Eduardo is on holiday right now, but he pinged me asking for this. It
makes some sense, and using half the *failed value* covers the case where
it's less than half what we expected.
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
It now looks like (for test_hardmpp):
```
# we have computed a set of 1 flows with probability 0.328, fees 0msat and delay 23
# Flow 1: amount=1800000000msat prob=0.328 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0->-103x3x0/1->
# Flow 1: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 1: Failure of 1800000000msat for 103x5x0/0 capacity [0msat,3000000000msat] -> [0msat,1799999999msat]
# we have computed a set of 2 flows with probability 0.115, fees 0msat and delay 23
# Flow 2: amount=500000000msat prob=0.475 fees=0msat delay=12 path=-103x6x0/0(min=max=4294967295msat)->-103x1x0/1->-103x4x0/1->
# Flow 3: amount=1300000000msat prob=0.242 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1799999999msat)->-103x3x0/1->
# Flow 3: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 3: Failure of 1300000000msat for 103x5x0/0 capacity [0msat,1799999999msat] -> [0msat,1299999999msat]
# we have computed a set of 2 flows with probability 0.084, fees 0msat and delay 23
# Flow 4: amount=260000000msat prob=0.467 fees=0msat delay=12 path=-103x6x0/0(500000000msat in 1 htlcs,min=max=4294967295msat)->-103x1x0/1(500000000msat in 1 htlcs)->-103x4x0/1(500000000msat in 1 htlcs)->
# Flow 5: amount=1040000000msat prob=0.179 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1299999999msat)->-103x3x0/1->
# Flow 5: Failed at node #1 (WIRE_TEMPORARY_CHANNEL_FAILURE): failed: WIRE_TEMPORARY_CHANNEL_FAILURE (reply from remote)
# Flow 5: Failure of 1040000000msat for 103x5x0/0 capacity [0msat,1299999999msat] -> [0msat,1039999999msat]
# we have computed a set of 2 flows with probability 0.052, fees 0msat and delay 23
# Flow 6: amount=120000000msat prob=0.494 fees=0msat delay=12 path=-103x6x0/0(760000000msat in 2 htlcs,min=max=4294967295msat)->-103x1x0/1(760000000msat in 2 htlcs)->-103x4x0/1(760000000msat in 2 htlcs)->
# Flow 7: amount=920000000msat prob=0.105 fees=0msat delay=12 path=-103x2x0/1(min=max=4294967295msat)->-103x5x0/0(max=1039999999msat)->-103x3x0/1->
# Flow 7: Success
```
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
I am doing to add more more debugging, but sent here is 0.
Document that clearly, and put a real value in sent.
Also: since we already sub 1 msat from x, amount_msat_less_eq should
be amount_msat_less (it may be equal to our min, in theory).
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
There's a fascinating bug report which suggests this happens on local channels,
implying spendable_msat is wrong?
See-also: #6567
Signed-off-by: Rusty Russell <rusty@rustcorp.com.au>
- adopt "const <type> *"convention
- remove use_shadow option for some pyln tests
- show prob. information of flows into paynotes
- show prob. of success of entire payment flow in paynotes
- minflow: We were not releasing the memory of flow arrays when replacing
them with a new canditate.
- use memleak_scan_obj in memleak_check
- replace u64 with size_t
Signed-off-by: Lagrang3 <eduardo.quintana@pm.me>
