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Adapted from
Project Euler problem 370
Quote:
Let us define a geometric triangle as an integer sided triangle with sides a <= b <= c so that its sides form a geometric progression, i.e. b^2 = a*c.
An example of such a geometric triangle is the triangle with sides a = 144, b = 156 and c = 169.
There are 42 geometric triangles with perimeter <= 100. Appropriately for this problem F(100)=42.
Write a program (42S or platform of your choice) optimized for speed then for size that can solve F(1000) on a real 42S or F(1e8) on PC emulator in a reasonable time.
Edited: 11 Mar 2012, 1:05 a.m.
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Unoptimized and very slow, I fear:
00 { 71Byte Prgm }
01>LBL 00
02 STO 00
03 3
04 BASE÷
05 1E3
06 +
07 STO 01
08 IP
09 STO 02
10>LBL 01
11 RCL 01
12 1
13 BASE
14 STO 03
15>LBL 02
16 RCL 01
17 IP
18 X^2
19 RCL÷ 03
20 STO 04
21 IP
22 RCL 03
23 RCL+ 01
24 X<Y?
25 GTO 04
26 RCL 04
27 FP
28 X!=0?
29 GTO 03
30 RCL 00
31 RCL 04
32 RCL 01
33 BASE+
34 RCL+ 03
35 X>Y?
36 GTO 03
37 1
38 STO+ 02
39>LBL 03
40 1
41 STO 03
42 GTO 02
43>LBL 04
44 DSE 01
45 GTO 01
46 RCL 02
47 .END.
100 XEQ 00 > 42 (1 min 19 sec, real HP42S)
1000 XEQ 00 > 532 (~1 second (Free42 Decimal)
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Quote:
There are 42 geometric triangles with perimeter <= 100.
I get a significant different number of 'goemetric integer traiangle' !
I may have miss samething, or badly interpret the conditions.
I count 83 integer sided triangles (a,b,c) with a <= b <=c and perimeter P less to 100 who satisfed b^{2} = a.c.
Are you sure you are not counting integer sided triangles with different side's length restrictions (such as 1 < a < b < c ) ?
Edited: 11 Mar 2012, 5:45 p.m. after one or more responses were posted
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Quote:
I count 83 integer sided triangles (a,b,c) with a <= b <=c and perimeter P less to 100.
Remember it must also satisfy b^2 = a*c
Are you checking that condition also?
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What is step "24 X<Y?" testing for? It looks like a+b < c as a loop termination condition. Why is this necessary?
Apart from the trivial cases of a=b=c this program would seem to produce (I converted it to C to play with so could be wrong here) these nine extras:
25 30 36
20 30 45
18 24 32
16 24 36
16 20 25
12 18 27
9 12 16
8 12 18
4 6 9
My next question is where are triples like 1 3 9 or 14 28 56? These look to be in geometric progression and satisfy a c = b^{2}. In fact I get fifty of these which include the nine above:
25 30 36
20 30 45
18 30 50
16 28 49
14 28 56
13 26 52
18 24 32
16 24 36
12 24 48
9 24 64
11 22 44
9 21 49
7 21 63
16 20 25
10 20 40
8 20 50
12 18 27
9 18 36
6 18 54
8 16 32
4 16 64
9 15 25
5 15 45
3 15 75
7 14 28
4 14 49
9 12 16
8 12 18
6 12 24
4 12 36
3 12 48
2 12 72
5 10 20
4 10 25
2 10 50
3 9 27
1 9 81
4 8 16
2 8 32
1 8 64
1 7 49
4 6 9
3 6 12
2 6 18
1 6 36
1 5 25
2 4 8
1 4 16
1 3 9
1 2 4
Add to this the 33 a=b=c cases and I'm at 83.
What am I missing or is the problem stated incorrectly?
 Pauli
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I'm getting 83 too (see above). They all satisfy the condition as stated.
 Pauli
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I was getting 83 also, but then I realized we are looking for triangles. In a triangle the length of one side cannot exceed the sum of the lengths of the other two sides, as we know. That's what step 24 is testing for.
Gerson.
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Quote:
What is step "24 X<Y?" testing for? It looks like a+b < c as a
How do you know this is a triangle:
1 9 81
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Yes, all the 83 I count satisfy b^{2}=ac condition.
I also get 50 by considering only a < b < c.
a b c P=a+b+c a b c P=a+b+c
1 2 4 7 nat 1 1 1 3
1 3 9 13 nat 2 2 2 6
2 4 8 14 nat 3 3 3 9
4 6 9 19 4 4 4 12
1 4 16 21 nat 5 5 5 15
3 6 12 21 nat 6 6 6 18
2 6 18 26 nat 7 7 7 21
4 8 16 28 nat 8 8 8 24
1 5 25 31 nat 9 9 9 27
5 10 20 35 nat 10 10 10 30
9 12 16 37 11 11 11 33
8 12 18 38 12 12 12 36
3 9 27 39 nat 13 13 13 39
4 10 25 39 nat 14 14 14 42
2 8 32 42 nat 15 15 15 45
6 12 24 42 nat 16 16 16 48
1 6 36 43 nat 17 17 17 51
7 14 28 49 nat 18 18 18 54
9 15 25 49 nat 19 19 19 57
4 12 36 52 nat 20 20 20 60
8 16 32 56 nat 21 21 21 63
1 7 49 57 nat 22 22 22 66
12 18 27 57 23 23 23 69
16 20 25 61 24 24 24 72
2 10 50 62 nat 25 25 25 75
3 12 48 63 nat 26 26 26 78
9 18 36 63 nat 27 27 27 81
5 15 45 65 nat 28 28 28 84
4 14 49 67 nat 29 29 29 87
10 20 40 70 nat 30 30 30 90
1 8 64 73 nat 31 31 31 93
18 24 32 74 32 32 32 96
16 24 36 76 33 33 33 99
11 22 44 77 nat
6 18 54 78 nat
8 20 50 78 nat
9 21 49 79 nat
4 16 64 84 nat
12 24 48 84 nat
2 12 72 86 nat
1 9 81 91 nat
7 21 63 91 nat
13 26 52 91 nat
25 30 36 91
3 15 75 93 nat
16 28 49 93 nat
20 30 45 95
9 24 64 97 nat
14 28 56 98 nat
18 30 50 98 nat
Edit : nat = not a triangle
Edited: 12 Mar 2012, 11:46 a.m. after one or more responses were posted
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Quote:
I'm getting 83 too (see above). They all satisfy the condition as stated.
Yes, you're on the right track. Your list satisfies every condition except one very important one... See my question above.
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Quote:
Yes, all the 83 I count satisfy b^{2}=ac condition.
I also get 50 by considering only a < b < c.
Ok, I assume you also get the 1,2,4 shape as above? What is the largest angle in a triangle with sides 1,2 and 4?
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Quote:
I'm getting 83 too (see above). They all satisfy the condition as stated.
 Pauli
As has already been said, the original statement explicitly says we're looking for triangles, not mere triplets of integers, and in every triangle you have that none of the sides can equal or exceed the sum of the others.
So, if you've got alleged triangle sides A, B, and C, then they must satisfy:
A+B > C
A+C > B
B+C > A
else you can't construct a triangle having those sides. If in doubt, just try to draw a triangle having sides 1, 2, and 4, and see if you succeed. You won't.
With that restriction in place you get exactly 42 integer triplets which can indeed form a triangle whose perimeter doesn't exceed 100.
Best regards from V.
Edited to correct a typo, as kindly pointed out below
Edited: 11 Mar 2012, 7:38 p.m. after one or more responses were posted
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OK.
You ask the good question.
I found 83 triplets (a,b,c), but I have not check at all if there are triangles !
In fact (1,2,4) is not a triangle, I found no way to draw it. One of the edge is too short.
That what I miss in the problem; a triplet (a,b,c) is not necessery a triangle.
Thanks.
I have to found a way to check for triangularity and to adjust my algorithm.
'
00 { 59Byte Prgm }
01>LBL "FTRI"
02 STO 01 @ Store perimeter in R01
03 3
04 ÷
05 IP @ Initate a to INT(P/3)
06 0 @ Initiate counter R00 to zero
07 STO 00
08 Rv
09>LBL 00 @  Main loop (over a in stack x:)
10 RCL ST X
11>LBL 01 @  Loop over b from a to ...
12 RCL ST X
13 RCL+ ST Z @ place a+b in stack
14 RCL ST Y
15 X^2
16 RCL÷ ST T @ estimate c = b.b/a
17 X>Y?
18 GTO 02 @ test triangle if ( c > a+b ) exit
19 +
20 RCL 01
21 X<Y?
22 GTO 02 @ test perimeter if ( a+b+c > P ) exit
23 +
24 FP
25 X=0? @ test c integer if c interger count triangle
26 ISG 00
27 NEG
28 Rv
29 1 @ increase b
30 +
31 GTO 01
32>LBL 02 @  exit
33 R^
34 DSE ST X
35 GTO 00 @ loop while a>0
36 RCL 00 @  recall counter
37 END
Register:
R00 store n (number of geometric triangles)
R01 store perimeter P
usage:
100 EXQ FTRI > 42
1000 EXQ FTRI > 532
10000 EXQ FTRI > 6427
Edited: 11 Mar 2012, 9:01 p.m.
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Quote:
So, if you've got alleged triangle sides A, B, and C, then they must satisfy:
A+B < C
A+C < B
B+C < A
If you change all 3 '<' to '>', then you're right. ;)
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Yes, of course, thank you.
Unless I incur in some other typo, I get (in reasonable times):
F(100) = 42
F(1000) = 532
F(10000) = 6427
Best regards from V.
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Valentin,
Greetings, What platform are you using? The trusty 71b?
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I knew I was missing something obvious, I feel silly now :)
I've not been thinking well recently :(
 Pauli
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An alternative to the usual triangle inequality arises from Heron's formula, which makes it clear that the semiperimeter of an actual triangle is greater than all three side lengths.
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Quote:
10000 EXQ FTRI > 6427
A very compact and small size solution. (Thank you for commenting the various loops and code) Very Nice!
Now as to the speed... How long does it take your emulator to calculate F(1e6)? What about F(1e8)?
Is there a clever way to speed up your program by a factor of 1000 or more?
Here are some target times for different machines:
F(N) 41C 32sii 42s Free42(D) Result

