details

property	value
status	complete
benchmark	18.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n023.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.437021017075 seconds
cpu usage	0.454769671
max memory	1.0072064E7

stage attributes

key	value
output-size	3072
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#(x, y) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y] eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1] eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2] eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0] eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0] eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0] R = eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y] eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1] eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2] eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0] eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0] eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0] The dependency graph for this problem is: 0 -> 1 -> 2 -> 2 3 -> 4 -> 2, 4, 5 5 -> 2, 4, 5 Where: 0) eval#(x, y) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y] 1) eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1] 2) eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2] 3) eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0] 4) eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0] 5) eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0] We have the following SCCs. { 4, 5 } { 2 } DP problem for innermost termination. P = eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2] R = eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y] eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1] eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2] eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0] eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0] eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z2 + -1 * 0 This gives the following inequalities: I3 > 0 && 0 >= I2 ==> I3 + -1 * 0 > I3 - 1 + -1 * 0 with I3 + -1 * 0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0] eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0] R = eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y] eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1] eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2] eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0] eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0] eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z1 This gives the following inequalities: I7 > 0 && I6 > 0 ==> I6 > I6 - 1 with I6 >= 0 I8 > 0 ==> I8 > I8 - 1 with I8 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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

job log

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actions

all output return to Integer TRS Innermost