### Table 1: Phylogenetic parsimony criteria.

"... In PAGE 31: ... The hypothesis encoded in this tree is preferred because it explains as much of the observed character distributions as possible by character-state transitions in a common ancestor, and invokes the fewest ad hoc hypotheses of subsequent character-state change [Far83]. There are several phylogenetic parsimony criteria, each of which encodes a di erent model of evolution by placing di erent restrictions on the types and numbers of character-state transitions allowable in a tree (see Table1 ). The Wagner Linear [KF69], Wagner General, and Fitch [Fit71] criteria assume the simplest model of evolution, in which character-state change is reversible.... In PAGE 33: ....2.1 Phylogenetic Parsimony Each of these problems is given as input a discrete character matrix for m taxa and d char- acters, and operates on an implicit graph G whose vertices are the set of all d-dimensional points de ned by the states of the given characters and whose edges are speci ed by the allow- able transitions between the states in these characters. Each phylogenetic parsimony problem seeks the evolutionary tree in G of minimum length that includes the given taxa, subject to the restrictions on character-state transitions that are particular to that problem apos;s criterion (see Table1 ). The given characters can be restricted in various ways to generate a family of... In PAGE 43: ... Question: Does the collection of characters C have a polarization such that there is a com- patible collection C0 C such that jC0j B? Unconstrained Qualitative Compatibility (UQC) Instance: Collection C of d qualitative characters de ned on a set of m objects; a positive integer B d. Question: Does the collection of characters C have a polarization and an ordering such that there is a compatible collection C0 C such that jC0j B? Table1 0: Character compatibility decision problems (adapted from [DS86]).... In PAGE 44: ... B0 = B BQC p m BCC [DS86] d0 = d m0 = m X0 = [x0i;j]; 1 i d0; 1 j m0 where a character apos;s most frequently occurring state becomes that charac- ter apos;s ancestral state in X0. B0 = B Table1 1: Reductions for character compatibility decision problems.... In PAGE 45: ... Question: Does there exist an additive tree T 2 Ad n such that X(D; A(T)) B? Fitting Unconstrained Matrices to Graph-Based Dominant Additive Trees (FUGT[ ]) Instance: Complete graph G = (V; E), jV j = n; semimetric D 2 Mn de ned on all pairs of vertices in G; set of taxa S V ; and a positive integer B. Question: Is there a subtree T of G that includes S such that Pfx;yg2T D(x; y) B and [ A(T)]S DS? Table1 2: Distance matrix tting decision problems (adapted from [Day83, KM86, Day87, Kri88]).... In PAGE 47: ... Question: Does there exist an ultrametric tree U 2 Un;2 such that X(D; U(U)) B? Fitting Binary Matrices to Dominant Ultrametric Trees of Height 2 VIA STATISTIC X (FBUT2[X, ]) [X 2 fF1; F2g] Instance: Set S of n taxa; semimetric D 2 Bn; and a positive integer B. Question: Does there exist an ultrametric tree U 2 Un;2 such that X(D; U(U)) B and U(U) D? Table1 3: Auxiliary decision problems for NP-hardness proofs of distance matrix tting decision problems (adapted from [KM86, Day87, Kri88]). .... In PAGE 48: ... S0 = S + yi; 1 i apos; D0 = [d0 i;j] = quot; D M M0 1 #, where M = [mi;j], mi;j = for all 1 i