F(100) 9s 2s 6s 0s 42
F(1000) 72s 15s 45s 0s 532
F(1e4) 651s 145s 426s 0s 6,427
F(1e5)    0s DD,DDD
F(1e6)    1.5s CCC,CCC
F(1e7)    5s B,BBB,BBB
F(1e8)    60s AAA,AAA,AAA

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Quote:
Valentin,
Greetings, What platform are you using? The trusty 71b?
Hi, Allen.
Yes, I'm using the ole trusty HP71B in its JF Garnier's Emu71 incarnation, as usual. Using the naïve approach you can't do better than this:
>LIST
10 DESTROY ALL @ INPUT K @ L=K DIV 3 @ N=L @ F=(1+SQR(5))/2
20 SETTIME 0 @ FOR A=1 TO L @ M=(KA)*A @ FOR B=A+1 TO MIN(A*F,L)
30 IF MOD(B*B,A) THEN 40 ELSE IF B*(A+B)>M THEN 50 ELSE N=N+1
40 NEXT B
50 NEXT A @ DISP N,TIME
>RUN
? 100
42 .02
>RUN
? 1000
532 .52
>RUN
? 10000
6427 51.57
which uses two nested loops and minimizes the number of arithmetic operations and comparisons in the inner one to speed execution up.
Nevertheless, using two loops implies increasing the running time by two orders of magnitude (x100) when the input increases by just one (x10), as was bound to be the case and as the timings above clearly demonstrate thus making this naïve approach utterly hopeless for large inputs.
What's needed is a subtler approach that keeps the running time approximately linear on the input, not quadratic as shown here, but I won't spoil the fun for now ... XD
Nice challenge, Allen, thanks for it and best regards from V.
Edited: 12 Mar 2012, 8:33 a.m.
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Quote:
F=(1+SQR(5))/2
I came up with the constraint A > (SQR(5)1)/2 * B, using a pencil and paper approach. Here are some thoughts:
 Loop over B (from 1 to N/3, maybe one more and check A+B+C)
 Add one to the count for the case A=B=C.
 Loop over A from CEIL((SQR(5)1)/2 * B) to B1 and check if B*B/A is an integer.
The last step can be omitted if B is prime or a power of a prime if this can be easily determined.
Edited: 12 Mar 2012, 10:16 a.m.
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I made the same mistake too, but I prefer to think of it as a temporary trip into a nonEuclidean geometry realm :)
Gerson.
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I did almost include a comment about the problem not specifying the triangles live in a Euclidean space. Riemannian spaces are so much more fun and correspond better to the word we live in.
 Pauli
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Quote:
Here are some thoughts:
 Loop over B...
 Loop over A...
Yes, this is in the right direction, Keep in mind, when dealing with
Big O Notation reducing the limits may be slightly faster, but not exponentially faster, since you have to loop over BOTH A and B to solve F(N).
As N grows, your solve time will still grow quadratically O(N^2) pronounced "Order N squared".
I have studied this problem for over a month and I am convinced you could teach a large part of an undergraduate computer science course based on solving this problem. In the initial Problem 370, the writers ask for F(2.5e13). The 14digit answer is impractical on a pocket calculator. But smaller limits like F(1e8) are well within "reasonable" times for emulated calculators, and F(10,000) within reasonable times on a real calculator.
After a month of study, I've seen/found/broken four different approaches to this problem:
 Naive approach O(N^2) with two loops (easy to write but slow)  like Valentin's nice 71b solution
 Reduced O(N^2) approach with two loops, with more mathematical limits and assumptions See Gerson and C.Ret's short and well posed solution I like them both for different reasons.
 An O(N) approach (I will post shortly)
 The most common approach posted in the Euler Discussion forum, which I will not discuss here.
Except for one program, all of the posted approaches in the Project Euler discussion forum for Problem 370 are in category 4 (due to the large size of 2e13). I believe category 4 approach is quite impractical on a pocked calculator for reasons I won't discuss here. But there is at least 1 category 3 approach that is worth discussing in more detail....
Edited: 12 Mar 2012, 9:09 p.m.
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Quote:
Apart from the trivial cases of a=b=c this program would seem to produce (I converted it to C to play with so could be wrong here) these nine extras:
25 30 36
20 30 45
18 24 32
16 24 36
16 20 25
12 18 27
9 12 16
8 12 18
4 6 9
Your conversion to C is perfect. Here is the BASIC program I wrote to test my second algorithm. The RPN code is just a quick & dirt conversion from it.
10 CLS
15 INPUT L: PRINT
20 N = INT(L / 3)
25 B = N
30 A = B  1
35 C = B * B / A
40 IF INT(C) > (A + B) THEN 65
45 IF C  INT(C) <> 0 THEN 55
50 IF C + A + B <= L THEN N = N + 1: PRINT USING "###"; A; B; C
55 A = A  1
60 GOTO 35
65 B = B  1
70 IF B > 2 THEN 30
75 PRINT : PRINT N
? 100
25 30 36
20 30 45
18 24 32
16 24 36
16 20 25
12 18 27
9 12 16
8 12 18
4 6 9
42
Would you mind posting your C program? Yesterday I tried to convert it to C by I quit after a while (literally after a while :). Looks like my BASIC program was not well structured enough. I changed it a bit later but still had trouble with the inner loop when trying to convert it to C:
10 CLS
15 INPUT L
20 N = INT(L / 3)
25 FOR B = N TO 2 STEP 1
30 A = B  1
35 C = B * B / A
40 IF INT(C) > (A + B) THEN 65
45 IF C  INT(C) = 0 THEN IF C + A + B <= L THEN N = N + 1
55 A = A  1
60 GOTO 35
65 '
70 NEXT B
75 PRINT N
Thanks,
Gerson.
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Seems the code had disappeared, but here is a second stab which is pretty close to the original.
 Pauli
#include <stdio.h>
const int p = 100;
int main() {
int a, b;
const int n = p / 3;
int count = n;
for (b=n; b>0; b)
for (a=b1; a>0; a) {
const int c = b*b/a;
if (b*b == a*c && c <= a+b && a+b+c <= p)
count++;
}
printf("%d\n", count);
return 0;
}
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Thank you!
F(100 000) = 75243 in 10 seconds. F(1000 0000) will take a while more * :)
Gerson.
* this would require int_64, I think.
Edited: 12 Mar 2012, 10:27 p.m.
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Yes it will. (10,000,000 / 3)^{2} exceeds the size of a 32 bit integer. Even 1,000,000 will be too large.
Given that the process is O(n^{2}), I'd expect 1,000,000 to take 100 times as long and 10,000,000 about 10,000 times as long (very roughly).
I've been trying to think of a way of iterating over c and determining possible values of a & b to scan over. No luck thus far. We know that a.x^{2} = b.x = c for some values of x.
 Pauli
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Hi, Allen:
Quote:
After a month of study, I've seen/found/broken four different approaches to this problem:
After a couple' hours of study I came up with a O(n) solution (or O(n)^{1+e}), or O(n*ln(n)), difficult to assess exactly ...) which can find the solution for reasonably large values of N in reasonable times on Emu71, about 1,000 times faster in a compiled highlevel language.
I suggest we all post and comment our solutions next Friday so as not to spoil the fun for people working right now on their very own approach. Deal ? :D
Best regards from V.
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Quote:
I came up with the constraint A > (SQR(5)1)/2 * B, using a pencil and paper approach.
I'll try to follow your steps:
Since
b^2 = a*c
and
c <= a + b
then
b^2 <= a*(a + b)
b^2 <= a^2 + a*b
b^2/a^2 <= 1 + b/a
(b/a)^2  b/a  1 <= 0
Solving for b/a:
b/a <= (sqrt(5) + 1)/2
a/b > (sqrt(5)  1)/2
a > b*(sqrt(5)  1)/2
I've modified my algorithm to use this constraint. I thought there was no need to check the sum of the sides of the triangles because that's what this constraint was supposed to, but it had to stay. Here are the corresponding C program and RPN code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int a, b, len, n;
scanf("%d", &len);
double c, k;
n = len/3;
k = (sqrt(5)  1) / 2;
for (b = n; b >= 2; b)
{
a = b  1;
while (a > b * k )
{
c = 1.0 * b * b / a;
if (c  int(c) == 0)
if (a + b + c <= len)
n++;
a;
}
}
printf("n = %d\n",n);
system("pause");
return 0;
}
00 { 85Byte Prgm }
01>LBL "GT"
02 STO 00 R00 < len
03 3
04 BASE÷
05 1E3
06 +
07 STO 01 R01 < b
08 IP
09 STO 02 R02 < n = int(len/3)
10 5
11 SQRT
12 1
13 
14 2
15 ÷
16 STO 04 R04 < k = (sqrt(5)  1)/2
17>LBL 01 for b = n down to 2
18 RCL 01
19 1
20 BASE
21 STO 03 R03 < a = b  1
22>LBL 02
23 RCL 01
24 IP
25 RCL× 04
26 RCL 03
27 X<=Y? a <= b*k ? (while a > b*k)
28 GTO 04
29 RCL 01
30 IP
31 X^2
32 RCL÷ 03 c = b^2/a
33 FP
34 X!=0? frac(c) != 0 ?
35 GTO 03
36 LASTX goto 03
37 RCL 01 else
38 BASE+
39 RCL+ 03
40 RCL 00
41 X<=Y? c + b + a >= len?
42 GTO 03 goto 03
43 1 else
44 STO+ 02 n = n + 1
45>LBL 03
46 1
47 STO 03 a = a  1
48 GTO 02 endwhile
49>LBL 04
50 DSE 01
51 GTO 01 endfor
52 RCL 02 disp n
53 .END.
100 XEQ GT > 42 (1 min 10 sec, real HP42S)
1000 XEQ GT > 532 (~ 0.5 seconds, Free42 Decimal)
10000 XEQ GT > 6427 ( 20.8 seconds, Free42 Decimal)
100000 XEQ GT > 75243 (~ 41 minutes, Free42 Decimal)
Still an O(n ^{2}) process. I've been busy these days and haven't given it much thought, if this can be an excuse. Anyway, no progress after three days. Perhaps I have to upgrade my hardware (brainware) :)
Gerson.
_{Edited to fix a typo as per Valentin's observation below.}
Edited: 14 Mar 2012, 12:14 p.m. after one or more responses were posted
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This is faster, but still an O(n^{2}) process:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int a, b, len, n;
scanf("%d", &len);
double c, k;
n = len/3;
k = (sqrt(5)  1) / 2;
for (b = n; b >= 2; b)
{
a = b  1;
while (a > b * k )
{
c = 1.0 * b * b / a;
if (c  int(c) == 0)
if (a + b + c <= len)
n++;
a;
}
}
printf("n = %d\n",n);
system("pause");
return 0;
}
100000
n = 75243 (7 seconds @ 1.86 GHz)
Gerson.
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Hi, Gerson:
Sorry but methinks you've got a typo (unless it's a program bug, hopefully not):
Quote:
10000 XEQ GT > 6437 ( 20.8 seconds, Free42 Decimal)
The correct value is 6427, not 6437.
Best regards from V.
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My paper and pencil approach was just what you found out: find the zero of a+b>c with ac=b^2. This doesn't make sure that a+b+c is bound in any way, that's just another inequality to check for.
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That was indeed a typo. The program does return 6427. I will fix it presently. Thank you for pointing it out.
Best regards,
Gerson.
Edited: 14 Mar 2012, 12:14 p.m.
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I am really glad to learn that my first approach doesn’t below to the first naïve category and is consider as ‘reduce naïve’.
Because, in my point of view, my first code is really a “brute force scan”, since very few optimization in the search strategy is done.
Two loops imbedded.
By luck the outer one scan step by step on a which is fortunately the shortest side of the triangle, make it walk trough it as short as possible because limited to (P/3) ! How Lucky I am here!
On the other hand, the inner loop (which will weight much more of the time to find the solution) is far for a good optimization. Despite the two exit tests which appropriately shortcut the long b walk through, the worst in my code is that at b is scan step by step (only increase by one). Here I am bad.
This bad strategy is perhaps not a shame for small perimeter, but will be a disaster for larger perimeter.
As a lot of Euler problems, here a more tricky strategy have to be take into account, to avoid this very long systematic scans of “brute force like” algorithm.
There must be a way to scan over a only, not over b !
What are the constraints:
(1) Sorted Sides: a <= b <= c
(2) Being a triangle c <= a + b
Since other inequalities are both verified:
Due to a < b we get a <= b + n whatever is the natural integer n.
Due to initial condition (1) b <= c so we get b <= c + n for any natural integer n, thus b <= c +a.
As long as initial condition (1) is respected, we only have to care about c <= a + b
(3) Limited perimeter a+b+c <= P
As Marcus Von Cube pointed it out, this constraint has to be tested. But generally it is less constringent than (2).
(4) Specific condition: b^{2} = a.c
Here is the key to reduce
Suppose that b decompose into the following product of primary factors : b_{1}, b_{2}, b_{3}…
With b = b_{1}. b_{2}. b_{3}… then (4) implies that a and c are both a subset of the primary factors of b.
To illustrate that, investigate all the triplets (a,b,c) that verify (4). Only a fraction of this list of triplets corresponds to constraints (1) (2) and (3):
Looking for triangle P<=100
b = 2 x 2 x 3 = 12 scanning all (a,c) pairs built up table 1:
TABLE 1:
========
a b c P=a+b+c Violate constraints
    
