display("CIGAR: %s",cigar[i].name());
            end
        foreach (pet[i])
            begin
                $display("PET : %s",pet[i].name());
            end
        foreach (pet[i])
            if (pet[i] == fish)
                $display("THE ANSWER TO THE RIDDLE : The %s has %s ", nationality[i].name(), pet[i].name());
    
    endtask // end display
    
    
endclass




program main ;

    initial
        begin
            Einstein_problem ep;
            ep = new();
            if(!ep.randomize())
                $display("ERROR");
            ep.display();
        end
endprogram // end of main

        

On May 7, 2013 Cadence announced a 30% productivity gain in the June 2013 Incisive Enterprise Simulator 13.1 release.  Advanced debug visualization, faster turn-around time, and the extension of eight years of low-power verification innovation to IEEE 1801/UPF are the key capabilities in the release.

When we talk about low-power verification its easy to equate it with simulation.  For certain, simulation is the heart of a low-power verification solution. Simulation enables engineers to run their design in the context of power intent.  The challenge is that a simulation-only approach is inadequate. For example, if engineers could achieve SoC quality by verifying the individual function of each power control module (PCM), then simulation could be enough.  For a single power domain, simulation can be enough. 

However, when the SoC has multiple power domains -- and we have seen SoCs with hundreds of them -- engineers have to check the PCMs and all of the arcs between the power modes.  These SoCs often synchronize some of the domain switching to reduce overall complexity, creating the potential for signal skew errors on the control signals for the connected domains.  Managing these complexities requires verification methodologies including advanced debug, verification planning, assertion-based verification, Universal Verification Methodology - Low Power (UVM-LP), and more (see Figure 1).

Figure 1:  Comprehensive Low-Power Verification 

But even advanced verification methodologies on top of simulation aren't enough.  For example, the state machine that defines the legal and illegal power mode transitions is often written in software. The speed and capacity of the Palladium emulation platform is ideal to verify in this context, and it is integrated with simulation sharing debug, UVM acceleration, and static checks for low-power. And, it reports verification progress into a holistic plan for the SoC.  Another example is the ability to compare the design in the implementation flow with the design running in simulation to make sure that what we verify is what we intend to build.

Taken together, verification across multiple engines provides the comprehensive low-power verification needed for today's advanced node SoCs.  That's the heart of this low-power verification announcement. 

Another point you may have noticed is the extension of the Common Power Format (CPF) based power-aware support in the Incisive Enterprise Simulator to IEEE 1801.  We chose to bring IEEE 1801 to simulation first because users like you sometimes need to mix vendors for regression flows.  Over time, Cadence will extend the low-power capabilities throughout its product suite to IEEE 1801.

If you are using CPF today, you already have the best low-power solution. The evidence is clear:  the upcoming IEEE 1801-2013 update includes many of the CPF features contributed to 1801/UPF to enable methodology convergence.  Since you already have those features in the CPF flow, any migration before you have a mature IEEE 1801-2013 tool flow would reduce the functionality you have today.

If you are using Unified Power Format (UPF) 1.0 today, you want to start planning your move toward the IEEE 1801-2013 standard.  A good first step would be to move to the IEEE 1801-2009 standard.  It fills holes in the earlier UPF 1.0 definition.  While it does lack key features in -2013, it is an improvement that will make the migration to -2013 easier. The Incisive 13.1 release will run both UPF 1.0 and IEEE 1801-2009 power intent today.

Over the next few weeks you'll see more technical blogs about the low-power capabilities coming in the Incisive 13.1 release.  You can also join us on June 19 for a webinar that will introduce those capabilities using the reference design supplied with the Incisive Enterprise Simulator release.

=Adam "The Jouler" Sherer

(Yes, "Sherilog" is still here.  :-) )

ST Microelectronics reported their success with IEEE 1801 / UPF low-power simulation using Incisive Enterprise Simulator at CDNLive India in November 2013. We were able to meet with Mohit Jain just after his presentation and recorded this video that explains the key points in his paper.

With eight years of experience and pioneering technology in native low-power simulation, Mohit was able to apply Incisive Enterprise Simulator to a low-power demonstrator in preparation for use with a production set-top box chip.  Mohit was impressed with the ease in which he was able to reuse his existing IEEE 1801 / UPF code successfully, including the power format files and the macro models coded in his Liberty files. Mohit also discusses how he used the power-aware Cadence SimVision debugger.

