details

property	value
status	complete
benchmark	csharp2.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.78212499619 seconds
cpu usage	1.893165182
max memory	1.767424E7

stage attributes

key	value
output-size	3615
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) R = b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 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 = b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] R = b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 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 -> 3 1 -> 2 -> 0 3 -> 4, 1 4 -> 0 Where: 0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 4) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] We have the following SCCs. { 0, 3, 4 } DP problem for innermost termination. P = b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b10#(I6, I7, I8) -> b14#(I6, I7, I8) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] R = b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) b10(I6, I7, I8) -> b14(I6, I7, I8) We use the extended value criterion with the projection function NU: NU[b14#(x0,x1,x2)] = x1 - 2 NU[b10#(x0,x1,x2)] = x1 - 2 NU[b15#(x0,x1,x2)] = -x0 + x1 - 2 This gives the following inequalities: ==> -sv14_14 + sv23_37 - 2 >= (sv23_37 - sv14_14) - 2 ==> I7 - 2 >= I7 - 2 I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1 ==> I1 - 2 > -I0 + I1 - 2 with I1 - 2 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b10#(I6, I7, I8) -> b14#(I6, I7, I8) R = b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 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 -> 3 3 -> Where: 0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) We have the following SCCs. popout

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

job log

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actions

all output return to Integ TRS Inner 43557