details

property	value
status	complete
benchmark	csharp1.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n016.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.08837294579 seconds
cpu usage	2.111261874
max memory	1.8317312E7

stage attributes

key	value
output-size	3994
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 = b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), 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 && (true && (0 < sv24_38 && 1 < sv23_37))), 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 && (true && (0 < sv24_38 && 1 < sv23_37))), 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 && (true && (0 < sv24_38 && 1 < sv23_37)))] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), 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 && (true && (0 < sv24_38 && 1 < sv23_37))), 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 && (true && (0 < sv24_38 && 1 < sv23_37)))] 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 && (true && (0 < sv24_38 && 1 < sv23_37)))] R = b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))), 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) 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: ==> -I0 + I1 - 2 >= (I1 - I0) - 2 ==> I7 - 2 >= I7 - 2 sv23_37 >= sv14_14 && (1 < sv14_14 && (true && (0 < sv24_38 && 1 < sv23_37))) ==> sv23_37 - 2 > -sv14_14 + sv23_37 - 2 with sv23_37 - 2 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 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 && (true && (0 < sv24_38 && 1 < sv23_37))), 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: 1 -> 3 3 -> Where: 1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 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