n and 1 j apos;, M0 is the transpose of M, and 1 is a square matrix with zeros on, but ones o , the main diagonal. B0 = B VC p m FUGT[ ] V = f g S f vi j 1 i jVVCjg S f ej j 1 j jEV Cjg D = [di;j], where d( ; vi) = 1 d( ; ej) = 2 d(vi; vj) = 4 d(vi; ej) = 1 if ej = fvi; xg 2 EV C, d(vi; ej) = 3 otherwise d(ei; ej) = 2 S = f g S f ej j 1 j jEVCjg B = K + jEV Cj Table1 4: Reductions for distance matrix tting decision problems.... In PAGE 52: ...Thesis Literature Phylogenetic UBfC,QgCS fC,QgCS [DJS86] Parsimony UBfC,QgDo fC,QgDO [DJS86] UBfC,QgCI fC,QgCI [DJS86] UBW SPQ [GF82, Day83] UUW SPP [GF82, Day83] WUOWL WTP [Day83] Character BfQ,CgC BfQ,CgC [DS86] Compatibility UfQ,CgC UfQ,CgC [DS86] Distance Matrix FBUT[F1] bHICy [KM86] Fitting FBUT2[F1] bHIC3y [KM86], 2 1y [Kri86], FUT[1] [Day87] FBUT2[F2] 2 2y [Kri86], FUT[2] [Day87] FUUT[F1] 1y [Kri86], HICy [KM86] FUUT[F2] 2y [Kri86] FUUT[F1; ] P4 [Kri88] FUDT[ ], 2 fF1; F2g FAT[ ], 2 f1; 2g [Day87] FUGT[ ] AET [Day83] Table1 5: Correspondence between phylogenetic inference problems in this thesis and problems in the literature. All solution problems are marked with daggers (y); all other problems are decision problems.... In PAGE 74: ...Unweighted Weighted Given-Cost Given-Limit Decision - NP-complete Evaluation FPNP[O(logn)]-C FPNP jj -hard y - Solution FPNP jj -hard, properly FPNP[O(logn)]-hard, 2 NPMVg FPNP 2 NPMVg Spanning 2 Span(NPMVg FPNP ) 2 SpanP Enumeration 2 FP#P Random 2 FRP p 2 Generation Table1 6: Computational complexities of phylogenetic inference functions. y Most weighted distance matrix tting evaluation problems are only known to be properly FPNP[O(logn)]-hard (see Corollary 26).... In PAGE 78: ... Formula: maxT jfcj9x[(P(c; x) ^ x 2 T) _ (N(c; x) ^ :(x 2 T))]gj where P, N, and T are as de ned for SAT. Table1 7: Formulations of SAT in rst-order logic (adapted from [KT90, PY91]). MAX NP.... In PAGE 79: ... Formula:maxT jf(x1; x2; x3)j [ ((x1; x2; x3) 2 C0 ! x1 2 T _ x2 2 T _ x3 2 T) ^ ((x1; x2; x3) 2 C1 ! x1 62 T _ x2 2 T _ x3 2 T) ^ ((x1; x2; x3) 2 C2 ! x1 62 T _ x2 62 T _ x3 2 T) ^ ((x1; x2; x3) 2 C3 ! x1 62 T _ x2 62 T _ x3 62 T) ] gj; where C0, C1, C2, C3, and T are as de ned for 3SAT. Table1 8: Formulations of SAT in rst-order logic (cont apos;d from Table 17).... In PAGE 82: ... 4. Given a solution W of cost c to an instance of SOL-MIN-FBUT2[F1] derived by the reduction from X3C given in [KM86] (See Table1 4), in polynomial time we can nd a canonical solution W0 with cost c0 c. 5.... In PAGE 82: ... 5. Given a solution W of cost c to an instance of SOL-MIN-FUDT[F ] ( 2 f1; 2g) derived by the reductions from FBUT2[ ] given in [Day87] (see Table1 4), in polynomial time we can nd a canonical solution W0 of cost c0 c.... In PAGE 84: ... As the Generalized parsimony criterion can simulate any ordered phylogenetic parsimony problem, (5) can be proved by a variant on any of the proofs for (1 - 4). Proofs of (6 { 7): By the reductions given in Table1 1, solutions to SOL-MAX-BCC (SOL- MAX-BQC) yield solutions to SOL-MAX-CLIQUE (SOL-MAX-BCC) of the same cost. Hence, these reductions yield L-reductions with = = 1.... In PAGE 84: ... Hence, this reduction is an L-reduction. Proof of (10): Consider the reduction from FBUT2[F ] to FUDT[F ] 2 f1; 2g given in [Day87] (see Table1 4). As OPTFBUT2[F ] = OPTFUDT[F ], condition (L1) is satis ed with = 1.... ..."