1 12 144 157 (2)(3) Not a triangle + P Too large
2 12 72 86 (2) Not a triangle
3 12 48 63 (2) Not a triangle
4 12 36 52 (2) Not a triangle
6 12 24 42 (2) Not a triangle
8 12 18 38 Ok n°1
9 12 16 37 Ok n°2
12 12 12 36 Ok n°3
16 12 9 37 (1) a b c not sorted
18 12 8 38 (1) a b c not sorted
24 12 6 42 (1) a b c not sorted
36 12 4 52 (1) a b c not sorted
48 12 3 63 (1) a b c not sorted
72 12 2 86 (1) a b c not sorted
144 12 1 157 (1)(3) a b c not sorted + Too large P
    
We can observe :
 the symmetry role of a and c, scanning over a or c is equivalent.
 half of the triplet have a>b (i.e. c<b), scan can be reduce considerably (for example by only scanning over a from 1 to b)
 not every avalue have to be scan.
In fact, only value of a composed using a subset of the primary factor from b^{2} :
Table 2 :
=========
b = 2 x 2 x 3 b.b = 2 x 2 x 2 x 2 x 3 x 3
 
a = 1 c = 2 x 2 x 2 x 2 x 3 x 3 = 144
a = 2 c = 2 x 2 x 2 x 3 x 3 = 72
a = 3 c = 2 x 2 x 2 x 2 x 3 = 48
a = 2 x 2 c = 2 x 2 x 3 x 3 = 36
a = 2 x 3 c = 2 x 2 x 2 x 3 = 24
a = 2 x 2 x 2 c = 2 x 3 x 3 = 18
a = 3 x 3 c = 2 x 2 x 2 x 2 = 16
a = 2 x 2 x 3 c = 2 x 2 x 3 = 12
...
This prove that scanning step by step over a or b is not the trickiest way to look for such triangle.
This suggest that an algorithm based on combination of a prime factors decomposition of bcan be more adapt to this WeekEnd challenge.
On the other hands, having to scan over the smallest side a will be much simpler and allow the uses of shortcut tests. I am on the way thinking that investigating how to built the correct b side from a given a under constrains (1) to (4) will be an elegant and surely efficient method.
Of course, the overhead need to analyze a and built up b will only be benefic for scanning over large Perimeter and large side. But the great interest would be that no more inner loop over b will be needed.
To investigate how to construct b from a given a, please consider a first example:
Looking for geometric triangle responding to constraints (1) to (4) with perimeter less or egal to P = 100.
Considering a prime :
Table 3: prime a = 7 , P <=100
=======
a b c P=a+b+c Violate constraints
    
7 7 7 21 none OK
7 14 28 49 (2) not a triangle (c>a+b)
7 21 63 91 (2) not a triangle
7 28 112 147 (2)(3) not a triangle & too large
7 35 175 217 (2)(3) not a triangle & too large
and so on...
    
Only first triplet (7,7,7) correspond to the researched geometric triangle.
As soon as the second line, triplets violate one or more constrains, so I may have stop the scan as soon.
Putting more values in the table shows how b value are build.
As a is prim, there is no combination of factors is possible, and we observe that b is multiple of a
The triplets are of the general formulae (a , k.a , a.k^{2}) with k=1, 2, 3, ...
Now, what if a is a simple composite: a = 8
Table 4: simple composite a = 8 , P <=100
=======
a b c P=a+b+c Violate constraints
    
8 8 8 24 none OK
8 12 18 38 none OK
8 16 32 56 (2) not a triangle
8 20 50 78 (2) not a triangle
8 24 72 104 (2)(3) not a triangle & too large
8 28 98 134 (2)(3) not a triangle & too large
and so on...
    
