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MathGroup Archive 2001

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RE: MatrixTemplate (was SpecialMatrix)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28451] RE: [mg28414] MatrixTemplate (was SpecialMatrix)
  • From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.de>
  • Date: Thu, 19 Apr 2001 03:26:39 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Allan, dear Carl,

these are most miraculous discoveries you both made. 
The Question arises, what makes CreateMatrix such fast?
and what did you really get? a fully laid-out matrix?
This is not so, I believe. For my investigation 
please see below...

> -----Original Message-----
> From: Allan Hayes [mailto:hay at haystack.demon.co.uk]
To: mathgroup at smc.vnet.net
> Sent: Tuesday, April 17, 2001 5:54 AM
> To: mathgroup at smc.vnet.net
> Subject: [mg28451] [mg28414] MatrixTemplate (was SpecialMatrix)
> 
> 
> The problem of creating a matrix with all the entries the 
> same, say all
> zero, has cropped up in recent postings.
> The following seems to be very quick way of doing this:
> 
> Clear["`*"]
> CreateMatrix[x_,m_,n_]:=
>   PadRight[{#},m,{#}]&[PadRight[{x},{n},{x}]]
> 
> Check
> 
> CreateMatrix[0,3,5]
> 
>         {{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0}}
> 
> Timings
> 
> CreateMatrix[0,5000,5000];//Timing
> CreateMatrix[0,5000,2500];//Timing
> CreateMatrix[0,2500,5000];//Timing
> CreateMatrix[0,5000,250];//Timing
> CreateMatrix[0,250,5000];//Timing
> 
>         {0. Second,Null}
> 
>         {0.11 Second,Null}
> 
>         {0. Second,Null}
> 
>         {0.06 Second,Null}
> 
>         {0.16 Second,Null}
> 
> For comparison:
> 
> IdentityMatrix[5000];//Timing
> 
>         {10.27 Second,Null}
> 
> and. using Table,
> 
> Clear["`*"]
> CreateMatrix[x_,m_,n_]:=
>   Table[Evaluate[Table[x,{n}]],{m}]
> 
> CreateMatrix[0,3,5]
> 
>         {{0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0}}
> 
> CreateMatrix[0,5000,5000];//Timing
> CreateMatrix[0,5000,2500];//Timing
> CreateMatrix[0,2500,5000];//Timing
> CreateMatrix[0,5000,250];//Timing
> CreateMatrix[0,250,5000];//Timing
> 
>         {10.22 Second,Null}
> 
>         {5. Second,Null}
> 
>         {4.67 Second,Null}
> 
>         {0.44 Second,Null}
> 
>         {0.55 Second,Null}
> 
> --
> Allan
> ---------------------
> Allan Hayes
> Mathematica Training and Consulting
> Leicester UK
> www.haystack.demon.co.uk
> hay at haystack.demon.co.uk
> Voice: +44 (0)116 271 4198
> Fax: +44 (0)870 164 0565
> 
> 
> 
> 
[Hartmut Wolf]

In[1]:= $Version
Out[1]= "4.0 for Microsoft Windows (April 21, 1999)"
In[2]:= mem0 = MemoryInUse[]
Out[2]= 1129872
In[3]:= k0 = 80554 K ;

...this is the memory used as read off Microsoft Task Manager 
at my computer (Windows 2000).
 
Now we go through a sequence of steps, which are obvious:

In[4]:=
m1 = PadRight[{#}, 2000, {#}] &@PadRight[{}, 3000, 0]; // Timing
Out[4]= {0.01 Second, Null}
In[5]:= mem1 = MemoryInUse[]
Out[5]= 1158200
In[6]:= k1 = 80532 K;

In[7]:=
m2 = PadRight[{{}}, {2000, 3000}, 0]; // Timing
Out[7]= {1.752 Second, Null}
In[8]:= mem2 = MemoryInUse[]
Out[8]= 25217776
In[9]:= k2 = 104100 K; 

In[10]:=
m3 = Table[0, {2000}, {3000}]; // Timing
Out[10]= {2.965 Second, Null}
In[11]:= mem3 = MemoryInUse[]
Out[11]= 49221952
In[12]:= k3 = 127560 K;


...just for comparison, a Matrix of same size:

In[13]:=
m4 = Partition[Range[2000*3000], 3000]; // Timing
Out[13]= {0.17 Second, Null}
In[14]:= mem4 = MemoryInUse[]
Out[14]= 73225392
In[15]:= k4 = 151024 K;

Now let's look at the memory used (per element):
 
In[16]:=
Subtract @@@ Partition[{mem4, mem3, mem2, mem1, mem0}, 2, 1]/(2000.*3000.)
Out[16]=
{4.00057, 4.0007, 4.00993, 0.00472133}

We see m2, m3, m4 use up four bytes/element each, which is what we 
expect. However m1 is much more compact!

In[17]:=
Subtract @@@ Partition[{k4, k3, k2, k1, k0}, 2, 1]
Out[17]=
{23464 K, 23460 K, 23568 K, -22 K}

This is memory consumption as read off the MS TaskManager. 
This is roughly consistent with recorded MemoryInUse.

In[18]:= MemoryInUse[]
Out[18]= 73231408
In[19]:= kk0 = 151412 K;
 
Now we work with our "objects"

In[20]:= m3 == m4 // Timing
Out[20]= {0. Second, False}

It takes no time, to find out the discrepancy, ok.

In[21]:= MemoryInUse[]
Out[21]= 73234112
In[22]:= kk1 = 151424 K;

...and the comparison takes no memory, ok.
However...

In[23]:= m1 == m4 // Timing
Out[23]= {4.797 Second, False}

... the comparison with m1 takes up more time, than was 
gained in first place! Not ok!

In[24]:= MemoryInUse[]
Out[24]= 73236368
In[25]:= kk2 = 275144 K;

...what happend here? Mathematica doesn't show additional
memory used, however much had been reclaimed from the OS!

In[26]:= m2 == m4 // Timing
Out[26]= {8.372 Second, False}
In[27]:= MemoryInUse[]
Out[27]= 73238624
In[28]:= kk3 = 275144 K;

..also much time for m2 needed! 

In[29]:= m1 == m2 // Timing
Out[29]= {0.521 Second, True}
In[30]:= MemoryInUse[]
Out[30]= 73240896

...this is fast

In[31]:= kk4 = 275144 K;
In[32]:= m2 == m3 // Timing
Out[32]= {5.217 Second, True}
In[33]:= MemoryInUse[]
Out[33]= 73243144

In[34]:= kk5 = 275544 K;
In[35]:= m1 == m3 // Timing
Out[35]= {4.887 Second, True}
In[36]:= MemoryInUse[]
Out[36]= 73245496
In[37]:= kk6 = 275168 K;

I admit that these last timings are not fully conclusive,
and might be somewhat dependend on the structure of the hardware 
used (This is a notebook computer from my company:
hp OmniBook 4150, and possibly there is a performance 
step down at 128 MB due to chip set, furthermore there 
could be some influence of paging for the last operations 
(paging times are not recorded by Timing[] though), 
my real memory was 192 MB).

I will refrain from fantasizing about what is going on,
whowever someone familiar with "Mathematica inside" 
is asked to comment. Esp. to the question: what must be 
paid for "rapid matrix creation", and when?

Kind regards, Hartmut



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