details

property	value
status	complete
benchmark	mc91test.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	34.5603 seconds
cpu usage	35.1285
user time	19.5207
system time	15.6078
max virtual memory	706980.0
max residence set size	15432.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) f7#(I0, I1, I2, I3, I4) -> f1#(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5, 7, 9 2 -> 3, 5, 7, 9 3 -> 2 4 -> 3, 5, 7, 9 5 -> 4 6 -> 3, 5, 7, 9 7 -> 6 8 -> 3, 5, 7, 9 9 -> 8 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f1#(0, 1, rnd3, I3, I4) [rnd3 = rnd3] 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] 4) f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] 6) f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 7) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] 8) f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) 9) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9 } DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f4#(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3#(I35, I36, I37, I38, I39) -> f1#(I35, I36, I37, I38, I39) f1#(I40, I41, I42, I43, I44) -> f3#(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(0, 1, rnd3, I3, I4) [rnd3 = rnd3] f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f6(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, 11 + I22, I23, I24) [I22 <= 100 /\ 1 <= I21] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f4(1, -1 + I31, -10 + I32, I31, I32) [101 <= I32 /\ 1 <= I31 /\ I30 <= 0] f3(I35, I36, I37, I38, I39) -> f1(I35, I36, I37, I38, I39) f1(I40, I41, I42, I43, I44) -> f3(1, 1 + I41, 11 + I42, I41, I42) [I42 <= 100 /\ 1 <= I41 /\ I40 <= 0] f1(I45, I46, I47, I48, I49) -> f2(I45, I46, I47, I48, I49) [I47 <= I49 /\ I48 <= I46 /\ 1 <= I45] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f4#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f5#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f1#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 NU[f6#(z1,z2,z3,z4,z5)] = 0 + -1 * z1 This gives the following inequalities: ==> 0 + -1 * I5 >= 0 + -1 * I5 101 <= I12 /\ 1 <= I11 ==> 0 + -1 * I10 >= 0 + -1 * I10 ==> 0 + -1 * I15 >= 0 + -1 * I15 I22 <= 100 /\ 1 <= I21 ==> 0 + -1 * I20 >= 0 + -1 * I20 ==> 0 + -1 * I25 >= 0 + -1 * I25 101 <= I32 /\ 1 <= I31 /\ I30 <= 0 ==> 0 + -1 * I30 > 0 + -1 * 1 with 0 + -1 * I30 >= 0 ==> 0 + -1 * I35 >= 0 + -1 * I35 I42 <= 100 /\ 1 <= I41 /\ I40 <= 0 ==> 0 + -1 * I40 > 0 + -1 * 1 with 0 + -1 * I40 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f6#(I10, -1 + I11, -10 + I12, I13, I14) [101 <= I12 /\ 1 <= I11] f5#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to ITS