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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983

This file explains the functions of the library routines.
Note:
Type FiniteField is defined as unsigned long
Type Double is defined as double
extern long nullspaceLong(const long n,
const long m,
const long *A,
mpz_t * *mp_N_pass);
/*
* Calling Sequence:
* nullspaceLong(n, m, A, mp_N_pass)
*
* Summary:
* Compute the right nullspace of A.
*
* Input: n: long, row dimension of A
* m: long, column dimension of A
* A: 1dim signed long array length n*m, representing n x m matrix
* in row major order
*
* Output:
*  *mp_N_pass: points to a 1dim mpz_t array of length m*s, where s is the
* dimension of the right nullspace of A
*  the dimension s of the nullspace is returned
*
* Notes:
*  The matrix A is represented by onedimension array in row major order.
*  Space for what mp_N_points to is allocated by this procedure: if the
* nullspace is empty, mp_N_pass is set to NULL.
*/
extern void nonsingSolvMM (const enum SOLU_POS solupos, const long n,
const long m, const long *A, mpz_t *mp_B,
mpz_t *mp_N, mpz_t mp_D);
/*
* Calling Sequence:
* nonsingSolvMM(solupos, n, m, A, mp_B, mp_N, mp_D)
*
* Summary:
* Solve nonsingular system of linear equations, where the left hand side
* input matrix is a signed long matrix.
*
* Description:
* Given a n x n nonsingular signed long matrix A, a n x m or m x n mpz_t
* matrix mp_B, this function will compute the solution of the system
* 1. AX = mp_B
* 2. XA = mp_B.
* The parameter solupos controls whether the system is in the type of 1
* or 2.
*
* Since the unique solution X is a rational matrix, the function will
* output the numerator matrix mp_N and denominator mp_D respectively,
* such that A(mp_N) = mp_D*mp_B or (mp_N)A = mp_D*mp_B.
*
* Input:
* solupos: enumerate, flag to indicate the system to be solved
*  solupos = LeftSolu: solve XA = mp_B
*  solupos = RightSolu: solve AX = mp_B
* n: long, dimension of A
* m: long, column or row dimension of mp_B depending on solupos
* A: 1dim signed long array length n*n, representing the n x n
* left hand side input matrix
* mp_B: 1dim mpz_t array length n*m, representing the right hand side
* matrix of the system
*  solupos = LeftSolu: mp_B a m x n matrix
*  solupos = RightSolu: mp_B a n x m matrix
*
* Output:
* mp_N: 1dim mpz_t array length n*m, representing the numerator matrix
* of the solution
*  solupos = LeftSolu: mp_N a m x n matrix
*  solupos = RightSolu: mp_N a n x m matrix
* mp_D: mpz_t, denominator of the solution
*
* Precondition:
* A must be a nonsingular matrix.
*
* Note:
*  It is necessary to make sure the input parameters are correct,
* expecially the dimension, since there is no parameter checks in the
* function.
*  Input and output matrices are row majored and represented by
* onedimension array.
*  It is needed to preallocate the memory space of mp_N and mp_D.
*
*/
extern void nonsingSolvLlhsMM (const enum SOLU_POS solupos, const long n,
const long m, mpz_t *mp_A, mpz_t *mp_B,
mpz_t *mp_N, mpz_t mp_D);
/*
* Calling Sequence:
* nonsingSolvLlhsMM(solupos, n, m, mp_A, mp_B, mp_N, mp_D)
*
* Summary:
* Solve nonsingular system of linear equations, where the left hand side
* input matrix is a mpz_t matrix.
*
* Description:
* Given a n x n nonsingular mpz_t matrix A, a n x m or m x n mpz_t
* matrix mp_B, this function will compute the solution of the system
* 1. (mp_A)X = mp_B
* 2. X(mp_A) = mp_B.
* The parameter solupos controls whether the system is in the type of 1
* or 2.
*
* Since the unique solution X is a rational matrix, the function will
* output the numerator matrix mp_N and denominator mp_D respectively,
* such that mp_Amp_N = mp_D*mp_B or mp_Nmp_A = mp_D*mp_B.