The Cadence low-power verification solution for IEEE 1801 / UPF also incorporates the patent-pending Power Supply Network visualization in the SimVision debugger.  You can learn more about that in the Incisive low-power verification Rapid Adoption Kit for IEEE 1801 / UPF here in Cadence Online Support.

Just another happy Cadence low-power verification user!

Regards,

 Adam "The Jouler" Sherer 

Hi Team,

I would like to download "Incisive Metrics Center User Guide", I could not find in the cadence/support/manuals. Can you please provide me the link or path to download the same ? I am doing functional coverage with IMC. 

Thank You,

Mahesh

Hi,

I am facing this elaboration error when using Xcelium:

Command>

    xmverilog -v200x +access+r +xm64bit -f vlist -reflib plib -timescale 1ns/1ps

Log>

    xmelab: *E,CUVMUR (<name>.v,538|18): instance 'LUTP0.C GLAT3' of design unit 'tlatntscad12' is unresolved in 'worklib.LUTP0:v'.

I guess the plib was not referred to as the simulation configuration because the tlatntscad12 is included in plib.

The plib is compiled by INCISIVE 13.20 and I am using the Xcelium 19.30.

Please tell me the correct command on how to refer to the library directory compiled by different versions.

Thank you,

Hello All,

Following is the einstein's puzzle solved by cadence Incisive92  (solved in less than 3 seconds -> FAST!!!!!!)

Thanks,

Vinay Honnavara

Verification engineer at Keyu Tech

vinayh@keyutech.com

 // Author: Vinay Honnavara

// Einstein formulated this problem : he said that only 2% in the world can solve this problem
// There are 5 different parameters each with 5 different attributes
// The following is the problem

// -> In a street there are five houses, painted five different colors (RED, GREEN, BLUE, YELLOW, WHITE)

// -> In each house lives a person of different nationality (GERMAN, NORWEGIAN, SWEDEN, DANISH, BRITAIN)

// -> These five homeowners each drink a different kind of beverage (TEA, WATER, MILK, COFFEE, BEER),

// -> smoke different brand of cigar (DUNHILL, PRINCE, BLUE MASTER, BLENDS, PALL MALL)

// -> and keep a different pet (BIRD, CATS, DOGS, FISH, HORSES)


///////////////////////////////////////////////////////////////////////////////////////
// *************** Einstein's riddle is: Who owns the fish? ***************************
///////////////////////////////////////////////////////////////////////////////////////

/*
Necessary clues:

1. The British man lives in a red house.
2. The Swedish man keeps dogs as pets.
3. The Danish man drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the center house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The Blends smoker lives next to the one who drinks water.
*/




typedef enum bit [2:0]  {red, green, blue, yellow, white} house_color_type;
typedef enum bit [2:0]  {german, norwegian, brit, dane, swede} nationality_type;
typedef enum bit [2:0]  {coffee, milk, water, beer, tea} beverage_type;
typedef enum bit [2:0]  {dunhill, prince, blue_master, blends, pall_mall} cigar_type;
typedef enum bit [2:0]  {birds, cats, fish, dogs, horses} pet_type;




class Einstein_problem;

    rand house_color_type house_color[5];
    rand nationality_type nationality[5];
    rand beverage_type beverage[5];
    rand cigar_type cigar[5];
    rand pet_type pet[5];
        rand int arr[5];
    
    constraint einstein_riddle_solver {
    
        
    
        foreach (house_color[i])
            foreach (house_color[j])
               if (i != j)
                house_color[i] != house_color[j];
        foreach (nationality[i])
            foreach (nationality[j])
               if (i != j)
                nationality[i] != nationality[j];
        foreach (beverage[i])
            foreach (beverage[j])
               if (i != j)
                beverage[i] != beverage[j];
        foreach (cigar[i])
            foreach (cigar[j])
               if (i != j)
                cigar[i] != cigar[j];
        foreach (pet[i])
            foreach (pet[j])
               if (i != j)
                pet[i] != pet[j];
    
    
        //1) The British man lives in a red house.
        foreach(nationality[i])
                (nationality[i] == brit) -> (house_color[i] == red);
                
        
        //2) The Swedish man keeps dogs as pets.
        foreach(nationality[i])
                (nationality[i] == swede) -> (pet[i] == dogs);
                
                
        //3) The Danish man drinks tea.        
        foreach(nationality[i])
                (nationality[i] == dane) -> (beverage[i] == tea);
        