### Table 1: Phylogenetic parsimony criteria.

"... In PAGE 26: ... The hypothesis encoded in this tree is preferred because it explains as much of the observed character distributions as possible by character-state transitions in a common ancestor, and invokes the fewest ad hoc hypotheses of subsequent character-state change [Far83]. There are several phylogenetic parsimony criteria, each of which encodes a di erent model of evolution by placing di erent restrictions on the types and numbers of character-state transitions allowable in a tree (see Table1 ). The Wagner Linear [KF69], Wagner General, and Fitch [Fit71] criteria assume the simplest model of evolution, in which character-state change is reversible.... In PAGE 28: ....2.1 Phylogenetic Parsimony Each of these problems is given as input a discrete character matrix for m taxa and d char- acters, and operates on an implicit graph G whose vertices are the set of all d-dimensional points de ned by the states of the given characters and whose edges are speci ed by the allow- able transitions between the states in these characters. Each phylogenetic parsimony problem seeks the evolutionary tree in G of minimum length that includes the given taxa, subject to the restrictions on character-state transitions that are particular to that problem apos;s criterion (see Table1 ). The given characters can be restricted in various ways to generate a family of... In PAGE 38: ... Question: Does the collection of characters C have a polarization such that there is a com- patible collection C0 C such that jC0j B? Unconstrained Qualitative Compatibility (UQC) Instance: Collection C of d qualitative characters de ned on a set of m objects; a positive integer B d. Question: Does the collection of characters C have a polarization and an ordering such that there is a compatible collection C0 C such that jC0j B? Table1 0: Character compatibility decision problems (adapted from [DS86]).... In PAGE 39: ... B0 = B BQC p m BCC [DS86] d0 = d m0 = m X0 = [x0i;j]; 1 i d0; 1 j m0 where a character apos;s most frequently occurring state becomes that charac- ter apos;s ancestral state in X0. B0 = B Table1 1: Reductions for character compatibility decision problems.... In PAGE 40: ... Question: Does there exist an additive tree T 2 Ad n such that X(D; A(T)) B? Fitting Unconstrained Matrices to Graph-Based Dominant Additive Trees (FUGT[ ]) Instance: Complete graph G = (V; E), jV j = n; semimetric D 2 Mn de ned on all pairs of vertices in G; set of taxa S V ; and a positive integer B. Question: Is there a subtree T of G that includes S such that Pfx;yg2T D(x; y) B and [ A(T)]S DS? Table1 2: Distance matrix tting decision problems (adapted from [Day83, KM86, Day87, Kri88]).... In PAGE 42: ... Question: Does there exist an ultrametric tree U 2 Un;2 such that X(D; U(U)) B? Fitting Binary Matrices to Dominant Ultrametric Trees of Height 2 VIA STATISTIC X (FBUT2[X, ]) [X 2 fF1; F2g] Instance: Set S of n taxa; semimetric D 2 Bn; and a positive integer B. Question: Does there exist an ultrametric tree U 2 Un;2 such that X(D; U(U)) B and U(U) D? Table1 3: Auxiliary decision problems for NP-hardness proofs of distance matrix tting decision problems (adapted from [KM86, Day87, Kri88]). .... In PAGE 43: ... S0 = S + yi; 1 i apos; D0 = [d0 i;j] = quot; D M M0 1 #, where M = [mi;j], mi;j = for all 1 i