This file is indexed.

/usr/share/pyshared/mpmath/libmp/libintmath.py is in python-mpmath 0.18-1.

This file is owned by root:root, with mode 0o644.

The actual contents of the file can be viewed below.

  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
"""
Utility functions for integer math.

TODO: rename, cleanup, perhaps move the gmpy wrapper code
here from settings.py

"""

import math
from bisect import bisect

from .backend import xrange
from .backend import BACKEND, gmpy, sage, sage_utils, MPZ, MPZ_ONE, MPZ_ZERO

def giant_steps(start, target, n=2):
    """
    Return a list of integers ~=

    [start, n*start, ..., target/n^2, target/n, target]

    but conservatively rounded so that the quotient between two
    successive elements is actually slightly less than n.

    With n = 2, this describes suitable precision steps for a
    quadratically convergent algorithm such as Newton's method;
    with n = 3 steps for cubic convergence (Halley's method), etc.

        >>> giant_steps(50,1000)
        [66, 128, 253, 502, 1000]
        >>> giant_steps(50,1000,4)
        [65, 252, 1000]

    """
    L = [target]
    while L[-1] > start*n:
        L = L + [L[-1]//n + 2]
    return L[::-1]

def rshift(x, n):
    """For an integer x, calculate x >> n with the fastest (floor)
    rounding. Unlike the plain Python expression (x >> n), n is
    allowed to be negative, in which case a left shift is performed."""
    if n >= 0: return x >> n
    else:      return x << (-n)

def lshift(x, n):
    """For an integer x, calculate x << n. Unlike the plain Python
    expression (x << n), n is allowed to be negative, in which case a
    right shift with default (floor) rounding is performed."""
    if n >= 0: return x << n
    else:      return x >> (-n)

if BACKEND == 'sage':
    import operator
    rshift = operator.rshift
    lshift = operator.lshift

def python_trailing(n):
    """Count the number of trailing zero bits in abs(n)."""
    if not n:
        return 0
    t = 0
    while not n & 1:
        n >>= 1
        t += 1
    return t

if BACKEND == 'gmpy':
    if gmpy.version() >= '2':
        def gmpy_trailing(n):
            """Count the number of trailing zero bits in abs(n) using gmpy."""
            if n: return MPZ(n).bit_scan1()
            else: return 0
    else:
        def gmpy_trailing(n):
            """Count the number of trailing zero bits in abs(n) using gmpy."""
            if n: return MPZ(n).scan1()
            else: return 0

# Small powers of 2
powers = [1<<_ for _ in range(300)]

def python_bitcount(n):
    """Calculate bit size of the nonnegative integer n."""
    bc = bisect(powers, n)
    if bc != 300:
        return bc
    bc = int(math.log(n, 2)) - 4
    return bc + bctable[n>>bc]

def gmpy_bitcount(n):
    """Calculate bit size of the nonnegative integer n."""
    if n: return MPZ(n).numdigits(2)
    else: return 0

#def sage_bitcount(n):
#    if n: return MPZ(n).nbits()
#    else: return 0

def sage_trailing(n):
    return MPZ(n).trailing_zero_bits()

if BACKEND == 'gmpy':
    bitcount = gmpy_bitcount
    trailing = gmpy_trailing
elif BACKEND == 'sage':
    sage_bitcount = sage_utils.bitcount
    bitcount = sage_bitcount
    trailing = sage_trailing
else:
    bitcount = python_bitcount
    trailing = python_trailing

if BACKEND == 'gmpy' and 'bit_length' in dir(gmpy):
    bitcount = gmpy.bit_length

# Used to avoid slow function calls as far as possible
trailtable = [trailing(n) for n in range(256)]
bctable = [bitcount(n) for n in range(1024)]

# TODO: speed up for bases 2, 4, 8, 16, ...

def bin_to_radix(x, xbits, base, bdigits):
    """Changes radix of a fixed-point number; i.e., converts
    x * 2**xbits to floor(x * 10**bdigits)."""
    return x * (MPZ(base)**bdigits) >> xbits

stddigits = '0123456789abcdefghijklmnopqrstuvwxyz'

def small_numeral(n, base=10, digits=stddigits):
    """Return the string numeral of a positive integer in an arbitrary
    base. Most efficient for small input."""
    if base == 10:
        return str(n)
    digs = []
    while n:
        n, digit = divmod(n, base)
        digs.append(digits[digit])
    return "".join(digs[::-1])

def numeral_python(n, base=10, size=0, digits=stddigits):
    """Represent the integer n as a string of digits in the given base.
    Recursive division is used to make this function about 3x faster
    than Python's str() for converting integers to decimal strings.

