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RE: syntax error

• To: mathgroup at smc.vnet.net
• Subject: [mg37905] RE: [mg37901] syntax error
• From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
• Date: Tue, 19 Nov 2002 03:51:16 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

>-----Original Message-----
>From: David Afonso [mailto:david-afonso at iol.pt]
>Sent: Monday, November 18, 2002 6:53 AM
>To: mathgroup at smc.vnet.net
>Subject: [mg37905] [mg37901] syntax error
>
>
>there must be a syntax error..
>
>Could you help me find it?
>
>f = Function[pp,
>    Module[{r, aux},
>      r = new;
>      aux = pp;
>      While[emptyQ[aux] == False,
>          While[emptyQ[top[aux]] == False,
>            r = push[top[top[aux]], r]; 
>            aux = push[pop[top[aux]] , pop[aux]]
>            ];
>          aux = pop[aux]
>          ]
>        r]]
>
>
>
>it gives back something estrange when applied to a list of 
>lists. Like: {3Null,7Null...etc}
>
>

(1) Not a syntax error, but most probably (at least!) a missing semicolon
after the outermost While (before you return r). This is kind of standard
typo whenever results of this sort show up.

(2) It is, however, difficult to answer a question of this sort, I mean,
without being told what new etc. is. So the answer certainly must be "who
knows?"

(3) Such, BTW, you leave out a chance for better advice!


Now, if I make a guess and construe stacks as lists, and make up the most
simple implementation of push, pop etc. I get with your code:

In[29]:= pp = Partition[Range[9], 3]
Out[29]= {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}

In[30]:= f[pp]
Out[30]=
{9 Null, 8 Null, 7 Null, 6 Null, 5 Null, 4 Null, 3 Null, 2 Null, Null}

Having added that semicolon mentioned it becomes

In[33]:= f[pp]
Out[33]=
{9, 8, 7, 6, 5, 4, 3, 2, 1}

Perhaps you will have been lucky too.

In[34]:= Quit[]


However I'd not implement stacks that way, 
and rather code your function more like this:


f[ss : stack[__stack]] :=
  Module[{r, s = ss, t}, r = stack[new];
    While[! stackEmptyQ[s],
      t = top[s]; pop[s];
      While[! stackEmptyQ[t],
        push[r, top[t]]; pop[t]]
      ]; r]
 

Let it run:

In[45]:= pp = stack[new]
Out[45]=
\[SkeletonIndicator]empty stack\[SkeletonIndicator]

In[46]:=
With[{s = Unique[p]}, s = stack[new]; Scan[push[s, #] &, #]; s] & /@ 
  Partition[Range[9, 1, -1], 3]
Out[46]=
{\[SkeletonIndicator]stack\[SkeletonIndicator], \[SkeletonIndicator]stack\
\[SkeletonIndicator], \[SkeletonIndicator]stack\[SkeletonIndicator]}

In[47]:= Scan[push[pp, #] &, %]

In[48]:= Apply[List, pp, {0, 1}]
Out[48]= {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}

In[49]:= f[pp]
Out[49]= \[SkeletonIndicator]stack\[SkeletonIndicator]
In[50]:= List @@ %
Out[50]= {9, 8, 7, 6, 5, 4, 3, 2, 1}

In[51]:= Apply[List, pp, {0, 1}]
Out[51]= {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}

This is consistent with your function, however I had to create and
initialize stack objects using their methods.

You might like to guess what this implementation of push, pop etc. was.


--
Hartmut Wolf