details

property	value
status	complete
benchmark	n-14.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	4.64281 seconds
cpu usage	4.64338
user time	2.31831
system time	2.32507
max virtual memory	732508.0
max residence set size	8572.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f10#(x1, x2, x3) -> f1#(x1, x2, x3) f9#(I0, I1, I2) -> f2#(I0, I1, I2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19] f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0] f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27] f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f1#(I40, I41, I42) -> f2#(I40, I41, I42) R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42) The dependency graph for this problem is: 0 -> 11 1 -> 5, 8, 10 2 -> 1 3 -> 2 4 -> 2 5 -> 3, 4 6 -> 7 -> 8 -> 6, 7 9 -> 5, 8, 10 10 -> 9 11 -> 5, 8, 10 Where: 0) f10#(x1, x2, x3) -> f1#(x1, x2, x3) 1) f9#(I0, I1, I2) -> f2#(I0, I1, I2) 2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] 3) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] 4) f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] 5) f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] 6) f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19] 7) f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0] 8) f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27] 9) f4#(I28, I29, I30) -> f2#(I28, I29, I30) 10) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] 11) f1#(I40, I41, I42) -> f2#(I40, I41, I42) We have the following SCCs. { 1, 2, 3, 4, 5, 9, 10 } DP problem for innermost termination. P = f9#(I0, I1, I2) -> f2#(I0, I1, I2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42) We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z3 NU[f7#(z1,z2,z3)] = z3 NU[f8#(z1,z2,z3)] = z3 NU[f2#(z1,z2,z3)] = z3 NU[f9#(z1,z2,z3)] = z3 This gives the following inequalities: ==> I2 (>! \union =) I2 0 <= -1 + -1 + I5 ==> I5 >! -1 + I5 1 <= I7 ==> I8 (>! \union =) I8 1 + I10 <= 0 ==> I11 (>! \union =) I11 rnd2 = rnd2 ==> I14 (>! \union =) I14 popout

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job log

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actions

all output return to ITS