*
* Input:
* solupos: enumerate, flag to indicate the system to be solved
*  solupos = LeftSolu: solve XA = mp_B
*  solupos = RightSolu: solve AX = mp_B
* n: long, dimension of A
* m: long, column or row dimension of mp_B depending on solupos
* mp_A: 1dim mpz_t array length n*n, representing the n x n left hand
* side input matrix
* mp_B: 1dim mpz_t array length n*m, representing the right hand side
* matrix of the system
*  solupos = LeftSolu: mp_B a m x n matrix
*  solupos = RightSolu: mp_B a n x m matrix
*
* Output:
* mp_N: 1dim mpz_t array length n*m, representing the numerator matrix
* of the solution
*  solupos = LeftSolu: mp_N a m x n matrix
*  solupos = RightSolu: mp_N a n x m matrix
* mp_D: mpz_t, denominator of the solution
*
* Precondition:
* mp_A must be a nonsingular matrix.
*
* Note:
*  It is necessary to make sure the input parameters are correct,
* expecially the dimension, since there is no parameter checks in the
* function.
*  Input and output matrices are row majored and represented by
* onedimension array.
*  It is needed to preallocate the memory space of mp_N and mp_D.
*
*/
extern void nonsingSolvRNSMM (const enum SOLU_POS solupos, const long n,
const long m, const long basislen,
const FiniteField *basis, Double **ARNS,
mpz_t *mp_B, mpz_t *mp_N, mpz_t mp_D);
/*
* Calling Sequence:
* nonsingSolvRNSMM(solupos, basislen, n, m, basis, ARNS, mp_B, mp_N, mp_D)
*
* Summary:
* Solve nonsingular system of linear equations, where the left hand side
* input matrix is represented in a RNS.
*
* Description:
* Given a n x n nonsingular matrix A represented in a RNS, a n x m or m x n
* mpz_t matrix mp_B, this function will compute the solution of the system
* 1. AX = mp_B
* 2. XA = mp_B.
* The parameter solupos controls whether the system is in the type of 1
* or 2.
*
* Since the unique solution X is a rational matrix, the function will
* output the numerator matrix mp_N and denominator mp_D respectively,
* such that A(mp_N) = mp_D*mp_B or (mp_N)A = mp_D*mp_B.
*
* Input:
* solupos: enumerate, flag to indicate the system to be solved
*  solupos = LeftSolu: solve XA = mp_B
*  solupos = RightSolu: solve AX = mp_B
* basislen: long, dimension of RNS basis
* n: long, dimension of A
* m: long, column or row dimension of mp_B depending on solupos
* basis: 1dim FiniteField array length basislen, RNS basis
* ARNS: 2dim Double array, dimension basislen x n*n, representation of
* n x n input matrix A in RNS, where ARNS[i] = A mod basis[i]
* mp_B: 1dim mpz_t array length n*m, representing the right hand side
* matrix of the system
*  solupos = LeftSolu: mp_B a m x n matrix
*  solupos = RightSolu: mp_B a n x m matrix
*
* Output:
* mp_N: 1dim mpz_t array length n*m, representing the numerator matrix
* of the solution
*  solupos = LeftSolu: mp_N a m x n matrix
*  solupos = RightSolu: mp_N a n x m matrix
* mp_D: mpz_t, denominator of the solution
*
* Precondition:
*  A must be a nonsingular matrix.
*  Any element p in RNS basis must satisfy 2*(p1)^2 <= 2^531.
*
* Note:
*  It is necessary to make sure the input parameters are correct,
* expecially the dimension, since there is no parameter checks in the
* function.
*  Input and output matrices are row majored and represented by
* onedimension array.
*  It is needed to preallocate the memory space of mp_N and mp_D.
*
*/
extern long certSolveLong (const long certflag, const long n, const long m,
const long *A, mpz_t *mp_b, mpz_t *mp_N,
mpz_t mp_D, mpz_t *mp_NZ, mpz_t mp_DZ);
/*
*
* Calling Sequence:
* 1/2/3 < certSolveLong(certflag, n, m, A, mp_b, mp_N, mp_D,
* mp_NZ, mp_DZ)
*
* Summary:
* Certified solve a system of linear equations without reducing the
* solution size, where the left hand side input matrix is represented
* by signed long integers
*
* Description:
* Let the system of linear equations be Av = b, where A is a n x m matrix,
* and b is a n x 1 vector. There are three possibilities:
*
* 1. The system has more than one rational solution
* 2. The system has a unique rational solution
* 3. The system has no solution
*
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is really the minimal denominator.
