Talk:Law of cosines - triples
Showing the sides in order?
I see the samples show the 60 degree triangles with the sides in ascending order, however ( a, b, c ) = ( 5, 7, 8 ) isn't a solution of
a^2 + b^2 - ab = c^2.
Where the task says "Find all integer solutions to the three specific cases, in order;", presumably this should be in order of ascending first (lowest) side?
--Tigerofdarkness (talk) 13:28, 23 September 2018 (UTC)
- You are right. Thanks for bringing this up. I fixed my Factor submission. --Chunes (talk) 14:28, 23 September 2018 (UTC)
- I changed the wording of that task's requirement (hopefully, it is more clearer). In any case, that's what I took it to mean, that the triangles are to be shown in increasing (ascending) order of the first side found. -- Gerard Schildberger (talk) 06:56, 24 September 2018 (UTC)
How to verify the number of triangles for the extra credit?
Now that there're several computer programming examples that have different output for the optional extra credit requirement, how does one verify which one is correct? -- Gerard Schildberger (talk) 11:07, 24 September 2018 (UTC)
- How about we each create a temporary "Law of cosines - triples/tmp/<language>" page containing just a list of the thousands of triples ?
- One space-separated triple per line
- Any line starting '#' treated as a comment, no other comments allowed.
- The tmp/... pages all to be deleted on 1st October 2018.
- We would have a week to examine each others results and refine the task. Paddy3118 (talk) 12:25, 24 September 2018 (UTC)
- Paddy, I suspect there may be a problem with the expression: int(c2**0.5) which occurs 3 times in your Python entry. If the power operator produces a value which is slightly less than the 'correct' integer value, then the 'int' function will round it down and the wrong tuple will be added to the set.
- Thanks to user Hout for the extra Python solution. I ran it then extracted the generation of the list for the extended credit solution and then filtered out occurrences of the same triangle to get a still differing result:
- <lang python>
t60u = triangles(f60unequal, 10000)
len(t60u) Out[4]: 18394
t60u[0] Out[5]: '[3, 7, 8]'
t60new = set([tuple(sorted(tri)) for tri in t60u])
len(t60new) Out[7]: 16161 </lang>
- P.S. I had actually developed a python
method2based on dicts (and the fact that dicts are ordered by default in Python3.6). I debugged the two implementations together and they agreed when tested against each other but the Python set version was slightly faster so I only published that. I'll invetstigate further tonight. Paddy3118 (talk) 09:25, 25 September 2018 (UTC)
- P.S. I had actually developed a python
The two different python methods that I developed together
Here you go: <lang python>N = 13
def method1(N=N):
squares = [x**2 for x in range(0, N+1)]
sqrset = set(squares)
tri90, tri60, tri120 = (set() for _ in range(3))
for a in range(1, N+1):
a2 = squares[a]
for b in range(1, a + 1):
b2 = squares[b]
c2 = a2 + b2
if c2 in sqrset:
tri90.add(tuple(sorted((a, b, int(c2**0.5)))))
continue
ab = a * b
c2 -= ab
if c2 in sqrset:
tri60.add(tuple(sorted((a, b, int(c2**0.5)))))
continue
c2 += 2 * ab
if c2 in sqrset:
tri120.add(tuple(sorted((a, b, int(c2**0.5)))))
return sorted(tri90), sorted(tri60), sorted(tri120)
- %%
if __name__ == '__main__':
print(f'Integer triangular triples for sides 1..{N}:')
for angle, triples in zip([90, 60, 120], method1(N)):
print(f' {angle:3}° has {len(triples)} solutions:\n {triples}')
_, t60, _ = method1(10_000)
notsame = sum(1 for a, b, c in t60 if a != b or b != c)
print('Extra credit:', notsame)
def method2(N=N): # Python 3.6+
sqr2root = {x**2: x for x in range(1, N+1)}
tri90, tri60, tri120 = (set() for _ in range(3))
for a2, a in sqr2root.items():
for b2, b in sqr2root.items():
if b > a:
break
c2 = a2 + b2
if c2 in sqr2root:
tri90.add(tuple(sorted((a, b, sqr2root[c2]))))
continue
ab = a * b
c2 -= ab
if c2 in sqr2root:
tri60.add(tuple(sorted((a, b, sqr2root[c2]))))
continue
c2 += 2 * ab
if c2 in sqr2root:
tri120.add(tuple(sorted((a, b, sqr2root[c2]))))
return sorted(tri90), sorted(tri60), sorted(tri120)
- %%
if __name__ == '__main__':
tt1 = method1()
tt2 = method2()
assert tt1 == tt2
#
method1_t60 = t60
_, method2_t60, _ = method2(10_000)
assert method1_t60 == method2_t60
notsame2 = sum(1 for a, b, c in method2_t60 if a != b or b != c)
print('Extra credit:', notsame2)
assert notsame == notsame2</lang>
- Output:
Integer triangular triples for sides 1..13:
90° has 3 solutions:
[(3, 4, 5), (5, 12, 13), (6, 8, 10)]
60° has 15 solutions:
[(1, 1, 1), (2, 2, 2), (3, 3, 3), (3, 7, 8), (4, 4, 4), (5, 5, 5), (5, 7, 8), (6, 6, 6), (7, 7, 7), (8, 8, 8), (9, 9, 9), (10, 10, 10), (11, 11, 11), (12, 12, 12), (13, 13, 13)]
120° has 2 solutions:
[(3, 5, 7), (7, 8, 13)]
Extra credit: 17806
Extra credit: 17806
- method1 uses sets and is slightly faster than method2 which is why I only posted it.
- method 2 uses `sqr2root` mapping squares to their square root, and furthermore, does not use floating point fractional exponation in its generation.
- the seconf `if __name__...` block computes the same stats using method2 and asserts that the answers are equivalent.
Looking again at my code, Ithink it's right, but, maybe I'm not seeing an off by one error? I am not sure.
I guess the next thing to do is compare Houts dict based solution and look at the differences. Paddy3118 (talk)
- Some particular 60 degree matches which I see in my output but not (I think) in yours include:
- (8, 15, 13), (16, 30, 26), (24, 45, 39), (32, 60, 52) etc ... up to (4704, 8820, 7644) (all of which, as far as I can see match the pattern ((a^2) + (b^2)) - (a * b) == c ^ 2) Hout (talk) 20:05, 26 September 2018 (UTC)