## Legendre’s Symbol

### May 1, 2012

Here are our new versions of the jacobi and legendre symbols, which combine the rules given above into a single function:

```(define (jacobi a m)   (if (not (integer? a)) (error 'jacobi "must be integer")     (if (not (and (integer? m) (positive? m) (odd? m)))         (error 'jacobi "modulus must be odd positive integer")         (let loop1 ((a (modulo a m)) (m m) (t 1))           (if (zero? a) (if (= m 1) t 0)             (let ((z (if (member (modulo m 8) (list 3 5)) -1 1)))               (let loop2 ((a a) (t t))                 (if (even? a) (loop2 (/ a 2) (* t z))                   (loop1 (modulo m a) a                          (if (and (= (modulo a 4) 3)                                   (= (modulo m 4) 3))                              (- t) t))))))))))```

Although we’ve rearranged things, this code faithfully executes the six rules; it is interesting to trace from the rules to the code. The time complexity is about the same as taking a gcd, so it’s very fast. The legendre symbol is shown below:

```(define (legendre a m)   (if (prime? m) (jacobi a m)     (error 'legendre "modulus must be prime")))```

And here are some tests; the first two failed on the old version of `jacobi`:

```> (jacobi -1 7) -1 > (jacobi 3 3) 0 > (jacobi 127 703) -1```

You can run the program at http://programmingpraxis.codepad.org/6uVnD0RR.

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### 13 Responses to “Legendre’s Symbol”

1. ardnew said

I first tried translating this to a completely iterative algorithm, but it was becoming a mess.. I also couldn’t think of a better way to reduce the rule 3 logic.

```use strict;
use warnings;

sub reduce
{
my (\$p, \$a, \$m) = (1, @_);

if (\$a > 2)
{
# rule 4
\$p *=  reduce(2, \$m) * reduce(\$a / 2, \$m) unless \$a & 1;

# rule 6
\$p *= -reduce(-\$m, \$a) if \$a % 4 == 3 and \$m % 4 == 3;
}

# rules 1 and 2
\$p *= \$a unless \$a >> 1;

# rule 3
if (2 == \$a)
{
my \$mod8 = \$m % 8;
\$p *= -1 if 3 == \$mod8 or 5 == \$mod8;
\$p *=  1 if 1 == \$mod8 or 7 == \$mod8;
}

return \$p;
}

sub legendre(\$)
{
my (\$a, \$m) = @{(shift)};

die "modulus must be prime positive integer\$/"
unless int(\$m) == \$m and \$m > 0 and \$m & 1;
# need to also verify \$m is prime, but meh

# rule 5
my \$n = \$a % \$m;

print "(\$a / \$m) = ".reduce(\$n, \$m).\$/;
}

legendre(\@ARGV);
```
2. programmingpraxis said

Ardnew: the usual iterative solution looks like this:

```a = a mod m; t = 1; while (a != 0) {     while (a even) {         a = a / 2;         if (m mod 8 in {3,5}) t = -t;     }     (a,m) = (m,a) // swap variables     if (a === m === 3 (mod 4)) t = -t     a = a mod m; } if (m == 1) return t; return 0;```

3. ardnew said

very nice! always seems so simple once you see the answer…

also, there’s a bug in my code that doesn’t reapply rule 5 when rule 6 is evaluated. I’m not familiar enough with the math to prove this behavior is incorrect though

4. programmingpraxis said

In a private email, Mike pointed out that I missed the `else` clause in Rule 6. It is now fixed.

5. Mike said

An iterative Python version:

```def jacobi(n, m):
j = 1

# rule 5
n %= m

while n:
# rules 3 and 4
t = 0
while not n & 1:
n /= 2
t += 1
if t&1 and m%8 in (3, 5):
j = -j

# rule 6
if (n % 4 == m % 4 == 3):
j = -j

# rules 5 and 6
n, m = m % n, n

return j if m == 1 else 0

```
6. ardnew said

a) I love that 3-way equality condition in rule 6. Never seen that before :)
b) Can you explain the behavior of the expression: “if t&1 and m%8 in (3, 5)”

And Mr. programmingpraxis (i assume that is your real name), a question:
Is rule 5 even a necessary reduction? I’m beginning to think its there merely for computational optimization

7. ardnew said

Oops, I was caught up in that equality expression, disregard comment b), Mike.

