My Sunday night hobby for the past couple of months is to try and programatically solve the NPR Sunday Puzzle. Each Sunday morning on Weekend Edition Sunday Will Shortz and Liane Hansen offer a puzzle. I have been a fan of this for well over ten years, but recently started using it as a way to have fun programming. Most of the recent puzzles have involved words of a certain type (such as job titles or animals). Since I don't have a complete database of words in a particular category, my resolution of the Sunday puzzle has not been complete, (i.e. to the point of saying "The Answer is: x"). After 5 or 6 weeks of getting a range of possible answers they finally gave me a puzzle that could easily be solved.
I missed last weeks puzzle because I didn't wake up in time, but remembered on Saturday night just in time to solve it before waking up to the answer tomorrow morning.
Last weeks puzzle is:
From Ed Pegg, Jr., who runs the Web site mathpuzzle.com: The numbers 2, 4, 6 and 30 are the first four numbers whose names lack the letter "E." What is the 23rd number whose name lacks an "E?"
I had wondered what the significance of the 23rd number was, if I was more quick witted, that alone should have been enough of a clue to solve the first puzzle of the new year.
Instead, I decided to use T-SQL as the language for this, in part because I think that the .Net Framework has a function that does what I did in the function below, taking the fun out of it.
So, here is the T-SQL code. I know it is choppy and clunky, and the variable names don't make sense, but it works really well (for numbers up to 4 digits long) and was very easy to write. In addition, it is a nice example of recursion, which makes me feel good after reading this post by Joel Spolsky. I noticed as I pasted the code into this post that I have done a number of implicit variable conversions, this code would not hold up well in a code review, but it got the job done.
create
function fn_NumberToText(@value int) returns varchar(100)
as
BEGIN
-- ********************************************************
-- * CODE COPYRIGHT 2006, Shawn Swaner *
-- * http://www.shawnswaner.com *
-- * Published under *
-- * Creative Commons Attribution Share-Alike 2.5 License *
-- * No warranty, express or implied is provide by author *
-- ********************************************************
declare
@name varchar(100), @valueText varchar(9),
@Left1
char(1), @RightOrdinal char(1), @RightPair char(2),
@RightTrio
varchar(3), @Left2 char(2)
set
@valueText = Convert(varchar(5), @value)
if
LEN(@valueText) = 4
BEGIN
set @Left1 = Left(@valueText, 1)
set @name = dbo.fn_NumberToText(@Left1) + ' ' + 'Thousand'
set @RightTrio = RIGHT(@valueText, 3)
set @name = @name + ' ' + dbo.fn_NumberToText(@RightTrio)
END
if
LEN(@valueText) = 3
BEGIN
set
@Left1 = LEFT(@valueText, 1)
set
@name = dbo.fn_NumberToText(@Left1) + ' ' + 'Hundred'
set
@RightPair = Convert(varchar(2), Convert(int, right(@valueText, 2)))
set
@name = @name + ' ' + dbo.fn_NumberToText(@RightPair)
END
if
LEN(@valueText) = 2 and LEFT(@valueText, 1) <> '1'
BEGIN
set
@Left1 = LEFT(@valueText, 1)
set
@RightOrdinal = Right(@valueText, 1)
select
@name = case @Left1
when
2 then 'Twenty'
When
3 then 'Thirty'
when
4 then 'Forty'
when
5 then 'Fifty'
when
6 then 'Sixty'
when
7 then 'Seventy'
when
8 then 'Eighty'
when
9 then 'Ninety'
end
select
@name = @name + ' ' + dbo.fn_NumberToText(@RightOrdinal)
END
if
LEN(@valueText) = 2 and LEFT(@valueText, 1) = '1'
select
@name = case @value
when
10 then 'Ten'
when
11 then 'Eleven'
when
12 then 'Twelve'
When
13 then 'Thirteen'
when
14 then 'Fourteen'
when
15 then 'Fifteen'
when
16 then 'Sixteen'
when
17 then 'Seventeen'
when
18 then 'Eighteen'
when
19 then 'Nineteen'
end
if
LEN(Convert(varchar(4), @value)) = 1
select
@name = case @value
when
1 then 'One'
when
2 then 'Two'
When
3 then 'Three'
when
4 then 'Four'
when
5 then 'Five'
when
6 then 'Six'
when
7 then 'Seven'
when
8 then 'Eight'
when
9 then 'Nine'
when
0 then ''
end
return
@name
end
go
-- test statement
-- select dbo.fn_NumberToText(1111)
-- script to solve the problem
declare
@iter int, @match int, @text varchar(100)
set
@iter = 1
set
@match = 0
while
@match < 23
BEGIN
set
@text = dbo.fn_NumberToText(@iter)
if
charindex('e', @text) = 0
BEGIN
set @match = @match + 1
if @match = 23
print 'The Answer is: ' + @Text
-- TESTING CODE -- print Convert(char(2), @match) + ': ' + @text
END
set
@iter = @iter + 1
END
This work is licensed under a
Creative Commons Attribution-ShareAlike 2.5 License.