*
* The 1 x n certificate vector z satisfies that z.A is an integer vector
* and z.b has the same denominator as the solution vector v.
* In this case, the function will output the solution with minimal
* denominator and optional certificate vector z (if certflag = 1).
*
* Note: if choose not to compute the certificate vector z, the solution
* will not garantee, but with high probability, to be the minimal
* denominator solution, and the function will run faster.
*
* In the second case, the function will only compute the unique solution
* and the contents in the space for certificate vector make no sense.
*
* In the third case, there exists a certificate vector q to certify that
* the system has no solution. The 1 x n vector q satisfies q.A = 0 but
* q.b <> 0. In this case, the function will output this certificate vector
* q and store it into the same space for certificate z. The value of
* certflag also controls whether to output q or not.
*
* Note: if the function returns 3, then the system determinately does not
* exist solution, no matter whether to output certificate q or not.
*
* Input:
* certflag: 1/0, flag to indicate whether or not to compute the certificate
* vector z or q.
*  If certflag = 1, compute the certificate.
*  If certflag = 0, not compute the certificate.
* n: long, row dimension of the system
* m: long, column dimension of the system
* A: 1dim signed long array length n*m, representation of n x m
* matrix A
* mp_b: 1dim mpz_t array length n, representation of n x 1 vector b
*
* Return:
* 1: the first case, system has more than one solution
* 2: the second case, system has a unique solution
* 3: the third case, system has no solution
*
* Output:
* mp_N: 1dim mpz_t array length m,
*  numerator vector of the solution with minimal denominator
* in the first case
*  numerator vector of the unique solution in the second case
*  make no sense in the third case
* mp_D: mpz_t,
*  minimal denominator of the solutions in the first case
*  denominator of the unique solution in the second case
*  make no sense in the third case
*
* The following will only be computed when certflag = 1
* mp_NZ: 1dim mpz_t array length n,
*  numerator vector of the certificate z in the first case
*  make no sense in the second case
*  numerator vector of the certificate q in the third case
* mp_DZ: mpz_t,
*  denominator of the certificate z if in the first case
*  make no sense in the second case
*  denominator of the certificate q in the third case
*
* Note:
*  The space of (mp_N, mp_D) is needed to be preallocated, and entries in
* mp_N and integer mp_D are needed to be initiated as any integer values.
*  If certflag is specified to be 1, then also needs to preallocate space
* for (mp_NZ, mp_DZ), and initiate integer mp_DZ and entries in mp_NZ to
* be any integer values.
* Otherwise, set mp_NZ = NULL, and mp_DZ = any integer
*
*/
extern long certSolveRedLong (const long certflag, const long nullcol,
const long n, const long m, const long *A,
mpz_t *mp_b, mpz_t *mp_N, mpz_t mp_D,
mpz_t *mp_NZ, mpz_t mp_DZ);
/*
*
* Calling Sequence:
* 1/2/3 < certSolveRedLong(certflag, nullcol, n, m, A, mp_b, mp_N, mp_D,
* mp_NZ, mp_DZ)
*
* Summary:
* Certified solve a system of linear equations and reduce the solution
* size, where the left hand side input matrix is represented by signed
* long integers
*
* Description:
* Let the system of linear equations be Av = b, where A is a n x m matrix,
* and b is a n x 1 vector. There are three possibilities:
*
* 1. The system has more than one rational solution
* 2. The system has a unique rational solution
* 3. The system has no solution
*
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is really the minimal denominator.
*
* The 1 x n certificate vector z satisfies that z.A is an integer vector
* and z.b has the same denominator as the solution vector v.
* In this case, the function will output the solution with minimal
* denominator and optional certificate vector z (if certflag = 1).
*
* Note: if choose not to compute the certificate vector z, the solution
* will not garantee, but with high probability, to be the minimal
* denominator solution, and the function will run faster.
*
* Lattice reduction will be used to reduce the solution size. Parameter
* nullcol designates the dimension of kernal basis we use to reduce the
* solution size as well as the dimension of nullspace we use to compute
* the minimal denominator. The heuristic results show that the solution
* size will be reduced by factor 1/nullcol.
*
* To find the minimum denominator as fast as possible, nullcol cannot be
* too small. We use NULLSPACE_COLUMN as the minimal value of nullcol. That
* is, if the input nullcol is less than NULLSPACE_COLUMN, NULLSPACE_COLUMN
* will be used instead. However, if the input nullcol becomes larger, the
* function will be slower. Meanwhile, it does not make sense to make
* nullcol greater than the dimension of nullspace of the input system.
