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MathGroup Archive 1997

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Re: Sequence is funny !

  • To: mathgroup at smc.vnet.net
  • Subject: [mg8952] Re: [mg8852] Sequence is funny !
  • From: Allan Hayes <hay at haystack.demon.co.uk>
  • Date: Mon, 6 Oct 1997 01:59:20 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

Gilles BARBIER <Gilles.Barbier at der.edfgdf.fr>
[mg8852] Sequence is funny !
writes

> For some people who like to understand the internal mechanism
> of Mathematica, I give them this little problem a friend sent to me > :
>  Why Table[i,{i,Sequence[1,3]}] gives {1,2,3,4} instead of {1,2,3} > !!
> That's not a bug, I even have an explanation,if you are interested.

Giles,
Very interesting.
Before giving my own explanation here is another teaser:

1/Sequence[2,3,4,5]
	
	 2^3^2^(2/5)
	
Now for Table[i,{i,Sequence[1,3]}].

Table[expr,{i,b}] evaluates as follows
a) Find the number of entries required in the table:
	n =  Floor[1+(b-1)] (Floor allows for b's like 7/2 and 7.8). 

b) Compute n entries starting from 1 and incrementing by 1:
		1, 1+1,       ,1+(n-1)

The example Table[i,{i,Sequence[1,3]}] evaluates as follows.
a) Since table has attribute HoldAll, Sequence[1,3] is passed into  
the calculation as b (Sequence is *not* stripped off immediately to  
give Table[
i,{i,1,3}] ).
So we have
n = Floor[1+(Sequence[1,3]-1)]
	-->Floor[1 + Plus[1,3,-1]]   (*note this*)
	--> Floor[1+3]
	-->4

b) entries =  1, 2, 3, 4

NOTES:
(1) More generally, Table[expr, {i,a,b,s}] evaluates as follows
Find the number of entries required in the table:
n =  Floor[1+((b-a)/s)]
Compute n entries starting from a incrementing by s:
		a, a+s,       ,a+(n-1)s
(default a is 1 and default s  is 1)

(2) We can show that HoldAll is crucial in the example:

ClearAttributes[Table,HoldAll]

Table[i,{i,Sequence[1,3]}]
	
	{1,2,3}

Explanation:  Table[i,{i,Sequence[1,3]}] is changed first to  
Table[i,{i,1,3}]
 .
SetAttributes[Table,HoldAll] (*restore HoldAll*)

(3) The use of parentheses is important in Floor[1+(Sequence[1,3]-1)]
With them we have

Floor[1+(Sequence[1,3]-1)]
	
	4

but without them we get

Floor[1+Sequence[1,3]-1]
	Floor::"argx":
    	"\!\(Floor\) called with \!\(2\) arguments; 1 argument is 	
	expected."

	Floor[1,3]

(4) We can follow the steps in all these calculations:
On[]
On::"trace": "\!\(On[]\) --> \!\(Null\)."
Now evaluate the following:

Floor[1+(Sequence[1,3]-1)]

Table[i,{i,Sequence[1,3]}]

Off[]

Allan Hayes
hay at haystack.demon.co.uk
http://www.haystack.demon.co.uk/training.html
voice:+44 (0)116 2714198
fax: +44 (0)116 2718642
Leicester,  UK


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