details

property	value
status	complete
benchmark	florian_pldi.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	24.8684 seconds
cpu usage	25.2503
user time	13.0611
system time	12.1891
max virtual memory	716952.0
max residence set size	11964.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] The dependency graph for this problem is: 0 -> 1 1 -> 8 2 -> 5 3 -> 5 4 -> 6 5 -> 6 6 -> 7 7 -> 10, 11 8 -> 9 9 -> 7 10 -> 2, 3, 4 11 -> 8 Where: 0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 1) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 9) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 } DP problem for innermost termination. P = f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] We use the extended value criterion with the projection function NU: NU[f4#(x0,x1,x2,x3)] = -x1 + x3 + 1 NU[f2#(x0,x1,x2,x3)] = -x1 + x3 + 1 NU[f1#(x0,x1,x2,x3)] = -x1 + x3 NU[f6#(x0,x1,x2,x3)] = -x1 + x3 NU[f7#(x0,x1,x2,x3)] = -x1 + x3 NU[f8#(x0,x1,x2,x3)] = -x1 + x3 NU[f3#(x0,x1,x2,x3)] = -x1 + x3 This gives the following inequalities: 1 + I4 <= 0 ==> -I5 + I7 >= -I5 + I7 1 <= I8 ==> -I9 + I11 >= -I9 + I11 popout

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