*
* As a result, the parameter nullcol will not take effect unless
* NULLSPACE_COLUMN < nullcol < dimnullspace is satisfied, where
* dimnullspace is the dimension of nullspace of the input system. If the
* above condition is not satisfied, the boundary value NULLSPACE_COLUMN or
* dimnullspace will be used.
*
* In the second case, the function will only compute the unique solution
* and the contents in the space for certificate vector make no sense.
*
* In the third case, there exists a certificate vector q to certify that
* the system has no solution. The 1 x n vector q satisfies q.A = 0 but
* q.b <> 0. In this case, the function will output this certificate vector
* q and store it into the same space for certificate z. The value of
* certflag also controls whether to output q or not.
*
* Note: if the function returns 3, then the system determinately does not
* exist solution, no matter whether to output certificate q or not.
*
* Input:
* certflag: 1/0, flag to indicate whether or not to compute the certificate
* vector z or q.
*  If certflag = 1, compute the certificate.
*  If certflag = 0, not compute the certificate.
* nullcol: long, dimension of nullspace and kernel basis of conditioned
* system,
* if nullcol < NULLSPACE_COLUMN, use NULLSPACE_COLUMN instead
* n: long, row dimension of the system
* m: long, column dimension of the system
* A: 1dim signed long array length n*m, representation of n x m
* matrix A
* mp_b: 1dim mpz_t array length n, representation of n x 1 vector b
*
* Return:
* 1: the first case, system has more than one solution
* 2: the second case, system has a unique solution
* 3: the third case, system has no solution
*
* Output:
* mp_N: 1dim mpz_t array length m,
*  numerator vector of the solution with minimal denominator
* in the first case
*  numerator vector of the unique solution in the second case
*  make no sense in the third case
* mp_D: mpz_t,
*  minimal denominator of the solutions in the first case
*  denominator of the unique solution in the second case
*  make no sense in the third case
*
* The following will only be computed when certflag = 1
* mp_NZ: 1dim mpz_t array length n,
*  numerator vector of the certificate z in the first case
*  make no sense in the second case
*  numerator vector of the certificate q in the third case
* mp_DZ: mpz_t,
*  denominator of the certificate z if in the first case
*  make no sense in the second case
*  denominator of the certificate q in the third case
*
* Note:
*  The space of (mp_N, mp_D) is needed to be preallocated, and entries in
* mp_N and integer mp_D are needed to be initiated as any integer values.
*  If certflag is specified to be 1, then also needs to preallocate space
* for (mp_NZ, mp_DZ), and initiate integer mp_DZ and entries in mp_NZ to
* be any integer values.
* Otherwise, set mp_NZ = NULL, and mp_DZ = any integer
*
*/
extern long certSolveMP (const long certflag, const long n, const long m,
mpz_t *mp_A, mpz_t *mp_b, mpz_t *mp_N,
mpz_t mp_D, mpz_t *mp_NZ, mpz_t mp_DZ);
/*
*
* Calling Sequence:
* 1/2/3 < certSolveMP(certflag, n, m, mp_A, mp_b, mp_N, mp_D,
* mp_NZ, mp_DZ)
*
* Summary:
* Certified solve a system of linear equations without reducing the
* solution size, where the left hand side input matrix is represented
* by mpz_t integers
*
* Description:
* Let the system of linear equations be Av = b, where A is a n x m matrix,
* and b is a n x 1 vector. There are three possibilities:
*
* 1. The system has more than one rational solution
* 2. The system has a unique rational solution
* 3. The system has no solution
*
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is really the minimal denominator.
*
* The 1 x n certificate vector z satisfies that z.A is an integer vector
* and z.b has the same denominator as the solution vector v.
* In this case, the function will output the solution with minimal
* denominator and optional certificate vector z (if certflag = 1).
*
* Note: if choose not to compute the certificate vector z, the solution
* will not garantee, but with high probability, to be the minimal
* denominator solution, and the function will run faster.
*
* In the second case, the function will only compute the unique solution
* and the contents in the space for certificate vector make no sense.
*
* In the third case, there exists a certificate vector q to certify that
* the system has no solution. The 1 x n vector q satisfies q.A = 0 but
* q.b <> 0. In this case, the function will output this certificate vector
* q and store it into the same space for certificate z. The value of
* certflag also controls whether to output q or not.
