## Fenderbender’s Square Reckoner

### November 29, 2016

There’s not much to talk about. Here’s the code:

(define (square? n) (let ((m (modulo n 128))) (if (positive? (bitwise-and (* m #x8bc40d7d) (* m #xa1e2f5d1) #x14020a)) #f (let ((large-mod (modulo n 3989930175))) ; (* 63 25 11 17 19 23 31) (and (let ((m (modulo large-mod 63))) (zero? (bitwise-and (* m #x3d491df7) (* m #xc824a9f9) #x10f14008))) (let ((m (modulo large-mod 25))) (zero? (bitwise-and (* m #x1929fc1b) (* m #x4c9ea3b2) #x51001005))) (let ((m (* #xd10d829a (modulo large-mod 31)))) (zero? (bitwise-and m (+ m #x672a5354) #x21025115))) (let ((m (modulo large-mod 23))) (zero? (bitwise-and (* m #x7bd28629) (* m #xe7180889) #xf8300))) (let ((m (modulo large-mod 19))) (zero? (bitwise-and (* m #x1b8bead3) (* m #x4d75a124) #x4280082b))) (let ((m (modulo large-mod 17))) (zero? (bitwise-and (* m #x6736f323) (* m #x9b1d499) #xc0000300))) (let ((m (modulo large-mod 11))) (zero? (bitwise-and (* m #xabf1a3a7) (* m #x2612bf93) #x45854000))) (let ((root (isqrt n))) (if (= (* root root) n) root #f)))))))

> (let ((p (- (expt 2 89) 1)) (q (- (expt 2 89) 21))) (square? (* p q))) #f > (let ((p (- (expt 2 89) 1))) (square? (* p p))) 618970019642690137449562111

You can run the program at http://ideone.com/hbxEG5.

@programmingpraxis: I converted the code to Python. It works about twice as fast compared to a method where you just use the 2 last digits for a first screening. Maybe I found an example where the method does not work. 152415787532388367504942236884722755800955129 is a square, but fenderbender’s method tells me not; it fails on the test using mod19. Could you check with your code?

@Paul: Works fine for me: http://ideone.com/sqY3zf.

@programmingparaxis: thanks. Apparently some checking to do.

Does anyone know why fenderbender chose mod x with x = 128? I have tried other x. By trial and error, 5! (= 120) gives a rejection rate of 85% and 6! a rejection rate of 93.3%.

Would either of these be more appropriate?

@Ernie: I suspect it is due to the computational efficiency of bit-shifting rather than performing a big-integer modulo.

An interesting method. In Python it seems to be more effective to use sets i.s.o. Bloom filters. The version with sets is about a factor of 1.7 times faster. For the sets version I used mod64 i.s.o. mod128, because I could not find the quadratic residues for mod128.

@Paul: It’s easy to compute the quadratic residues of a modulus, let’s say

n= 11. For each number from 0 ton−1, square the number and take it modn; thus, (0*0)%11=0, (1*1)%11=1, . . . , (5*5)%11=3, . . . , (10*10)%11=1. Those numbers are the quadratic residues of 11: 0, 1, 3, 4, 5, and 9. For the quadratic residues of 128 I get@programmingpraxis: thanks, I found out in the meantime (and get the same numbers as you). Using mod128 did not give significant different results from using mod64. I am still a little working on the problem, what would be the best mod’s to use.

I’ve implemented 4 square finding routines and timed them.

> Timings (Standard Python):

3.6116340733525862

6.418930000824364

9.941192323010673

11.628690812502668

> Timings (Pypy):

0.3549592833256139

2.8981952367599213

0.24083208547439217

0.14977518674845003

As you can see it seems to matter with which interpreter you run your script. Feel free to correct me if any of that source code is wrong.

To make the function more general purpose, it is useful to handle negative n properly (it should return false) and n = 0 (it should return 0).