pythagorean triples exercise
MRAB
python at mrabarnett.plus.com
Fri Oct 22 19:42:18 CEST 2010
On 22/10/2010 13:33, Baba wrote:
> On Oct 22, 8:07 am, Dennis Lee Bieber<wlfr... at ix.netcom.com> wrote:
>> On Thu, 21 Oct 2010 03:51:07 -0700 (PDT), Baba<raoul... at gmail.com>
>> declaimed the following in gmane.comp.python.general:
>>
>>> Hi everyone
>>
>>> i need a hint regarding the following exercise question:
>>
>>> "Write a program that generates all Pythagorean triples whose small
>>> sides are no larger than n.
>>> Try it with n<= 200."
>>
>>> what is "n" ? i am guessing that it is a way to give a bound to the
>>> triples to be returned but i can't figure out where to fit in "n".
>>
>>> a^a + b^b = c^c is the condition to satisfy and i need to use loops
>>
>> Well, I'd interpret it to mean that
>>
>> a<= 200
>> AND
>> b<= 200
>>
>> since c is the hypotenuse, which by definition is longer than either of
>> the sides.
>>
>> The brute force approach is a nested set of "for" loops, running
>> 1-200 (remember that (x)range(200) runs 0-199<G>).
>>
>> The alternative is to study the information at
>>
>> http://www.math.uic.edu/~fields/puzzle/triples.html
>>
>> and filtering out entries where the first two components are>200...
>> Looking at the middle term "2*m*n" would have to be less than 201, and
>> the first term "n*n-m*m"< 201
>>
>> In Python, this can all be done in a one-liner...
>>
>> [(n*n - m*m, 2*m*n, n*n + m*m) for n in xrange(2,101) for m in
>> xrange(1,n) if 2*m*n< 201 and n*n-m*m< 201]
>>
>> Converting this to a clean set of nested loops and nice lines of
>> output is a different matter.
>>
>> Oh, and DO study the link I gave so you can cite it when you turn in
>> this intriguing formulation... after all, no need for math.sqrt<G>
>>
>> --
>> Wulfraed Dennis Lee Bieber AF6VN
>> wlfr... at ix.netcom.com HTTP://wlfraed.home.netcom.com/
>
> Hi Wulfraed,
>
> only a has an upper limit of 200
>
Really? The quote you gave included "whose small sides are no larger
than n". Note: "sides", plural.
> the program needs to output the following triple for a == 200: (200 ,
> 609,641)
>
> so at this stage my problem is: how to set the upper limit for b in a
> smart way?
>
> My solution below is not efficient i believe.
>
> import math
> for a in range (190,200):
> for b in range (a,a*a):
> csqrd = a * a + b * b
> csqrt = math.sqrt(csqrd)
> for c in range (1, csqrd):
> if c * c == a * a + b * b and math.floor(csqrt) == csqrt:
> print (a,b,c)
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