*
* Note: if the function returns 3, then the system determinately does not
* exist solution, no matter whether to output certificate q or not.
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is the minimal denominator.
*
* Input:
* certflag: 1/0, flag to indicate whether or not to compute the certificate
* vector z or q.
*  If certflag = 1, compute the certificate.
*  If certflag = 0, not compute the certificate.
* n: long, row dimension of the system
* m: long, column dimension of the system
* mp_A: 1dim mpz_t array length n*m, representation of n x m matrix A
* mp_b: 1dim mpz_t array length n, representation of n x 1 vector b
*
* Return:
* 1: the first case, system has more than one solution
* 2: the second case, system has a unique solution
* 3: the third case, system has no solution
*
* Output:
* mp_N: 1dim mpz_t array length m,
*  numerator vector of the solution with minimal denominator
* in the first case
*  numerator vector of the unique solution in the second case
*  make no sense in the third case
* mp_D: mpz_t,
*  minimal denominator of the solutions in the first case
*  denominator of the unique solution in the second case
*  make no sense in the third case
*
* The following will only be computed when certflag = 1
* mp_NZ: 1dim mpz_t array length n,
*  numerator vector of the certificate z in the first case
*  make no sense in the second case
*  numerator vector of the certificate q in the third case
* mp_DZ: mpz_t,
*  denominator of the certificate z if in the first case
*  make no sense in the second case
*  denominator of the certificate q in the third case
*
* Note:
*  The space of (mp_N, mp_D) is needed to be preallocated, and entries in
* mp_N and integer mp_D are needed to be initiated as any integer values.
*  If certflag is specified to be 1, then also needs to preallocate space
* for (mp_NZ, mp_DZ), and initiate integer mp_DZ and entries in mp_NZ to
* be any integer values.
* Otherwise, set mp_NZ = NULL, and mp_DZ = any integer
*
*/
extern long certSolveRedMP (const long certflag, const long nullcol,
const long n, const long m, mpz_t *mp_A,
mpz_t *mp_b, mpz_t *mp_N, mpz_t mp_D,
mpz_t *mp_NZ, mpz_t mp_DZ);
/*
*
* Calling Sequence:
* 1/2/3 < certSolveRedMP(certflag, nullcol, n, m, mp_A, mp_b, mp_N, mp_D,
* mp_NZ, mp_DZ)
*
* Summary:
* Certified solve a system of linear equations and reduce the solution
* size, where the left hand side input matrix is represented by signed
* mpz_t integers
*
* Description:
* Let the system of linear equations be Av = b, where A is a n x m matrix,
* and b is a n x 1 vector. There are three possibilities:
*
* 1. The system has more than one rational solution
* 2. The system has a unique rational solution
* 3. The system has no solution
*
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is really the minimal denominator.
*
* The 1 x n certificate vector z satisfies that z.A is an integer vector
* and z.b has the same denominator as the solution vector v.
* In this case, the function will output the solution with minimal
* denominator and optional certificate vector z (if certflag = 1).
*
* Note: if choose not to compute the certificate vector z, the solution
* will not garantee, but with high probability, to be the minimal
* denominator solution, and the function will run faster.
*
* Lattice reduction will be used to reduce the solution size. Parameter
* nullcol designates the dimension of kernal basis we use to reduce the
* solution size as well as the dimension of nullspace we use to compute
* the minimal denominator. The heuristic results show that the solution
* size will be reduced by factor 1/nullcol.
*
* To find the minimum denominator as fast as possible, nullcol cannot be
* too small. We use NULLSPACE_COLUMN as the minimal value of nullcol. That
* is, if the input nullcol is less than NULLSPACE_COLUMN, NULLSPACE_COLUMN
* will be used instead. However, if the input nullcol becomes larger, the
* function will be slower. Meanwhile, it does not make sense to make
* nullcol greater than the dimension of nullspace of the input system.
*
* As a result, the parameter nullcol will not take effect unless
* NULLSPACE_COLUMN < nullcol < dimnullspace is satisfied, where
* dimnullspace is the dimension of nullspace of the input system. If the
* above condition is not satisfied, the boundary value NULLSPACE_COLUMN or
* dimnullspace will be used.
*
* In the second case, the function will only compute the unique solution
* and the contents in the space for certificate vector make no sense.
