details

property	value
status	complete
benchmark	a.11.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	6.79333782196 seconds
cpu usage	7.2337054
max memory	2.8704768E7

stage attributes

key	value
output-size	2629
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_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] eval_2#(I0, I1, I2) -> eval_1#(I0, I1, I2 + 1) [I0 > I2] eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7] R = eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] We use the reverse value criterion with the projection function NU: NU[eval_1#(z1,z2,z3)] = z1 + -1 * z3 NU[eval_2#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: z >= x ==> x + -1 * z >= x - 1 + -1 * z I0 > I2 ==> I0 + -1 * I2 > I0 + -1 * (I2 + 1) with I0 + -1 * I2 >= 0 I3 > I5 ==> I3 + -1 * I5 >= I3 + -1 * I5 I6 > I7 ==> I6 + -1 * I8 >= I6 + -1 * I8 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] eval_1#(I6, I7, I8) -> eval_2#(I6, I7, I8) [I6 > I7] R = eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] We use the reverse value criterion with the projection function NU: NU[eval_1#(z1,z2,z3)] = z1 + -1 * z2 NU[eval_2#(z1,z2,z3)] = z1 - 1 + -1 * z2 This gives the following inequalities: z >= x ==> x - 1 + -1 * y >= x - 1 + -1 * y I3 > I5 ==> I3 - 1 + -1 * I4 >= I3 + -1 * (I4 + 1) I6 > I7 ==> I6 + -1 * I7 > I6 - 1 + -1 * I7 with I6 + -1 * I7 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] R = eval_2(x, y, z) -> eval_1(x - 1, y, z) [z >= x] eval_2(I0, I1, I2) -> eval_1(I0, I1, I2 + 1) [I0 > I2] eval_2(I3, I4, I5) -> eval_1(I3, I4 + 1, I5) [I3 > I5] eval_1(I6, I7, I8) -> eval_2(I6, I7, I8) [I6 > I7] The dependency graph for this problem is: 0 -> 2 -> Where: 0) eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [z >= x] 2) eval_2#(I3, I4, I5) -> eval_1#(I3, I4 + 1, I5) [I3 > I5] We have the following SCCs. popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557