details

property	value
status	complete
benchmark	csharp3.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.43451786041 seconds
cpu usage	1.47136189
max memory	1.5015936E7

stage attributes

key	value
output-size	2514
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) The dependency graph for this problem is: 0 -> 1 -> 3 2 -> 1 3 -> 4, 0 4 -> 1 Where: 0) b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 4) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] We have the following SCCs. { 1, 3, 4 } DP problem for innermost termination. P = b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) popout

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

job log

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actions

all output return to Integer TRS Innermost