details

property	value
status	complete
benchmark	A07.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.17100596428 seconds
cpu usage	1.209941946
max memory	1.2423168E7

stage attributes

key	value
output-size	3465
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = g#(true, x, y, z) -> f#(x > z, x, y, z + 1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) f#(true, I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) R = g(true, x, y, z) -> f(x > z, x, y, z + 1) g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = g#(true, x, y, z) -> f#(x > z, x, y, z + 1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4] f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4] f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2] g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z] R = g(true, x, y, z) -> f(x > z, x, y, z + 1) g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) The dependency graph for this problem is: 0 -> 1 -> 2 -> 7, 6, 5, 4, 3 3 -> 4 -> 5 -> 7, 6, 5, 3, 4 6 -> 7 -> 7, 3, 4, 5, 6 8 -> 3, 4, 5, 6, 7 9 -> 3, 4, 5, 6, 7 Where: 0) g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 2) f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 3) f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 4) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4] 5) f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] 6) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4] 7) f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] 8) g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2] 9) g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z] We have the following SCCs. { 5, 7 } DP problem for innermost termination. P = f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] R = g(true, x, y, z) -> f(x > z, x, y, z + 1) g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) We use the reverse value criterion with the projection function NU: NU[f_1#(z1,z2,z3)] = z1 + -1 * z2 This gives the following inequalities: I3 > I4/\I3 > I5 ==> I3 + -1 * I4 > I3 + -1 * (I4 + 1) with I3 + -1 * I4 >= 0 I3 > I4/\I3 > I5 ==> I3 + -1 * I4 >= I3 + -1 * I4 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] R = g(true, x, y, z) -> f(x > z, x, y, z + 1) g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) We use the reverse value criterion with the projection function NU: NU[f_1#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: I3 > I4/\I3 > I5 ==> I3 + -1 * I5 > I3 + -1 * (I5 + 1) with I3 + -1 * I5 >= 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