details

property	value
status	complete
benchmark	19.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.604934 seconds
cpu usage	0.615621
user time	0.342493
system time	0.273128
max virtual memory	576992.0
max residence set size	8720.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = eval#(x, y, z) -> eval#(x, y, z) [y > z && z >= x && z >= y] eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1] eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8] eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14] R = eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y] eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1] eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8] eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14] 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, z) -> eval#(x, y, z) [y > z && z >= x && z >= y] 1) eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1] 2) eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] 3) eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8] 4) eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] 5) eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14] We have the following SCCs. { 4, 5 } { 2 } DP problem for innermost termination. P = eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] R = eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y] eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1] eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8] eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z2 + -1 * z3 This gives the following inequalities: I4 > I5 && I5 >= I3 ==> I4 + -1 * I5 > I4 - 1 + -1 * I5 with I4 + -1 * I5 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14] R = eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y] eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1] eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3] eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8] eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11] eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: I10 > I11 && I9 > I11 ==> I9 + -1 * I11 > I9 - 1 + -1 * I11 with I9 + -1 * I11 >= 0 I12 > I14 ==> I12 + -1 * I14 > I12 - 1 + -1 * I14 with I12 + -1 * I14 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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all output return to Integer TRS Innermost