    The 'size' parameters specifies the number of digits in n; this
    number is only used to determine splitting points and need not be
    exact."""
    if n <= 0:
        if not n:
            return "0"
        return "-" + numeral(-n, base, size, digits)
    # Fast enough to do directly
    if size < 250:
        return small_numeral(n, base, digits)
    # Divide in half
    half = (size // 2) + (size & 1)
    A, B = divmod(n, base**half)
    ad = numeral(A, base, half, digits)
    bd = numeral(B, base, half, digits).rjust(half, "0")
    return ad + bd

def numeral_gmpy(n, base=10, size=0, digits=stddigits):
    """Represent the integer n as a string of digits in the given base.
    Recursive division is used to make this function about 3x faster
    than Python's str() for converting integers to decimal strings.

    The 'size' parameters specifies the number of digits in n; this
    number is only used to determine splitting points and need not be
    exact."""
    if n < 0:
        return "-" + numeral(-n, base, size, digits)
    # gmpy.digits() may cause a segmentation fault when trying to convert
    # extremely large values to a string. The size limit may need to be
    # adjusted on some platforms, but 1500000 works on Windows and Linux.
    if size < 1500000:
        return gmpy.digits(n, base)
    # Divide in half
    half = (size // 2) + (size & 1)
    A, B = divmod(n, MPZ(base)**half)
    ad = numeral(A, base, half, digits)
    bd = numeral(B, base, half, digits).rjust(half, "0")
    return ad + bd

if BACKEND == "gmpy":
    numeral = numeral_gmpy
else:
    numeral = numeral_python

_1_800 = 1<<800
_1_600 = 1<<600
_1_400 = 1<<400
_1_200 = 1<<200
_1_100 = 1<<100
_1_50 = 1<<50

def isqrt_small_python(x):
    """
    Correctly (floor) rounded integer square root, using
    division. Fast up to ~200 digits.
    """
    if not x:
        return x
    if x < _1_800:
        # Exact with IEEE double precision arithmetic
        if x < _1_50:
            return int(x**0.5)
        # Initial estimate can be any integer >= the true root; round up
        r = int(x**0.5 * 1.00000000000001) + 1
    else:
        bc = bitcount(x)
        n = bc//2
        r = int((x>>(2*n-100))**0.5+2)<<(n-50)  # +2 is to round up
    # The following iteration now precisely computes floor(sqrt(x))
    # See e.g. Crandall & Pomerance, "Prime Numbers: A Computational
    # Perspective"
    while 1:
        y = (r+x//r)>>1
        if y >= r:
            return r
        r = y

def isqrt_fast_python(x):
    """
    Fast approximate integer square root, computed using division-free
    Newton iteration for large x. For random integers the result is almost
    always correct (floor(sqrt(x))), but is 1 ulp too small with a roughly
    0.1% probability. If x is very close to an exact square, the answer is
    1 ulp wrong with high probability.

    With 0 guard bits, the largest error over a set of 10^5 random
    inputs of size 1-10^5 bits was 3 ulp. The use of 10 guard bits
    almost certainly guarantees a max 1 ulp error.
    """
    # Use direct division-based iteration if sqrt(x) < 2^400
    # Assume floating-point square root accurate to within 1 ulp, then:
    # 0 Newton iterations good to 52 bits
    # 1 Newton iterations good to 104 bits
    # 2 Newton iterations good to 208 bits
    # 3 Newton iterations good to 416 bits
    if x < _1_800:
        y = int(x**0.5)
        if x >= _1_100:
            y = (y + x//y) >> 1
            if x >= _1_200:
                y = (y + x//y) >> 1
                if x >= _1_400:
                    y = (y + x//y) >> 1
        return y
    bc = bitcount(x)
    guard_bits = 10
    x <<= 2*guard_bits
    bc += 2*guard_bits
    bc += (bc&1)
    hbc = bc//2
    startprec = min(50, hbc)
    # Newton iteration for 1/sqrt(x), with floating-point starting value
    r = int(2.0**(2*startprec) * (x >> (bc-2*startprec)) ** -0.5)
    pp = startprec
    for p in giant_steps(startprec, hbc):
        # r**2, scaled from real size 2**(-bc) to 2**p
        r2 = (r*r) >> (2*pp - p)
        # x*r**2, scaled from real size ~1.0 to 2**p
        xr2 = ((x >> (bc-p)) * r2) >> p
        # New value of r, scaled from real size 2**(-bc/2) to 2**p
        r = (r * ((3<<p) - xr2)) >> (pp+1)
        pp = p
    # (1/sqrt(x))*x = sqrt(x)
    return (r*(x>>hbc)) >> (p+guard_bits)

