details

property	value
status	complete
benchmark	24.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.343121051788 seconds
cpu usage	0.307879002
max memory	9441280.0

stage attributes

key	value
output-size	2831
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = eval_3#(x, y) -> eval_1#(x, y) [0 >= y] eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0] eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2] eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0] eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6] eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9] R = eval_3(x, y) -> eval_1(x, y) [0 >= y] eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0] eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2] eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0] eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6] eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9] The dependency graph for this problem is: 0 -> 1 -> 0, 1 2 -> 3 -> 2, 3 4 -> 1 5 -> 3 Where: 0) eval_3#(x, y) -> eval_1#(x, y) [0 >= y] 1) eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0] 2) eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2] 3) eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0] 4) eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6] 5) eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9] We have the following SCCs. { 1 } { 3 } DP problem for innermost termination. P = eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0] R = eval_3(x, y) -> eval_1(x, y) [0 >= y] eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0] eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2] eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0] eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6] eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9] We use the reverse value criterion with the projection function NU: NU[eval_2#(z1,z2)] = z1 This gives the following inequalities: I4 > 0 ==> I4 > I4 - 1 with I4 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0] R = eval_3(x, y) -> eval_1(x, y) [0 >= y] eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0] eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2] eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0] eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6] eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9] We use the reverse value criterion with the projection function NU: NU[eval_3#(z1,z2)] = z2 + -1 * 0 This gives the following inequalities: I1 > 0 ==> I1 + -1 * 0 > I1 - 1 + -1 * 0 with I1 + -1 * 0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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job log

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actions

all output return to Integer TRS Innermost