*
* In the third case, there exists a certificate vector q to certify that
* the system has no solution. The 1 x n vector q satisfies q.A = 0 but
* q.b <> 0. In this case, the function will output this certificate vector
* q and store it into the same space for certificate z. The value of
* certflag also controls whether to output q or not.
*
* Note: if the function returns 3, then the system determinately does not
* exist solution, no matter whether to output certificate q or not.
* In the first case, there exist a solution vector v with minimal
* denominator and a rational certificate vector z to certify that the
* denominator of solution v is the minimal denominator.
*
* Input:
* certflag: 1/0, flag to indicate whether or not to compute the certificate
* vector z or q.
*  If certflag = 1, compute the certificate.
*  If certflag = 0, not compute the certificate.
* nullcol: long, dimension of nullspace and kernel basis of conditioned
* system,
* if nullcol < NULLSPACE_COLUMN, use NULLSPACE_COLUMN instead
* n: long, row dimension of the system
* m: long, column dimension of the system
* mp_A: 1dim mpz_t array length n*m, representation of n x m matrix A
* mp_b: 1dim mpz_t array length n, representation of n x 1 vector b
*
* Return:
* 1: the first case, system has more than one solution
* 2: the second case, system has a unique solution
* 3: the third case, system has no solution
*
* Output:
* mp_N: 1dim mpz_t array length m,
*  numerator vector of the solution with minimal denominator
* in the first case
*  numerator vector of the unique solution in the second case
*  make no sense in the third case
* mp_D: mpz_t,
*  minimal denominator of the solutions in the first case
*  denominator of the unique solution in the second case
*  make no sense in the third case
*
* The following will only be computed when certflag = 1
* mp_NZ: 1dim mpz_t array length n,
*  numerator vector of the certificate z in the first case
*  make no sense in the second case
*  numerator vector of the certificate q in the third case
* mp_DZ: mpz_t,
*  denominator of the certificate z if in the first case
*  make no sense in the second case
*  denominator of the certificate q in the third case
*
* Note:
*  The space of (mp_N, mp_D) is needed to be preallocated, and entries in
* mp_N and integer mp_D are needed to be initiated as any integer values.
*  If certflag is specified to be 1, then also needs to preallocate space
* for (mp_NZ, mp_DZ), and initiate integer mp_DZ and entries in mp_NZ to
* be any integer values.
* Otherwise, set mp_NZ = NULL, and mp_DZ = any integer
*
*/
extern void RowEchelonTransform (const FiniteField p, Double *A, const long n,
const long m, const long frows,
const long lrows, const long redflag,
const long eterm, long *Q, long *rp,
FiniteField *d);
/*
* Calling Sequence:
* RowEchelonTransform(p, A, n, m, frows, lrows, redflag, eterm, Q, rp, d)
*
* Summary:
* Compute a mod p rowechelon transform of a mod p input matrix
*
* Description:
* Given a n x m mod p matrix A, a rowechelon transform of A is a 4tuple
* (U,P,rp,d) with rp the rank profile of A (the unique and strictly
* increasing list [j1,j2,...jr] of column indices of the rowechelon form
* which contain the pivots), P a permutation matrix such that all r leading
* submatrices of (PA)[0..r1,rp] are nonsingular, U a nonsingular matrix
* such that UPA is in rowechelon form, and d the determinant of
* (PA)[0..r1,rp].
*
* Generally, it is required that p be a prime, as inverses are needed, but
* in some cases it is possible to obtain an echelon transform when p is
* composite. For the cases where the echelon transform cannot be obtained
* for p composite, the function returns an error indicating that p is
* composite.
*
* The matrix U is structured, and has last nr columns equal to the last nr
* columns of the identity matrix, n the row dimension of A.
*
* The first r rows of UPA comprise a basis in echelon form for the row
* space of A, while the last nr rows of U comprise a basis for the left
* nullspace of PA.
*
* For efficiency, this function does not output an echelon transform
* (U,P,rp,d) directly, but rather the expression sequence (Q,rp,d).
* Q, rp, d are the form of arrays and pointers in order to operate inplace,
* which require to preallocate spaces and initialize them. Initially,
* Q[i] = i (i=0..n), rp[i] = 0 (i=0..n), and *d = 1. Upon completion, rp[0]
* stores the rank r, rp[1..r] stores the rank profile. i<=Q[i]<=n for
* i=1..r. The input Matrix A is modified inplace and used to store U.