Here, two triangles of small side 8 exist (8,8,8) and (8,12,18). Both candidates respect all the constraints.
If we consider the b set , we now observe an arithmetic suite a, a+4, a+2x4, a+3x4,…
At each rwo, we have to increase b by 4. Where come this value from?
Is there any chance that 4 is built from half of the prime factors of a ?
a = 1x2x2x2 = 8 , the delta D to use to built the set of b is D = 2x2 = 4
The triplets are of the form (a, a+D.k, c)., where k = 0, 1, 2, ....
Now, what if a is a[italic] square: [italic]a = 9
Table 5: simple composite a = 25 , P <= 100
=======
a b c P=a+b+c Violate constraints
    
25 25 25 75 none OK
25 30 36 91 none OK
25 35 49 109 (3) perimeter too long
25 40 64 129 (3) perimeter too long
25 45 81 151 (2)(3) not a triangle & too long
25 50 100 175 (2)(3) not a triangle & too long
25 55 121 201 (2)(3) not a triangle & too long
and so on...
    
At each row, b have to be increase by D = 5 (not 25). So we have to construct D as small as possible with “half” of the factors of a.
Triplets are of the form ( a^{2}, a^{2}+k.a, (a+k) ^{2}).
In fact, D has to be built from each factor of a, but using only half of the multiple factors (i.e; with half to power of each factor).
Table 6: simple composite a = 42 , P <= 500
=======
a b c P=a+b+c Violate constraints
    
42 42 25 126 none OK
42 84 168 294 (2) not a triangle
42 126 378 546 (2)(3) nat & perimeter too long
42 168 672 882 (2)(3) nat & perimeter too long
    
Here a = 1x2x3x7 = 42 , and b = a + 42.k with k = 0, 1, 2, ....
This confirming that D has to be built from each factor of a, but using only half of the multiple factors (i.e; with half to power of each factor).
Table 7: Samples of D from factor decomposition of a
=======
a D  a D  a D 
         
1 1x1 1  31 1x31 31  61 1x61 61 
2 1x2 2  32 1x2x2x2x2x2 8 = 2x2x2  62 1x2x31 62 = 2x31 
3 1x3 3  33 1x33 33  63 1x63 63 
4 1x2x2 2 = 2  34 1x2x17 34 = 2x17  64 1x2x2x2x2x2x2 8 = 2x2x2 
5 1x5 5  35 1x5x7 35 = 5x7 
6 1x2x3 6 = 2x3  36 1x2x13 36 = 2x2x2 
7 1x7 7  37 1x37 37 
8 1x2x2x2 4 = 2x2  38 1x2x19 38 = 2x19 
9 1x3x3 3 = 3  39 1x3x13 39 = 3x13 
10 1x2x5 10 = 2x5  40 1x2x2x2x5 20 = 2x2x5 
11 1x11 11  41 1x41 41 
12 1x2x2x3 6 = 2x3  42 1x2x3x7 42 = 2x3x7 
13 1x13 13  43 1x42 43 
14 1x2x7 14 = 2x7  44 1x2x2x11 22 = 2x11 
15 1x3x5 15 = 3x5  45 1x3x3x5 15 = 3x5 
16 1x2x2x2x2 4 = 2x2  46 1x2x23 46 = 2x23 
17 1x17 17  47 1x47 47 
18 1x2x3x3 6 = 2x3  48 1x2x2x2x2x3 12 = 2x2x3 
19 1x19 19  49 1x7x7 7 = 7 
20 1x2x2x5 10 = 2x5  50 1x2x5x5 10 = 2x25 
21 1x3x7 21 = 3x7  51 1x3x17 51 = 3x17 
22 1x2x11 22 = 2x11  52 1x2x2x13 26 = 2x13 
23 1x23 23  53 1x53 53 
24 1x2x2x2x3 12 = 2x2x3  54 1x2x3x3x3 18 = 2x3x3 
25 1x5x5 5 = 5  55 1x5x11 55 = 5x11 
26 1x2x13 26 = 2x13  56 1x2x2x2x7 28 = 2x2x7 
27 1x3x3x3 9 = 3x3  57 1x3x19 57 = 3x19 
28 1x2x2x7 14 = 2x7  58 1x2x29 58 = 2x29 
29 1x29 29  59 1x59 59 
30 1x2x3x5 30 = 2x3x5  60 1x2x2x3x5 30 = 2x3x5 
      
