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02 March 2022 02:54:59 PM SUBSET_TEST C version Test the SUBSET library. ASM_ENUM_TEST ASM_ENUM returns the number of alternating sign matrices of a given order. 0 1 1 1 2 2 3 7 4 42 5 429 6 7436 7 218348 ASM_TRIANGLE_TEST ASM_TRIANGLE returns a row of the alternating sign matrix triangle. 0 1 1 1 1 2 2 3 2 3 7 14 14 7 4 42 105 135 105 42 5 429 1287 2002 2002 1287 429 6 7436 26026 47320 56784 47320 26026 7436 7 218348 873392 1813968 2519400 2519400 1813968 873392 218348 BELL_TEST BELL computes Bell numbers. N exact C(I) computed C(I) 0 1 1 1 1 1 2 2 2 3 5 5 4 15 15 5 52 52 6 203 203 7 877 877 8 4140 4140 9 21147 21147 10 115975 115975 CATALAN_TEST CATALAN computes Catalan numbers. N exact C(I) computed C(I) 0 1 1 1 1 1 2 2 2 3 5 5 4 14 14 5 42 42 6 132 132 7 429 429 8 1430 1430 9 4862 4862 10 16796 16796 CATALAN_ROW_NEXT_TEST CATALAN_ROW_NEXT computes a row of the Catalan triangle. First, compute row 7: 7 1 7 27 75 165 297 429 429 Now compute rows consecutively, one at a time: 0 1 1 1 1 2 1 2 2 3 1 3 5 5 4 1 4 9 14 14 5 1 5 14 28 42 42 6 1 6 20 48 90 132 132 7 1 7 27 75 165 297 429 429 8 1 8 35 110 275 572 1001 1430 1430 9 1 9 44 154 429 1001 2002 3432 4862 4862 10 1 10 54 208 637 1638 3640 7072 11934 16796 16796 CFRAC_TO_RAT_TEST CFRAC_TO_RAT continued fraction => fraction. Regular fraction is 4096/15625 Continued fraction coefficients: 0 0 1 3 2 1 3 4 4 2 5 1 6 1 7 11 8 13 The continued fraction convergents. The last row contains the value of the continued fraction, written as a common fraction. I, P(I), Q(I), P(I)/Q(I) 0 0 1 0 1 1 3 0.333333 2 1 4 0.25 3 5 19 0.263158 4 11 42 0.261905 5 16 61 0.262295 6 27 103 0.262136 7 313 1194 0.262144 8 4096 15625 0.262144 CFRAC_TO_RFRAC_TEST CFRAC_TO_RFRAC: continued fraction to ratio; Rational polynomial fraction coefficients: P: 1 1 2 Q: 1 3 1 1 Continued fraction coefficients: 0 1.000000 1 0.500000 2 1.333333 3 -0.500000 4 -1.500000 5 2.000000 Recovered rational polynomial: P: 1 1 2 Q: 1 3 1 1 CHANGE_GREEDY_TEST CHANGE_GREEDY makes change using the biggest coins first. The total for which change is to be made: 73 The available coins are: 1 5 10 25 50 100 The number of coins in change is: 6 4 2 2 0 0 0 73 50 10 10 1 1 1 CHANGE_NEXT_TEST CHANGE_NEXT displays the next possible way to make change for a given total The total for which change is to be made: 50 The available coins are: 1 5 10 25 50 100 1 50 2 25 25 3 25 10 10 5 4 25 10 10 1 1 1 1 1 5 25 10 5 5 5 6 25 10 5 5 1 1 1 1 1 7 25 10 5 1 1 1 1 1 1 1 1 1 1 8 25 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 9 25 5 5 5 5 5 10 25 5 5 5 5 1 1 1 1 1 CHINESE_CHECK_TEST CHINESE_CHECK checks a set of moduluses for use with the Chinese Remainder representation. Modulus set #1: 0 1 1 3 2 8 3 25 IERROR = 0 Modulus set #2: 0 1 1 3 2 -8 3 25 IERROR = 1 Modulus set #3: 0 1 1 3 2 1 3 25 IERROR = 2 Modulus set #4: 0 1 1 3 2 8 3 24 IERROR = 3 CHINESE_TO_I4_TEST CHINESE_TO_I4 computes an integer with the given Chinese Remainder representation. The moduli: 0 3 1 4 2 5 3 7 The number being analyzed is 37 The remainders: 0 1 1 1 2 2 3 2 The reconstructed number is 37 The remainders of the reconstructed number are: 0 1 1 1 2 2 3 2 COMB_NEXT_TEST COMB_NEXT produces combinations. Combinations of size K = 1 0 1 2 3 4 5 Combinations of size K = 2 0 1 0 2 0 3 0 4 0 5 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 Combinations of size K = 3 0 1 2 0 1 3 0 1 4 0 1 5 0 2 3 0 2 4 0 2 5 0 3 4 0 3 5 0 4 5 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5 Combinations of size K = 4 0 1 2 3 0 1 2 4 0 1 2 5 0 1 3 4 0 1 3 5 0 1 4 5 0 2 3 4 0 2 3 5 0 2 4 5 0 3 4 5 1 2 3 4 1 2 3 5 1 2 4 5 1 3 4 5 2 3 4 5 Combinations of size K = 5 0 1 2 3 4 0 1 2 3 5 0 1 2 4 5 0 1 3 4 5 0 2 3 4 5 1 2 3 4 5 COMB_ROW_NEXT_TEST COMB_ROW_NEXT computes the next row of the Pascal triangle. 0 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 7 1 7 21 35 35 21 7 1 8 1 8 28 56 70 56 28 8 1 9 1 9 36 84 126 126 84 36 9 1 10 1 10 45 120 210 252 210 120 45 10 1 COMB_UNRANK_TEST COMB_UNRANK returns a combination of N things out of M, given the lexicographic rank. The total set size is M = 10 The subset size is N = 5 The number of combinations of N out of M is 252 Rank Combination 1 1 2 3 4 5