def sqrtrem_python(x):
    """Correctly rounded integer (floor) square root with remainder."""
    # to check cutoff:
    # plot(lambda x: timing(isqrt, 2**int(x)), [0,2000])
    if x < _1_600:
        y = isqrt_small_python(x)
        return y, x - y*y
    y = isqrt_fast_python(x) + 1
    rem = x - y*y
    # Correct remainder
    while rem < 0:
        y -= 1
        rem += (1+2*y)
    else:
        if rem:
            while rem > 2*(1+y):
                y += 1
                rem -= (1+2*y)
    return y, rem

def isqrt_python(x):
    """Integer square root with correct (floor) rounding."""
    return sqrtrem_python(x)[0]

def sqrt_fixed(x, prec):
    return isqrt_fast(x<<prec)

sqrt_fixed2 = sqrt_fixed

if BACKEND == 'gmpy':
    if gmpy.version() >= '2':
        isqrt_small = isqrt_fast = isqrt = gmpy.isqrt
        sqrtrem = gmpy.isqrt_rem
    else:
        isqrt_small = isqrt_fast = isqrt = gmpy.sqrt
        sqrtrem = gmpy.sqrtrem
elif BACKEND == 'sage':
    isqrt_small = isqrt_fast = isqrt = \
        getattr(sage_utils, "isqrt", lambda n: MPZ(n).isqrt())
    sqrtrem = lambda n: MPZ(n).sqrtrem()
else:
    isqrt_small = isqrt_small_python
    isqrt_fast = isqrt_fast_python
    isqrt = isqrt_python
    sqrtrem = sqrtrem_python


def ifib(n, _cache={}):
    """Computes the nth Fibonacci number as an integer, for
    integer n."""
    if n < 0:
        return (-1)**(-n+1) * ifib(-n)
    if n in _cache:
        return _cache[n]
    m = n
    # Use Dijkstra's logarithmic algorithm
    # The following implementation is basically equivalent to
    # http://en.literateprograms.org/Fibonacci_numbers_(Scheme)
    a, b, p, q = MPZ_ONE, MPZ_ZERO, MPZ_ZERO, MPZ_ONE
    while n:
        if n & 1:
            aq = a*q
            a, b = b*q+aq+a*p, b*p+aq
            n -= 1
        else:
            qq = q*q
            p, q = p*p+qq, qq+2*p*q
            n >>= 1
    if m < 250:
        _cache[m] = b
    return b

MAX_FACTORIAL_CACHE = 1000

def ifac(n, memo={0:1, 1:1}):
    """Return n factorial (for integers n >= 0 only)."""
    f = memo.get(n)
    if f:
        return f
    k = len(memo)
    p = memo[k-1]
    MAX = MAX_FACTORIAL_CACHE
    while k <= n:
        p *= k
        if k <= MAX:
            memo[k] = p
        k += 1
    return p

def ifac2(n, memo_pair=[{0:1}, {1:1}]):
    """Return n!! (double factorial), integers n >= 0 only."""
    memo = memo_pair[n&1]
    f = memo.get(n)
    if f:
        return f
    k = max(memo)
    p = memo[k]
    MAX = MAX_FACTORIAL_CACHE
    while k < n:
        k += 2
        p *= k
        if k <= MAX:
            memo[k] = p
    return p

if BACKEND == 'gmpy':
    ifac = gmpy.fac
elif BACKEND == 'sage':
    ifac = lambda n: int(sage.factorial(n))
    ifib = sage.fibonacci

def list_primes(n):
    n = n + 1
    sieve = list(xrange(n))
    sieve[:2] = [0, 0]
    for i in xrange(2, int(n**0.5)+1):
        if sieve[i]:
            for j in xrange(i**2, n, i):
                sieve[j] = 0
    return [p for p in sieve if p]

if BACKEND == 'sage':
    # Note: it is *VERY* important for performance that we convert
    # the list to Python ints.
    def list_primes(n):
        return [int(_) for _ in sage.primes(n+1)]

small_odd_primes = (3,5,7,11,13,17,19,23,29,31,37,41,43,47)
small_odd_primes_set = set(small_odd_primes)

def isprime(n):
    """
    Determines whether n is a prime number. A probabilistic test is
    performed if n is very large. No special trick is used for detecting
    perfect powers.