The larger is D, the more rapid is the scan. As we can see, the process (involving the two embedded loops) has a great chance to be speedup, since most of the a value leads to D = a. Especially when a is prime.
The inner loop (the one over b) can be suppress if we found a way of forecasting at which k we have to stop because the triplet violate one (or more) constraints.
To test for constraint (2) : ( a, b, c) being a triangle, we have to test c >= a + b
From the given a value, we may get D (see table 7 above) or find a way to built D from a factors.
With such a D, we know that the triplet (a, a+k.D, c) respect (4) b^{2} = a.c when
b = a + k.D , therefore c = b^{2} / a
To test for constrinat (2), we have to express c >= a + b as a function of variable k and parameters a and D.
The inequality c >= a + b can be transform into k^{2}.D^{2} + k.D.(2a)  a^{2} <= 0. Resolving the corresponding quadratic equation, allow to indicate the domain of validity:
The variable have to be set from 0 up to k_{2} = (a/2D).(1V5)
To test for constraint (3) : a+b+c <=P , we may found the limit for k the same way, by expression the inequality as a function of variable k and parameters a, P and D
a+b+c <=P is transform into k^{2}.D^{2} + k.D.(2+a) + 3a^{2}aP <= 0.
Solving the corresponding quadratic equation for k indicate that k may varie from 0 upto k_{3} = (a/2D).(3V(4P/a  3))
This give you the way to code for a more efficient algorithm using only one loop over a :
Input P upper limit of perimeter
Initiate Sum = 0
For each integer a from 1 to P/3
Determine D ( 1 <= D <= a ) from the prime factor decomposition of a.
In most of the cases D will be equal to a
or less than a in specific cases
IF P > (3+SQR(5))*a
THEN ' Limited by triangle (2)
kl = (a/2D)*(1SQR(5))
ELSE ' Limited by perimeter (3)
kl = (a/2D)*(3SQR(4P/a  3))
End if
There is n = INT( 1 + kl ) geometric triangles of small side [italic]a[/italic] since k = 0,1, 2, ..., kl
Sum = Sum + n Add integer n into the Sum
At end, display Sum that contains number of researched triangle of perimeter less or equal to P.
In conclusion, I leave you a few to compose you own code on your favorite calculator or system.
I curious to know how much time this new algorithm loose on shot perimeter (P = 100, 1000), and what it is win for larger size (P> 10^4).
Edited: 15 Mar 2012, 7:26 p.m. after one or more responses were posted
Posts: 2,761
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Quote:
I am really glad to learn that my first approach doesn’t below to the first naïve category and is consider as ‘reduce naïve’. Because, in my point of view, my first code is really a “brute force scan”, since very few optimization in the search strategy is done.
That is exactly my feelings, regarding mine. I haven't timed your first approach, but mine isn't any better than Valentin's, quite the contrary:
10 DESTROY ALL @ INPUT L @ N= L DIV 3 @ K=(SQR(5)1)/2
20 SETTIME 0 @ FOR B = N TO 2 STEP 1 @ A = B  1
30 IF A < B * K THEN 90
40 C = B * B / A
50 IF FP(C) <> 0 THEN 70
60 IF A + B + C <= L THEN N = N + 1
70 A = A  1
80 GOTO 30
90 NEXT B @ PRINT N,TIME
>RUN
? 100
42 .05
>RUN
? 1000
532 2.51
>RUN
? 10000
6427 255.22
Valentin's times in my notebook running EMU71 @ 1.86 GHz are:
>RUN
? 100
42 .05
>RUN
? 1000
532 1.2
>RUN
? 10000
6427 114.67
Gerson.
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Joined: Oct 2008
You may try this last version :
(Microsoft QBASIC)
DEFLNG AZ '  All variables long integer except k and t (single precision)
DIM k AS SINGLE
DIM t AS SINGLE
PRINT
INPUT "ENTER perimeter"; P
IF P < 1 THEN END
'
' Initiate Sum and timer
'
Sum = 0
t0 = TIMER
'
'  MAIN LOOP OVER a
'
FOR a = 1 TO P / 3
'
'  Compute step D
'
x = a: i = 2: j = 1: D = 1
'
'           Isprime? Test Loop
DO UNTIL i * i > x
n = 0
DO WHILE x MOD i = 0: n = n + 1: x = x / i: LOOP
IF n > 0 THEN D = D * i ^ INT(.5 + n / 2)
i = i + j: j = 2
LOOP
D = D * x
'
'  test Limit for k
'
IF P < (3 + SQR(5)) * a THEN
k = SQR(4 * P / a  3)  3 ' Limited by perimeter
ELSE
k = SQR(5)  1 ' Limited by triangle
END IF
'
'  Add current number to sum
'
Sum = Sum + 1 + INT(a / 2 / D * k)
NEXT a
'
'  Display result and time
'
PRINT : PRINT
PRINT "There is "; Sum; "triangles of perimeter <="; P; " with b²=a.c"
PRINT "("; INT((TIMER  t0) * 10) / 10; "s )"
PRINT
Approximetely translate into HP71 BASIC : not sure about correct syntax of MOD CEIL and power (^).
10 DESTROY ALL @ INPUT P @ SETTIME 0
REM  MAIN LOOP OVER A
20 S = 0 @ FOR A = 1 TO P DIV 3
REM  DETERMINE STEP D
30 X = A @ I = 2 @ D = 1 @ J = 1
REM        ISPRIME TEST LOOP
40 N = 0 @ IF I * I > X THEN 90
50 IF X MOD I = 0 THEN N = N + 1 @ X = X DIV I @ GOTO 50
60 IF N > 0 THEN D = D * I ^ CEIL( N / 2 )
70 I = I + J @ J = 2
80 GOTO 40
REM  COUNT SOLUTIONS
90 D = D * X @ K = 1  SQRT 5
100 IF P < (3 + SQRT 5) * A THEN K = 3  SQRT(4 * P / A  3)
110 S = S + 1 + INT(  A / 2 / D * K)
120 NEXT A @ PRINT S,TIME
Edit : Replace N by S at line 120 in HP71 listing.
Edited: 15 Mar 2012, 6:53 p.m. after one or more responses were posted
Posts: 2,761
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Joined: Jul 2005
It looks like there is an error in your conversion from QBASIC to HP71 BASIC or I have introduced one myself:
10 DESTROY ALL @ INPUT P @ SETTIME 0
20 S=0 @ FOR A=1 TO P DIV 3
23 REM
25 REM  determine step d
27 REM
30 X=A @ I=2 @ D=1 @ J=1
33 REM
35 REM        isprime test loop
37 REM
40 N=0 @ IF I*I>X THEN 90
45 N=0
50 IF MOD(X,I)=0 THEN N=N+1 @ X=X DIV I @ GOTO 50
60 IF N>0 THEN D=D*I^CEIL(N/2)
70 I=I+J @ J=2
80 GOTO 40
83 REM
85 REM  count solutions
87 REM
90 D=D*X @ K=1SQR(5)
100 IF P<(3+SQR(5))*A THEN K=3SQR(4*P/A3)
110 S=S+1+INT(A/2/D*K)
120 NEXT A @ PRINT N,TIME
>RUN
? 1000
0 .32
>RUN
? 10000
0 5.1
>RUN
? 100000
0 102.21
The QBASIC times are:
F(100000) > 1.4 s
F(1000000) > 30.6 s
F(10000000) > 567.8 s
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Yes, I made a mistake in the last line where N stand for S.
Sorry.
10 DESTROY ALL @ INPUT P @ SETTIME 0 @ S=0 @ FOR A=1 TO P DIV 3 @ X=A @ I=2 @ D=1 @ J=1
40 N=0 @ IF I*I>X THEN 90
50 IF MOD(X,I)=0 THEN N=N+1 @ X=X DIV I @ GOTO 50
60 IF N>0 THEN D=D*I^CEIL(N/2)
70 I=I+J @ J=2 @ GOTO 40
90 D=D*X @ K=SQR(5)1 @ IF P<(3+SQR(5))*A THEN K=SQR(4*P/A3)3
110 S=S+1+INT(A/2/D*K) @ NEXT A @ PRINT S,TIME
Here the same code for "old" RPL aka HP28C/S.
"New PRL" may take advantage of ISPRIME? and NEXTPRIME instructions !
« > P
« 0 @ Initiate Sum=0
1 P 3 / FOR a
1 SF
2 @ Initiate D = 2
a @ x = a
2 @ i = 2
WHILE DUP2 SQ >= @ while i²<=x
REPEAT
0 3 ROLLD @ Initate n = 0
WHILE DUP2 MOD NOT @ while x MOD i = 0
REPEAT
SWAP OVER / SWAP @ x = x/i
ROT 1 + 3 ROLLD @ n = n+1
END
4 ROLL OVER 5 ROLL 2 /
CEIL ^ * 3 ROLLD @ D = D*i^CEIL(n/2)
1 + @ Inc i
IF 1 FC?C THEN 1 + END @ Inc i when i>3
END
DROP * @ D = D*x
a SWAP /
IF P a 3 5 SQRT + * <
THEN 4 P * a / 3  SQRT 3  @ limited by perimeter
ELSE 5 SQRT 1  @ limited by triangle side
END
* IP 1 + + @ Sum = Sum + 1 + INT(...)
NEXT
»
» 'B2AC' STO
Execution Times HP28S
100 B2AC > 42 (12s.)
1000 B2AC > 532 (3min45s)
Edited: 15 Mar 2012, 8:04 p.m. after one or more responses were posted
Posts: 2,761
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Here are your times up to 10^5:
Emu71: HP71B & HPIL system emulator JF GARNIER 1996, 2006
>list
10 DESTROY ALL @ FOR E=2 TO 5 @ P=10^E @ SETTIME 0 @ S=0
20 FOR A=1 TO P DIV 3 @ X=A @ I=2 @ D=1 @ J=1
40 N=0 @ IF I*I>X THEN 90
50 IF MOD(X,I)=0 THEN N=N+1 @ X=X DIV I @ GOTO 50
60 IF N>0 THEN D=D*I^CEIL(N/2)
70 I=I+J @ J=2 @ GOTO 40
90 D=D*X @ K=SQR(5)1 @ IF P<(3+SQR(5))*A THEN K=SQR(4*P/A3)3
110 S=S+1+INT(A/2/D*K) @ NEXT A @ PRINT P,S,TIME @ NEXT E
>run
100 42 .05
1000 532 .3
10000 6427 4.81
100000 75243 94.45
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Thank you.
It's look like this version is not as fast as Valentin Albillo's code (see post #023) who is using machine ROM PRIM function.
My code loose much of his time to test for primility, where the PRIM function spare a lot of time.
As as explain above, most od the time, D equal a. In most of the case my code test all factors of a where a simple D = a will do the job.
Following a transcription of the above code into HP32S RPN program :
00 { 141Byte Prgm } : :
01>LBL "B2AC" : 31 1 : 61 SQRT
02 STO 01 : 32 + : 62 +
03 3 : 33 GTO 02 : 63 ×
04 ÷ : 34>LBL 03< : 64 X<=Y?
05 IP : 35 Rv : 65 GTO 06
06 STO 02 : 36 X=0? : 66 Rv
07 0 : 37 GTO 04 : 67 4
08 STO 00 : 38 2 : 68 ×
09>LBL 00<: 39 1/X : 69 RCL÷ 02
10 2 : 40 STO× ST Y : 70 3
11 STO 03 : 41 + : 71 STO ST Y
12 RCL 02 : 42 IP : 72 X<>Y
13 2 : 43 RCL ST Y : 73 GTO 07
14 SF 00 : 44 X<>Y : 74>LBL 06<
15>LBL 01 : 45 Y^X : 75 1
16 X^2 : 46 STO× 03 : 76 5
17 X>Y? : 47>LBL 04< : 77>LBL 07<
18 GTO 05 : 48 CLX : 78 SQRT
19 X<> ST L : 49 1 : 79 X<>Y
20 0 : 50 FC?C 00 : 80 
21>LBL 02< : 51 STO+ ST Y : 81 RCL× 02
22 RCL ST Z : 52 + : 82 RCL÷ 03
23 RCL÷ ST Z : 53 GTO 01 : 83 IP
24 FP : 54>LBL 05<: 84 1
25 X>0? : 55 Rv : 85 +
26 GTO 03 : 56 STO× 03 : 86 STO+ 00
27 Rv : 57 RCL 01 : 87 DSE 02
28 X<>Y : 58 RCL 02 : 88 GTO 00
29 STO÷ ST Z : 59 3 : 89 RCL 01
30 X<>Y : 60 5 : 90 RCL 00
: : 91 END
REgisters
R00 : Sum ,number of triangles
R01 : P ,perimeter
R02 : a ,small side
R03 : D ,
>LBL 00 : Main loop (For a = INT(P/3) downto 1
Factors analysis of a
>LBL 01 : Look for factors of a
>LBL 02 : search for multiple factor of a
>LBL 03 : Built D by adding each factor of a (but half of each multiple factor number).
>LBL 04 : next factor f=2,3,5,...
Compute number of triangle of small side a
>LBL 05 : Nbr of triangle limited by perimeter a+b+c < P
>LBL 06 : Nbr of triangle limited by side c < a+b
>LBL 07 : Add number of traingle of side a to global sum (R00)
Edited: 16 Mar 2012, 5:56 p.m.
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Hi, all
We're already on Friday so time to post some nonnaïve solutions to this challenge.
When I created my above naïve solution I fully realized that having two nested loops, though simple and immediately coded, was not the way to go as you'll inevitably get times of order O(n^{2}). Some clever tricks with the nested loop limits did significantly improve the timing but it still was O(n^{2}).
Regrettably I couldn't dedicate more than a few hours to this but I decided to make the most of the available time and last Tuesday I concocted this much improved 6line, 234byte solution for the HP71B which I'll presently comment:
1 DESTROY ALL @ INPUT K @ P=4*K @ N=0 @ F=(1+SQR(5))/2 @ SETTIME 0
2 FOR A=1 TO K/3 @ N=N+FNT(FNF(A)) @ NEXT A @ DISP N;TIME
3 DEF FNT(N)=MIN(A*F,A*((SQR(P/A3)1)/2)) DIV NA DIV N+NOT MOD(A,N)
4 DEF FNF(N) @ M=N @ E=0
5 D=PRIM(N) @ IF D<2 THEN FNF=M/(E*(N=E)+(N#E)) @ END ELSE N=N/D
6 IF E=D THEN M=M/E @ E=0 @ GOTO 5 ELSE E=D @ GOTO 5
First, some timings:
Perimeter Triangles Time (sec). Time increase

