Talk:Pascal's triangle: Difference between revisions
Line 26:
for i in range(len(row)):
row[i] = str(format(row[i], '6d'))
print(tabs+''.join( str(row[i]) for i in range(len(row)) )
for i in range(len(row)):
row[i] = int(row[i])
|
Revision as of 04:41, 14 March 2012
Right Output Format?
The task bdescription says the triangle looks like this:
1 1 1 1 2 1 1 3 3 1
And yet, some examples are showing the easier to construct:
1 1 1 1 2 1 1 3 3 1
I think that maybe all example output should follow the task description format of an isosceles triangle. --Paddy3118 08:59, 27 December 2009 (UTC)
- That's not always easy to do. I think the important part of the task is the generation of each row. We don't need to complicate it with output formatting that isn't important to the theory involved. --Mwn3d 18:37, 27 December 2009 (UTC)
- Need it or not, the Python~2 code as available strays from theory by virtue of the return statement and status as a function rather than a complete program. I submit that completeness and conformity trumps raw theory in the case of code examples and would serve a broader public with formatting applied. I further submit the following additions as alternative and addition to present code under the potential heading of "Traditionally Formatted, Python~3":
- <lang python>def pascal(n):
# Prints out n rows of Pascal's triangle, traditionally formatted. # Good for 1 <= n <= 20 rows. row = [1] k = [0] z = n-1 for x in range(max(n,0)): tabs = ' '*z for i in range(len(row)): row[i] = str(format(row[i], '6d')) print(tabs+.join( str(row[i]) for i in range(len(row)) ) ) for i in range(len(row)): row[i] = int(row[i]) row=[l+r for l,r in zip(row+k,k+row)] z -= 1</lang>Likely, there are more elegant ways. This is to my ability. --Jnever1 04:35, 14 March 2012 (UTC)
J Explanation
<lang j>!/~@i. N</lang>
The triangle itself is simply the table of number-of-combinations, for the first N non-negative integers.
<lang J> !/~i.5 1 1 1 1 1 0 1 2 3 4 0 0 1 3 6 0 0 0 1 4 0 0 0 0 1</lang>
That is, C(n,k) for all n,k in [0 .. n). J's notation for C(n,k) is k ! n
(mnemonic: combinations are closely related to factorials, which are denoted by ! in math).
So, for example, the number of ways to choose a poker hand (5 cards from the deck of 52): <lang j> 5!52 2598960</lang>
So !
is the mathematical choose function. As for /~@i.
, the /~
would be "table of" and i.
"the first N non-negative integers (i.e. 0 .. N-1)". (And @
is just glue.)
But, formatting the thing takes a bit more effort:
<lang j> (-@|. |."_1 [: ;:inv [: ":@-.&0&.>@|: !/~)@i. 5
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1</lang>
Here, we have refactored the operation slightly:
<lang j>(stuff)@i.5</lang> makes the sequence 0 1 2 3 4 available every odd verb (counting from the right) in stuff
The "stuff" are (from right to left) <lang j>!/~</lang> build our table of combinations <lang j>":@-.&0&.>@|:</lang> transpose that array, and for each number remove it if it is zero, format it as a string and put it in a box <lang j>[:</lang> placeholder meaning "no verb here" the verb to the right does not get a left argument <lang j>;:inv</lang> combine the strings in rows of boxes as word (with spaces between them <lang j>[:</lang> placeholder meaning "no verb here" the verb to the right does not get a left argument <lang j>|."_1</lang> rotate each item (each row of characters) by the indicated amount (rotation would be moving contents left except that the amounts are negative). <lang j>-@|.</lang> reverse and negate the argument (resulting in the negative sequence _4 _3 _2 _1 0)