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:: 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. -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk: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? -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk: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. [[User:Paddy3118|Paddy3118]] ([[User talk: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.
::I'd see if you get a different answer for the extra credit if you use instead: int(round(c2**0.5)). --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 14:46, 24 September 2018 (UTC)
::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>
::I will try removing the **.5 tonight. Thanks.<br> [[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 09:15, 25 September 2018 (UTC)
::P.S. I had actually developed a python <code>method2</code> based 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. [[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 09:25, 25 September 2018 (UTC)
===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>
{{out}}
<pre>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</pre>
# 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.
<br>
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. [[User:Paddy3118|Paddy3118]] ([[User talk: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) [[User:Hout|Hout]] ([[User talk:Hout|talk]]) 20:05, 26 September 2018 (UTC)
:: PS I notice that you are braver than me and mutating the referent of the name '''c2'''. Is it possible that an earlier mutation is (in a subset of cases) shadowing a later test ? You have ''c2 -= ab'' and then ''c2 += 2 * ab'' [[User:Hout|Hout]] ([[User talk:Hout|talk]]) 20:18, 26 September 2018 (UTC)
:::Yep, that's the reason for the discrepancy, Hout :)
:::I just ran it through my Go version and there's apparently 588 cases where both (a*a + b*b) and (a*a + b*b -a*b) are perfect squares. So they are being filtered out by the 90° case.
:::As 18394 - 588 = 17806 that explains Paddy's results. --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 21:20, 26 September 2018 (UTC)
:::: Aha ! Well spotted :-) Mutation is a tricky business ... [[User:Hout|Hout]] ([[User talk:Hout|talk]]) 21:43, 26 September 2018 (UTC)
====I woke in the night...====
With an idea that the continue statements might be at fault.<br>
(I am disowning them in my use of English, but it was me).<br>
I commented-out all the continue statements in method1 and method2 as I literally had that lightbulb moment of thinking that in my loops, I didn't allow for one triple to be in more than one of the three groups.
Before sleeping I had added Houts case as method3; seen that hehad captured cases that I had not; but was unable to work out why and had gonne to bed with the problem
<lang python># Houts code with a slight modification to his `main` function so that it returned `triangles(f60unequal, 10000)`
# ...
#%%
method3_t60_uneven_strings = main() # Houts'
method3_t60_uneven = [tuple(eval(triple_list_str))
for triple_list_str in method3_t60_uneven_strings]
method2_t60_uneven = [(a, b, c) for a, b, c in method2_t60 if a != b or b != c]
method1_t60_uneven = [(a, b, c) for a, b, c in method1_t60 if a != b or b != c]
#%%
methods_t60_uneven = [method1_t60_uneven, method2_t60_uneven, method3_t60_uneven]
whos_methods = "Paddys set, Paddys dict, Houts dict".split(", ")
print("\n# Stated extra credit answers")
for who, t60u in zip(whos_methods, methods_t60_uneven):
print(f" {who:12s}: {len(t60u)}")
print("\n# Filtered (again in some cases) for unequal sides")
for who, t60u in zip(whos_methods, methods_t60_uneven):
t60uf = [(a, b, c) for a, b, c in t60u if a != b or b != c]
print(f" {who:12s}: {len(t60uf)}")
print("\n# Filtered, ordered, and duplicates removed: size changes")
for who, t60u in zip(whos_methods, methods_t60_uneven):
t60ufod = set([tuple(sorted([a, b, c]))
for a, b, c in t60u if a != b or b != c])
print(f" {who:12s}: From {len(t60u)} to {len(t60ufod)}")
diff13 = sorted(set(method1_t60_uneven) - set(method3_t60_uneven))
diff31 = sorted(set(method3_t60_uneven) - set(method1_t60_uneven))
print(f'\n# I have {len(diff13)} triples that Hout does not have')
print(f'# Hout has {len(diff31)} triples that I do not have')</lang>
;Original output:
<pre>60 degrees - uneven triangles of maximum side 10000. Total:
18394
# Stated extra credit answers
Paddys set : 17806
Paddys dict : 17806
Houts dict : 18394
# Filtered (again in some cases) for unequal sides
Paddys set : 17806
Paddys dict : 17806
Houts dict : 18394
# Filtered, ordered, and duplicates removed: size changes
Paddys set : From 17806 to 17806
Paddys dict : From 17806 to 17806
Houts dict : From 18394 to 18394
# I have 0 triples that Hout does not have
# Hout has 588 triples that I do not have</pre>
;Output when those continue statements in method1 and method2 are removed:
<pre># Stated extra credit answers
Paddys set' : 18394
Paddys dict': 18394
Houts dict : 18394
# Filtered (again in some cases) for unequal sides
Paddys set' : 18394
Paddys dict': 18394
Houts dict : 18394
# Filtered, ordered, and duplicates removed: size changes
Paddys set' : From 18394 to 18394
Paddys dict': From 18394 to 18394
Houts dict : From 18394 to 18394
# I have 0 triples that Hout does not have
# Hout has 0 triples that I do not have</pre>
'''Yay!!!''' From the discussions of sharper minds above, I found that you had already found the answer, but I thought I would go ahead and finish my public debugging session. <br>
Thanks people :-)<br>
[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 01:18, 27 September 2018 (UTC)
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