* Let A' denote the state of A on completion. Then U is obtained from the
* identity matrix by replacing the first r columns with those of A', and P
* is obtained from the identity matrix by swapping row i with row Q[i], for
* i=1..r in succession.
*
* Parameters flrows, lrows, redflag, eterm control the specific operations
* this function will perform. Let (U,P,rp,d) be as constructed above. If
* frows=0, the first r rows of U will not be correct. If lrows=0, the last
* nr rows of U will not be correct. The computation can be up to four
* times faster if these flags are set to 0.
*
* If redflag=1, the rowechelon form is reduced, that is (UPA)[0..r1,rp]
* will be the identity matrix. If redflag=0, the rowechelon form will not
* be reduced, that is (UPA)[1..r,rp] will be upper triangular and U is unit
* lower triangular. If frows=0 then redflag has no effect.
*
* If eterm=1, then early termination is triggered if a column of the
* input matrix is discovered that is linearly dependant on the previous
* columns. In case of early termination, the third return value d will be 0
* and the remaining components of the echelon transform will not be correct.
*
* Input:
* p: FiniteField, modulus
* A: 1dim Double array length n*m, representation of a n x m input
* matrix
* n: long, row dimension of A
* m: long, column dimension of A
* frows: 1/0,
*  if frows = 1, the first r rows of U will be correct
*  if frows = 0, the first r rows of U will not be correct
* lrows: 1/0,
*  if lrows = 1, the last nr rows of U will be correct
*  if lrows = 0, the last nr rows of U will not be correct
* redflag: 1/0,
*  if redflag = 1, compute rowechelon form
*  if redflag = 0, not compute reowechelon form
* eterm: 1/0,
*  if eterm = 1, terminate early if not in full rank
*  if eterm = 0, not terminate early
* Q: 1dim long array length n+1, compact representation of
* permutation vector, initially Q[i] = i, 0 <= i <= n
* rp: 1dim long array length n+1, representation of rank profile,
* initially rp[i] = 0, 0 <= i <= n
* d: pointer to FiniteField, storing determinant of the matrix,
* initially *d = 1
*
* Precondition:
* ceil(n/2)*(p1)^2+(p1) <= 2^531 = 9007199254740991 (n >= 2)
*
*/
extern Double * mAdjoint (const FiniteField p, Double *A, const long n);
/*
* Calling Sequence:
* Adj < mAdjoint(p, A, n)
*
* Summary:
* Compute the adjoint of a mod p square matrix
*
* Description:
* Given a n x n mod p matrix A, the function computes adjoint of A. Input
* A is not modified upon completion.
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no error
* message is returned
* A: 1dim Double array length n*n, representation of a n x n mod p matrix.
* The entries of A are casted from integers
* n: long, dimension of A
*
* Return:
* 1dim Double matrix length n*n, repesentation of a n x n mod p matrix,
* adjoint of A
*
* Precondition:
* n*(p1)^2 <= 2^531 = 9007199254740991
*
*/
extern long mBasis (const FiniteField p, Double *A, const long n,
const long m, const long basis, const long nullsp,
Double **B, Double **N);
/*
* Calling Sequence:
* r/1 < mBasis(p, A, n, m, basis, nullsp, B, N)
*
* Summary:
* Compute a basis for the rowspace and/or a basis for the left nullspace
* of a mod p matrix
*
* Description:
* Given a n x m mod p matrix A, the function computes a basis for the
* rowspace B and/or a basis for the left nullspace N of A. Row vectors in
* the r x m matrix B consist of basis of A, where r is the rank of A in
* Z/pZ. If r is zero, then B will be NULL. Row vectors in the nr x n
* matrix N consist of the left nullspace of A. N will be NULL if A is full
* rank.
*
* The pointers are passed into argument lists to store the computed basis
* and nullspace. Upon completion, the rank r will be returned. The
* parameters basis and nullsp control whether to compute basis and/or
* nullspace. If set basis and nullsp in the way that both basis and
* nullspace will not be computed, an error message will be printed and
* instead of rank r, 1 will be returned.