100 42 0.03 
1000 532 0.16 5.2x
10000 6427 1.8 11.3x
100000 75243 19.52 10.8x
1000000 861805 219.62 11.3x
10000000 9712598 2589.75 11.8x
Now for the comments:
 As you can see, the main program is just 2line long and all it does is to initialize some needed constants so that the ensuing loop will go as fast as possible by removing invariant operations within the loop, then the single loop on the shortest side (A) is executed, which simply calls a userdefined function FNT which returns the number of valid triangles for that value of A and adds this up to the tally, eventually returning the total and the timing when the loop is done for. Simple as can be.
 Line 3 is a nifty userdefined singleline function, FNT, that duly returns the number of valid triangles within limits given by the maximum and minimum possible values for side B, which are cleverly computed so that no tests about triangularity and/or perimeter are necessary within the loop at all, this way:
 the A*F term, where F is the golden ratio, ensures that A, B, and C form inded a triangle, as C is guaranteed to be less than A+B.
 the A*((SQR(P/A3)1)/2)) term ensures that the sum of A+B+C is guaranteed to be less than or equal to the given perimeter.
 Lines 45 constitute another userdefined function, FNF, that given the value of A computes the periodicity with which B*B is divisible by A. This way there's no need for another nested loop that examines each B in turn to check for divisibility but we can simply tell at once just how many values of B do meet the divisibility condition by simply using this computed periodicity as input for the trianglecounting function FNT.
Thus, this approach does allow us to get over the dreaded O(n ^{2}) to achive something much more like O(n*log(n)), as the results above clearly show ( ~ 11.3x increase in time for every 10x increase in the input value).
Note:
FNF uses the PRIM function from the HP71B JPC ROM for faster execution. If this function were unavailable the functionality can easily be achieved using plainvanilla HP71B BASIC code, which would affect speed by a constant factor (~3x slower) but would not change the order O(n*log(n)) so timing would ultimately be much faster than any O(n^{2}) approach for sufficiently large input values of the perimeter.
That's all on my side, thanks to Allen for an interesting challenge.
Best regards from V.
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Valentin, C.Ret, Gerson thank you for your interesting solutions. I hope I was clear in saying there is nothing wrong with the naive approach only an observation that it is the easiest solution to develop and program. Your solutions are certainly well thought out and not "naive" as in a bad.
I admit my approach is not the fastest I've seen (or even close) but I propose a method that neither counts a nor b, but only focuses on counting similar triangles with ratio of b/a. In so doing, There are no perimeter checks, no triangle inequality, and no loops over a or b in fact there is only 1 loop to generate a series of fractions, and 1 check to make sure that fraction b/a does not pass the golden ratio. This allows execution to proceed in O(N) time  requiring roughly N/16 cycles for F(N).
How?
F(100) can be calculated in 7 iterations of fraction loop that creates coprime numerator and denominator pairs called a Farey Sequence (A close cousin of the SternBrocot tree). Using Integer Division, these coprime pairs can be used to quickly sum how many similar triangles are less than the perimeter. A more lengthy description can be found in this PDF file: Solving Project Euler.net Problem 370 on an HP calculator. This PDF also includes Bar codes for HP41, .RAW file if you want to use it in an Free42S emulator, commented HP42S Code, and a C program if you wish to compile on a PC.
Some example Run Times:
F(N) 41C 32sii 42s Free42(D) Result Iterations Time increase
F(100) 9s 2s 6s 0s 42 7 
F(1000) 72s 15s 45s 0s 532 64 9.142857143
F(1e4) 651s 145s 426s 0s 6427 619 9.671875
F(1e5)    0s 75243 6262 10.11631664
F(1e6)    1.5s 861805 62710 10.0143724
F(1e7)    5s 9712598 626180 9.985329294
F(1e8)    60s 108073540 6262634 10.00133189
Note: in MATLAB or C the sum of times for F(2) to F(8)) is around 0.35 seconds
The time increase approaches 10X for a 10X increase in F(N) so it's almost exactly O(N) all the way up to higher orders.
Here is the commented HP2S code. It uses 10 registers: (1 for the Total, 1 for the order of the sequence, 1 for Phi, the golden ratio, 1 temporary register to simplify calculations, and 6 registers to iterate the Farey Sequence.)
 Lines 122 set up 10 registers for Main loop
 Lines 2434 find the smallest similar Triangle with sides a and b
 Lines 3662 Creates the next Farey sequence.
 Lines 6366 Check to exit if m2/n2> golden ratio
00 { 72Byte Prgm } Comment
01>LBL 01 Lines 122 set up 10 registers for Main loop
02 STO 00 p=Perimeter
03 3
04 ÷
05 SQRT Set Order for Recursive Farey Series
06 IP
07 STO 01 =floor(sqrt(P/3))
08 SIGN save a byte to get 1 on the stack
09 STO 03 mt=1  Initial Farey upper bound mt/nt=1/1
10 STO 04 nt=1
11 STO 05 m2=current side a=1
12 STO 06 n2=current side b=1
13 STO 07 n1=1
14 5
15 SQRT
16 +
17 2
18 ÷
19 STO 02 Phi Golden Ratio (the upper limit for the ratio of mt/nt)
20 CLX
21 STO 08 m1=0 Initial Farey lower bound m1/n1=0/1
22 STO 09 Total triangles= 0
23>LBL 02 Main loop
24 RCL 00 Lines 2434 find the smallest similar Triangle with sides a and b
25 RCL 05
26 RCL 06
27 +
28 X^2 The smallest perimeter is ps=(a+b)^2ab
29 RCL 05
30 RCL 06
31 ×
32 
33 ÷
34 IP
35 STO+ 09 Total=total+floor(p/ps) [if ps>p this adds 0 no branching test needed]
36 RCL 07 Lines 3662 Create the next Farey sequence.
37 RCL 01 This guarantees the numbers mt and nt are coprime and thus similar triangles are not
38 + counted twice. Example: For the trivial case where the perimeter p=100,
39 RCL 06 the order 5 Farey series is: 0/1, 1/1, 6/5, 5/4, 4/3, 7/5, 3/2, 8/5…
40 ÷ Contributing 0+33+1+1+2+0+5+0= 42 respectively.
41 IP h=floor((n1+d)/n2)
42 ENTER
43 ENTER
44 RCL 05
45 ×
46 RCL 08
47 
48 STO 03 mt=h*m2m1
49 X<>Y
50 RCL 06
51 ×
52 RCL 07
53 
54 STO 04 nt=h*n2n1
55 RCL 06
56 STO 07 n1=n2
57 RCL 05
58 STO 08 m1=m2
59 RCL 03
60 STO 05 m2=mt
61 RCL 04
62 STO 06 n2=nt
63 ÷
64 RCL 02
65 X>Y? Lines 6366 Check to exit if m2/n2> golden ratio
66 GTO 02 Next Farey sequence
67 RCL 09 Otherwise recall total (END)
68 .END.
For F(2.5e13) the C program will get the solution to Euler Problem 370 eventually, but since it's O(N) this takes a long time. Nonetheless, because of the relatively small size program and minimal data requirements, finding Similar triangles using the Farey Sequence is reasonable for smaller orders of N. For these reasons, I believe this approach is is moderately suited for many values of N using a real or emulated HP calculator.
Many thanks to those who contributed solutions, you have some clever and compact solutions, all of which I studied and learned from.
Edited: 16 Mar 2012, 9:37 p.m.
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Congrats, Allen, this is a really beautiful solution!
For fun I coded it up in RPL+ and ended up with this:
\<< =n
0 =:m1 =sum
1 =:m2 =:n1 =n2
floor(sqrt(n/3)) =d
goldenRatio EVAL =gr
WHILE m2/n2<=gr
REPEAT
floor(n/((m2+n2)*(m2+n2)m2*n2)) =+sum
floor((n1+d)/n2) =h
h*m2m1 =mt
h*n2n1 =nt
n2 =n1 nt =n2
m2 =m1 mt =m2
END
sum
\>>
goldenRatio is a builtin continued fraction. It's EVAL'ed in order to yield a real.
Runtimes:
ND1 (iPhone): F(1000): <0.1s; F(1e6): 35s
CalcPad (Mac): F(1000): 0.007s; F(1e6): 1.8s
I used your C code to arrive at this in JavaScript:
function (n) { /*as is*/
var sum=0, maxp = n;
var mt=1, nt=1, m2=1, n1=1, n2=1, h, m1=0;
var d = Math.floor(Math.sqrt(maxp/3));
var gr = 1.6180339887498948482;
while (m2/n2 <= gr ) {
sum += Math.floor(maxp/((m2+n2)*(m2+n2)m2*n2));
h = Math.floor((n1+d)/n2);
mt = h * m2  m1;
nt = h * n2  n1;
n1 = n2;
n2 = nt;
m1 = m2;
m2 = mt;
}
return sum;
}
Runtimes:
ND1 (iPhone): F(1000): 0.003s; F(1e8): 32s
CalcPad (Mac): F(1000): instant; F(1e8): 3.6s
My goal for RPL+ is to eventually achieve roughly half of JavaScript speed.
Edited: 17 Mar 2012, 8:03 a.m.
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Nice code as always, C.Ret!
Quote:
Execution Times HP28S 100 B2AC > 42 (12s.) 1000 B2AC > 532 (3min45s)
I ran it (unchanged) in ND1, yielding the following runtimes:
B2AC(100): 0.2s; B2AC(1000): 2.8s
It's not obvious to me how to change the code to use NEXTPRIME and/or ISPRIME?. Can you show me?
The 50g or ND1 also have FACTORS, which can be used to derive your "D". It's straightforward but not terribly elegant, as you'd still need a loop to do an integer divide on every second element in the factors array (which interleaves factors and their exponents).
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Hi again, Allen:
Quote:
Valentin, C.Ret, Gerson thank you for your interesting solutions. I hope I was clear in saying there is nothing wrong with the naive approach only an observation that it is the easiest solution to develop and program. Your solutions are certainly well thought out and not "naive" as in a bad.
Well, I wouldn't say my 6line HP71B solution above is that naïve in any sense good or bad but I get your point. As kinda proofofconcept, this is the exact same solution directly converted to just 5 lines of interpreted UBASIC code almost verbatim:
10 word 3 : point 2 : for I=2 to 8 : K=10^I : P=4*K : S=0 : F=(1+sqrt(5))/2 : clr time
20 for A=1 to K\3 : N=A : M=N : E=0
30 D=prmdiv(N) : if D>1 then N\=D : if E=D then M\=D : E=0 : goto 30 else E=D : goto 30
40 S+=int(min(A*F,A*((sqrt(P/A3)1)/2))/M)A\M+(res=0) : next
50 print I,S,time : next : print "Yasta"
Upon running it, the results and execution times are, in the same hardware:
run
2 42 0:00:00
3 532 0:00:00
4 6427 0:00:00
5 75243 0:00:00
6 861805 0:00:01
7 9712598 0:00:23
8 108073540 0:04:47
Yasta
which produces all the results you asked for in reasonable times, about two full orders of magnitude (~100x) faster than Emu71, and of course still O(n*log(n)) [ 4:47 / 00:23 = 12.5x ]. A conversion to compiled C# code produces almost two additional orders of magnitude of speed increase.
Quote:
Many thanks to those who contributed solutions, you have some clever and compact solutions, all of which I studied and learned from.
You're welcome and again, thanks for both your interesting challenge and your detailed, welldocumented solution.
Best regads from V.
Edited: 17 Mar 2012, 12:08 p.m.
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Nice chALLENge and even nicer solution!
While I was watching a BBC documentary about Henry the Meek, I decided to try F(100000) on the real HP42S. The calculator bravely accomplished the task just before the second episode was over.
Let me present an alternative G(N) function for those of you who are in a hurry and can tolerate some error:
00 { 119Byte Prgm }
01>LBL "T"
02 STO 00
03 3
04 BASE÷
05 RCL 00
06 2
07 ÷
08 SQRT
09 BASE+
10 RCL 00
11 LN
12 RCL× ST L
13 4.76115738459E2
14 ×
15 1.29705734564E1
16 RCL× 00
17 
18 47.8815
19 
20 +
21 RCL 00
22 SQRT
23 LASTX
24 LN
25 8.60090977995E2
26 ×
27 1.50150179296
28 
29 ×
30 53.456
31 +
32 BASE+
33 .END.
absolute percent
N F(N) G(N) error error
1E+02 42 43 +1 2.381
1E+03 532 530 2 0.376
1E+04 6427 6426 1 0.016
1E+05 75243 75244 +1 0.001
1E+06 861805 861805 0 0.000
1E+07 9712598 9712232 366 0.004
1E+08 108073540 108074423 +883 0.001
1E+09 1190212172 1190326147 +113975 0.010
1E+10 12996874312 12999364614 +2490302 0.019
Less than one second on the real HP42S :)
Gerson.
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I suspect there was some unintended interpretations of the naïve method noted above. Just to be clear when referring to programming and algorithms, "naïve method" referrs to the following definition from http://en.wikipedia.org/wiki/Na%C3%AFve_algorithm#Classification
Quote:
Bruteforce or exhaustive search. This is the naïve method method of trying every possible solution to see which is best.
In programming this does not carry any negative connotation only an acknowledgement that is is a bruteforce approach, unoptimized.
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Hi, Gerson:
Quote:
Let me present an alternative G(N) function for those of you who are in a hurry and can tolerate some error:
Good idea, here's my take on the subject for the HP71B:
1 DESTROY ALL @ OPTION BASE 1 @ DIM C(8) @ READ C @ INPUT K @ FIX 0
2 DATA 2.23467244246E8,1.76140742602E6,5.78605330267E5,1.04128697469E3
3 DATA 1.14617819796E2,8.37891596549E2,2.79672680182,1.59739297094
4 P=LGT(K) @ N=C(1) @ FOR I=2 TO 8 @ N=N*P+C(I) @ NEXT I @ DISP EXP(N)
The results are:
P Exact value Computed val. Error % error

