details

property	value
status	complete
benchmark	weakness.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n111.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	3.03805 seconds
cpu usage	3.09471
user time	1.51072
system time	1.58399
max virtual memory	722824.0
max residence set size	8700.0

stage attributes

key	value
starexec-result	MAYBE

output

3.01/3.03 MAYBE 3.01/3.03 3.01/3.03 DP problem for innermost termination. 3.01/3.03 P = 3.01/3.03 f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 3.01/3.03 f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3] 3.01/3.03 f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 3.01/3.03 f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9] 3.01/3.03 R = 3.01/3.03 f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 3.01/3.03 f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3] 3.01/3.03 f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 3.01/3.03 f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9] 3.01/3.03 3.01/3.03 The dependency graph for this problem is: 3.01/3.03 0 -> 1 3.01/3.03 1 -> 3 3.01/3.03 2 -> 3 3.01/3.03 3 -> 2 3.01/3.03 Where: 3.01/3.03 0) f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 3.01/3.03 1) f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3] 3.01/3.03 2) f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 3.01/3.03 3) f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9] 3.01/3.03 3.01/3.03 We have the following SCCs. 3.01/3.03 { 2, 3 } 3.01/3.03 3.01/3.03 DP problem for innermost termination. 3.01/3.03 P = 3.01/3.03 f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 3.01/3.03 f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9] 3.01/3.03 R = 3.01/3.03 f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 3.01/3.03 f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3] 3.01/3.03 f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 3.01/3.03 f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9] 3.01/3.03 3.01/6.01 EOF popout

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

job log

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54