n and 1 j apos;, M0 is the transpose of M, and 1 is a square matrix with zeros on, but ones o , the main diagonal. B0 = B VC p m FUGT[ ] V = f g S f vi j 1 i jVVCjg S f ej j 1 j jEV Cjg D = [di;j], where d( ; vi) = 1 d( ; ej) = 2 d(vi; vj) = 4 d(vi; ej) = 1 if ej = fvi; xg 2 EV C, d(vi; ej) = 3 otherwise d(ei; ej) = 2 S = f g S f ej j 1 j jEVCjg B = K + jEV Cj Table1 4: Reductions for distance matrix tting decision problems.... In PAGE 47: ...Thesis Literature Phylogenetic UBfC,QgCS fC,QgCS [DJS86] Parsimony UBfC,QgDo fC,QgDO [DJS86] UBfC,QgCI fC,QgCI [DJS86] UBW SPQ [GF82, Day83] UUW SPP [GF82, Day83] WUOWL WTP [Day83] Character BfQ,CgC BfQ,CgC [DS86] Compatibility UfQ,CgC UfQ,CgC [DS86] Distance Matrix FBUT[F1] bHICy [KM86] Fitting FBUT2[F1] bHIC3y [KM86], 2 1y [Kri86], FUT[1] [Day87] FBUT2[F2] 2 2y [Kri86], FUT[2] [Day87] FUUT[F1] 1y [Kri86], HICy [KM86] FUUT[F2] 2y [Kri86] FUUT[F1; ] P4 [Kri88] FUDT[ ], 2 fF1; F2g FAT[ ], 2 f1; 2g [Day87] FUGT[ ] AET [Day83] Table1 5: Correspondence between phylogenetic inference problems in this thesis and problems in the literature. All solution problems are marked with daggers (y); all other problems are decision problems.... In PAGE 68: ...Unweighted Weighted Given-Cost Given-Limit Decision - NP-complete Evaluation FPNP[O(logn)]-C FPNP jj -hard y - Solution FPNP jj -hard, properly FPNP[O(logn)]-hard, 2 NPMVg FPNP 2 NPMVg Spanning 2 Span(NPMVg FPNP ) 2 SpanP Enumeration 2 FP#P Random 2 FRP p 2 Generation Table1 6: Computational complexities of phylogenetic inference functions. y Most weighted distance matrix tting evaluation problems are only known to be properly FPNP[O(logn)]-hard (see Corollary 26).... In PAGE 71: ... Formula: maxT jfcj9x[(P(c; x) ^ x 2 T) _ (N(c; x) ^ :(x 2 T))]gj where P, N, and T are as de ned for SAT. Table1 7: Formulations of SAT in rst-order logic (adapted from [KT90, PY91]). MAX NP.... In PAGE 72: ... Formula:maxT jf(x1; x2; x3)j [ ((x1; x2; x3) 2 C0 ! x1 2 T _ x2 2 T _ x3 2 T) ^ ((x1; x2; x3) 2 C1 ! x1 62 T _ x2 2 T _ x3 2 T) ^ ((x1; x2; x3) 2 C2 ! x1 62 T _ x2 62 T _ x3 2 T) ^ ((x1; x2; x3) 2 C3 ! x1 62 T _ x2 62 T _ x3 62 T) ] gj; where C0, C1, C2, C3, and T are as de ned for 3SAT. Table1 8: Formulations of SAT in rst-order logic (cont apos;d from Table 17).... In PAGE 75: ... 4. Given a solution W of cost c to an instance of SOL-MIN-FBUT2[F1] derived by the reduction from X3C given in [KM86] (See Table1 4), in polynomial time we can nd a canonical solution W0 with cost c0 c. 5.... In PAGE 75: ... 5. Given a solution W of cost c to an instance of SOL-MIN-FUDT[F ] ( 2 f1; 2g) derived by the reductions from FBUT2[ ] given in [Day87] (see Table1 4), in polynomial time we can nd a canonical solution W0 of cost c0 c.... In PAGE 77: ... As the Generalized parsimony criterion can simulate any ordered phylogenetic parsimony problem, (5) can be proved by a variant on any of the proofs for (1 - 4). Proofs of (6 { 7): By the reductions given in Table1 1, solutions to SOL-MAX-BCC (SOL- MAX-BQC) yield solutions to SOL-MAX-CLIQUE (SOL-MAX-BCC) of the same cost. Hence, these reductions yield L-reductions with = = 1.... In PAGE 77: ... Hence, this reduction is an L-reduction. Proof of (10): Consider the reduction from FBUT2[F ] to FUDT[F ] 2 f1; 2g given in [Day87] (see Table1 4). As OPTFBUT2[F ] = OPTFUDT[F ], condition (L1) is satis ed with = 1.... ..."