        
        //4) The Green house is next to, and on the left of the White house.
        foreach(house_color[i])        
                 if (i<4)
                    (house_color[i] == green) -> (house_color[i+1] == white);
        
        
        //5) The owner of the Green house drinks coffee.
        foreach(house_color[i])
                (house_color[i] == green) -> (beverage[i] == coffee);
                
        
        //6) The person who smokes Pall Mall rears birds.
        foreach(cigar[i])
                (cigar[i] == pall_mall) -> (pet[i] == birds);
        
        
        //7) The owner of the Yellow house smokes Dunhill.
        foreach(house_color[i])
                (house_color[i] == yellow) -> (cigar[i] == dunhill);
        
        
        //8) The man living in the center house drinks milk.
        foreach(house_color[i])
                if (i==2) // i==2 implies the center house (0,1,2,3,4) 2 is the center
                    beverage[i] == milk;
        
        
        
        //9) The Norwegian lives in the first house.
        foreach(nationality[i])        
                if (i==0) // i==0 is the first house
                    nationality[i] == norwegian;
        
        
        
        //10) The man who smokes Blends lives next to the one who keeps cats.
        foreach(cigar[i])        
                if (i==0) // if the man who smokes blends lives in the first house then the person with cats will be in the second
                    (cigar[i] == blends) -> (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i>0 && i<4) // if the man is not at the ends he can be on either side
                    (cigar[i] == blends) -> (pet[i-1] == cats) || (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i==4) // if the man is at the last
                    (cigar[i] == blends) -> (pet[i-1] == cats);
        
        foreach(cigar[i])        
                if (i==4)
                    (pet[i] == cats) -> (cigar[i-1] == blends);
        
        
        //11) The man who keeps horses lives next to the man who smokes Dunhill.
        foreach(pet[i])
                if (i==0) // similar to the last case
                    (pet[i] == horses) -> (cigar[i+1] == dunhill);
        
        foreach(pet[i])        
                if (i>0 & i<4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill) || (cigar[i+1] == dunhill);
                    
        foreach(pet[i])        
                if (i==4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill);
                    


        //12) The man who smokes Blue Master drinks beer.
        foreach(cigar[i])
                (cigar[i] == blue_master) -> (beverage[i] == beer);
        
        
        //13) The German smokes Prince.
        foreach(nationality[i])        
                (nationality[i] == german) -> (cigar[i] == prince);
        

        //14) The Norwegian lives next to the blue house.
        foreach(nationality[i])
                if (i==0)
                    (nationality[i] == norwegian) -> (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i>0 & i<4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue) || (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i==4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue);
        

        //15) The Blends smoker lives next to the one who drinks water.            
        foreach(cigar[i])            
                if (i==0)
                    (cigar[i] == blends) -> (beverage[i+1] == water);
        
        foreach(cigar[i])        
                if (i>0 & i<4)
                    (cigar[i] == blends) -> (beverage[i-1] == water) || (beverage[i+1] == water);
                    
        foreach(cigar[i])        
                if (i==4)
                    (cigar[i] == blends) -> (beverage[i-1] == water);
        
    } // end of the constraint block
    


    // display all the attributes
    task display ;
        foreach (house_color[i])
            begin
                $display("HOUSE : %s",house_color[i].name());
            end
        foreach (nationality[i])
            begin
                $display("NATIONALITY : %s",nationality[i].name());
            end
        foreach (beverage[i])
            begin
                $display("BEVERAGE : %s",beverage[i].name());
            end
        foreach (cigar[i])
            begin
                $display("CIGAR: %s",cigar[i].name());
            end
        foreach (pet[i])
            begin
                $display("PET : %s",pet[i].name());
            end
        foreach (pet[i])
            if (pet[i] == fish)
                $display("THE ANSWER TO THE RIDDLE : The %s has %s ", nationality[i].name(), pet[i].name());
    
    endtask // end display
    
    
endclass




program main ;

    initial
        begin
            Einstein_problem ep;
            ep = new();
            if(!ep.randomize())
                $display("ERROR");
            ep.display();
        end
endprogram // end of main

        

Just in time for the holidays, inside the posted tar ball is some code to solve 9x9 Sudoku puzzles with the Assertion-Driven Simulation (ADS) capability of Incisive Enterprise Verifier (IEV). Enjoy! Joerg Mueller Solutions Engineer for Team Verify

Online Resource