Created
November 12, 2011 21:43
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euler 9
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| <html> | |
| <body> | |
| <script type="text/javascript"> | |
| for (a=1;a<=1000;a++) | |
| { | |
| for (b=a+1;b<=1000;b++) | |
| { | |
| c = Math.sqrt(Math.pow(a,2) + Math.pow(b,2)); | |
| sum = a+b+c; | |
| product = a*b*c; | |
| if (sum==1000) | |
| { | |
| document.write(product); | |
| } | |
| } | |
| } | |
| </script> | |
| </body> | |
| </html> |
function is_int(value) {
return (isNaN(value));
}
a = 333;
sum = 0;
product = 0;
while (!(sum == 1000)) {
a--;
b=(1000-2*a);
while(b>10 && !(sum==1000)) {
b--;
c = Math.sqrt(Math.pow(a,2) + Math.pow(b,2));
if (Math.floor(c) == c) {
// console.log(a, b, Math.floor(c), Math.sqrt(Math.pow(c,2)));
sum = a+b+c;
product = a*b*c;
}
}
}
console.log(sum, product, a, b, c);
This is faster, saved some math.
Right- use the m and n relationship to limit scope of a and b, count down
instead of up. that all makes sense.
jake
Jake Dobkin
Gothamist.com
(p) 917 627 6915
(f) 646 349 3893
AIM, YIM, GTalk: jakedobkin
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…On Sun, Nov 13, 2011 at 11:19 AM, David Jacobs < ***@***.*** > wrote:
```
function is_int(value) {
return (isNaN(value));
}
a = 333;
sum = 0;
product = 0;
while (!(sum == 1000)) {
a--;
b=(1000-2*a);
while(b>10 && !(sum==1000)) {
b--;
c = Math.sqrt(Math.pow(a,2) + Math.pow(b,2));
if (Math.floor(c) == c) {
// console.log(a, b, Math.floor(c),
Math.sqrt(Math.pow(c,2)));
sum = a+b+c;
product = a*b*c;
}
}
}
console.log(sum, product, a, b, c);
```
This is faster, saved some math.
---
Reply to this email directly or view it on GitHub:
https://gist.github.com/1361162
OK, a couple other minor nits (I never finished writing that is_int function, which didn't work) and of course while b>a is better than b > 10.
a = 293;
sum = 0;
product = 0;
while (sum != 1000) {
a--;
b=(1000-2*a);
while(b>a && !(sum==1000)) {
b--;
c = Math.sqrt(Math.pow(a,2) + Math.pow(b,2));
if (Math.floor(c) == c) {
// console.log(a, b, Math.floor(c), Math.sqrt(Math.pow(c,2)));
sum = a+b+c;
product = a*b*c;
}
}
}
console.log(sum, product, a, b, c);
a = 293;
sum = 0;
while (sum != 1000) {
a--;
b=(1000-2*a);
while(b>a && !(sum==1000)) {
b--;
c = Math.sqrt(Math.pow(a,2) + Math.pow(b,2));
if (Math.floor(c) == c) {
// console.log(a, b, Math.floor(c), Math.sqrt(Math.pow(c,2)));
sum = a+b+c;
}
}
}
product = a*b*c;
console.log(sum, product, a, b, c);
and computing product can be outside the loop.
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Here's mine - I agree that using natural number sequences would be much faster. The other thing is to count down a from 333 (since we know the numbers sum to 1000 and is the smallest of them).