details

property	value
status	complete
benchmark	practical1.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n184.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.66772890091 seconds
cpu usage	2.818740202
max memory	1.4385152E7

stage attributes

key	value
output-size	2108
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#(i, j) -> eval_1#(i - 1, j) [j > i - 1] eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 >= 0] R = eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] We use the reverse value criterion with the projection function NU: NU[eval_1#(z1,z2)] = z1 + -1 * 0 NU[eval_2#(z1,z2)] = z1 - 1 + -1 * 0 This gives the following inequalities: j > i - 1 ==> i - 1 + -1 * 0 >= i - 1 + -1 * 0 I1 <= I0 - 1 ==> I0 - 1 + -1 * 0 >= I0 - 1 + -1 * 0 I2 >= 0 ==> I2 + -1 * 0 > I2 - 1 + -1 * 0 with I2 + -1 * 0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1] eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] R = eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] The dependency graph for this problem is: 0 -> 1 -> 0, 1 Where: 0) eval_2#(i, j) -> eval_1#(i - 1, j) [j > i - 1] 1) eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I1 <= I0 - 1] R = eval_2(i, j) -> eval_1(i - 1, j) [j > i - 1] eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I1 <= I0 - 1] eval_1(I2, I3) -> eval_2(I2, 0) [I2 >= 0] We use the reverse value criterion with the projection function NU: NU[eval_2#(z1,z2)] = z1 - 1 + -1 * z2 This gives the following inequalities: I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 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