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	n143.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	8.91551 seconds
cpu usage	9.05914
user time	4.62598
system time	4.43316
max virtual memory	724780.0
max residence set size	11992.0

stage attributes

key	value
starexec-result	YES

output

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 Integer TRS Innermost