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Which NHL team has the deepest farm system? How about the weakest? Which are trending up or down? We reset the pipelines with prospect system superlatives.

Hi,

Is it possible to use a SystemVerilog package which is compiled into a library and then use it in a VHDL-2008 design file? Is such mixed-language flow supported?

I'm considering the latest versions of Incisive / Xcelium available today (Oct 2019).

Thank you,

Michal

I have been working on a an ASA 17.2 project for the last 6 months.

When I go to Project --> Settings, the settings window does not open. 

The tool indicates that a window is open, as I cannot click on anything else in the project. But it does not show the Settings window.

This has been happening only for the last 2 months. Before that it was working fine.

If I send the project to my colleague, the settings window shows up for him.

It’s now official: Perspec System Verifier is rated the #1 product in the #1 category of Portable Stimulus, according to the 2017 EDA User Survey published on Deepchip.com. There were 33 user responses in favor of Perspec as the #1 tool, and dr...(read more)

Cadence continues to be a leader in SoC verification and has expanded our industry investment in Accellera portable stimulus language standardization. Some customers have expressed reservations that portable stimulus requires the effort of learn...(read more)

Hi All,

I've just about to finish my first PCB layout, and I want to understand some 2 issues better:

1. Soldermask layer: when exactly do we want to define for some SMT pad (say at Pad Designer tool), to have soldermask top and when we want to define solermask bottom? 

if it's a TH pad, I guess we always want to define both layers soldermask (top, bottom), because the pads are crossing all the layers. However, if it's a SMT pad, which SM we want?

2. Pastemask layer: is this layer necessary for the gerber files generation, when we have SMT components in our circuit?

And again, when we define for a TH/SMT pads pastemask top, and when Pastemask bottom?

Thanks!

Hi, I am uploading a register class, which can be used for modeling hardware registers. I am uploading the source code and examples on how to run it. I also have a user guide which has all the APIs listed and explained. The user guide is ARV.pdf in the attached tar file. I have named the class ARV, which stands for Architect's Register View. It has got very good randomization and coverage features. Users have told me that its better than RAL. You can download it from http://verisilica.info/ARV.php
. There is a limit of 750KB in this cadence website. The ARV file is 4MB. That is why, I am uploading it at this site. I have a big pdf documentation and a doxygen documentation there. That is the reason for the bigger file size. The password to open the ZIP file is ovm_arv. I hope, everyone will use these classes.

Please contact me for any help.
Regards ANil

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

        

Hi Everyone,

Many of you would have heard about the Cadence Stylus Common UI and are wondering what it is and what the advantages might be to use it versus legacy UI.

The webinar answers the following questions:

• Why did Cadence develop Stylus UI and what is Stylus Common UI?
• How does someone invoke and use the Stylus Common UI?
• What are some of the important and useful features of the Stylus Common UI?
• What are the key ways in which the Stylus Common UI is different from the default UI?​

If you want to learn more about Stylus UI in the context of implementation, view the 45-minute recorded webinar on the Cadence support site.

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The Mandos system allows computers to have encrypted root file systems and at the same time be capable of remote or unattended reboots. The computers run a small client program in the initial RAM disk environment which will communicate with a server over a network. All network communication is encrypted using TLS. The clients are identified by the server using an OpenPGP key that is unique to each client. The server sends the clients an encrypted password. The encrypted password is decrypted by the clients using the same OpenPGP key, and the password is then used to unlock the root file system.