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more fun with decimal arithmetic
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| epc8@juno.com 2006-05-19, 3:55 am |
| A number of years ago a small university in western Canada tried a
publicity stunt to recruit prospective students. It offered a
scholarship to the first group of students who could solve a puzzle.
The last part of the puzzle instructed the contestants to add up the
decimal digits in the numbers from one to one million and send the
answer in for a chance at the prize.
One way to find this answer is to use some math. I'll skip the math
:-). Another way is to use brute force and write a computer program to
find the answer.
Just for fun, along the lines of the "99 bottles of beer program", see
how easy this is in COBOL.
[For math history buffs, this is not along the same lines as Gauss
adding the numbers from 1 to 100 in grade school.]
-- Elliot
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| Alistair 2006-05-20, 7:55 am |
| For anybody interested in such activities, Personal Computer World
magazine published a series of articles (back in the 90s) for idle
minds to occupy themselves. I believe they were authored by Mike Mudge
and may be available in a collected edition.
epc8@juno.com wrote:
> A number of years ago a small university in western Canada tried a
> publicity stunt to recruit prospective students. It offered a
> scholarship to the first group of students who could solve a puzzle.
> The last part of the puzzle instructed the contestants to add up the
> decimal digits in the numbers from one to one million and send the
> answer in for a chance at the prize.
>
> One way to find this answer is to use some math. I'll skip the math
> :-). Another way is to use brute force and write a computer program to
> find the answer.
>
> Just for fun, along the lines of the "99 bottles of beer program", see
> how easy this is in COBOL.
>
> [For math history buffs, this is not along the same lines as Gauss
> adding the numbers from 1 to 100 in grade school.]
>
> -- Elliot
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| Howard Brazee 2006-05-22, 7:55 am |
| On 18 May 2006 21:15:31 -0700, epc8@juno.com wrote:
>A number of years ago a small university in western Canada tried a
>publicity stunt to recruit prospective students. It offered a
>scholarship to the first group of students who could solve a puzzle.
>The last part of the puzzle instructed the contestants to add up the
>decimal digits in the numbers from one to one million and send the
>answer in for a chance at the prize.
So we're all thinking Gauss now.
>One way to find this answer is to use some math. I'll skip the math
>:-). Another way is to use brute force and write a computer program to
>find the answer.
Not much force is required when we have a computer and very easy code.
>Just for fun, along the lines of the "99 bottles of beer program", see
>how easy this is in COBOL.
>
>[For math history buffs, this is not along the same lines as Gauss
>adding the numbers from 1 to 100 in grade school.]
How is it different?
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| Peter Lacey 2006-05-22, 6:55 pm |
| Howard Brazee wrote:
>
> On 18 May 2006 21:15:31 -0700, epc8@juno.com wrote:
>
>
So: if "sigma" = n(n + 1)/2 to calculate the sum of 1, 2, ..., n, then
the result would be
sigma (1000000) + (sigma(1000000) - sigma(100000)) (for 100000 to
1000000) etc. etc.?
Or are we calcualting the sum of each column and adding the six totals
together?
PL
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| epc8@juno.com 2006-05-22, 6:55 pm |
|
Howard Brazee wrote:
> On 18 May 2006 21:15:31 -0700, epc8@juno.com wrote:
>
>
> So we're all thinking Gauss now.
>
>
> Not much force is required when we have a computer and very easy code.
>
>
> How is it different?
The digits have no positional value. 10 contributes a value of 1 not a
value of 10.
99 has a value of 18, for example.
-- Elliot
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| epc8@juno.com 2006-05-22, 6:55 pm |
|
Peter Lacey wrote:
> Howard Brazee wrote:
>
>
> So: if "sigma" = n(n + 1)/2 to calculate the sum of 1, 2, ..., n, then
> the result would be
>
> sigma (1000000) + (sigma(1000000) - sigma(100000)) (for 100000 to
> 1000000) etc. etc.?
>
> Or are we calcualting the sum of each column and adding the six totals
> together?
>
> PL
Yes. Exactly. The digits in this problem have no positional (place)
value at all. So 99 contributes 18 to the sum, for example.
-- Elliot
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| Howard Brazee 2006-05-22, 6:55 pm |
| On 22 May 2006 10:38:42 -0700, epc8@juno.com wrote:
>The digits have no positional value. 10 contributes a value of 1 not a
>value of 10.
>99 has a value of 18, for example.
OK. It's still a trivial CoBOL task. It might be more interesting
to have a mathematical short-cut, but that won't make for a better
program.
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| epc8@juno.com 2006-05-22, 6:55 pm |
|
Howard Brazee wrote:
> On 22 May 2006 10:38:42 -0700, epc8@juno.com wrote:
>
>
> OK. It's still a trivial CoBOL task. It might be more interesting
> to have a mathematical short-cut, but that won't make for a better
> program.
How might you solve the problem in other languages? One way is to take
each number in binary and convert it to base 10 using quotient and
remainder operations. A second way is to convert each number to a
string, then add up the digits.
Just for fun, here is a version in Microsoft BASIC:
10 T#=0
20 FOR I=1 TO 1000000!
30 S$=STR$(I)
40 FOR J=2 TO LEN(S$)
50 T#=T#+ASC(MID$(S$,J,1))-ASC("0")
60 NEXT J
70 NEXT I
80 PRINT T#
90 SYSTEM
[J starts at 2 to skip the leading space in S$.]
In CoBOL it's a trivial program becasue there is no type conversion
involved at all.
-- Elliot
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| Oliver Wong 2006-06-17, 7:55 am |
|
<epc8@juno.com> wrote in message
news:1148320969.215163.4070@i40g2000cwc.googlegroups.com...
>
> Howard Brazee wrote:
>
> How might you solve the problem in other languages? One way is to take
> each number in binary and convert it to base 10 using quotient and
> remainder operations. A second way is to convert each number to a
> string, then add up the digits.
>
> Just for fun, here is a version in Microsoft BASIC:
>
> 10 T#=0
> 20 FOR I=1 TO 1000000!
> 30 S$=STR$(I)
> 40 FOR J=2 TO LEN(S$)
> 50 T#=T#+ASC(MID$(S$,J,1))-ASC("0")
> 60 NEXT J
> 70 NEXT I
> 80 PRINT T#
> 90 SYSTEM
>
> [J starts at 2 to skip the leading space in S$.]
>
> In CoBOL it's a trivial program becasue there is no type conversion
> involved at all.
Sorry for the late reply. It's also trivial in any language which has
the addition, division and modulus operator, and does truncation for integer
division, and has a conditional looping construct.
int sum = 0;
for (int i = 1 to 1000000) {
int temp = i;
while (temp > 0) {
sum += temp % 10;
temp = temp / 10;
}
}
- Oliver
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