details

property	value
status	complete
benchmark	p-10.t2.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)	16.1867 seconds
cpu usage	16.4211
user time	8.19751
system time	8.22364
max virtual memory	744716.0
max residence set size	10576.0

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) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, I1, I2, I3, I4, I5, rnd7, rnd8, rnd9, rnd10) [rnd7 = rnd10 /\ rnd8 = rnd9 /\ rnd9 = rnd9 /\ rnd10 = rnd10] f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, rnd4, rnd5, rnd6, I26, I27, I28, I29) [y1 = I21 /\ y2 = I22 /\ 0 <= -1 - y1 + y2 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ y3 = I21 /\ rnd6 = rnd6] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, I1, I2, I3, I4, I5, rnd7, rnd8, rnd9, rnd10) [rnd7 = rnd10 /\ rnd8 = rnd9 /\ rnd9 = rnd9 /\ rnd10 = rnd10] f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20, I21, I22, rnd4, rnd5, rnd6, I26, I27, I28, I29) [y1 = I21 /\ y2 = I22 /\ 0 <= -1 - y1 + y2 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ y3 = I21 /\ rnd6 = rnd6] f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(rnd1, I31, I32, I40, I41, I35, I36, I37, I38, I39) [I42 = I31 /\ I43 = I32 /\ -1 * I42 + I43 <= 0 /\ I40 = I40 /\ I41 = I41 /\ rnd1 = rnd1] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, I1, I2, I3, I4, I5, rnd7, rnd8, rnd9, rnd10) [rnd7 = rnd10 /\ rnd8 = rnd9 /\ rnd9 = rnd9 /\ rnd10 = rnd10] 2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 3) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, rnd4, rnd5, rnd6, I26, I27, I28, I29) [y1 = I21 /\ y2 = I22 /\ 0 <= -1 - y1 + y2 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ y3 = I21 /\ rnd6 = rnd6] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, rnd4, rnd5, rnd6, I26, I27, I28, I29) [y1 = I21 /\ y2 = I22 /\ 0 <= -1 - y1 + y2 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ y3 = I21 /\ rnd6 = rnd6] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, I1, I2, I3, I4, I5, rnd7, rnd8, rnd9, rnd10) [rnd7 = rnd10 /\ rnd8 = rnd9 /\ rnd9 = rnd9 /\ rnd10 = rnd10] f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20, I21, I22, rnd4, rnd5, rnd6, I26, I27, I28, I29) [y1 = I21 /\ y2 = I22 /\ 0 <= -1 - y1 + y2 /\ rnd4 = rnd4 /\ rnd5 = rnd5 /\ y3 = I21 /\ rnd6 = rnd6] f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(rnd1, I31, I32, I40, I41, I35, I36, I37, I38, I39) [I42 = I31 /\ I43 = I32 /\ -1 * I42 + I43 <= 0 /\ I40 = I40 /\ I41 = I41 /\ rnd1 = rnd1] popout

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