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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995051
details
property
value
status
complete
benchmark
a.03.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n035.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
29.6699 seconds
cpu usage
29.992
user time
15.887
system time
14.1051
max virtual memory
749484.0
max residence set size
14696.0
stage attributes
key
value
starexec-result
YES
output
29.92/29.66 YES 29.92/29.66 29.92/29.66 DP problem for innermost termination. 29.92/29.66 P = 29.92/29.66 eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 29.92/29.66 eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] 29.92/29.66 eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 29.92/29.66 eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 29.92/29.66 eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 29.92/29.66 eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 29.92/29.66 eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 29.92/29.66 eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 29.92/29.66 eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34] 29.92/29.66 eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2] 29.92/29.66 R = 29.92/29.66 eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 29.92/29.66 eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] 29.92/29.66 eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 29.92/29.66 eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 29.92/29.66 eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 29.92/29.66 eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 29.92/29.66 eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 29.92/29.66 eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 29.92/29.66 eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] 29.92/29.66 eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] 29.92/29.66 29.92/29.66 The dependency graph for this problem is: 29.92/29.66 0 -> 8 29.92/29.66 1 -> 8 29.92/29.66 2 -> 29.92/29.66 3 -> 0, 1 29.92/29.66 4 -> 2, 3, 4, 5, 6, 7 29.92/29.66 5 -> 2, 3, 4, 5, 6, 7 29.92/29.66 6 -> 0, 1 29.92/29.66 7 -> 0, 1 29.92/29.66 8 -> 2, 3, 4, 5, 6, 7 29.92/29.66 9 -> 8 29.92/29.66 10 -> 8 29.92/29.66 Where: 29.92/29.66 0) eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 1) eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 29.92/29.66 2) eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] 29.92/29.66 3) eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 29.92/29.66 4) eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 29.92/29.66 5) eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 29.92/29.66 6) eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 29.92/29.66 7) eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 29.92/29.66 8) eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 29.92/29.66 9) eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34] 29.92/29.66 10) eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2] 29.92/29.66 29.92/29.66 We have the following SCCs. 29.92/29.66 { 0, 1, 3, 4, 5, 6, 7, 8 } 29.92/29.66 29.92/29.66 DP problem for innermost termination. 29.92/29.66 P = 29.92/29.66 eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 29.92/29.66 eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 29.92/29.66 eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 29.92/29.66 eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 29.92/29.66 eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 29.92/29.66 eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 29.92/29.66 eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 29.92/29.66 R = 29.92/29.66 eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 29.92/29.66 eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] 29.92/29.66 eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 29.92/29.66 eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 29.92/29.66 eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 29.92/29.66 eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 29.92/29.66 eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 29.92/29.66 eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 29.92/29.66 eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] 29.92/29.66 eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] 29.92/29.66 29.92/29.66 We use the reverse value criterion with the projection function NU: 29.92/29.66 NU[eval_3#(z1,z2,z3,z4,z5)] = z3 29.92/29.66 NU[eval_2#(z1,z2,z3,z4,z5)] = z3 29.92/29.66 NU[eval_4#(z1,z2,z3,z4,z5)] = z3 29.92/29.66 29.92/29.66 This gives the following inequalities: 29.92/29.66 2 > l && l >= 1 && r >= 2 ==> l >= l 29.92/29.66 I2 >= 2 && I2 >= 1 && I3 >= 2 ==> I2 > I2 - 1 with I2 >= 0 29.92/29.66 I11 >= I9 && I9 > I11 - 1 ==> I10 >= I10 29.92/29.66 I15 >= I13 && I15 - 1 >= I13 && I13 >= 1 ==> I14 >= I14 29.92/29.66 I19 >= I17 && I19 - 1 >= I17 && I17 >= 1 ==> I18 >= I18 29.92/29.66 I23 >= I21 && I23 - 1 >= I21 ==> I22 >= I22 29.92/29.66 I27 >= I25 && I27 - 1 >= I25 ==> I26 >= I26 29.92/29.66 I31 >= 2 ==> I30 >= I30 29.92/29.66 29.92/29.66 We remove all the strictly oriented dependency pairs. 29.92/29.66 29.92/29.66 DP problem for innermost termination. 29.92/29.66 P = 29.92/29.66 eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 29.92/29.66 eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
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return to Integer_TRS_Innermost 2019-03-21 04.53