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-1 char, bugfix
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hlt
hlt
  • 191
  • 5

Python - 121121 120 chars

def p(x,s=2):
 while s<abss*s<=abs(x)**.5:yield x%s;s+=1
f=lambda a,b:(all(p(a*a+b*b))if b else f(b,a))if a else(b%4>2)&all(p(b))

p checks whether abs(x) is prime by iterating over all numbers from 2 to abs(x)**.5 (which is sqrt(abs(x))). It does so by yielding x % s for each s. all then checks whether all the yielded values are non-zero and stops generating values once it encounters a divisor of x. In f, f(b,a) replaces the case for b==0, inspired by @killmous'@killmous' Haskell answer.

-1 char and bugfix from @PeterTaylor

Python - 121 chars

def p(x,s=2):
 while s<abs(x)**.5:yield x%s;s+=1
f=lambda a,b:(all(p(a*a+b*b))if b else f(b,a))if a else(b%4>2)&all(p(b))

p checks whether abs(x) is prime by iterating over all numbers from 2 to abs(x)**.5 (which is sqrt(abs(x))). It does so by yielding x % s for each s. all then checks whether all the yielded values are non-zero and stops generating values once it encounters a divisor of x. In f, f(b,a) replaces the case for b==0, inspired by @killmous' Haskell answer.

Python - 121 120 chars

def p(x,s=2):
 while s*s<=abs(x):yield x%s;s+=1
f=lambda a,b:(all(p(a*a+b*b))if b else f(b,a))if a else(b%4>2)&all(p(b))

p checks whether abs(x) is prime by iterating over all numbers from 2 to abs(x)**.5 (which is sqrt(abs(x))). It does so by yielding x % s for each s. all then checks whether all the yielded values are non-zero and stops generating values once it encounters a divisor of x. In f, f(b,a) replaces the case for b==0, inspired by @killmous' Haskell answer.

-1 char and bugfix from @PeterTaylor

Source Link
hlt
hlt
  • 191
  • 5

Python - 121 chars

def p(x,s=2):
 while s<abs(x)**.5:yield x%s;s+=1
f=lambda a,b:(all(p(a*a+b*b))if b else f(b,a))if a else(b%4>2)&all(p(b))

p checks whether abs(x) is prime by iterating over all numbers from 2 to abs(x)**.5 (which is sqrt(abs(x))). It does so by yielding x % s for each s. all then checks whether all the yielded values are non-zero and stops generating values once it encounters a divisor of x. In f, f(b,a) replaces the case for b==0, inspired by @killmous' Haskell answer.