1E02 42 42 0 0.001
1E03 532 532 0 +0.001
1E04 6427 6427 0 0.001
1E05 75243 75242 +1 +0.001
1E06 861805 861813 8 0.001
1E07 9712598 9712504 +94 +0.001
1E08 108073540 108074589 1049 0.001
1E09 1190212172 1190200618 +11554 +0.001
1E10 12996874312 12997000479 126167 0.001
and of course it runs almost instantaneously.
Best regards from V.
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Hi, Valentin!
Great fit! What about fitting H(N) = F(N)  (N DIV 3 + INT(SQR(N/2))) and then making G(N) = N DIV 3 + INT(SQR(N/2)) + H(N)? Would this result in an even better fit?
Best regards,
Gerson.
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Not only a 42S program! It runs with almost no change on my HP67.
The only thing I needed to change was to replace SIGN on line 8 with 1.
It calculated F(1000) in 2 minutes 26 sec!
**vp
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Thank you for this nice challenge.
And felicitations for your brilliant solution. Using Farey series is a much more elegant solution that the crude and brut way to determine coprimes factors from a or b the way I made it!
I am studying your solution with great interest. The use of the coprime factors use from Farey the fractions is really stricky. But I still miss why or how to select range of the serie. Why are FLOOR(SQRT(P)) appropriate ? Is it a kind of magic ? lol
I can't figure out how to relay this with the problem!
I can't resist to present here an oldstyle userRPL version using your clever Farey's tricks: It is closelooking to RPL+ version above. Except that I have use more stack manipulations to spare local variables (and speed up runs on my poor old HP28S).
42 (6s) 532 (20s) 6427 (1min30s)
« DUP @ Preserve P
3 / SQRT FLOOR @ Set Order for Recursive Farey Series
1 5 SQRT + 2/ @ golden ratio
> P d gr
« 0 @ Initiate Sum
0 1 1 1 @ m1 n1 m2 n2 leave and treated in stack
WHILE DUP2 / gr < @ test m2/n2 < gr
REPEAT
DUP2 * LAST + SQ SWAP 
P SWAP / FLOOR @ FLOOR(P/((a+b)²a*b))
6 ROLL + 5 ROLLD @ add to sum and ROLLD back to top
3 PICK d + OVER / FLOOR @ > h
3 PICK OVER * 6 ROLL  @ compute new mt (and remove m1 from stack)
SWAP 3 PICK * 5 ROLL  @ compute new nt (and remove n1 from stack)
END
4 DROPN @ remove m1 n1 m2 n2 from stack (Sum remain on top)
»
»
And again, thank to you for this nice challenge and for other forumers to participate. Great Time !
EDIT: remove swapped >> from code listing.
Edited: 19 Mar 2012, 12:00 p.m. after one or more responses were posted
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You are most welcome I enjoyed working on this problem with what time I had.
You raise a good point about how to get the order of the Farey sequence. I had two challenges, first the Sequence normally covers {0..1},but I needed {1..phi} for reasons already discussed.
The second challenge, determining the order = floor(sqrt(p/3)) was not as trivial. In examining the data for the smaller O(N^2) runs, the irreducible coprime factors came up. I noticed that in every case F(N), the boundary closest to unity: k+1/k always stopped at a certain value of k_max (and that no greater denominator existed anywhere in the set):
N k_max

