details

property	value
status	complete
benchmark	p-1a.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	15.8029 seconds
cpu usage	16.0462
user time	8.10892
system time	7.93732
max virtual memory	749084.0
max residence set size	10580.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) f2#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f4#(I10, I11, I12, I13, I14, rnd6, rnd7, rnd8, I18, I19) [y2 = I13 /\ y3 = I14 /\ 0 <= -1 - y2 + y3 /\ rnd7 = rnd7 /\ rnd8 = rnd8 /\ y1 = I13 /\ rnd6 = rnd6] f1#(I34, I35, I36, I37, I38, I39, I40, I41, I42, I43) -> f2#(I34, rnd2, rnd3, I37, I38, I39, I40, I41, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ B0 = B0 /\ rnd3 = rnd3 /\ B1 = B1 /\ rnd2 = rnd2] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f1(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f2(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) f2(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f4(I10, I11, I12, I13, I14, rnd6, rnd7, rnd8, I18, I19) [y2 = I13 /\ y3 = I14 /\ 0 <= -1 - y2 + y3 /\ rnd7 = rnd7 /\ rnd8 = rnd8 /\ y1 = I13 /\ rnd6 = rnd6] f2(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(rnd1, I21, I22, I23, I24, I25, I30, I31, I28, I29) [I32 = I23 /\ I33 = I24 /\ -1 * I32 + I33 <= 0 /\ I30 = I30 /\ I31 = I31 /\ rnd1 = rnd1] f1(I34, I35, I36, I37, I38, I39, I40, I41, I42, I43) -> f2(I34, rnd2, rnd3, I37, I38, I39, I40, I41, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ B0 = B0 /\ rnd3 = rnd3 /\ B1 = B1 /\ rnd2 = rnd2] The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) 2) f2#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f4#(I10, I11, I12, I13, I14, rnd6, rnd7, rnd8, I18, I19) [y2 = I13 /\ y3 = I14 /\ 0 <= -1 - y2 + y3 /\ rnd7 = rnd7 /\ rnd8 = rnd8 /\ y1 = I13 /\ rnd6 = rnd6] 3) f1#(I34, I35, I36, I37, I38, I39, I40, I41, I42, I43) -> f2#(I34, rnd2, rnd3, I37, I38, I39, I40, I41, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ B0 = B0 /\ rnd3 = rnd3 /\ B1 = B1 /\ rnd2 = rnd2] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) f2#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f4#(I10, I11, I12, I13, I14, rnd6, rnd7, rnd8, I18, I19) [y2 = I13 /\ y3 = I14 /\ 0 <= -1 - y2 + y3 /\ rnd7 = rnd7 /\ rnd8 = rnd8 /\ y1 = I13 /\ rnd6 = rnd6] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f1(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f2(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) f2(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f4(I10, I11, I12, I13, I14, rnd6, rnd7, rnd8, I18, I19) [y2 = I13 /\ y3 = I14 /\ 0 <= -1 - y2 + y3 /\ rnd7 = rnd7 /\ rnd8 = rnd8 /\ y1 = I13 /\ rnd6 = rnd6] f2(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(rnd1, I21, I22, I23, I24, I25, I30, I31, I28, I29) [I32 = I23 /\ I33 = I24 /\ -1 * I32 + I33 <= 0 /\ I30 = I30 /\ I31 = I31 /\ rnd1 = rnd1] f1(I34, I35, I36, I37, I38, I39, I40, I41, I42, I43) -> f2(I34, rnd2, rnd3, I37, I38, I39, I40, I41, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ B0 = B0 /\ rnd3 = rnd3 /\ B1 = B1 /\ rnd2 = rnd2] popout

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actions

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