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no
* error message is returned
* A: 1dim Double array length n*m, representation of a n x m mod p
* matrix. The entries of A are casted from integers
* n: long, row dimension of A
* m: long, column dimension of A
* basis: 1/0, flag to indicate whether to compute basis for rowspace or
* not
*  basis = 1, compute the basis
*  basis = 0, not compute the basis
* nullsp: 1/0, flag to indicate whether to compute basis for left nullspace
* or not
*  nullsp = 1, compute the nullspace
*  nullsp = 0, not compute the nullspace
*
* Output:
* B: pointer to (Double *), if basis = 1, *B will be a 1dim r*m Double
* array, representing the r x m basis matrix. If basis = 1 and r = 0,
* *B = NULL
*
* N: pointer to (Double *), if nullsp = 1, *N will be a 1dim (nr)*n Double
* array, representing the nr x n nullspace matrix. If nullsp = 1 and
* r = n, *N = NULL.
*
* Return:
*  if basis and/or nullsp are set to be 1, then return the rank r of A
*  if both basis and nullsp are set to be 0, then return 1
*
* Precondition:
* n*(p1)^2 <= 2^531 = 9007199254740991
*
* Note:
*  In case basis = 0, nullsp = 1, A will be destroyed inplace. Otherwise,
* A will not be changed.
*  Space of B and/or N will be allocated in the function
*
*/
extern long mDeterminant (const FiniteField p, Double *A, const long n);
/*
* Calling Sequence:
* det < mDeterminant(p, A, n)
*
* Summary:
* Compute the determinant of a square mod p matrix
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no error
* message is returned
* A: 1dim Double array length n*n, representation of a n x n mod p matrix.
* The entries of A are casted from integers
* n: long, dimension of A
*
* Output:
* det(A) mod p, the determinant of square matrix A
*
* Precondition:
* ceil(n/2)*(p1)^2+(p1) <= 2^531 = 9007199254740991 (n >= 2)
*
* Note:
* A is destroyed inplace
*
*/
extern long mInverse (const FiniteField p, Double *A, const long n);
/*
* Calling Sequence:
* 1/0 < mInverse(p, A, n)
*
* Summary:
* Certified compute the inverse of a mod p matrix inplace
*
* Description:
* Given a n x n mod p matrix A, the function computes A^(1) mod p
* inplace in case A is a nonsingular matrix in Z/Zp. If the inverse does
* not exist, the function returns 0.
*
* A will be destroyed at the end in both cases. If the inverse exists, A is
* inplaced by its inverse. Otherwise, the inplaced A is not the inverse.
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no error
* message is returned
* A: 1dim Double array length n*n, representation of a n x n mod p matrix.
* The entries of A are casted from integers
* n: long, dimension of A
*
* Return:
*  1, if A^(1) mod p exists
*  0, if A^(1) mod p does not exist
*
* Precondition:
* ceil(n/2)*(p1)^2+(p1) <= 2^531 = 9007199254740991 (n >= 2)
*
* Note:
* A is destroyed inplace
*
*/
extern long mRank (const FiniteField p, Double *A, const long n, const long m);
/*
* Calling Sequence:
* r < mRank(p, A, n, m)
*
* Summary:
* Compute the rank of a mod p matrix
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no
* error message is returned
* A: 1dim Double array length n*m, representation of a n x m mod p
* matrix. The entries of A are casted from integers
* n: long, row dimension of A
* m: long, column dimension of A
*
* Return:
* r: long, rank of matrix A
*
* Precondition:
* ceil(n/2)*(p1)^2+(p1) <= 2^531 = 9007199254740991 (n >= 2)
*
* Note:
* A is destroyed inplace
*
*/
extern long * mRankProfile (const FiniteField p, Double *A,
const long n, const long m);
/*
* Calling Sequence:
* rp < mRankProfile(p, A, n, m)
*
* Summary:
* Compute the rank profile of a mod p matrix
*
* Input:
* p: FiniteField, prime modulus
* if p is a composite number, the routine will still work if no
* error message is returned
* A: 1dim Double array length n*m, representation of a n x m mod p
* matrix. The entries of A are casted from integers
* n: long, row dimension of A
* m: long, column dimension of A
*
* Return:
* rp: 1dim long array length n+1, where
*  rp[0] is the rank of matrix A
*  rp[1..r] is the rank profile of matrix A
*
* Precondition:
* ceil(n/2)*(p1)^2+(p1) <= 2^531 = 9007199254740991 (n >= 2)
*
* Note:
* A is destroyed inplace
*
*/