100 5
1000 18
10000 57
And so on. Since I already came to the same conclusion many others had about the importance of "P/3" in determining the upper limit for side A, It did not take much guess work to identify SQRT(P/3) as an interesting number in the Farey sequence, and the FLOOR function simply ensured the remainder of the denominators were integers.
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Quote:
« DUP @ Preserve P
3 / SQRT FLOOR @ Set Order for Recursive Farey Series
1 5 SQRT + 2/ @ golden ratio
> P d gr
« 0 @ Initiate Sum
0 1 1 1 @ m1 n1 m2 n2 leave and treated in stack
WHILE DUP2 / gr < @ test m2/n2 < gr
REPEAT
DUP2 * LAST + SQ SWAP 
P SWAP / FLOOR @ FLOOR(P/((a+b)²a*b))
6 ROLL + 5 ROLLD @ add to sum and ROLLD back to top
3 PICK d + OVER/ FLOOR @ > h
3 PICK OVER * 6 ROLL  @ compute new mt (and remove m1 from stack)
SWAP 3 PICK * 5 ROLL  @ compute new nt (and remove n1 from stack)
»
END
4 DROPN @ remove m1 n1 m2 n2 from stack (Sum remain on top)
»
Are you sure this code will work? Aren't >> and END swapped over?
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You are right, the \>> swap here from a bad copypaste operation!
Corrected listing:
« DUP @ Preserve P
3 / SQRT FLOOR @ Set Order for Recursive Farey Series
1 5 SQRT + 2/ @ golden ratio
> P d gr
« 0 @ Initiate Sum
0 1 1 1 @ m1 n1 m2 n2 leave and treated in stack
WHILE DUP2 / gr < @ test m2/n2 < gr
REPEAT
DUP2 * LAST + SQ SWAP 
P SWAP / FLOOR @ FLOOR(P/((a+b)²a*b))
6 ROLL + 5 ROLLD @ add to sum and ROLLD back to top
3 PICK d + OVER / FLOOR @ > h
3 PICK OVER * 6 ROLL  @ compute new mt (and remove m1 from stack)
SWAP 3 PICK * 5 ROLL  @ compute new nt (and remove n1 from stack)
END
4 DROPN @ remove m1 n1 m2 n2 from stack (Sum remain on top)
»
»
Sorry.
Edited: 19 Mar 2012, 12:03 p.m.
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If a <= b <= c
and n/d is the irreducible fraction of b/a,
then the smallest triangle you can construct (subject to b^2 = a*c) is
(a,b,c) = (d^2,n*d,n^2)
because
b = k*n
a = k*d
for k a positive integer, then
b^2 = a*c
k^2 * n^2 = k*d*c
c = k*n^2/d
Since n and d are coprime, k >= d
Since a <= P/3, it follows that
d <= SQRT(P/3)
Werner
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Werner, A very nice derivation, thank you!
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Here is what the code using ISPRIME? and NEXTPRIME instruction may look at.
where ISPRIME? returns true (1) for prime numbers only.
and NEXTPRIME converts its argument to the next prime number (ex. 1 NEXTPRIME is 2, 2 NEXTPRIME is 3, 3 NEXTPRIME is 5, 5 NEXTPRIME is 7 and so on...)
PROGRAM: @ Comments @ Stack 5: 4: 3: 2: 1:
@ @ P
« > P
« 0 @ Initiate Sum=0 @ 0 : 1 : P/3
1 P 3 / FOR a @ Main Loop @ Sum
2 @ Initiate D < 2
a @ x < a
1 @ i < 1 @ Sum : D : x : i
WHILE OVER ISPRIME? NOT @ loop until x is prime
REPEAT
NEXTPRIME @ i' < nextprime ( i ) @ Sum : D : x : i
0 3 ROLLD @ Initiate n < 0 @ Sum : D : n : x : i
WHILE DUP2 MOD NOT @ while x MOD i = 0
REPEAT
SWAP OVER / SWAP @ x < x/i @ Sum : D : n : x/i : i
ROT 1 + 3 ROLLD @ n < n+1 @ Sum : D : n+1 : x/i : i
END
4 ROLL OVER 5 ROLL 2 /
CEIL ^ * 3 ROLLD @ D = D*i^CEIL(n/2) @ Sum : D' : x : i
END
DROP * @ D = D*x @ Sum : D*x
a SWAP / @ Sum : a/2D
IF P a 3 5 SQRT + * <
THEN 4 P * a / 3  SQRT 3  @ limited by perimeter @ Sum : a/2D : SQRT(4P/a3)3
ELSE 5 SQRT 1  @ limited by triangle side @ Sum : a/2D : SQRT(5)1
END
* IP 1 + + @ Sum' = Sum + 1 + INT(...) @ Sum'
NEXT
»
» @ Sum
The code is not far from the original one without ISPRIME? and NEXTPRIME.
Time is spare since for large arguments, fewer tests or loops are needed to detemine all cofactor. Especely, no more loop are waste to test for composite (ex. i=9 is always negative because i=3 was alreday tested, but no way !)
Most of the time, D is close to a. With a large a great number of loops are no more waste. And when a is large and prim, no loop at all, immediately D=a is detected.
Time may be lost when ISPRIME? and NEXTPRIME are long time runs. This would be the case with my HP28S if I have to programm ISPRIME? or NEXTPRIME as USERRPL. Without builtin instruction using internal speed machine code, no gain.
But still, as demonstrate by Allen, finding cofactor using Farey Series is much more tricky and faster than determining D for each a value, since there is no more loops for cofactors determination or prime testing. The Farey Series directly give coprime b / a pairs and a few scale and endtest is only need :
« DUP @ Preserve P
3 / SQRT FLOOR @ Set Order for Recursive Farey Series
1 5 SQRT + 2/ @ golden ratio
> P d gr
« 0 @ Initiate Sum
0 1 1 1 @ m1 n1 m2 n2 leave and treated in stack
WHILE DUP2 / gr < @ test m2/n2 < gr
REPEAT
DUP2 * LAST + SQ SWAP 
P SWAP / FLOOR @ FLOOR(P/((a+b)²a*b))
6 ROLL + 5 ROLLD @ add to sum and ROLLD back to top
3 PICK d + OVER/ FLOOR @ > h
3 PICK OVER * 6 ROLL  @ compute new mt (and remove m1 from stack)
SWAP 3 PICK * 5 ROLL  @ compute new nt (and remove n1 from stack)
END
4 DROPN @ remove m1 n1 m2 n2 from stack (Sum remain on top)
»
»
Edited: 21 Mar 2012, 8:03 a.m.
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Thanks for this, C.Ret!
Your points about there being no speed gain if the functions aren't accelerated is understood. I was really just curious if I missed something obvious and the code would simplify significantly.
I tried running it but it gets stuck in an endless loop, with ISPRIME? getting 1 as arguments apparently indefinitely.
I'll try to follow your stack diagram and see where the problem is.
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You have to check, 1 ISPRIME? my have return 1 as expected.
You have to check also for 2: 2 ISPRIME? may have return 1 (2 is prime).
Perhaps, an inifinite loop result from NEXTPRIME fonction. I have no idea of how it is working.
My assumption is that :
1 NEXTPRIME returns 2, 2 NEXTPRIME returns 3 etc.
But, depending of impletation, maybe NEXTPRIME returns the closest prime number, resulting in the catastrophic chain reaction:
1 NEXTPRIME returns 1 (as 1 is the closest prime number),
4 NEXTPRIME return 5 (as 5 is the closest prime number  greatest or equal to the argument).
In this case, the modified version:
PROGRAM: @ Comments @ Stack 5: 4: 3: 2: 1:
@ @ P
« > P
« 0 @ Initiate Sum=0 @ 0 : 1 : P/3
1 P 3 / FOR a
2 @ Initiate D < 2
a @ x < a
2 @ i < 2 @ Sum : D : x : i
WHILE OVER ISPRIME? NOT @ loop until x is prime
REPEAT
NEXTPRIME @ i' < nextprime ( i ) @ Sum : D : x : i
0 3 ROLLD @ Initiate n < 0 @ Sum : D : n : x : i
WHILE DUP2 MOD NOT @ while x MOD i = 0
REPEAT
SWAP OVER / SWAP @ x < x/i @ Sum : D : n : x/i : i
ROT 1 + 3 ROLLD @ n < n+1 @ Sum : D : n+1 : x/i : i
END
4 ROLL OVER 5 ROLL 2 /
CEIL ^ * 3 ROLLD @ D = D*i^CEIL(n/2) @ Sum : D' : x : i
1 + @ inc i @
END
DROP * @ D = D*x @ Sum : D*x
a SWAP / @ Sum : a/2D
IF P a 3 5 SQRT + * <
THEN 4 P * a / 3  SQRT 3  @ limited by perimeter @ Sum : a/2D : SQRT(4P/a3)3
ELSE 5 SQRT 1  @ limited by triangle side @ Sum : a/2D : SQRT(5)1
END
* IP 1 + + @ Sum' = Sum + 1 + INT(...) @ Sum'
NEXT
»
» @ Sum
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Hi C.Ret.
Believe it or not, 1 is not a prime... ;)
You're not alone in thinking that it should be one. Here's what the Wikipedia page about primes has to say about the matter:
Quote:
Most early Greeks did not even consider 1 to be a number,[4] so did not consider it a prime. In the 19th century however, many mathematicians did consider the number 1 a prime. For example, Derrick Norman Lehmer's list of primes up to 10,006,721, reprinted as late as 1956,[5] started with 1 as its first prime.[6] Henri Lebesgue is said to be the last professional mathematician to call 1 prime.[7] Although a large body of mathematical work is also valid when calling 1 a prime, the above fundamental theorem of arithmetic does not hold as stated. For example, the number 15 can be factored as 3 · 5 or 1 · 3 · 5. If 1 were admitted as a prime, these two presentations would be considered different factorizations of 15 into prime numbers, so the statement of that theorem would have to be modified. Furthermore, the prime numbers have several properties that the number 1 lacks, such as the relationship of the number to its corresponding value of Euler's totient function or the sum of divisors function.[8][9]
Anyway, I had suspected that and changed my ISPRIME? function to return true for "1" but that led to stack underrun. I also tried starting the FOR loop with 2.
I shall debug this a little later when I have time.
The 50g (and ND1) behavior for these two functions is:
ISPRIME?: the first int for which 1 is returned is 2
NEXTPRIME: returns 2 for any int smaller than 2; any int is valid input, the next closest prime will be returned
This is obviously hard to develop if you don't actually have the functions builtin.