### Table 3: Number of trees for the Parsimony computation, compared to the number of computations for TCS.

"... In PAGE 4: ...grows rapidly as the number of taxa increases, the problem is not nearly as difficult as total exploration of the tree space. Table3 shows the number of computations for each problem. Execution time for TCS varies significantly depending on the data set.... ..."

### Table 4 Computing times (msec) of training (ensembles of M = 100 trees)

2006

"... In PAGE 9: ... Thus, from a practical point of view, the increase in complexity is detrimental only in terms of memory requirements. Regarding computing times, Table4 shows that Extra-Trees training is systematically faster than that of Random Forests and Tree Bagging. In classification problems, the average 3 Pruning by ten-fold cross-validation only slightly affects accuracy (see Table 8), but leads to learning times about ten-times higher than for unpruned trees.... ..."

### Table 2. Percent Reduction in Loss of Ensemble Selection Over the Best Models From Any Learning Algorithm.

2004

"... In PAGE 5: ... Normalized scores do not have this bias. Table2 shows the percent reduction in loss for ensem- ble selection on the 7 test problems and 10 metrics, compared to the best models selected for each problem and metric. As with normalized scores, final perfor- mances are estimated on large final test sets not used for training and is the average over two trials with each problem.... ..."

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### Table 2 Maximum Parsimony and Likelihood Results

"... In PAGE 12: ...1 General sequencing and tree estimation results All ILD tests showed no significant incongruence between any of the partitions. All individual genes contained informative indels; the number of informative indels are shown in Table2 . All indels supported clades with high maximum parsimony boostrap (MP BP) support and did not contribute support to any weakly supported clades.... In PAGE 12: ... All indels supported clades with high maximum parsimony boostrap (MP BP) support and did not contribute support to any weakly supported clades. The aligned length of each gene partition is shown in Table2 . A 204 base pair (bp) insertion in all canid species in GHR was removed from the alignment, as was a 219 bp insertion in the red panda in RHO1.... In PAGE 12: ... In these cases, sequence was not included in the final data matrix (Table 1). For all partitions, multiple most parsimonious trees were obtained ( Table2 ). Both the GHR and complete data set MP searches were stopped due to computer memory constraints ... In PAGE 13: ... 12 retention index (RI) are reported for each partition in Table2 . Individual gene topologies (MP BP) contained polytomies but did not differ significantly from each other or from the concatenated total data MP BP topology.... In PAGE 13: ... Differences between trees were only found in areas of weak support; no hard incongruencies (opposing topologies supported by 80% bootstrap support) were observed. All maximum likelihood searches yielded a single most likely tree (log likelihoods in Table2 ), except FES, which recovered six equally likely trees. All FES trees were identical in topology and all parameters except the transition:transversion (Ti:Tv) ratio, Ti:Tv kappa, and gamma shape.... ..."

### Table 2. The reduction (in percent) in the parsimony score of trees when using inversion medians, for two genome sizes, under three different evolutionary scenarios.

"... In PAGE 11: ... While parsimony scores (the sum of the minimum inversion distances along the edges of the tree) do not have direct topological signifi- cance, the goal of the GRAPPA optimization is to obtain a most parsimonious tree, so that any reduction in the score is an indication that the algorithm is behaving more ef- fectively. Table2 shows percent reductions for a variety of settings; note that reductions... ..."

### TABLE I RESULTS OF INDIVIDUAL AND ENSEMBLE LEARNING

2006

Cited by 1

### Table 3: Comparative Performance of Minority amp; Majority Classes (Unbalanced Data Sets)

2001

"... In PAGE 12: ...1 Learning from Unbalanced Data Sets Classifiers were generated and evaluated on the 25 data sets using the naturally occurring class distributions. The results are summarized in Table3 . The first column in Table 3 identifies the data set and the second column specifies the natural class distribution, which was also displayed in Table 2.... In PAGE 13: ...ority class. In section 2.2.1 we stated one reason why minority-labeled leaves have a higher error rate than the majority-labeled leaves is because of the test distribution effect of having more majority-class than minority-class test examples. Table3 suggests a second reason, which is related to the fact that minority-labeled leaves tend to be formed from fewer training examples as a consequence of having fewer minority-class examples in the training set. Small disjuncts, which are disjuncts (i.... In PAGE 13: ... Consequently, we expect the rules/leaves labeled with the minority class to have a higher error rate because they suffer more from this problem of small disjuncts. The last three columns of Table3 provide additional class-specific information about the classifiers. The Leaf ER column specifies the error rates for leaves labeled with the minority and majority classes, based on the performance of these leaves at classifying the test examples.... In PAGE 13: ... This information from the last three columns is presented graphically in Figure 3. 0 20 40 60 80 100 2040608010 Minority Class Ma jo r i ty C l a s s Leaf ER Example ER Recall Figure 3: Comparison of Minority and Majority Class Measurements (Unbalanced Data Sets) According to the results in Table3 and Figure 3, the classifiers perform much worse on the minority-class test examples than on the majority-class test examples. Over the 24 data sets (coding is excluded from the analysis), the average error rate for the minority-class test examples is 42.... In PAGE 15: ...15 0 20 40 60 80 100 2040608010 Minority Class Majority Clas s Leaf ER Example ER Recall Figure 4: Comparison of Minority and Majority Class Error Rates for Balanced Data Sets By comparing Table 4 with Table3 and Figure 4 with Figure 3, we see that, as one would ex- pect, the minority and majority classes behave much more similarly when the classifiers are built from the balanced data sets. However, the classifiers built from the balanced data sets do not show identical, or even symmetric behavior for the two classes and there are some consistent, and surprising, differences.... ..."

Cited by 46