        >>> sum(list_primes(100000))
        454396537
        >>> sum(n*isprime(n) for n in range(100000))
        454396537

    """
    n = int(n)
    if not n & 1:
        return n == 2
    if n < 50:
        return n in small_odd_primes_set
    for p in small_odd_primes:
        if not n % p:
            return False
    m = n-1
    s = trailing(m)
    d = m >> s
    def test(a):
        x = pow(a,d,n)
        if x == 1 or x == m:
            return True
        for r in xrange(1,s):
            x = x**2 % n
            if x == m:
                return True
        return False
    # See http://primes.utm.edu/prove/prove2_3.html
    if n < 1373653:
        witnesses = [2,3]
    elif n < 341550071728321:
        witnesses = [2,3,5,7,11,13,17]
    else:
        witnesses = small_odd_primes
    for a in witnesses:
        if not test(a):
            return False
    return True

def moebius(n):
    """
    Evaluates the Moebius function which is `mu(n) = (-1)^k` if `n`
    is a product of `k` distinct primes and `mu(n) = 0` otherwise.

    TODO: speed up using factorization
    """
    n = abs(int(n))
    if n < 2:
        return n
    factors = []
    for p in xrange(2, n+1):
        if not (n % p):
            if not (n % p**2):
                return 0
            if not sum(p % f for f in factors):
                factors.append(p)
    return (-1)**len(factors)

def gcd(*args):
    a = 0
    for b in args:
        if a:
            while b:
                a, b = b, a % b
        else:
            a = b
    return a


#  Comment by Juan Arias de Reyna:
#
#  I learn this method to compute EulerE[2n] from van de Lune.
#
#  We apply the formula   EulerE[2n] = (-1)^n 2**(-2n) sum_{j=0}^n a(2n,2j+1)
#
#  where the numbers a(n,j) vanish for  j > n+1 or j <= -1  and satisfies
#
#  a(0,-1) = a(0,0) = 0;  a(0,1)= 1; a(0,2) = a(0,3) = 0
#
#  a(n,j) = a(n-1,j)                              when n+j is even
#  a(n,j) = (j-1) a(n-1,j-1) + (j+1) a(n-1,j+1)   when n+j is odd
#
#
#  But we can use only one array unidimensional a(j) since to compute
#  a(n,j) we only need to know a(n-1,k) where k and j are of different parity
#  and we have not to conserve the used values.
#
#  We cached up the values of Euler numbers to sufficiently high order.
#
#  Important Observation: If we pretend to use the numbers
#     EulerE[1], EulerE[2], ... , EulerE[n]
#     it is convenient to compute first EulerE[n], since the algorithm
#     computes first all
#     the previous ones, and keeps them in the CACHE

MAX_EULER_CACHE = 500

def eulernum(m, _cache={0:MPZ_ONE}):
    r"""
    Computes the Euler numbers `E(n)`, which can be defined as
    coefficients of the Taylor expansion of `1/cosh x`:

    .. math ::

        \frac{1}{\cosh x} = \sum_{n=0}^\infty \frac{E_n}{n!} x^n

    Example::

        >>> [int(eulernum(n)) for n in range(11)]
        [1, 0, -1, 0, 5, 0, -61, 0, 1385, 0, -50521]
        >>> [int(eulernum(n)) for n in range(11)]   # test cache
        [1, 0, -1, 0, 5, 0, -61, 0, 1385, 0, -50521]

    """
    # for odd m > 1, the Euler numbers are zero
    if m & 1:
        return MPZ_ZERO
    f = _cache.get(m)
    if f:
        return f
    MAX = MAX_EULER_CACHE
    n = m
    a = [MPZ(_) for _ in [0,0,1,0,0,0]]
    for  n in range(1, m+1):
        for j in range(n+1, -1, -2):
            a[j+1] = (j-1)*a[j] + (j+1)*a[j+2]
        a.append(0)
        suma = 0
        for k in range(n+1, -1, -2):
            suma += a[k+1]
            if n <= MAX:
                _cache[n] = ((-1)**(n//2))*(suma // 2**n)
        if n == m:
            return ((-1)**(n//2))*suma // 2**n

def stirling1(n, k):
    """
    Stirling number of the first kind.
    """
    if n < 0 or k < 0:
        raise ValueError
    if k >= n:
        return MPZ(n == k)
    if k < 1:
        return MPZ_ZERO
    L = [MPZ_ZERO] * (k+1)
    L[1] = MPZ_ONE
    for m in xrange(2, n+1):
        for j in xrange(min(k, m), 0, -1):
            L[j] = (m-1) * L[j] + L[j-1]
    return (-1)**(n+k) * L[k]

def stirling2(n, k):
    """
    Stirling number of the second kind.
    """
    if n < 0 or k < 0:
        raise ValueError
    if k >= n:
        return MPZ(n == k)
    if k <= 1:
        return MPZ(k == 1)
    s = MPZ_ZERO
    t = MPZ_ONE
    for j in xrange(k+1):
        if (k + j) & 1:
            s -= t * MPZ(j)**n
        else:
            s += t * MPZ(j)**n
        t = t * (k - j) // (j + 1)
    return s // ifac(k)