details

property	value
status	complete
benchmark	simple_control_on_input.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)	0.988629 seconds
cpu usage	1.00751
user time	0.534342
system time	0.473164
max virtual memory	720344.0
max residence set size	8688.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f7#(x1, x2) -> f6#(x1, x2) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4#(I2, I3) -> f3#(I2, I3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) The dependency graph for this problem is: 0 -> 1 1 -> 3, 4 2 -> 5, 6 3 -> 2 4 -> 5 -> 2 6 -> Where: 0) f7#(x1, x2) -> f6#(x1, x2) 1) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 2) f4#(I2, I3) -> f3#(I2, I3) 3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] 4) f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] 5) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] 6) f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] We have the following SCCs. { 2, 5 } DP problem for innermost termination. P = f4#(I2, I3) -> f3#(I2, I3) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2)] = z1 + -1 * (1 + z2) NU[f4#(z1,z2)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I2 + -1 * (1 + I3) >= I2 + -1 * (1 + I3) 1 + I9 <= I8 ==> I8 + -1 * (1 + I9) > I8 + -1 * (1 + (1 + I9)) with I8 + -1 * (1 + I9) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I2, I3) -> f3#(I2, I3) R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) The dependency graph for this problem is: 2 -> Where: 2) f4#(I2, I3) -> f3#(I2, I3) We have the following SCCs. popout

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job log

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actions

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