details

property	value
status	complete
benchmark	small21.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.80619 seconds
cpu usage	2.84459
user time	1.52911
system time	1.31548
max virtual memory	359208.0
max residence set size	8572.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f1#(I0, I1) f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0] f1#(I8, I9) -> f2#(I8, I8) [1 <= I8] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5] f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0] f1(I8, I9) -> f2(I8, I8) [1 <= I8] The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 3, 4 3 -> 2 4 -> 5 5 -> 3 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f1#(I0, I1) 2) f3#(I2, I3) -> f2#(I2, I3) 3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] 4) f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0] 5) f1#(I8, I9) -> f2#(I8, I8) [1 <= I8] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0] f1#(I8, I9) -> f2#(I8, I8) [1 <= I8] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5] f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0] f1(I8, I9) -> f2(I8, I8) [1 <= I8] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1)] = x0 + 1 NU[f2#(x0,x1)] = x0 NU[f3#(x0,x1)] = x0 This gives the following inequalities: ==> I2 >= I2 1 <= I5 ==> I4 >= I4 I7 <= 0 ==> I6 >= (-1 + I6) + 1 1 <= I8 ==> I8 + 1 > I8 with I8 + 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5] f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0] f1(I8, I9) -> f2(I8, I8) [1 <= I8] The dependency graph for this problem is: 2 -> 3, 4 3 -> 2 4 -> Where: 2) f3#(I2, I3) -> f2#(I2, I3) 3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] 4) f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5] f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0] f1(I8, I9) -> f2(I8, I8) [1 <= I8] popout

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job log

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actions

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