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Integer_TRS_Innermost 2019-03-21 04.53 pair #429994991
details
property
value
status
complete
benchmark
A08.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
2.62047 seconds
cpu usage
2.65928
user time
1.46054
system time
1.19874
max virtual memory
723688.0
max residence set size
8648.0
stage attributes
key
value
starexec-result
YES
output
2.61/2.61 YES 2.61/2.61 2.61/2.61 DP problem for innermost termination. 2.61/2.61 P = 2.61/2.61 f#(true, x, y, z) -> f#(x > y && x > z, x, y, z + 1) 2.61/2.61 f#(true, I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 R = 2.61/2.61 f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 2.61/2.61 f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 2.61/2.61 This problem is converted using chaining, where edges between chained DPs are removed. 2.61/2.61 2.61/2.61 DP problem for innermost termination. 2.61/2.61 P = 2.61/2.61 f#(true, x, y, z) -> f_2#(x, y, z) 2.61/2.61 f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 2.61/2.61 f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 2.61/2.61 f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 2.61/2.61 f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.61 f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 R = 2.61/2.61 f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 2.61/2.61 f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 2.61/2.61 The dependency graph for this problem is: 2.61/2.61 0 -> 5, 4, 3 2.61/2.61 1 -> 7, 6, 2 2.61/2.61 2 -> 2.61/2.61 3 -> 2.61/2.61 4 -> 7, 6, 2 2.61/2.61 5 -> 5, 3, 4 2.61/2.61 6 -> 7, 6, 2 2.61/2.61 7 -> 3, 4, 5 2.61/2.61 Where: 2.61/2.61 0) f#(true, x, y, z) -> f_2#(x, y, z) 2.61/2.61 1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 2.61/2.61 2) f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 3) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 2.61/2.61 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 2.61/2.61 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.61 6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 2.61/2.61 We have the following SCCs. 2.61/2.61 { 4, 5, 6, 7 } 2.61/2.61 2.61/2.61 DP problem for innermost termination. 2.61/2.61 P = 2.61/2.61 f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 2.61/2.61 f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.61 f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 2.61/2.61 R = 2.61/2.61 f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 2.61/2.61 f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.61 2.61/2.61 We use the reverse value criterion with the projection function NU: 2.61/2.62 NU[f_1#(z1,z2,z3)] = z1 + -1 * z2 2.61/2.62 NU[f_2#(z1,z2,z3)] = z1 + -1 * z2 2.61/2.62 2.61/2.62 This gives the following inequalities: 2.61/2.62 x > y && x > z ==> x + -1 * y >= x + -1 * y 2.61/2.62 x > y && x > z ==> x + -1 * y >= x + -1 * y 2.61/2.62 I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 2.61/2.62 I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 2.61/2.62 2.61/2.62 We remove all the strictly oriented dependency pairs. 2.61/2.62 2.61/2.62 DP problem for innermost termination. 2.61/2.62 P = 2.61/2.62 f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 2.61/2.62 f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.62 R = 2.61/2.62 f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 2.61/2.62 f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.62 2.61/2.62 The dependency graph for this problem is: 2.61/2.62 4 -> 2.61/2.62 5 -> 5, 4 2.61/2.62 Where: 2.61/2.62 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 2.61/2.62 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.62 2.61/2.62 We have the following SCCs. 2.61/2.62 { 5 } 2.61/2.62 2.61/2.62 DP problem for innermost termination. 2.61/2.62 P = 2.61/2.62 f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 2.61/2.62 R = 2.61/2.62 f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) 2.61/2.62 f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 2.61/2.62 2.61/2.62 We use the reverse value criterion with the projection function NU: 2.61/2.62 NU[f_2#(z1,z2,z3)] = z1 + -1 * z3 2.61/2.62 2.61/2.62 This gives the following inequalities: 2.61/2.62 x > y && x > z ==> x + -1 * z > x + -1 * (z + 1) with x + -1 * z >= 0
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return to Integer_TRS_Innermost 2019-03-21 04.53