8. Mike said

@ardnew,

The ability to chain relational operators is a nice feature of Python syntax. It compiles to a series of 2-way relational tests ‘and’ed together. “if n%4 == m%4 == 3:” compiles like “if n%4 == m%4 and m%4 == 3:”, but the m%4 is only evaluated once. It is handy when you want to check if a value is between two limits, like “if lowerbound <= x = y <= z:" in a previous exercise to determine that y was no larger than either x or z. You can also chain any number together: "a c == d …”

Rule 5 is the only way to reduce ‘a’ when it’s not even.

9. Mike said

Oops, wordpress ate some of my relational operators as html tags. Should be:

The ability to chain relational operators is a nice feature of Python syntax. It compiles to a series of 2-way relational tests ‘and’ed together. “if n%4 == m%4 == 3:” compiles like “if n%4 == m%4 and m%4 == 3:”, but the m%4 is only evaluated once. It is handy when you want to check if a value is between two limits, like “if lowerbound <= x < upperbound:". I used something like "if x >= y <= z:" in a previous exercise to determine that y was no larger than either x or z. You can also chain any number together: "a < b > c == d …”

Rule 5 is the only way to reduce ‘a’ when it’s not even.

10. Graham said

Since Mike already did an iterative Python version, here’s a straight translation of the rules to Haskell:

```jacobi :: Integer -> Integer -> Integer
jacobi a m
| even m        = error "m must be odd."
| a == 0        = 0
| a == 1        = 1
| a == 2        = if (m `mod` 8) `elem` [3, 5] then -1 else 1
| even a        = jacobi 2 m * jacobi (a `div` 2) m
| a >= m        = jacobi (a `mod` m) m
| otherwise     = let j = jacobi m a in
if (a `mod` 4 == 3) && (m `mod` 4 == 3) then -j else j
```

Not the most efficient solution, probably. Happy to see another math exercise, though! (even though it should be titled “Jacobi’s Symbol,” since we don’t know m is prime)

11. Mike said

As a preprocessing step, I think rule 5 needs to be applied if a (7 / -1) by rule 6 (the only rule that could apply
(7 / -1) => (0 / -1) by rule 5
(0 / -1) => 0 by rule 1

However, (-1 / 7) = -1 according to the programmingpraxis solution.

If rule 5 is applied first, then:
(-1 / 7) => (6 / 7) by rule 5
(6 / 7) => (2 / 7) * (3 / 7) => (3 / 7) => -(7 / 3) => -(1 / 3) => -1 by rules 4, 3, 6, 5, and 2, respectively

12. Mike said

I did it again. (Maybe I need an option to preview a post)

The previous post should be:

As a preprocessing step, I think rule 5 needs to be applied if a < 0.

For example:
(-1 / 7) => (7 / -1) by rule 6 (the only rule that could apply)
(7 / -1) => (0 / -1) by rule 5
(0 / -1) => 0 by rule 1

However, (-1 / 7) = -1 according to the programmingpraxis solution.

If rule 5 is applied first, then:
(-1 / 7) => (6 / 7) by rule 5
(6 / 7) => (2 / 7) * (3 / 7) => (3 / 7) => -(7 / 3) => -(1 / 3) => -1; by rules 4, 3, 6, 5, and 2, respectively.

13. programmingpraxis said

The jacobi symbol for (-1 7) is -1. Wolfram Alpha agrees.

In arithmetic mod m, the only numbers allowed are 0 to m-1 inclusive. There are no negative numbers. So any time you see a negative number you have to add m repeatedly until it is in the range 0 to m-1 inclusive. So yes, Rule 5 applies whenever a<0. I'll edit the rule.